Improper Riemann Integrals [2 ed.] 1032557982, 9781032557984

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Improper Riemann Integrals The scope of this book is the improper or generalized Riemann integral and infinite sum (series). The reader will study its convergence, principal value, evaluation and application to science and engineering. Improper Riemann integrals and infinite sums are interconnected. In the new edition, the author has involved infinite sums more than he did in the first edition. Apart from having computed and listed a large number of improper integrals and infinite sums, we have also developed the necessary theory and various ways of evaluating them or proving their divergence. Questions, problems and applications involving various improper integrals and infinite sums (series) of numbers emerge in science and application very often. Their complete presentations and all rigorous proofs would require taking the graduate-level courses on these subjects. Here their statements are adjusted to a level students of all levels can understand and use them efficiently as powerful tools in a large list of problems and applications. Professor Ioannis Markos Roussos was born on November 5, 1954, at the village Katapola of the island of Amorgos, Greece. After primary and secondary education, he studied mathematics at the National and Kapodistrian University of Athens and received his BSc Degree (1972–1977). Then, he studied graduate mathematics and computer sciences at the University of Minnesota and received his Masters and PhD degrees (1977–1986). His specialization in mathematics was in Differential Geometry and Analysis. He has taught mathematics at the University of Minnesota (1977–1987), University of South Alabama (1987–1990) and Hamline University (1990–2022). Besides this book, he has published 17 research papers, ten expository papers and the book Basic Lessons on Isometries, Similarities and Inversions in the Euclidean Plane. He has participated in meetings and has refereed papers and promotions of other professors. Other interests are classical music, history, international relations and travelling.

Improper Riemann Integrals Second Edition

Ioannis M. Roussos

Second edition published 2024 by CRC Press 6000 Broken Sound Parkway NW, Suite 300, Boca Raton, FL 33487-2742 and by CRC Press 4 Park Square, Milton Park, Abingdon, Oxon, OX14 4RN CRC Press is an imprint of Taylor & Francis Group, LLC © 2024 Ioannis Roussos First edition published by Taylor and Francis Group, LLC 2014 Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint. Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, access www.copyright. com or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. For works that are not available on CCC please contact mpkbookspermissions@ tandf.co.uk Trademark notice: Product or corporate names may be trademarks or registered trademarks and are used only for identification and explanation without intent to infringe. ISBN: 978-1-032-55798-4 (hbk) ISBN: 978-1-032-56035-9 (pbk) ISBN: 978-1-003-43347-7 (ebk) DOI: 10.1201/9781003433477 Typeset in CMR10 font by KnowledgeWorks Global Ltd. Publisher’s note: This book has been prepared from camera-ready copy provided by the authors.

Dedication To the memory of: My parents, Markos Ioannou Roussos and Margaro Nikita Grispou, and my first motivator and teacher in mathematics, my uncle, Michael Ioannou Roussos, and his wife, Evaggelia Louka Gavala.

Contents

Acknowledgments

ix

Prologue

xi

Additional Prologue

xv

Note for Readers

xvii

1 Improper Riemann Integrals, Definitions, Criteria 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10

Definitions and Examples . . . . . . . . . Applications . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . Cauchy Principal Value . . . . . . . . . . A Note on the Integration by Substitution Problems . . . . . . . . . . . . . . . . . . Some Criteria of Existence . . . . . . . . Problems . . . . . . . . . . . . . . . . . . Three Important Notes on Chapter 1 . . Uniformly Continuous Functions . . . . .

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2 Calculus Techniques 2.1 2.2 2.3

99

Normal Distribution Integral . . . . . . . . . . . . . . . Applications . . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . .

3 Real Analysis Techniques 3.1 3.2 3.3 3.4 3.5 3.6

Integrals Dependent on Parameters Problems . . . . . . . . . . . . . . Commuting Limits and Integrals . Commuting Limits and Derivatives Problems . . . . . . . . . . . . . . Double Integral Technique . . . .

1 18 23 32 37 41 47 82 89 93

99 103 107 119

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119 143 168 200 205 224 vii

viii 3.7 3.8 3.9 3.10 3.11 3.12 3.13 3.14 3.15

Problems . . . . . . . . . . Frullani Integrals . . . . . . Problems . . . . . . . . . . The Real Gamma Functions The Beta Function . . . . . Applications . . . . . . . . Problems . . . . . . . . . . Appendix . . . . . . . . . . Problems . . . . . . . . . .

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4 Laplace Transform 4.1 4.2 4.3 4.4 4.5

Laplace Transform, Definitions, Theory Problems . . . . . . . . . . . . . . . . . Inverse Laplace Transform . . . . . . . Applications . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . .

240 248 267 279 296 316 333 373 376 379

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380 400 414 416 424

Bibliography

431

Index

437

Acknowledgments

I would like to express my deep thanks to my colleague, Professor of Mathematics, Dr. Arthur Guetter. Without him, this work would not have been motivated, started and finished. He has also been my LaTeX teacher for the technical typing of this material. Secondly, I would like to thank my student, Nathan Davis, whose fluent expertise in LaTeX and figure making have been indispensable and very valuable in this work. He has spent a lot of time for this cause, and I feel indebted to him for all his work and help. I also wish to thank some of my friends for their moral support and encouragement.

ix

Prologue of the First Edition of Improper Riemann Integrals (2013)

This book is written at the masters level to help students of mathematics, statistics, applied sciences and engineering. Its scope is the improper or generalized Riemann integral, its convergence, principal value, evaluation and its application to science and engineering. Questions, problems and applications involving various improper integrals and series of numbers often emerge in these subjects. At the undergraduate level, results concerning useful improper integrals are mostly taken for granted, provided by an authority or obtained through tables and computer programs or packages. Here we try to give students sufficient knowledge and tools to enable them to answer these questions by themselves and acquire a deeper understanding of this matter and/or prepare them to do so with some further study of the matter. We try to achieve these goals by explaining the concepts involved, presenting sufficient theory and using a number of theorems, some with their proofs and others without. A complete, general and advanced exposition of this vast area of mathematics would contain a much greater number of theorems and proofs and involve advanced mathematical theories of real and complex analysis, integral transforms, special functions, etc., that lie far beyond the undergraduate and/or masters curriculum. We must add that the student, with this book at hand, is assumed to be fluent with the rules of antidifferentiation (indefinite integration and computing antiderivatives), the u-substitution, integration by parts and deriving recursive formulae, the change of variables with multiple integrals and the theorems of basic calculus, advanced calculus and mathematical analysis. Whenever possible, we present the material in a self-contained manner. We have proved many results, but not all. Sometimes our proofs are not established under the most general conditions that the more advanced theories can provide, but under conditions accessible to the undergraduate and sufficient for application. We also state and use a few advanced general theorems, results and tools from real and complex xi

xii

Improper Riemann Integrals

analysis without proofs. Their complete presentations and rigorous proofs would require taking the graduate-level courses on these subjects. Here their statements are adjusted to a level that students can understand and are interpreted in a way so that students can handle, manipulate and use them efficiently as powerful tools in our list of problems. In this way we avoid stating and proving a great number of criteria and partial results and thus avoid forcing the students into too much searching (a lot of times done by trial and error) for finding out the case they deal with each time and what criterion to apply. Thus, we try to render these advanced mathematical results and tools accessible and useful even to the undergraduate students with sufficient background so that they can use them in fairly straightforward manners in many pertinent problems they may come across in the subjects aforementioned. Moreover, our presentation and use of these advanced and general theorems and results give the undergraduate student a taste of the power of the graduate-level mathematics and motivate the interested one to take these courses at the graduate level in due time. We also expose a great number of detailed examples in order to illustrate the concepts and practice a lot with the tools that check convergence of improper integrals and evaluate their exact value when this is possible. We include many exercises and problems in every section. These are carefully chosen to serve both as practice and for further application. They are representative enough so that the student, on the basis of these, can solve many other exercises and problems not included in this book and also use them in many situations of application. We try to keep the number of exercises at a level so that, on the one hand, the student does not get lost in a vast sea of exercises and, on the other hand, the opportunity to practice and learn the material well and apply it is not compromised. A few problems that are lengthy have several questions and may be hard; these could be assigned as projects to an individual student or a group of students. Also, the input and help from the teacher or pertinent bibliography may be significant. Many examples are presented several times in different ways in order to see them from various points of view and see how different methods can give correct answers to the same questions. That is, their solutions are achieved in various ways depending on the context. We also repeat a few problems from section to section, and we seek their solutions within the new context. In this way we try to show students the interconnection of the whole matter, how a given question may be viewed in many ways and within various contexts, and that there are many ways to achieve a correct answer. This is something generally lacking in the undergraduate mathematics education.

Prologue

xiii

This book includes many theorems and methods for checking the convergence and the computation of most improper integrals encountered in applications. The content is sufficient to provide answers to the majority of them. We briefly examine the Laplace transform, Mellin transform and Fourier transform. Except for a few results, we do not develop the theory behind these integral transforms, but we mostly concentrate on their evaluation and some applications. We have omitted other integral transforms, such as the Hankel transform, Hilbert transform, etc. At this level, we did not include many special and hard integrals such as improper integrals in several variables, elliptic and hyper-elliptic integrals, integrals involving special functions such as Bessel functions, hypergeometric functions, asymptotic expansions, methods of steepest descents, etc., and some very special cases of contour integration (Cauchy, Legendre, Mellin, etc.). These are topics in the area of Special Functions at a more advanced level. However, a lot of concrete cases out of these special integrals can still be resolved by making appropriate use of the tools provided here. Also, in advanced mathematics, we encounter the singular integrals (especially in higher dimensions), which is a whole subject in itself, very important in mathematics and application. We must say that one will encounter several not fully explained points, indicated in the text by expressions like “justify this,” etc. All of them, however, can be justified by the versatile, studious and knowledgeable master-level student. An undergraduate could also clarify all of them with the help of the teacher. The proofs of theorems and results omitted in the text can be found in real analysis, complex analysis and applied mathematics literature. In the chapter of complex analysis methods, we have avoided the theory and the formulae that involve the index or winding number of a curve with respect to a point (or, of a point with respect to a curve). In this way, we do not get too far into the theory of complex variables and, at the same time, we do not lose anything much with respect to the computations of improper integrals. In an advanced complex analysis course, we see the local and global Cauchy theorems, the residue theorem and other theorems, and all the pertinent formulae stated and proved in the general context that involves some multiplications with the index (or rotation) number. This number assumes only integer values and in the development of our formulae we arrange the hypothesis so that its value is 1. The interested reader can consult a good book in complex analysis and study these topics in this generality. (A good number of such books have been listed in our bibliography.) In conclusion, the useful and practical material of this book is accessible to and can be mastered by any student who has finished a calculus sequence and has taken some multi-variable calculus, basic ordinary

xiv

Improper Riemann Integrals

differential equations, basic mathematical analysis, complex numbers and the basics of complex variables. Knowing this material, a student may not rely on authority, tables or computer packages to give and understand answers to questions related to this important material in theory and application. On account of all these and its whole content, this book can be used as the text for an undergraduate course or a supplementary text to other courses of mathematics, statistics, applied sciences and engineering. It can also become a very helpful manual and reference to students at the master level and even beyond. At the senior undergraduate level, this material can be used for a capstone course of a program and also serve for a good review of calculus and basic mathematical and complex analysis. At the starting graduate level, we find many illustrations of several strong tools of real and complex analysis with numerous examples and problems, a good many of which are quite involved. We use these tools, results and theorems not just in computing examples and solving problems but also in justifying that our methods of various computations are legitimate. A student who knows advanced calculus and has learnt the material and problems of this book must be able to verify at least all the integrals numbered 582–709 that appear in pages 448–455 of the CRC Standard Mathematical Tables and Formulae, by Daniel Zwillinger, 31st Edition, Chapman & Hall/CRC Press, 2003. Have a look at and practice with them after you have finished studying this book. At the end of this book, we have collected in a list all the major integrals evaluated one way or another in the text and the major finite and infinite sums in a different list. We did not go into computing and collecting infinite products. That would have been another chapter in the book. As expected we have included a sufficient bibliography, but far from all the bibliography that circulates in the world on these subjects. For the convenience of students and readers, an index of terms and names is also included. Finally, we thank all the people who study and use this book and we kindly ask them, if they encounter a typo or error that has escaped our attention, to communicate it to the author for correcting it in a prospective next edition. Also, suggesting new interesting and pertinent problems is highly appreciated. Dr. Ioannis M. Roussos, Professor of Mathematics Hamline University Saint Paul, Minnesota, 2013

Additional Prologue on the Second Edition: Improper Riemann Integrals

The present book is an improved second edition of the first half of the book with the title Improper Riemann Integrals. Even though this new book is conceptually similar to the previous book, it constitutes an extended, more complete and detailed exposition of the subject matter. More experience, a new bibliography, additional practice and knowledge with the subject matter along with readers’ suggestions, reports and critiques have motivated me to put forward this newly supplemented version. Apart from having corrected typos and errors and having taken care of some cosmetics, the old material has been extended to 1000 pages. This new material is dispersed within the old material almost uniformly. Every part of the old book has been touched and extended by something interesting or important. Apart from the fact of having computed and listed a large number of improper integrals and infinite sums, we have also developed various ways of evaluating them, in detail. A chapter of mathematics that deals with the infinite sums a lot is the Fourier series. We do not explore this chapter because the volume of this book would increase by much more and because the Fourier series is a subject well exposed in many books of the international literature. The new material includes ADDITIONAL: Theory Theorems Results and formulae on improper integrals and infinite sums Examples Applications, some of which are in partial differential equations Results on Laplace and Mellin transforms Problems Hints Inter-text references Footnotes xv

xvi

Improper Riemann Integrals Index references Bibliography

Also, we have corrected Problems 2.5.15, 3.7.55 and 3.7.81 from the old book. The new material is interesting, challenging, important, useful and applicable in mathematics, engineering and science. It supplements the material of the old book in a substantial and useful way. Some of these items were added at the request and suggestion of some readers of the first book, whom we thank greatly. This new edition will be much more helpful for the interested student, professor, scientist, engineer and reader. Given that the material collected was large, this book contains the chapters: (1) A preliminary chapter on Improper Riemann Integrals. (2) Real analysis techniques. (3) The bibliography. The complex analysis techniques and the necessary theory can be provided by the author. Professor Emeritus Ioannis Markos Roussos Hamline University Twin Cities, Minnesota, 2023

Note for Readers

For Part II of this work, which pertains to complex analysis and improper integrals, you can download it from the link imroussos/Improper Integrals Part2: Improper Riemann Integrals through Complex Analysis. Part II (github.com) Or, you can also write to the author to send you a free electronic copy at the emails: [email protected]

[email protected]

Given that the whole material was extended to 1000 pages, we have published just the real analysis and improper integrals material in this regular print. For the complex analysis material, send a message to the author. Complex analysis is a very powerful tool for computing improper integrals and infinite sums. We have also included the lists of the results obtained in the whole material. The inter-text references inside this book that begin with “II 1.” refer to the material of the chapter in complex analysis. All the references are written in bold numbers and or letters. Similarly, in the complex analysis material, the references that begin with “I 1.” refer to Chapter 1 and with “I 2.” refer to Chapters 2, 3 and 4 of this book, in which we expose the calculus and real analysis techniques in improper integrals.

xvii

Chapter 1 Improper Riemann Integrals, Definitions, Criteria

1.1

Definitions and Examples

Many theorems in Mathematics and many applications in science and technology depend on the evaluation and on the properties of improper Riemann integrals. Therefore, we are going to state the definitions of improper or generalized integrals and then discuss their properties. Subsequently, we discuss criteria for checking their existence (or non-existence) and then we develop methods and mathematical techniques we can use in order to evaluate them. Certainly, the answers to many important improper Riemann integrals have been tabulated in mathematical handbooks and can also be found with the help of various computer programs, which we can use if we can trust in them, of course. However, these means can never exhaust every interesting case. Hence, the good knowledge of the mathematical theory of how to understand, handle and compute improper integrals, at a higher level, will always remain very important for being able to deal with new cases and checking the accuracy of the answers provided in tables or found by computer programs or packages. In a regular undergraduate Calculus course, we study the Fundamental Theorem of Integral Calculus. This states: Theorem 1.1.1 [Fundamental Theorem of Integral Calculus] If a real function f : [a, b] → R (a < b are real numbers) is continuous, then it possesses antiderivatives F (x), i.e., functions that satisfy F 0 (x) = f (x) for every x ∈ [a, b] [at the endpoints we consider the appropriate side derivatives, F+0 (a) = f (a) (right derivative at a) and F−0 (b) = f (b) (left derivative at b)]. As continuously differentiable, any such antiderivative F (x) of f (x) is necessarily continuous in (a, b), right continuous at a, [i.e., F (a) = lim F (x) ], and left continuous at b, [i.e., x→a+

F (b) = lim− F (x) ]. x→b

DOI: 10.1201/9781003433477-1

1

2

Improper Riemann Integrals

All of these antiderivatives of the given continuous function f (x) on [a, b] are given by Z x F (x) = f (t) dt + C, ∀ x ∈ [a, b], a

where C is a real constant (constant of integration), satisfy and are characterized by the following properties: (a)

d [F (x)] = F 0 (x) = f (x), dx

(b) F (a) = C, Z b (c) f (x) dx = F (b) − F (a). a

We emphasize the three hypotheses that must hold in order for this Theorem to be valid: 1. [a, b] is a closed and bounded interval in the real line. 2. f (x) is continuous and therefore, by the extreme value Theorem, bounded on [a, b]. Z b 3. In the computation f (x) dx = F (b) − F (a), the function F (x) a

can be any fixed continuous antiderivative. We know that on [a, b], there are infinitely many antiderivatives of a continuous function f (x), but any two of them differ by a real constant C. Since they are differentiable they are continuous, and since their derivative is the continuous function f (x) they are continuously differentiable. Under these conditions, the integral Z b f (x) dx = F (b) − F (a) a

is called the proper Riemann1 integral of f (x) over the interval [a, b]. This is well defined and equal to the limit of the Riemann sums of f (x) over [a, b], as the maximum length max{∆xk := xk − xk−1 , k = 1, 2, . . . , n}, of the subintervals [xk−1 , xk ] into which we subdivide [a, b], where 1 Georg

Friedrich Bernhard Riemann, German mathematician, 1826–1866.

Improper Riemann Integrals, Definitions, Criteria

3

a = x0 < x1 < x2 < . . . < xn = b, in this well-known process, approaches zero. I.e., for any point selection x∗k , with xk−1 ≤ x∗k ≤ xk , for k = 1, 2, . . . , n, we have Z

b

f (x) dx = a

lim

max{∆xk }→0

n X

f (x∗k )∆xk = F (b) − F (a).

k=1

Example 1.1.1 Consider the function    x, if 0 ≤ x ≤ 1, f (x) =   1, if 1 < x ≤ 2, which is continuous on [0, 2]. (Check this!) An antiderivative of f (x) is  2 x   if 0 ≤ x ≤ 1,  Z x  2, F (x) = f (x)dx =  0    x − 1 , if 1 < x ≤ 2. 2 (Any other antiderivative of f (x) is given by G(x) = F (x) + C, with C an arbitrary real constant.) F (x) is continuously differentiable on [0, 2], F 0 (x) = f (x) and Z

2

 f (x)dx = F (2) − F (0) =

0

2−

1 2

 −0=

3 . 2

The function  2 x    , if 0 ≤ x ≤ 1, 2 H(x) =    x, if 1 < x ≤ 2, is not an antiderivative of f (x), even though for x 6= 1, we have    x, if 0 ≤ x < 1, H 0 (x) =   1, if 1 < x ≤ 2. (At x = 0 and x = 2, we consider the right and left derivative, respectively.) At x = 1, the left derivative of H(x) is 1, but the right derivative

4

Improper Riemann Integrals

does not exist. (Check this!) Hence at x = 1, the derivative does not exist. N We can go a bit beyond the undergraduate interpretation of the Fundamental Theorem of Integral Calculus and relax the above hypotheses as follows: We more generally consider f : [a, b] → R piecewise continuous and bounded. Then its Riemann integral exists. In such a case, we can also find F (x) antiderivative of f (x) which is continuous in (a, b), right continuous at a, left continuous at b and differentiable only at the points of continuity of f (x). At the points of discontinuity of f (x), F (x) may have a left or right derivative but not derivative. Sometimes f : [a, b] → R may be continuous in (a, b), right continuous at a, left continuous at b, but in order to obtain, by means of the usual methods and rules of antidifferentiation, an antiderivative F (x) of f (x) which is continuous in (a, b), right continuous at a, left continuous at b, we may have to make necessary adjustments, by adjusting certain constants, at certain points of the interval of integration [a, b]. Only then we can guarantee the result Z b f (x) dx = F (b) − F (a), a

in such cases. In fact, the Fundamental Theorem of Calculus proves that if f (x) is a bounded Riemann integrable function on the closed and bounded interval [a, b], then Z x f (t)dt is continuous in [a, b] F (x) = a

and satisfies d F (x)|x=w = f (w) at all points of continuity w of f (x) on [a, b]. dx So, if f (x) is continuous on [a, b], the function F (x) is an antiderivative of f (x) on [a, b] and is continuously differentiable. The anomaly we discuss here is not due to any deficiency of the Fundamental Theorem of Calculus, but it is created by the standard rules and methods of antidifferentiation. At times, the answers obtained by these rules are not defined at certain points and therefore are discontinuous at these points. To obtain the continuity as the Fundamental Theorem of Calculus guarantees and requires, we must adjust these answers appropriately. To understand this extraordinary situation and be

Improper Riemann Integrals, Definitions, Criteria

5

aware of its occurrence, let us study the following example (and see also Problem 1.3.4). Example 1.1.2 We consider the function f (x) =

3 . 5 − 4 cos(x)

This function is defined for every x ∈ R. It is continuous at every x ∈ R,  1 bounded ≤ f (x) ≤ 3 , periodic with period 2π and even. 3 When we integrate rational functionsof sine and cosine, we usually x . (See also Remark 2 of use the half angle substitution u = tan 2 Example II 1.7.14.) Then we find (work it out) Z h  x i 3 dx = 2 arctan 3 tan + C. 5 − 4 cos(x) 2 We let C = 0 (as we usually do in calculus when we evaluate definite integrals). So, we choose h  x i F (x) = 2 arctan 3 tan . 2 This function is defined for all real x 6= (2k + 1)π, with k ∈ Z, since at x x = (2k + 1)π, with k ∈ Z, tan is not defined. At these exceptional 2 points, however, we have lim

x→(2k+1)π −

F (x) = 2 ·

π =π 2

and

lim

x→(2k+1)π +

F (x) = 2 ·

−π = −π. 2

So, at each x = (2k + 1)π, with k ∈ Z, F (x) has a jump discontinuity with jump |π − (−π)| = 2π. As x crosses each (2k + 1)π, F (x) jumps to a different branch of the function 2×(arc-tangent). Notice also that F (x) is bounded and −π < F (x) < π. Therefore, to evaluate the definite integrals Z b 3 dx = F (b) − F (a), for any − π ≤ a, b ≤ π, 5 − 4 cos(x) a we can use the continuous antiderivative  −π, if x = −π + ,        h  x i ¯ , if − π < x < π, F (x) = 2 arctan 3 tan  2       π, if x = π − .

6

Improper Riemann Integrals Z

π

E.g., −π

3 dx = F¯ (π − ) − F¯ (−π + ) = π − (−π) = 2π. 5 − 4 cos(x)

(See also the second half of Example II 1.4.4.) But, we cannot use F (x) or F¯ (x) to evaluate definite integrals if a or b does not satisfy −π ≤ a, b ≤ π. For instance, if we use it with a = −2π and b = 4π, we find Z 4π 3 dx = F (4π) − F (−2π) = 0 − 0 = 0, 5 − 4 cos(x) −2π which is incorrect, since the continuous integrand function f (x) is positive, and therefore its definite integral should be positive. This error has occurred because the chosen antiderivative over the interval [−2π, 4π] is discontinuous at the exceptional points examined above. y

8

y=Fc (x)

6

f (x) = 4 y=F(x)

3 5-4 cos(x )

F(x) = 2 arctan 3 tan

y=F(x)

2

x 2



x

x +π

2

2π

Fc (x) = 2 arctan3 tan  + 2π



y=f (x) O

-2

2

4

6

8

x

y=Fc (x) -2

FIGURE 1.1: The three functions in Example 1.1.2.

The correct answer is obtained if we use the adjusted antiderivative    h  x i x+π   2 arctan 3 tan + 2π , if x 6= (2k + 1)π, k ∈ Z,  2 2π Fc (x) =    (2k + 1)π, if x = (2k + 1)π, k ∈ Z,  where

x+π 2π

 is the integer part or floor function of

x+π . 2π

Improper Riemann Integrals, Definitions, Criteria

7

This new Fc (x) is now continuous, differentiable and Fc0 (x) = f (x) at all points of [−2π, 4π] as the Fundamental Theorem of Calculus claims and requires. (In fact, this is true at every x ∈ R. See Problems 1.3.1 and 1.3.2.) Now, we get: Z 4π 3 dx = Fc (4π) − Fc (−2π) = 4π − (−2π) = 6π > 0. 5 − 4 cos(x) −2π This result is the correct one and was also expected since f (x) is 2π−periodic, with integral 2π over [−π, π] and we have integrated it over an interval of length 6π, i.e., three times its period. (About periodic functions, in general, see Problem 1.3.8.) N In this context we also state (without proof) the following nice Theorem2 because is related to the Fundamental Theorem of Calculus. Theorem 1.1.2 If a < b are real numbers, the real function f : [a, b] −→ R is differentiable at every point of [a, b] and Z b |f 0 (x)| dx < ∞, then a

Z ∀ x ∈ [a, b],

f (x) − f (a) =

x

f 0 (x) dx.

a

Note that differentiability is assumed to hold at every point of [a, b]. (In this context, study also bibliography, Bartle 1996.) Example 1.1.3 On the interval [−1, 1], we consider the function    1 2   if − 1 ≤ x 6= 0 ≤ 1, x sin x , f (x) =    0, if x = 0. By usual computation, we find that its derivative exists at every point and it is      1 1   2x sin − cos , if − 1 ≤ x 6= 0 ≤ 1,  x x 0 f (x) =    0, if x = 0. 2 For

the proof, see the bibliography, e.g., Rudin 1987, Theorem 7.21, p. 149.

8

Improper Riemann Integrals The derivative satisfies Z 1 Z |f 0 (x)|dx < −1

1

3 dx = 6.

−1

Then, by the above Theorem, we get     Z 1 Z 1 1 1 f 0 (x) dx = 2x sin − cos dx = x x −1 −1 f (1) − f (−1) = sin(1) − sin(−1) = 2 sin(1). (See also Example 1.1.11.) N Now we continue with the improper or generalized Riemann integrals. Definition 1.1.1 An integral of a piecewise continuous real function of a real variable is called an improper or generalized Riemann integral if at least one of the following three conditions occurs: 1. The integrated function is unbounded over the interval of integration. 2. The interval of integration is not closed. 3. The interval of integration is unbounded. In all the pertinent definitions that follow, an improper or generalized Riemann integral of a real piecewise continuous function of a real variable defined over a set I ⊆ R will be defined to be a certain limit of proper Riemann integrals. More concretely, we present four cases and definitions in our exposition, each of which may include two or more subcases, that generalize the proper Riemann integrals: Definition 1.1.2 Suppose y = f (x) is a real function continuous in [a, b) ⊂ R, then we define: For b < ∞ Z b Z ρ Z b− f (x) dx = lim− f (x) dx = lim+ f (x) dx. ρ→b

a

For b = ∞ Z

→0

a

∞

Z f (x) dx = lim

a

M →∞

a

M

f (x) dx. a

Improper Riemann Integrals, Definitions, Criteria

9

Examples Example 1.1.4 Z 0

 ρ Z ρ −1 dx 3 2 3 3 √ = lim x dx = lim = x 3 x ρ→0− −1 ρ→0− 2 −1 −1   2 2 3 3 3 3 2 3 3 3 lim ρ 3 − (−1) 3 = lim ρ 3 − = · 0 − = − . − − 2 2 2 ρ→0 2 2 2 2 ρ→0 N

Example 1.1.5 Z

∞

dx = lim [arctan(x)]M 1 = + 1 M →∞ 1 π π π π lim [arctan(M ) − arctan(1)] = lim arctan(M ) − = − = . M →∞ M →∞ 4 2 4 4 x2

[π.3 ] N Example 1.1.6 Z ∞ √ √ √ dx √ = lim [2 x]M 1 = lim 2 M − 2 1 = ∞ − 2 = ∞. M →∞ x M →∞ 1 N Example 1.1.7 Z 0

∞

M

sin(x) dx = lim [− cos(x)]0 = M →∞

lim [− cos(M ) + cos(0)] = − lim cos(M ) + 1 = does not exist.

M →∞

M →∞

This limit does not exist because cos(M ) oscillates between −1 and 1. N 3 Pi, π, $, Π the sixteenth letter of the Greek alphabet. International symbol of the transcendental number equal to the fixed ratio of the circumference of any circle divided by its diameter(∼ = 3.1415926536...). Its conceptualization and first good approximation was done by Archimedes of Syracuse (in Sicily, today in Italy), Greek mathematician, 287–212 B.C.E. The symbol π was introduced by the English mathematician William Jones (1675–1749) in 1706 and adopted by Euler in 1748. It is one of the most important numbers in mathematics, science, technology and applications. Archimedes is considered by the great majority of mathematicians to be the greatest mathematician of all times. He is the first who conceived the process of integration by which he found the area of the of radius r to be A = πr2 , the length of its R 1 circle circumference c = 2πr and also −1 x2 dx = 32 . His method was a limiting process of sums of areas or lengths that could be computed elementarily, similar to the Riemann or Riemann-Darboux or Darboux sums of the nineteenth century. (Jean Gaston Darboux, French mathematician, 1842–1917.)

10

Improper Riemann Integrals

Definition 1.1.3 Suppose y = f (x) is a real continuous function in (a, b] ⊂ R. Then we define: For −∞ < a Z b Z b Z b f (x) dx. f (x) dx = lim+ f (x) dx = lim+ σ→a

a

δ→0

σ

a+δ

For a = −∞ Z

b

b

Z f (x) dx =

−∞

lim

N →−∞

f (x) dx. N

Examples Example 1.1.8 Z 2 √ √ √ √ √ dx √ = lim [2 x]2σ = 2 2 − lim 2 σ = 2 2 − 0 = 2 2. + + x σ→0 σ→0 0 N Example 1.1.9 Z

0

0

ex dx = lim [ex ]N = N →−∞ −∞  0  N lim e − e = 1 − lim eN = 1 − 0 = 1.

N →−∞

N →−∞

[e.4 ] N Example 1.1.10 Z

1 2



x dx = −∞

lim

N →−∞

x3 3

1

 =

N

lim

N →−∞

 1 N3 − = 3 3

1 1 N3 − lim = − (−∞) = ∞. 3 N →−∞ 3 3 N 4e

is the symbol of the transcendental number lim

n→∞

1+


1 n n

=

∞ P n=0

1 . n!

It is one of

the most important numbers in mathematics, science, technology and applications. It is the base of the natural logarithms. It was in some sense known to the Scottish mathematician John Napier, 1550–1617, but the Swiss mathematicians Jakob (Jacques) Bernoulli, 1654–1705, and Leonhard Euler, 1707–1783, were the ones who recognized its highest significance to mathematics and applications. Its approximate value is: e ∼ = 2.718281828459045... .

Improper Riemann Integrals, Definitions, Criteria

11

Example 1.1.11 Consider the function    1   , if − 1 ≤ x 6= 0 ≤ 1, x sin  x f (x) =    0, if x = 0, on the interval [−1, 1]. This function is continuous and its derivative is      1 1 1   if − 1 ≤ x 6= 0 ≤ 1, sin x − x cos x , f 0 (x) =    does not exist, if x = 0. [Check this! Also, f 0 (x) is odd.] The derivative does not satisfy the conditions of Theorem 1.1.2 on [−1, 1]. It does not exist at x = 0, and the integral of its absolute value is not finite. (Check!) But it exists on any interval [−1, b] with −1 < b < 0 and on any interval [a, 1] with 0 < a < 1. Then, by Theorem 1.1.2 or Theorem 1.1.1, we obtain the improper integral   Z 1   1 1 1 sin − cos dx = x x x 0   Z 1   1 1 1 lim+ sin − cos dx = lim+ [f (1) − f (a)] = x x x a→0 a→0 a    1 lim sin(1) − a · sin = sin(1) − 0 = sin(1). a a→0+ Similarly, we obtain   Z 0   1 1 1 sin − cos dx = x x x −1   Z b    1 1 1 lim sin − cos dx = lim− [f (b) − f (−1)] = x x x b→0− −1 b→0     1 lim b · sin − [− sin(−1)] = 0 − sin(1) = − sin(1). b b→0− (See also Example 1.1.3.) N

12

Improper Riemann Integrals

Example 1.1.12 As in the previous Example, we consider the function    1 2   , if − 1 ≤ x 6= 0 ≤ 1, x sin  x3 g(x) =    0, if x = 0. This function is continuous and its derivative is      3 1 1   if − 1 ≤ x 6= 0 ≤ 1, 2x sin x3 − x2 cos x3 , 0 g (x) =    0, if x = 0. [Check this! Also, g 0 (x) is even.] The integral of the absolute value of the derivative is not finite, and so it does not satisfy that condition of Theorem 1.1.2 on [−1, 1]. The above function g(x) is an example of a differentiable function whose derivative is not properly Riemann integrable, but only in the generalized sense, as x −→ 0+ or x −→ 0− . In any neighborhood of x = 0, g 0 (x) is unbounded above and/or below. N Definition 1.1.4 Suppose y = f (x) is a real continuous function in (a, b) ⊆ R. Then we define: For −∞ < a < b < ∞ Z b Z ρ f (x) dx = lim f (x) dx. a

For a = −∞ and b = ∞ Z ∞ f (x) dx = −∞

For a = −∞ and b < ∞ Z b f (x) dx =

ρ→b− σ→a+

σ

M

Z lim

f (x) dx.

M →∞ N →−∞

N

Z

ρ

f (x) dx.

lim

ρ→b− N →−∞

N

For −∞ < a and b = ∞ Z ∞ Z f (x) dx = lim

M

−∞

a

M →∞ σ→a+

f (x) dx.

σ

In the above double limits, the two limiting processes are independent of each other in general.

Improper Riemann Integrals, Definitions, Criteria

13

Examples Example 1.1.13

Z lim−

ρ→1 σ→−1+

ρ

σ

1

 Z 1  dx 1 1 1 = − dx = 2 x−1 x+1 −1 x − 1 −1 2   1 1 1 1 ρ − dx = lim [ln |x − 1| − ln |x + 1|]σ = 2 x−1 x+1 2 ρ→1− Z

σ→−1+

1 lim [ln |ρ − 1| − ln |ρ + 1| − ln |σ − 1| + ln |σ + 1|] . 2 ρ→1− σ→−1+

We have that lim ln |ρ − 1| = −∞,

lim ln |ρ + 1| = ln(2),

ρ→1−

ρ→1−

lim ln |σ − 1| = ln(2),

lim ln |σ + 1| = −∞.

σ→−1+

σ→−1+

So, the above improper integral as double limit is Z 1 1 dx = [−∞ − ln(2) − ln(2) − ∞] = −∞. 2 2 −1 x − 1 N Example 1.1.14 Z ∞ dx = lim [arctan(x)]M N = 2+1 M →∞ x −∞ N →−∞

lim [arctan(M ) − arctan(N )] =

M →∞ N →−∞

(since both partial limits exist separately, we get) π  π = π. lim arctan(M ) − lim arctan(N ) = − − M →∞ N →−∞ 2 2 N Example 1.1.15 Z

0

−∞

 lim

ρ→0−

1 − ρ

dx = x2



 lim

ρ→0− N →−∞

 − lim

N →−∞

−1 N

−1 x

ρ = N

 = −(−∞) − 0 = ∞. N

14

Improper Riemann Integrals

Example 1.1.16 Z ∞ √ √ √ dx √ = lim [2 x]M σ = lim 2 M − lim+ 2 σ = ∞ − 0 = ∞. M →∞ x M →∞ σ→0 0 σ→0+

N Example 1.1.17 Z

∞

 x dx =

−∞

 lim

M →∞ N →−∞

lim

M →∞ N →−∞

x2 2

M = N

 N2 M2 − = ∞ − ∞ = does not exist. 2 2

√ In fact, if for instance we let M = N 2 + 2A, where A is any real number such that N 2 + 2A ≥ 0, then  2  N2 M 2A lim − = A. = lim M →∞ N →−∞ 2 2 2 N →−∞

So, this double limiting process may produce any real number as limit. Similarly, we can make this double limit equal to −∞ or ∞ or make it oscillate. (Find some limiting processes that produce these results.) N Definition 1.1.5 Suppose y = f (x) real function continuous defined in the set [a, c) ∪ (c, b] ⊂ R with a, b finite and at x = c, y = f (x) is unbounded, that is, it approaches ±∞ as x approaches c. Then we define: Z

b

Z

c

f (x) dx = a

Z

b

f (x) dx + a

f (x) dx c

where the two partial integrals have been defined in Definitions 1.1.2 and 1.1.3. Instead of [a, c) ∪ (c, b] we could have (a, c) ∪ (c, b) ⊂ R with a, b finite or infinite. Then Z

b

Z f (x) dx =

a

c

Z f (x) dx +

a

b

f (x) dx c

in where the two partial integrals have been defined in Definitions 1.1.2, 1.1.3 and 1.1.4.

Improper Riemann Integrals, Definitions, Criteria

15

Examples Example 1.1.18 Z 3

Z 1 Z 3 dx dx dx = + = 3 3 3 −2 (x − 1) −2 (x − 1) 1 (x − 1) Z ρ Z 3 dx dx lim− + lim = 3 ρ→1 σ→1+ σ (x − 1)3 −2 (x − 1) ρ 3   (x − 1)−2 (x − 1)−2 lim − + lim − = 2 2 ρ→1− σ→1+ −2 σ     −1 1 −1 1 lim− + + + lim . 2(ρ − 1)2 18 8 2(σ − 1)2 ρ→1 σ→1+

By manipulating the two limiting processes, this double limit may assume any possible value, finite or infinite. It follows that this improper integral does not exist. N Example 1.1.19 Now we examine the following integral which, we must notice, is improper at x = 1: Z 3 Z 1 Z 3 dx dx dx = + = x − 1 x − 1 x −1 −2 −2 1 Z ρ Z 3 dx dx lim + lim = − + ρ→1 σ→1 −2 x − 1 σ x−1 ρ

3

lim [ln |x − 1|]−2 + lim+ [ln |x − 1|]σ =

ρ→1−

σ→1

lim [ln |ρ − 1| − ln(3)] + lim+ [ln(2) − ln |σ − 1|] =

ρ→1− +

σ→1

+

[ln(0 ) − ln(3)] + [ln(2) − ln(0 )] = [−∞ − ln(3)] + [ln(2) − (−∞)] = −∞ + ∞ = does not exist! We could also write lim [ln |ρ − 1| − ln(3)] + lim [ln(2) − ln |σ − 1|] = σ→1+     2 |ρ − 1| + lim ln . ln 3 |σ − 1| ρ→1−

ρ→1−

σ→1+

 We can easily see that the double limit lim− ln ρ→1 σ→1+

any value as σ → 1+ and ρ → 1− independently. Z 3 dx Therefore, the improper integral −2 x − 1

|ρ − 1| |σ − 1|

 can assume

does not exist.

16

Improper Riemann Integrals

Notice that the point x = 1, at which this integral is improper, is an interior point of the interval of integration [−2, 3]. So, if we inadvertently write   Z 3 dx 2 3 = [ln |x − 1| ]−2 = ln(2) − ln(3) = ln , x − 1 3 −2 then we find a wrong answer, and we have made a bad mistake! 1 , x−1 namely F (x) = ln |x−1|, is not defined (let alone continuous) at x = 1, a point inside the interval of integration [−2, 3]. See also Example 1.1.2.] N [We must notice that the antiderivative of the function f (x) =

Important Remark and Note for Improper Integrals: As we have seen in the previous three examples, whenever the final evaluation of an improper integral (without splitting it) takes final formal form ∞ − ∞, then the improper integral does not exist. By manipulating the limiting processes, we may make it assume any possible value, finite or infinite, and so such an improper integral does not exist. This should not be confused with the limits of the indeterminate form ∞ − ∞. These limits may exist and are resolved by some mathematical manipulation and/or adjusting the well-known L’ Hˆ opital’s5 rule. Obviously, +∞ + ∞ = +∞, −∞ − ∞ = −∞, finite+∞ = +∞ and finite−∞ = −∞. Therefore, breaking an improper integral as a sum two integrals that yield one of these four forms, then this splitting is legitimate and the value of this improper integral is +∞ or −∞, accordingly. But, if in such a breaking one integral is ∞ and the other is −∞, this splitting is illegitimate (even if the improper integral does not exist finally). Otherwise, all sorts of mistakes can occur. Example 1.1.20 Similarly with the previous Example, we have: Z For ∞ ≤ a < 0 < b ≤ ∞, the integral a

b

dx does not exist. x

Verify this by work analogous to the work of the previous Example. (See also Problem 1.3.5 and compare with Example 1.4.5.) N Sometimes an integral seems to be improper, whereas it is proper. For instance, let us investigate the following two examples. 5 Guillaume Fran¸ cois Antoine Marquis de L’ Hˆ opital, French mathematician, 1661– 1704.

Improper Riemann Integrals, Definitions, Criteria Example 1.1.21 The integral Z

1

−1

17

sin(x) dx x

sin(x) is proper, even though the function f (x) = at x = 0 takes the x 0 indeterminate form . This is so because, as we know from calculus, 0 lim f (x) = lim

x→0

x→0

sin(x) = 1. x

sin(x) is bounded on the interval [−1, 1] and can be x continuously defined at x = 0, by assigning the value f (0) = 1. This integral can be evaluated, by means of power series, as a series of real numbers. By using the power series expansion of the function sin(x) we find that the power series of the function f (x) is:

Therefore, f (x) =

sin(x) f (x) = = x

P∞

n=0

2n+1

x (−1)n (2n+1)!

x

=

∞ X

(−1)n

n=0

x2n , ∀ x ∈ R. (2n + 1)!

Since we can integrate a power series, within its interval of convergence [here (−∞, ∞)], term by term, we get Z 1 sin(x) dx = x −1 " #  Z 1 X ∞ Z 1  ∞ X x2n x2n n n (−1) dx = (−1) dx = (2n + 1)! (2n + 1)! −1 n=0 n=0 −1 #1 "∞ ∞ 2n+1 X X 2 x = (−1)n . (−1)n (2n + 1)(2n + 1)! (2n + 1)(2n + 1)! n=0 n=0 −1

(See also Problem II 1.2.37.) N Important Remark: In the previous Example, we have used the fact that we can integrate power series term by term, which means Rb that we can commute the integral a with the infinite summation   P∞ Pk lim n=0 . (That is, we can switch the order of inten=0 = 0≤k→∞

gration and the limit process.) Whereas this is always legitimate with integrals of power series when the limits of integration a and b are inside their intervals of convergence, it does not hold in every situation with

18

Improper Riemann Integrals

limits of sequences or series of functions, even if the limits of integration are within the domain of definition of all functions involved. Serious mistakes may occur if such a commutation is performed while it is not valid! (For this see Section 3.3.) Example 1.1.22 As in the previous Example so the following integral Z 5 1 − cos(x) dx x2 −3 is proper. 1 − cos(x) Again, at the singular point x = 0, the function g(x) = x2 is bounded and can be continuously defined by assigning the value 1 g(0) = . This follows from the fact that under certain hypotheses we 2 0 can resolve a limit of type by using L’ Hˆ opital’s rule. Indeed, we have: 0 1 1 1 − cos(x) [1 − cos(x)]0 sin(x) = ·1= . = lim = lim 2 x→0 x→0 x→0 2x x (x2 )0 2 2

lim g(x) = lim

x→0

N Remark: Under certain necessary conditions, we can resolve limits 0 ±∞ of the types or by using L’ Hˆ opital’s rule. (Remember that not 0 ±∞ every such limit can be answered by this rule, but only those that satisfy the necessary conditions. Review this rule one more time from a calculus or mathematical analysis book!) These limits may assume any real value, or ±∞, or may not exist. If this problem arises at a point of a set over which we examine an integral and such a limit is equal to a real number, then the integral is proper with respect to this singular point. Otherwise, it is improper.

1.2

Applications

Application 1: In calculus, geometry, differential geometry and other areas we encounter the logarithmic spirals. In polar coordinates (r, θ) they are given by the formula r = aebθ , where a 6= 0 and b 6= 0 real constants.

Improper Riemann Integrals, Definitions, Criteria

19

y

15

10

5

-10

10

20

x

-5

-10

-15

-20

FIGURE 1.2: Logarithmic spiral r = a ebθ

For such a curve r = f (θ) and θ1 ≤ θ ≤ θ2 , as we learn in Calculus, the arc-length is given by s  2 Z θ2 dr L(θ1 , θ2 ) = dθ. r2 + dθ θ1 Applying this to the logarithmic spiral, we find √ Z θ2 p
1 + b2 bθ2 bθ 2 L(θ1 , θ2 ) = |a| 1 + b e dθ = |a| e − ebθ1 . b θ1 Now for b > 0 and any θ ∈ R the L(−∞, θ) is an improper integral but has finite value. Namely √ 1 + b2 bθ L(−∞, θ) = |a| e . b

20

Improper Riemann Integrals Similarly for b < 0 and θ ∈ R, we get √ 1 + b2 bθ L(θ, ∞) = |a| e . −b

Application 2: In physics we learn that the Earth creates around it a conservative gravitational field. If the mass of the Earth is M , then the force W (weight) exerted on a mass m located at distance r from the center of gravity of the Earth, by Newton’s6 law of gravitational attraction, has measure W = −G

Mm , r2

where G is the universal gravitational constant. The minus sign has the meaning that the force is directed toward the center of gravity of the Earth. The gravitational potential energy of m at a point P located at distance R from the center of gravity of the Earth O is equal to the work needed to move m from distance R to infinite distance. Then, Z ∞ E= W (r) dr. R

Since the gravitational field is conservative (i.e., this integral is independent of the path), we can evaluate E by moving on the straight line OP from R to ∞, where O is considered to be the origin. So,  ∞ Z ∞ Z ∞ − GM Mm 1 = E= W (r) dr = −G 2 dr = −GM m − m. r r R R R R The potential of the gravitational field of the Earth at any point P at distance R from O is defined to be the above energy E per unit-mass, and so it is E GM U= =− . m R

Application 3: The decaying law of a radioactive substance is m(t) = m0 ekt , where t is time, m(t) is the radioactive mass remaining after time t, 6 Sir Isaac Newton, English mathematician and physicist, one of the greatest mathematicians and scientists of all time, 1643–1727.

Improper Riemann Integrals, Definitions, Criteria

21

m0 = m(0) is the initial mass at time t = 0 and k is a negative constant representing the percentage rate of decay of the substance. The mean life of an atom of this substance is Z ∞ tekt dt. µ = −k 0

We can compute the improper integral and find that the mean life, in fact, is Z ∞ Z ∞  kt  e µ = −k tekt dt = −k td = k 0 0  kt ∞  kt ∞ Z ∞ kt 1 e e e +k =− . dt = −[0 − 0] + k −k t 2 k 0 k k k 0 0

[Notice that if k < 0, lim t ekt = 0 and lim ekt = 0.] t→∞

t→∞

Application 4: The plane curve given implicitly by 2

2

x3 + y3 = 1 has four cusps at the points {(1, 0), (0, 1), (−1, 0), (0, −1)}. It is symmetrical about either axis, the lines y = ±x and about the origin. (See Figure 1.3.) Then, by its symmetries (in the axes and the origin), its arc-length (by the well-known formula from calculus) is going to be Z

1

p

L=4

1 + (y 0 )2 dx.

0 1

By implicit differentiation, we find s p

1+

(y 0 )2

=

2

dy y3 = − 1 and so dx x3

2

x3 + y3 x

2 3

r =

1 x

2 3

=

1 x

1 3

=x

−1 3

.

So, even if the arc-length of this curve is finite, it is given by an improper integral as " 2 #1 Z 1p Z 1 −1 x3 3 L=4 = 4 · = 6. 1 + (y 0 )2 dx = 4 x 3 dx = 4 2 2 0 0 3 0

[See also Problem 3.13.52, (e).]

22

Improper Riemann Integrals y

1.0

0.5

-1.0

O

-0.5

0.5

1.0

x

-0.5

-1.0

2

2

FIGURE 1.3: Astroid x 3 + y 3 = 1

Application 5: If a company expects annual profits p(t), t years from now with interest compounded continuously at an annual interest rate r, then the present value for all future profits, also called present value of the income stream p(t), using appropriate Riemann sums, can be shown to be given by the improper integral  −rt  Z ∞ Z ∞ e −rt Present Value = e p(t)dt = p(t) d = −r 0 0  ∞ Z ∞ −rt Z e−rt e p(0) 1 ∞ −rt p(t) + dp(t) = + e dp(t). −r 0 r r r 0 0 e−r∞ = 0, which is a natural condition.] −r If we need to find the present value for a time interval 0 ≤ a ≤ b, then we compute the above integral from a to b. [We have assumed that p(∞)

Improper Riemann Integrals, Definitions, Criteria

1.3

23

Problems

1.3.1 Study the graphs of the functions f (x), F (x), F¯ (x) and Fc (x) of Example 1.1.2 by using the graphs and the information already provided and the information obtained by studying their first and second derivatives. Compare them with each other and observe the similarities and differences! 1.3.2 Prove that the function Fc (x) in Example 1.1.2, is: (a) Continuous at every x ∈ R. (b) Differentiable at every x ∈ R, by showing: At the points x = (2k + 1)π with k ∈ Z, use the definition of side derivatives and the help of L’ Hˆ opital’s rule to resolve the corresponding 1 0 limits to show Fc [(2k + 1)π] = f [(2k + 1)π] = . 3 Then show that at every x ∈ R Fc0 (x) = f (x), where f (x) is the function given in Example 1.1.2. 1.3.3 Using the transformation u = Z

∞

(a)

∞

Z f (x) dx =

f

0

Z

0

∞

(b)

  1 1 · 2 dx, f x x

∞

  1 1 f · 2 dx. x x

f (x) dx =

Z (c)

0

1

Z f (x) dx =

0

1

Z So,

∞

f (x) dx = 0

  1 1 · 2 dx, x x

1

Z

1

1 prove that x

1 2

Z 0

∞

    1 1 f (x) + f · 2 dx, x x

etc.

(These relations can be useful in some computations. For example, see Problems 2.3.23, II 1.7.135 and Example II 1.7.47, etc.)

24

Improper Riemann Integrals

1.3.4 (a) Check that for any real constant C     d 1 3x(1 − x2 ) x4 + 1 arctan 4 + C = 6 > 0, 2 dx 3 x − 4x + 1 x +1 which is a positive continuous function defined over all R. (b) Using (a), we find   1 Z 1 4 1 3x(1 − x2 ) x +1 = 0 − 0 = 0, dx = arctan 6 3 x4 − 4x2 + 1 0 0 x +1 and Z

∞

1

  ∞ x4 + 1 1 3x(1 − x2 ) dx = arctan = 0 − 0 = 0. x6 + 1 3 x4 − 4x2 + 1 1

An integral of a positive function is 0(?)! Explain what has happened. (Similar problem as in Example 1.1.2.) (c) Derive the antiderivative given in (a). [Hint: Derive and integrate the partial fractions x4 + 1 2 1 1 1 1 1 √ √ = · 2 + · + · . x6 + 1 3 x + 1 6 x2 − 3 x + 1 6 x2 + 3 x + 1 Putting the three integrated parts in one basket, by using the appropriate trigonometric formulae, derives the antiderivative given in (a), but it also introduces 5 branches of the arc-tangent over all R, two of which have domains intersecting the interval [0, 1]   1 π −π π and have a jump equal to − = . Similarly with the 3 2 2 3 interval [1, ∞]. Etc.] (d) Prove that Z ∞ 4 Z Z 1 4 x +1 π 1 ∞ x4 + 1 x +1 dx = dx = = dx, 6 x6 + 1 3 2 0 x6 + 1 1 0 x +1 and so

Z 0

∞

x4 + 1 2π dx = . 6 x +1 3

[See also Examples 3.1.6, (b) and 3.11.13 and Problem II 1.7.14, (c).]

Improper Riemann Integrals, Definitions, Criteria

25

1.3.5 Z

2

1 2 dx = [ln(|x|)]−2 = ln(2) − ln(2) = 0. x −2 But prove, this integral does not exist. (See Example 1.1.20.)

(a) Someone wrote

Explain the error that this someone made. (See also and compare with Example 1.4.5.) Z 1 ln(|x|) dx = −2 and this integral is improper. (b) Now prove that −1

Prove that an antiderivative of f (x) = ln(|x|) is   if x > 0, x ln(x) − x, F (x) =   x ln(−x) − x, if x < 0 Then, F (1) − F (−1) = −1 − 1 = −2, which is correct. Why now this antiderivative works correctly for this improper integral! 1.3.6 Let f (x) =

Z Then,

  1, if

0 ≤ x ≤ 1,

  2, if

1 < x ≤ 2.

2

f (x)dx = 3 and as an antiderivative of f (x) may be considered 0

the function F (x) =

  x,

if

0 ≤ x ≤ 1,

  2x, if

1 < x ≤ 2.

(a) But, F (2) − F (0) = 4 − 0 = 4 6= 3. Why has this happened? (b) Find an antiderivative G(x) of f (x) such that G(2) − G(0) = 3, which is the correct answer. 1.3.7 (a) If n ∈ N0 , find all continuous functions f : R −→ R, which satisfy the functional relation Z n Z f n (x) dx = f (x) dx ,

26

Improper Riemann Integrals with the arbitrary constants of the two indefinite integrations involved equal to zero. [Hint: Consider cases: n = 0, n = 1, and n ≥ 2. In the last case, Z let g(x) = f (x) dx and so g 0 (x) = f (x). Then, differentiate, find g(x), and eventually find that f (x) = c e bitrary constant.]

√ nx

n−1

, where c is an ar-

(b) What happens if one or both of the arbitrary constants of the two indefinite integrations involved are not equal to zero. (c) Examine also the case in which n is a negative integer, e.g., n = −1, n = −2, etc. 1.3.8 Project on periodic real functions In this project, without loss of generality, we consider real functions of a real variable y = f (x) : R −→ R. Such a function is called periodic if there is a real number q 6= 0 such that f (x) = f (x+q), ∀ x ∈ R. This number q is called a period of the real function y = f (x). Otherwise, y = f (x) is called non-periodic. Obviously, q = 0 satisfies this condition for every function. So q = 0 does not tell us anything about any function. We can call q = 0 trivial period for any function. Also, if y = f (x) is a constant function, then obviously any real number q ∈ R is a period of it. [Note: In integrals that we study in this text, at times, we use properties (7.) and (8.) below. With this opportunity, we try to present a more complete exposition of the periodic real functions. See also Problem 1.6.39.] 1. If functions y = f (x) and y = g(x) have a common period q and c ∈ R is a constant, then prove that the functions f + g, f − g, f c · f, f · g, , |f | and f ◦ g have q as a period. For the composition g f ◦ g we can relax one hypothesis. Which one and why? [See also items (6.) and (14.) below.] 2. For any q period of y = f (x), prove that −q and in general any kq with k ∈ Z (integer) is another period. If moreover r is any other period of y = f (x) (including the trivial q one), then q ± r is also a period, but qr and may not be periods r of y = f (x).

Improper Riemann Integrals, Definitions, Criteria

27

Also prove that for any a 6= 0 and b ∈ R, the function f (ax + b) is q periodic with a period . |a| 3. For any real numbers s 6= t, we define a so-called Dirichlet7 function f : R −→ R by   if x = rational, s, y = f (x) =   t, if x = irrational. In the literature, many times, we encounter such a function with s = 0 and t = 1 which is the characteristic function of the irrationals in R, χR−Q , or with s = 1 and t = 0 which is the characteristic function of the rationals in R, χQ . For any Dirichlet function y = f (x), prove: (a) It is nowhere continuous. (b) It is even [i.e., f (−x) = f (x)]. (c) It is periodic and any rational number r ∈ Q is a period. (d) Any irrational number w ∈ R − Q is not a period of y = f (x). 4. If y = f (x) possesses a point of continuity (i.e., there is an x0 ∈ R such that lim f (x) = f (x0 ) [= f ( lim x)] ) and a sequence of nonx→x0

x→x0

zero periods (qn 6= 0) with n ∈ N such that lim qn = 0, then n→∞

y = f (x) is identically constant. 5. If y = f (x) is periodic, non-constant and possesses a point of continuity, then it cannot have a sequence of distinct periods that converges to zero. Then prove that such a function has a minimum positive period and any other period is an integer multiple of it. I.e., if p := inf{ positive periods of y = f (x) }, then p > 0 and f (x) = f (x + p), ∀ x ∈ R. Hence, p := inf{ positive periods of y = f (x) } = min{ positive periods of y = f (x) }. Moreover, for any other period q of y = f (x), there is k ∈ Z such that q = kp. 7 Johann

Peter Gustav Lejeune Dirichlet, German mathematician, 1805–1859.

28

Improper Riemann Integrals In such a case, this number p is unique and we call it the period of the real function y = f (x). Then, the function y = f (x) is called p -periodic. Also prove that for any a = 6 0 and b ∈ R, the function f (ax + b) is p . periodic with period |a| 6. Give an example of two p -periodic functions y = f (x) and y = g(x) such that the period q of their sum f + g, or difference f − g, or f product f g and/or ratio is not p and so, by items (1.) and (5.) g above, q < p. In such a case, what are the possible values of the p ratio ? q 7. If a periodic function y = f (x) is Riemann integrable, then for any Z a+q a ∈ R and any of its periods q ∈ R the integral f (x) dx is a

fixed, that is, independent of a ∈ R. (For q = 0 this is trivially true regardless.) 8. Suppose y = f (x) is a real periodic function on R with period p > 0 which is Riemann integrable in the interval [0, p ]. Then prove that for any real numbers a < b, f (x) is Riemann integrable in the interval [a, b] and Z 1 u+p f (x) dx = f (x) dx, ∀ u ∈ R. p u 0 0   x x [Hint: For any x > 0 consider n := , the integer part of . p p # "n−1 Z Z Z x 1 x 1 X (k+1)p Then use: f (t) dt + f (t) dt = f (t) dt , x 0 x kp np 1 lim x→∞ x

Z

x

1 f (t) dt = p

Z

p

k=0

the previous result, the inequality

n n 1 ≥ > (prove it p x (n + 1)p

first), the Squeeze Lemma, etc.] 9. Let y = f (x) be a real, Riemann integrable, periodic function on R with period p > 0. Suppose that there are some numbers Z b Z ∞ 0 ≤ a < b < ∞ such that f (x) dx 6= 0. Prove that f (x) dx a

0

is either +∞, or −∞, or does not exist due to oscillation.

Improper Riemann Integrals, Definitions, Criteria

29

10. If y = f (x) is periodic and differentiable, then show that its derivative y 0 = f 0 (x) is also periodic with the same periods as y = f (x). Also prove that y 0 = f 0 (x) is zero at least one point in every interval of length greater than or equal to any positive period (in particular to its period p > 0). 11. Give anZexample of a periodic function y = f (x) whose integral x f (t) dt is not periodic. F (x) = 0

12. Give examples of periodic functions y = f (x) with two irrational a periods a and b, such that is rational. b 13. Give an example of two non-periodic functions whose composition is periodic. 14. Check if the compositions g ◦ f and f ◦ g of a periodic function f and a non-periodic function g are periodic? 15. If y = f (x) is a continuous and periodic function with an irrational period q, then prove that the set f (Z) := {f (n) | n ∈ Z} is dense in its range f (R) := {f (x) | x ∈ R}). That is, between any two different numbers in the range f (R), there is a number of the set f (Z). This is equivalent to the fact that any number in the range f (R) is the limit of a sequence in f (Z). (The latter statement is easier to prove.) 16. Using the previous result and the properties of the trigonometric functions y = cos(x) and y = sin(x), prove that the sets { cos(n) | n ∈ N} and { sin(n) | n ∈ N} are dense in the range [−1, 1]. a 17. If y = f (x) is periodic and has two periods a and b, such that is b irrational, then it has a sequence of different periods that converges to zero. In such a case, in order for y = f (x) to be non-constant [by (4.) above] it must be discontinuous everywhere. 18. Give (construct) an example of a non-constant function with two a periods a and b, such that is irrational. b 19. Read again the definition of “Riemann integrable function” and/or some criteria of “Riemann integrability” from appropriate books of Mathematical Analysis and prove that the Dirichlet function, defined in item (3.) above, restricted on any interval [a, b], with a < b real constants, is not Riemann integrable.

30

Improper Riemann Integrals

20. With the help of the previous item, give an example of a function f (x) on an interval [a, b], with a < b real constants, such that |f (x)| is Riemann integrable, but f (x) is not. [Hint: In Items (15.), (17.) and (18.) you can use the following fact due to Kronecker:8 “For any irrational number t the set Z + t · Z is dense in R. I.e., between any two different real numbers there is a number of the form k + t · l, with k and l integers.” You may provide a proof of this fact, but if you cannot, just use it readily in the above items.] 1.3.9 Project on the modified Dirichlet function. Part I: For any rational number r ∈ Q we consider two integers p ∈ Z and q ∈ N (q ≥ 1), such that: p and q have no common factors p except the trivial 1 [i.e., gcd(p, q) = 1] and r = . For any integer m q m (including m = 0) we have m = and so p = m and q = 1. We call such 1 a representation of the rational number r reduced representation. With this in mind we define the so-called modified Dirichlet or Riemann function g : R −→ R by:  1 p    q , if x = q , rational in reduced representation, y = g(x) =    0, if x = irrational. Now prove: 1. ∀ x ∈ R, 0 ≤ g(x) ≤ 1,

g −1 ({0}) = R − Q

and g −1 ({1}) = Z.

2. y = g(x) is even [i.e., g(−x) = g(x)] and also even about any k ∈ Z 1 and any k + with k ∈ Z. 2 3. ∀ w ∈ R, lim g(x) = 0. x→w x6=w

4. y = g(x) is continuous at every x irrational and discontinuous at every x rational.9 5. y = g(x) is periodic and the set of its periods is exactly Z. 6. y = g(x) is nowhere differentiable. 8 Leopold

Kronecker, German mathematician, 1823–1891. is an example of a real function which is continuous at exactly the irrational numbers. It can be proved that there is no real function which is continuous at exactly the rational numbers. Besides the manes of Dirichlet and Riemann, the name of Carl Johannes Thomae (German mathematician, 1840–1821) is also associated with this function. 9 This

Improper Riemann Integrals, Definitions, Criteria

31

Part II: We consider the function y = h(x) to be the restriction of y = g(x) on the closed interval [0, 1]. (In general we could consider any interval [a, b], where −∞ < a < b < ∞, but we use [0, 1] without loss of generality.) I.e., h : [0, 1] −→ R is defined by:  p 1    q , if x = q rational in reduced representation in [0, 1], y = h(x) =    0, if x = irrational in [0, 1]. Then: Z

1

1. Prove that y = h(x) is Riemann integrable and

h(x)dx = 0. 0

2. Define u : [0, 1] −→ R by

y = u(x) =

  1,

if 0 < x ≤ 1,

  0,

if x = 0. Z

Prove that y = u(x) is Riemann integrable and

1

u(x)dx = 1. 0

3. Prove that

y = (u ◦ h)(x) = χ[0,1]∩Q (x) =

  1,

if x = rational in [0, 1],

  0,

if x = irrational in [0, 1].

Then prove that the composition of these two Riemann integrable functions, u ◦ h = χ[0,1]∩Q , is not a Riemann integrable function. 4. However, u ◦ h = χ[0,1]∩Q is the point-wise limit of a sequence of Riemann integrable functions, defined as follows: Let [0, 1] ∩ Q = {r1 , r2 , r3 , . . . , } be an enumeration of the rational numbers in [0, 1]. Then ∀ n ∈ N define   if x ∈ {r1 , r2 , . . . , rn }, 1, y = vn (x) =   0, if x ∈ [0, 1] − {r1 , r2 , . . . , rn }. Now prove that: ∀ n ∈ N, y = vn (x) is a Riemann integrable

32

Improper Riemann Integrals Z function with lim vn (x) = function. n→∞

1

vn (x)dx = 0, but the point-wise limit 0 χ[0,1]∩Q (x), ∀ x ∈ [0, 1] is not a Riemann

5. Finally, prove that: lim

h

m→∞

integrable

i lim cos2n (m!πx) = χ[0,1]∩Q (x).

n→∞

That is, χ[0,1]∩Q (x) is an iterated limit of a double limit process of bounded continuous functions. (See also Theorem 3.3.12 and Examples 3.3.19 and 3.3.20.)

1.4

Cauchy Principal Value

In some cases we can define the so-called Cauchy10 principal value or simply principal value of an improper integral. This is a certain symmetrical limit and it is defined in the following four situations: Definition 1.4.1 If the integral is improper simply because the set of integration is R = (−∞, ∞), then we define its principal value to be: Z

∞

P.V.

Z

def

R

f (x) dx = lim

R→∞

−∞

f (x) dx. −R

Definition 1.4.2 If the set of integration is [a, c)∪(c, b], where a < c < b finite real numbers, and the integral becomes improper at c, then we define its principal value to be: "Z # Z Z b

P.V. a

c−

def

f (x) dx = lim+ →0

b

f (x) dx + a

f (x) dx . c+

Definition 1.4.3 If both situations of the previous two definitions occur, i.e., we have improper integrals over (−∞, c) ∪ (c, ∞), with c ∈ R, then we combine the two definitions and we define the principal value of this improper integral to be: Z

∞

P.V.

c−

f (x) dx = lim ∞

10 Augustin

"Z

def

→0+ R→∞

Z

R

f (x) dx + −R

# f (x) dx .

c+

Louis Cauchy, French mathematician, 1789–1857.

Improper Riemann Integrals, Definitions, Criteria

33

Definition 1.4.4 If the set of integration is the finite open interval (a, b) (a < b are finite real numbers), and the integral is improper just because the interval is open at both endpoints, then we define the principal value of this improper integral to be: b

Z P.V. a

Z

def

f (x) dx = lim+ →0

b−

f (x) dx. a+

Again we see that the principal values are obtained by symmetrical limiting processes and therefore are special. However, they turn out to be very useful in mathematics and applications. We will see applications of the principal value in many sections that follow.

Examples Example 1.4.1 Z P.V.

∞

Z

R

x dx = lim

R→∞

−∞

 lim

R→∞

 x dx = lim

R→∞

−R

x2 2

R = −R

 R2 (−R)2 − = lim 0 = 0. R→∞ 2 2 N

Example 1.4.2 Z ∞ Z 2 P.V. x dx = lim R→∞

−∞

 lim

R→∞

R

−R



2

x dx = lim

R→∞

x3 3

R = −R

 R3 (−R)3 2R3 − = lim = ∞. R→∞ 3 3 3 N

Example 1.4.3  Z 1− Z 3 Z 3 dx dx dx P.V. = = lim + 3 3 (x − 1)3 →0+ −2 (x − 1) −2 1+ (x − 1) ( 1−  3 ) −1 −1 lim + = 2(x − 1)2 −2 2(x − 1)2 1+ →0+   1 1 1 1 −5 1 1 lim − 2 + − + − = . = 2 18 8 22 18 8 72 →0+ N

34

Improper Riemann Integrals

Example 1.4.4 Z

∞

P.V. −∞

lim [arctan(x)]R −R R→∞

dx = +1

x2

= lim [arctan(R) − arctan(−R)] = R→∞

(since both of the two partial limits exist) π  π = π. lim arctan(R) − lim arctan(−R) = − − R→∞ R→∞ 2 2 N Example 1.4.5 (a) As we have seen in Example 1.1.20 the integral Z 1 dx does not exist, but −1 x Z −  Z 1 dx dx dx = lim + = x →0+ −1 x  −1 x n o − 1 lim+ [ln(|x|)]−1 + [ln(|x|)] =

Z P.V.

1

→0

lim+ [ln() − ln(1) + ln(1) − ln()] = lim+ 0 = 0.

→0

→0

Z

3

(b) Similarly −2

dx does not exist, but x

Z −  Z 3 o n dx dx dx − 3 = lim + = lim [ln(|x|)]−2 + [ln(|x|)] x →0+ →0+ −2 x −2 x    3 = lim+ [ln() − ln(2) + ln(3) − ln()] = lim+ [− ln(2) + ln(3)] = ln . 2 →0 →0 Z ∞ dx (c) Also does not exist, but −∞ x ! Z − Z ∞ Z R dx dx dx = lim + P.V. = x →0+ −R x  −∞ x R→∞ n o − R lim+ [ln(|x|)]−R + [ln(|x|)] =

Z P.V.

3

→0 R→∞

lim [ln() − ln(R) + ln(R) − ln()] = lim+ 0 = 0.

→0+ R→∞

(See also Problem 1.3.5.)

→0 R→∞

N

Improper Riemann Integrals, Definitions, Criteria

35

Example 1.4.6 By Example 1.1.11, we obtain   Z 1   1 1 1 P.V. sin − cos dx = x x x −1      Z − 1 1 1 lim sin − cos dx + + x x x →0 −1   Z 1   1 1 1 sin − cos dx = − sin(1) + sin(1) = 0. lim+ x x x →0  N By the above definitions and examples we conclude the following: (a) If the improper integral exists, then all limiting processes give the same answer which is the value of the improper integral, and so the principal value also exists and it is equal to the improper integral. (b) If the improper integral does not exist, then its principal value may or may not exist. (c) If the principal value does not exist, then the improper integral does not exist either, since the principal value is one of the limiting processes. Thus, the principal value of an improper integral constitutes a proper generalization of the improper integral. When we know ´a-priori that the improper integral exists, we can evaluate it by just computing its principal value, especially when the computation of this symmetric limit is easier than any other way. The following definitions and immediate results are also useful: (a) If y = f (x) is an odd function in R, i.e., by definition ∀ x ∈ R, f (−x) = −f (x) (its graph is symmetrical about the origin), then Z ∞ Z R P.V. f (x) dx = lim f (x) dx = 0. −∞

R→∞

−R

(b) If y = f (x) is an even function in R, i.e., by definition ∀ x ∈ R, f (−x) = f (x) (its graph is symmetrical in the y-axis), then Z ∞ Z R P.V. f (x) dx = lim f (x) dx = −∞ Z R

2 lim

R→∞

R→∞

−R

Z

0

f (x) dx = 2 lim 0

R→∞

f (x) dx. −R

36

Improper Riemann Integrals In general, a function y = f (x), where f : R −→ R, is odd about a point c ∈ R, if by definition ∀ u ∈ R, f (c − u) = −f (c + u), or ∀ x ∈ R, f (2c − x) = −f (x). Also, a function y = f (x), where f : R −→ R, is even about a point c ∈ R, if by definition ∀ u ∈ R, f (c − u) = f (c + u), or ∀ x ∈ R, f (2c − x) = f (x). Now in (c) and (d) below, we consider a function y = f (x) defined in (−∞, c) ∪ (c, ∞), with c ∈ R. Then we have:

(c) If y = f (x) is an odd function about c, that is, ∀ x 6= c, f (2c − x) = −f (x) [even if f (x) is not defined at x = c], then Z

∞

P.V.

Z

c−

f (x) dx = lim

→0+ R→∞

−∞

Z

R

f (x) dx + lim

→0+ R→∞

−R

f (x) dx = 0. c+

(d) If y = f (x) is an even function about c, that is, ∀ x 6= c, f (2c − x) = f (x) [even if f (x) is not defined at x = c], then Z

∞

P.V. −∞

Z f (x) dx = lim+

→0 R→∞

Z

−R

Z f (x) dx + lim+

R

2 lim

→0+ R→∞

c−

Z

→0 R→∞ c−

f (x) dx = 2 lim c+

→0+ R→∞

R

f (x) dx = c+

f (x) dx. −R

(e) Rule of translate or shift: Consider a function f : [a, a + r] −→ R,

with r > 0.

The translate or shift of y = f (x) at the interval [b, b + r] is given by y = f (x − b + a),

with b ≤ x ≤ b + r.

(f) Any function y = f (x), defined on a set symmetrical about x = 0, is written as the sum f (x) = fe (x) + fo (x), where fe (x) = f (x) − f (−x) f (x) + f (−x) is even and fo (x) = is odd (check!). 2 2 This decomposition into a sum of an even and an odd function is unique. fe is called the even part of f and fo the odd part of f . (See Problem 1.6.33.)

Improper Riemann Integrals, Definitions, Criteria

1.5

37

A Note on the Integration by Substitution

We must say a few words on the application of a substitution with proper or improper integrals. When applied casually, it may create mathematical and computational errors reminiscent to Example 1.1.2. In a calculus course, it is stated for indefinite and definite integral in the following way: Theorem 1.5.1 (Indefinite Integrals and u-Substitution.) Let f (u) be a continuous function with antiderivative F (u) and u = u(x) a function with continuous derivative. Then Z Z 0 f [u(x)]u (x) dx = f (u) du = F (u) + C = F [u(x)] + C, where C is an arbitrary real constant. [So, when we find F (u) + C, the indefinite integral with respect to u, we substitute u = u(x) into it to find the answer to the first integral.] Theorem 1.5.2 (Definite Integrals and u-Substitution.) Let u = u(x) be a function with continuous derivative on an interval [a, b] and let f (u) be a continuous function on u([a, b]), the range of u. Then Z

b

f [u(x)]u0 (x) dx =

Z

a

u(b)

f (u) du. u(a)

So, in applying the method of integration by substitution, we actually want to integrate a function g(x) which can be recognised as g(x) = f [u(x)]u0 (x). Then, if we apply the above two Theorems to g(x), we respectively obtain Z Z g(x) dx = f (u) du u=u(x) , and Z

b

Z

u(b)

g(x) dx = a

f (u) du. u(a)

For computational purposes, the substitution must make the righthand side integral easier than the left-hand side one, but in theoretical considerations this does not matter. Here is a simple example:

38

Improper Riemann Integrals


Example 1.5.1 We consider the function y := g(x) = 2x sin x2 + 3 , with −1 ≤ x ≤ 1. We see that g(x) = f [u(x)]u0 (x) with u(x) = x2 + 3 and f (u) = sin(u), both are continuously differentiable in R. So, Z 1 Z u(1) Z 4
2x sin x2 + 3 dx = sin(u) du = sin(u) du = 0, −1

u(−1)

4

as the limits of integration coincide. This agrees with the fact that g(x) is odd and so its integral is zero over any interval symmetrical about the origin. It also agrees with the Fundamental Theorem of Calculus,
1.1.1, with G(x) = − cos x2 + 3 a continuous antiderivative of g(x), Z 1

1 2x sin x2 + 3 dx = − cos x2 + 3 −1 = − cos(4) + cos(4) = 0. −1

Z

11π 2

sin(x) cos(x) dx

Similarly

f (u)=u=sin(x)

Z

=

0

−1

 u du =

0

u2 2

−1 = 0

1 . 2

This agrees with the computation through antidifferentiation   11π Z 11π Z 11π 2 2 sin(2x) − cos(2x) 2 1 sin(x) cos(x) dx = dx = = . 2 4 2 0 0 0 N But, most of the times, especially when g(x) cannot be directly recognised as f [u(x)]u0 (x), we apply the substitution method on integrals in the following alternative way. Without loss of generality, we consider the proper integral Z b I := g(x) dx a

of a Riemann integrable function g : [a, b] −→ R, usually continuous. We choose an appropriate u = u(x), usually continuously differentiable, so that du = u0 (x) dx with u0 (x) continuous and therefore Riemann integrable. Then, we make the direct substitution x = x(u) in the integral I, i.e, we use the inverse function of u = u(x). If x0 = x0 (u) exists and is continuous, then dx = x0 (u) du, we replace every x in I in terms of u and we write Z u(b) f (u) du, where f (u) = g[x(u)]x0 (u). I := u(a)

Now, we expose some problems that may arise when we casually apply the substitution method on definite (proper or improper) integrals in this alternative way, by examining the following simple examples.

Improper Riemann Integrals, Definitions, Criteria

39
Example 1.5.2 We consider the function y := g(x) = sin x2 + 3 , with −1 ≤ x ≤ 1. Its integral Z 1
sin x2 + 3 dx J := −1

can be estimated and J 6= 0. We pick as u = u(x) = x2 + 3 on [−1, 1] and so du = 2x dx, du du = √ , u(−1) = (−1)2 + 3 = 4 and u(1) = 12 + 3 = 4. dx = 2x 2 u−3 So, if we apply the last formula, J becomes Z 4 du J := sin(u) √ = 0, 2 u−3 4 in view of the fact that the limits of integration are equal. This result is obviously wrong (since J 6= 0) and we must explain why this mistake has happened. First of all, the range of u = u(x) is [3, 4] and not [4, 4]. Also, when du du we write dx = = √ , we tacitly use as inverse function of 2x 2 u−3 √ u = u(x) is the function x = x(u) = u−1 (u) = u − 3. This is defined on the range of u, [3, 4], but covers only [0, 1] in the domain of x, i.e., 1 half of the domain of x, [−1, 1]. Moreover, x0 (u)|u=3 = √ 2 u − 3 u=3 does not exists, contrary√to the hypothesis of the Theorem. So, with x = x(u) = u − 3 on [3, 4], we must write Z

1 2




Z

sin x + 3 dx = 0

3

4

du sin(u) √ . 2 u−3

sin(u) [Notice that lim+ √ = +∞, and the proper integral above beu→3 2 u − 3 comes improper, but the latter is also positive finite, since the power of 1 u − 3 in the denominator is .] 2 The other part √ of the integral J, on [−1, 0], is found if we use as x = x(u) = − u − 3, i.e., the other branch of the inverse of u = u(x) on [3, 4]. Then, u = x2 + 3 again and Z

0 2




Z

sin x + 3 dx = −1

4

3

du sin(u) √ = −2 u − 3

The addition of these two parts fixes the error.

Z 3

4

du sin(u) √ . 2 u−3

40

Improper Riemann Integrals

Another way to look at this integral that avoids this problem, given that f (x) is an even function, i.e., f (−x) = f (x), is the following: Z 1 Z 1

2 J := sin x + 3 dx = 2 sin x2 + 3 dx = −1

Z 2 3

4

0

Similarly, Z 1 Z
sin x2 + 3 dx = −1 2

4

Z

du sin(u) √ = 2 u−3

3

du sin(u) √ , u−3

0


sin x2 + 3 dx + −1 2

1 2

Z 2

which is correct.

Z

1


sin x2 + 3 dx =

0 1

Z

sin x2 + 3 dx +



sin x2 + 3 dx. 1 2

0

√ 2 In the final form, we can use u = x + 3 with x = + u − 3 and √ −1 u ,1 = [3, 4], etc. Otherwise, we must use x = − u − 3, when 2 √ −1 ≤ x ≤ 0, and x = + u − 3, when 0 ≤ x ≤ 1. 2 N Example 1.5.3 We would like to test the alternative method with the basic integral Z π sin(x) dx = 2, 0

by making the substitution u = sin(x). If we write Z π Z 0 u du √ = 0, sin(x)dx = ± 1 − u2 0 0 then this is obviously wrong. h π πi Because the inverse of sine maps [−1, 1] onto − , , we must write 2 2 Z π Z π2 Z π sin(x)dx + sin(x)dx = sin(x)dx =

0

1

u du √ + + 1 − u2

π 2

0

0

Z

Z 1

0

u du √ =2 − 1 − u2

Z 0

1


1 u du √ = −2 1 − u2 0 = 2, 1 − u2

which is the correct answer! N (Study also how the u-substitution is stated, treated and used in mathematical analysis books, e.g., Apostol 1974, Rudin 1976, etc., and some good calculus textbooks.)

Improper Riemann Integrals, Definitions, Criteria

1.6

41

Problems

1.6.1 Give all the reasons as to why the following six integrals are improper: Z ∞ Z ∞ sin(x) I1 = ln(x) dx, I2 = dx, x 0 −∞ Z ∞ Z ∞ cos(x) 1 I3 = dx, I4 = dx, x 2x −1 0 0 Z ∞ Z ∞ sin(5x) dx, I = xn e−x dx, n ∈ Z. I5 = 6 2x − 1 e 0 0 In Problems 1.6.2–1.6.25 compute the given improper integrals. (Prove that they are equal to −∞, or +∞, or the provided value and/or you find their values. In some of these problems you have to distinguish different cases depending on the values of the constants/parameters involved.) 1.6.2 Z ∞ e

−µx

∞

Z dx,

e

0

√ −µ x

Z

∞

dx,

−µ

e

0

√ 3 x

Z

∞

dx, and −∞

0

where µ ∈ R is a constant. 1.6.3 Z

∞

Z

α

1

x ln(x) dx and 1

xα ln(x) dx,

0

where α ∈ R is a constant. 1.6.4

Z

∞

−∞

1.6.5 Z 0

1.6.6

9

|x| dx x2 + 1

Z and

∞

P.V. −∞

Z

x dx (x − 3)2

and

P.V. 0

Z

∞

−∞

2

|x|e−x dx.

9

|x| dx. x2 + 1

x dx. (x − 3)2

e−µ

√ 3 x

dx,

42

Improper Riemann Integrals

1.6.7 Z ∞ a

dx xp

∞

Z and a

dx p, x [ln(x)]

where a and p are real constants.

(Investigate the possible cases depending on the values of a and p, as −∞ ≤ a < ∞ and ∞ < p < ∞.) 1.6.8

π 2

Z

Z tan(x) dx

and

tan(x) dx.

−π 2

1.6.9

−π 2

∞

Z 0

1.6.10

∞

Z

π 2

P.V.

1

dx (x − 2)3

1.6.11

Z

∞

3

1.6.12 Z

∞

dx . (x − 1)3

Z

∞

Z

dx (x + 1)3

and 0

and

P.V. 1

dx 1 = ln x2 + x − 2 3

dx . (x − 2)3

  5 . 2

1

xp−1 dx,

where p is a real constant.

0

(Investigate cases depending on the values of p, as −∞ < p < ∞.) 1.6.13 Z ∞ e−αx sin(βx) dx = 0

β , α2 + β 2

where α > 0 and β ∈ R constants.

(See also Problem II 1.7.58.) 1.6.14 Z ∞

e−αx sin(βx) dx,

where α > 0, β ∈ R and a ∈ R constants.

a

(See also Problem II 1.7.58.)

Improper Riemann Integrals, Definitions, Criteria 1.6.15 Z ∞ e−αx cos(βx) dx = 0

α , α2 + β 2

where α > 0 and β ∈ R constants.

(See also Problem II 1.7.58.) 1.6.16 Z ∞

e−αx cos(βx) dx,

where α > 0, β ∈ R and a ∈ R constants.

a

(See also Problem II 1.7.58.) 1.6.17 If α > 0 and −α < β < α constants, then Z ∞ β e−αx sinh(βx) dx = 2 . α − β2 0 1.6.18 If α > 0 and −α < β < α constants, then Z ∞ α . e−αx cosh(βx) dx = 2 α − β2 0 1.6.19 Z

∞

e−x cos(x) dx

43

Z

∞

and

−∞

e−x sin(x) dx.

−∞

1.6.20 Let a ≥ 0, b ≥ −1 and c > 0 be real constants. Prove:  +∞, if 0 ≤ a ≤ 1,    R ∞ −cx (a) 0 a dx =  1   , if a > 1. c · ln(a)  +∞, if 0 ≤ a ≤ 1,    R∞ 1 (b) 0 acx +b dx =    ln(1 + b) , if a > 1. bc · ln(a) (See also a case in Example II 1.7.26.) [Hint: In (b) use u = acx , when a > 1.]

44

Improper Riemann Integrals

1.6.21 If c > |b| real constants, prove Z ∞ dx π =√ . 2 + 2bx + c2 2 − b2 x c −∞ [See also Problem II 1.7.45, (a), (b), (c).] 1.6.22 If a > 0 constant, find the integrals Z a Z ∞ (a) ln(x) dx and ln(x) dx, 0

Z (b)

a a

Z

ln2 (x) dx

∞

and

0

ln2 (x) dx,

a

Z

1

1.6.23 Show that the improper integral −1

ex dx does not exist, −1

ex

but its principal value is 1. 1.6.24 By making two successive appropriate u-substitutions, prove that Z ∞ 1 dx 2 = ln [ln(3)] . x ln(x) [ ln [ln(x)] ] 3

1.6.25 By making two successive appropriate u-substitutions, prove that Z ∞ dx = ∞. x ln(x) ln [ln(x)] 3 Z

1

1.6.26 Show that the integral −1

ex − 1 dx is proper and find its value x

as a series of real numbers. Z 2 Z 1 Z 1 ln(x + 1) ln(x) ln(x) dx = dx = − dx. 1.6.27 Show that: x − 1 x x +1 0 0 1 1.6.28 If a > 0 and b ≥ 0 use the exponential power series to prove Z 0

1

b a(x ) dx =

∞ X [ln(a)]n . n!(bn + 1) n=0

Investigate this integral when: (1) b ∈ R and a = 0 or a = 1. (2) b < 0 and 0 < a < 1 or a > 1.

Improper Riemann Integrals, Definitions, Criteria

45

1.6.29 Use the power series of cos(x) to compute the following integrals as series of real numbers: Z 2 Z 5 1 − cos(x) 1 − cos(x) dx and dx. 2 x x2 −2 −3 1.6.30 Prove that Z ∀ a ∈ R, 0

a

2 ∞ X 1 − e−x (−1)n a2n+1 dx = . x2 (2n + 1) · (n + 1)! n=0

1.6.31 (a) Prove that with compositions of real functions in R we have the rules: f ◦ even = even, where f is any real function in R. even ◦ odd = even, odd ◦ odd = odd. (b) Prove that with multiplications of real functions in R we have the rules: even · even = even, even · odd = odd, odd · even = odd, odd · odd = even. (c) Prove that with addition and subtraction of real functions in R we have the rules: even ± even = even, odd ± odd = odd. 1 11 2 1.6.32 Show that the polynomial f (x) = x4 − 2x3 + x − 6x is a 4 2 even function about c = 2. 1.6.33 Consider a function f : R −→ R. Prove: (a) If y = f (x) is odd (about zero), then f (0) = 0. (Similarly, if y = f (x) is odd about c ∈ R, then f (c) = 0.) (b) If y = f (x) is both odd and even, then it is identically zero. (c) y = f (x) is odd about every c ∈ R, if and only if, y = f (x) is identically zero. (d) y = f (x) is even about every c ∈ R, if and only if, y = f (x) is identically constant. (e) The decomposition of f into even and odd parts is unique. (f) Give examples of real odd and / or even functions that are not defined at x = 0. 1.6.34 Consider a function f : R −→ R. Prove: (a) If y = f (x) is differentiable and odd, then f 0 (x) is even.

46

Improper Riemann Integrals

(b) If y = f (x) is differentiable and even, then f 0 (x) is odd. Z x f (t)dt is even. (c) If y = f (x) is integrable and odd, then F (x) = 0

Z

x

f (t)dt is odd.

(d) If y = f (x) is integrable and even, then F (x) = 0

(e) The inverse function of an odd function (if it exists) is odd. (What happens with an even function?) 1.6.35 Consider a function f : R −→ R. Then prove that any two of the following statements imply the third one: (a) f (x) is odd about x = 0, i.e., f (−x) = −f (x). (b) f (x) is odd about x = c, i.e., f (2c − x) = −f (x) or f (c − u) = −f (c + u). (c) f (x) is 2c−periodic, i.e., f (x + 2c) = f (x). 1.6.36 Consider a function f : R −→ R. Then prove that any two of the following statements imply the third one: (a) f (x) is even about x = 0, i.e., f (−x) = f (x). (b) f (x) is even about x = c, i.e., f (2c − x) = f (x) or f (c − u) = f (c + u). (c) f (x) is 2c−periodic, i.e., f (x + 2c) = f (x). 1.6.37 Consider a function f : R −→ R. Then prove: (a) If f (x) is odd about x = 0 and even about x = c, then f (x) is 4c−periodic, i.e., f (x + 4c) = f (x). (b) If f (x) is even about x = 0 and odd about x = c, then f (x) is 4c−periodic, i.e., f (x + 4c) = f (x). (c) Notice that here we cannot have results similar to the results of the two previous Problems. Why? 1.6.38 Consider any c > 0 and let f (x) = x(c − x), for 0 ≤ x ≤ c. Show that the extension of this function all over (−∞, ∞), such that the extended function is odd about both x = 0 and x = c, is given by: ∀ n ∈ Z, if nl ≤ x ≤ (n + 1)c, then f (x) = (−1)n (x − nc) [(n + 1)c − x] .

Improper Riemann Integrals, Definitions, Criteria

47

[Hint: Extend the function to [−c, 0] as odd and then use Problem 1.6.35 and the Rule of shift, (e), above, etc.] 1.6.39 For any a ∈ R and p > 0, consider any real function y = f (x) defined in the interval [a, a + p) or (a, a + p]. (a) Extend this function to the whole R periodically with a period equal to p. (b) Give an example in which the period of the extended function in (a) is less than p. (c) If we consider f (x) defined on the closed interval [a, a + p], then give an example of a function f (x) which cannot be extended as a periodic function. (d) Under what condition a real function f (x) defined on the closed interval [a, a + p] can be extended as a periodic function to the whole R and with a period equal to p? (e) If g is a periodic function with period p > 0, then for any function f the composition f ◦ g is also periodic with period p. But, the composition g ◦ f may or may not be periodic. (Give examples and counterexamples.) 1.6.40 Let f be a Riemann integrable function in R and a > 0. Prove the following five relations:
Ra (a) −a x · f x2 dx = 0.

Ra Ra (b) −a f x2 dx = 2 0 f x2 dx. Rπ Rπ f [cos(x)] dx = 02 f [sin(x)] dx = 12 0 f [sin(x)] dx.  R mπ  2  Rπ  (d) ∀ m ∈ N, f cos (x) dx = m 0 f cos2 (x) dx. 0  R mπ  2  Rπ  f sin (x) dx = m 0 f sin2 (x) dx. (e) ∀ m ∈ N, 0 (c)

1.7

R

π 2

0

Some Criteria of Existence

We have defined the improper integrals as certain limits. These limits may or may not exist. When such a limit exists, we say that the

48

Improper Riemann Integrals

improper integral exists or it is convergent. If the limit does not exist, then we say that the improper integral does not exist or it is divergent. In the previous definitions, for more generality, the real value function y = f (x) was considered to be piecewise continuous rather than continuous. (In the most general theory of integration developed in advanced real analysis, we deal with more general integrals of a “very large” class of functions, the class of the measurable functions. We study these in an advanced course of real analysis.) Necessary and sufficient conditions for the existence of improper integrals are developed in advanced calculus, mathematical analysis and real analysis. Most of these are beyond the scope of this book. So, we will content ourselves with the few criteria stated in this section, some of which are reminiscent to criteria for the convergence of infinite series in calculus. These criteria are sufficient and powerful enough to give answers about existence or non-existence (convergence or non-convergence) questions for almost all the interesting improper integrals of mathematics and scientific applications at this primary level. Definition 1.7.1 Non-standard Definition: In this book, we shall call a function to be a “nice function” if it is piecewise continuous in its domain of definition with finitely many discontinuities each of which is of the following four types: 1. Removable discontinuity. 2. Essential jump discontinuity with finite or infinite jump. 3. Essential discontinuity, such that the limit of the function, as x approaches the point of discontinuity, is ±∞. 4. Essential discontinuity, such that the limit of the function, as x approaches the point of discontinuity, does not exist and is not ±∞, i.e., the function oscillates without limit. (In this case, for the purposes of the related material, we may assume some extra condition that the function must satisfy in some interval containing the point of such an essential discontinuity, e.g., bounded, etc.) The continuous functions are of course a subset of this set of nice functions since they have zero discontinuities in their domain. We use this non-standard term of “nice functions” for short, so that we do not have to repeat these conditions whenever we need them throughout this book. So, from now on we must remember what we mean by this nonstandard term of “nice function” whenever we refer to it.

Improper Riemann Integrals, Definitions, Criteria

49

When a jump discontinuity has infinity jump or the limit of the function at the point of the discontinuity is ±∞, then the function is unbounded. In general, the discontinuity is essential if the limit of the function at the point of the discontinuity does not exist. Also the domain of definition of such a function is going to be denoted by a capital letter like A, where A ⊆ R is any set that we have already encountered in the definitions of the previous two sections and/or any nice set that we have already dealt with in an undergraduate calculus course (e.g., a bounded closed interval, a finite union of bounded closed intervals, etc.). Theorem 1.7.1 (Comparison Test with Non-negative Functions) Let f and g be two nice functions defined in a set A ⊆ R (as we have indicated in the previous paragraph) and satisfying the inequality 0 ≤ f (x) ≤ g(x). Then we have: Z Z (a) If g(x) dx exists, then f (x) dx exists. A

A

(For a non-negative function its “integral exists” means that it assumes a finite non-negative value.) In this case we have the inequality Z Z 0≤ f (x) dx ≤ g(x) dx < ∞. A

A

Z (b) If

Z f (x) dx does not exist, then

A

g(x) dx does not exist. A

[For a non-negative function its “integral does not exist” means that the integral is infinite, Zi.e., it has value +∞. So, in this case Z f (x) dx = ∞ =

we have A

g(x) dx.] A

Proof The proof of this criterion is rather obvious, since for any closed interval [p, q] ⊆ A and for any nice function satisfying the inequality 0 ≤ f (x) ≤ g(x), by basic calculus we have Z q Z q 0≤ f (x) dx ≤ g(x) dx p

p

and the limiting processes preserve the ≤ inequalities. Then, to prove claim (a) and claim (b) of the Theorem, we respectively use the facts Z Z g(x) dx < ∞ and f (x) dx = ∞. A

A



50

Improper Riemann Integrals

Note: This criterion is reminiscent of the Comparison Test for convergence of non-negative series, in calculus and mathematical analysis. Remark: Another result reminiscent to the convergence of monotonic and bounded sequences is the following: Suppose −∞ ≤ a < b ≤ ∞ and f : (a, b) −→ R is a non-negative function such that for any two real numbers σ < ρ in (a, b) we have, Z ρ f (x)dx < M, σ

for some fixed real constant M ≥ 0. Then, the improper integral of f (x) over (a, b) exists as a nonnegative real value ≤ M , i.e., b

Z a

Z f (x)dx = lim−

ρ→b σ→a+

ρ

f (x)dx ≤ M. σ

This is so because the above limit is, non-decreasing as ρ → b− and/or σ → a+ and is bounded above. (Otherwise, the integral is unbounded and is +∞.) Analogously, if a function is non-positive, then its integral either exists as a non-positive real value or is −∞.

Examples Example 1.7.1 Prove that Z ∞

2

e−x dx exists.

0

(In other words, it is convergent or equals a finite value.) 2 Consider the continuous function f (x) = e−x on [0, ∞) and define  −x2  , for 0 ≤ x ≤ 1, f (x) = e g(x) =   −x e , for 1 ≤ x < ∞. Then 0 < f (x) ≤ g(x), ∀ x ∈ [0, ∞) and Z

∞

Z g(x) dx =

0

0

1

2

e−x dx +

Z 1

∞

e−x dx =

Z 0

1

 ∞ 2 e−x dx + −e−x 1 =

(finite value) + [0 − (−e−1 )] = (finite value).

Improper Riemann Integrals, Definitions, Criteria

51

Therefore, Z

∞

∞

Z

Z

∞

g(x) dx < ∞ 0

0

0

Z

2

e−x dx
0 and if k ≤ x1 ≤ x2 < ∞ then f (x1 ) ≥ f (x2 ).] We let an = f (n) for n = k, k + 1, k + 2, .... Then Z ∞ f (x) dx converges (diverges) k

if and only if

∞ X

an

converges (diverges).

n=k

(The proof of this criterion can be found in any good calculus or mathematical analysis book.) Z ∞Note: For positive functions and positive series, respectively, the P∞ f (x) dx and n=k an diverges means it is equal to ∞. k

Remark: Whereas in calculus the Integral Test is mainly used to check the convergence or divergence of a positive series that satisfies the hypotheses of this criterion, here we use it in the converse way to check the convergence or divergence of an improper Riemann integral under these hypotheses. So we need to prove that the respective positive series converges or diverges. To this end we employ any different criterion that gives an answer for the series, among all those someone can find in books of advanced calculus or real analysis. Review these criteria one more time. For instance, we remark that under the conditions of the Integral Test the following three criteria are often very convenient. Theorem 1.7.3 (Cauchy Posit. Series Condensation Theorem) Suppose that a1 ≥ a2 ≥ a3 ≥ ... ≥ 0 is a decreasing sequence of nonnegative numbers. Then ∞ X

an converges (diverges)

if and only if

n=1 ∞ X

2k a2k = a1 + 2a2 + 4a4 + 8a8 + ... converges (diverges).

k=0

Theorem 1.7.4 (Absolute Root Test. (Cauchy)) Consider a se∞ X ries of real numbers an . Suppose the following limit exists or is ∞ n=k

0 ≤ lim

n→∞

p n

|an | = ρ ≤ ∞.

Improper Riemann Integrals, Definitions, Criteria

53

Then: (1) If 0 ≤ ρ < 1, the series

∞ X

an converges absolutely and therefore

n=k

it converges. (2) If 1 < ρ ≤ ∞, the series

∞ X

an diverges.

n=k

(3) If ρ = 1, the test is inconclusive. Theorem 1.7.5 (Absolute Ratio Test. (D’ Alembert))12 Consider ∞ X a series of real numbers an . Suppose the following limit exists or is n=k

∞

0 ≤ lim

n→∞

|an+1 | = ρ ≤ ∞. |an |

Then: (1) If 0 ≤ ρ < 1, the series

∞ X

an converges absolutely and therefore

n=k

it converges. (2) If 1 < ρ ≤ ∞, the series

∞ X

an diverges.

n=k

(3) If ρ = 1, the test is inconclusive. Remark: The root and ratio tests, as presented here, are not stated in the most general form, one can find them in a good mathematical analysis book. In such a book these tests are stated in terms of the liminf and limsup of sequences of real numbers for achieving more general results. Study this material from a good mathematical analysis book. (See also the useful and powerful Dini-Kummer criterion for the convergence or divergence of positive series, Section 3.12, Application 7, footnote.) Theorem 1.7.6 (Limit Comparison Test) Consider two “nice” non-negative real functions 0 ≤ f , 0 ≤ g : [a, b) −→ R where f (x) −∞ < a < b ≤ +∞. Suppose lim = l ∈ [0, +∞]. Then we have: x→b− g(x) 12 Jean

Le Rond d’ Alembert, French mathematician, 1717–1783.

54

Improper Riemann Integrals

(a) If 0 < l < ∞, then b

Z

b

Z f (x) dx

exists

⇐⇒

g(x) dx

exists.

a

a

Or equivalently, if one of these improper integrals does not exist, (= ∞), then the other improper integral does not exist, (= ∞), either. Z b Z b (b) If l = 0 and g(x) dx exists, then f (x) dx exists. a

a

Z

b

b

Z g(x) dx = ∞, then

(c) If l = ∞ and a

f (x) dx = ∞. a

Proof (a) By the hypotheses of this case, it follows that there exists N ≥ a in R, such that g(x) > 0

and

l f (x) 3l < < , 2 g(x) 2

∀ x ≥ N.

So, l 3l g(x) < f (x) < g(x), ∀ x ≥ N, 2 2 and the conclusion follows by a double application of the Comparison Test, Theorem 1.7.1. (b) As in (a), by the hypotheses of this case, it follows that there exists N ≥ a in R, such that g(x) > 0

and

f (x) < 1, g(x)

∀ x ≥ N.

So, 0 < f (x) < g(x),

∀ x ≥ N,

and the conclusion follows by applying the Comparison Test, Theorem 1.7.1. (c) By the hypotheses of this case, it follows that there exists N ≥ a in R, such that g(x) > 0

and

1
−1,

1

the integral 0

xp − 1 dx exists. ln(x)

xp − 1 = lim− p xp = p (by L’ Hˆopital’s x→1 ln(x) x→1 rule) and so the integral is essentially proper at x = 1. Also, for any p ≥ 0, the integral is essentially proper, since at x = 0 the value of the −1 integrand is = 0. So, the integral exists for all p ≥ 0. −∞ Singularity occurs at x = 0, when −1 < p < 0. In this case, we consider a q such that 0 < −p < q < 1 (and so p + q > 0) and we apply the above Limit Comparison Test, 1.7.6, with absolute values (Absolute Limit Comparison Test). We then have Notice that for any p, lim−

p

lim+

−1 | xln(x) |

x→0

x−q

p+q x − xq 0 − 0 = lim+ = = 0, ln(x) −∞ x→0

and for 0 < q < 1, Z 0

1

x−q dx =



x−q+1 −q + 1

1 = 0

1 , 1−q

exists. 1

p x − 1 Therefore, by the Limit Comparison Test, 1.7.6, ln(x) dx 0 converges. I.e., the initial integral converges absolutely and therefore it exists. N Z

56

Improper Riemann Integrals

Example 1.7.5 The results presented in this example are straightforward, but because they are very useful we find them at times under the name “p -Test.” When combined with other tests it can answer a lot of questions on convergence or divergence of integrals rather easily. Since ∀ p ∈ R the antiderivative of the function f (x) = x−p = is

1 , xp

x ∈ (0, ∞)

 −p+1 x    −p + 1 + c, if p 6= 1, F (x) =    ln (|x|) + c, if p = 1,

where c is an arbitrary constant, we obtain the following easy but useful results: Let 0 < k < ∞ be a constant. Then:  if p ≤ 1,   ∞, Z ∞ dx  (1) =  xp 1 k   , if p > 1. k p−1 (p − 1)  1−p k 1    = p−1 , if p < 1, Z k  1−p k (1 − p) dx (2) = p  0 x    ∞, if p ≥ 1. So, by both (1) and (2) ∞

Z ∀ p ∈ R we have: 0

dx = ∞. xp

We now obtain the two byproducts: (1) For any a < b and p < 1 real constants Z a

b

dx = (x − a)p

Z

b−a

dt = tp

0

Z

b

a

dx (b − a)1−p = . (b − x)p 1−p

(2) For any a, b and p real constants Z a

∞

dx = (x − a)p

Z 0

∞

dt = tp

Z

b

−∞

dx = ∞. (b − x)p

Improper Riemann Integrals, Definitions, Criteria

57

An example of using the p -Test is in proving the convergence of the integral Z 1 Z ∞ Z ∞ sin2 (x) sin2 (x) sin2 (x) dx = dx + dx. x2 x2 x2 0 1 0 (The a´-priori splitting of this integral is legitimate since the integrand function is positive.) For the part Z 1 sin2 (x) dx x2 0 we observe

sin2 (x) = 1, x→0 x2 a fact that makes the integral proper and therefore finite. For the second part, we observe Z ∞ Z ∞ sin2 (x) 1 dx < dx = 1 x2 x2 1 1 lim

and so this integral is convergent, by the Comparison Test 1.7.1 and the p -Test with p = 2 > 1. On the other hand, the integral Z ∞ Z 1 Z ∞ sin2 (x) sin2 (x) sin2 (x) dx = dx + dx x3 x3 x3 0 0 1 diverges because Z 0

1

sin2 (x) dx = ∞. x3

sin2 (x) = 1 and so we can find a constant x2 x→0+ sin2 (x) 1 1 sin2 (x) 1 0 < k < 1 such that = · > · for all 0 < x < k. 3 x x2 x 2 x Then  k Z 1 Z k sin2 (x) 1 1 ln(x) 1 dx > · dx = = · [ln(k) + ∞] = ∞. 3 x 2 0 2 0 0 2 x This is so because lim

N Example 1.7.6 In this example we apply the Limit Comparison Test, Theorem 1.7.6, which is reminiscent to the limit comparison test for

58

Improper Riemann Integrals

positive series. Here, we find the limit of the ratio of two positive functions as we approach a singularity or ±∞ and then we make an appropriate comparison of their integrals. Z ∞ xq e−x dx converges. (1) Prove that ∀ q ∈ R, 1

In [1, ∞) we compare the positive function f (x) := xq e−x with the 1 positive function g(x) := 2 by taking the limit x xq+2 f (x) = lim xq+2 e−x = lim x = 0 (by L’ Hˆopital’s rule, e.g.). x→∞ x→∞ e x→∞ g(x) lim

Therefore, there is a constant k > 1 such that

f (x) < 1 if k ≤ x < ∞ g(x)

or 0 < f (x) < g(x) < 1 if k ≤ x < ∞. But then  ∞ Z ∞ Z ∞ −1 1 1 q −x x e < dx = = . 2 x x k k k k Hence, Z

∞ q −x

x e

k

Z

q −x

dx =

1

x e

∞

Z

xq e−x dx < finite +

dx +

1

k

1 −1,

1

xq e−x dx

converges.

0

Notice that when q ≥ 0 the integral is proper and so it is finite (converges). When −1 < q < 0, the positive function f (x) := xq e−x has singularity at x = 0 (it tends to ∞ as x −→ 0+ ). So, for −1 < q < 0, we compare f (x) with the positive function g(x) := xq by taking the limit f (x) = lim e−x = 1. g(x) x→0+

lim

x→0+

f (x) < 2 if g(x) q 0 < x ≤ k, i.e., 0 < f (x) < 2 g(x) = 2x if 0 < x ≤ k. But, in this case q + 1 > 0 and Therefore, there is a constant 0 < k < 1 such that

Z

k q −x

x e 0

Z dx < 0

k

xq+1 2 x dx = 2 q+1 q



k = 0

2k q+1 . q+1

Improper Riemann Integrals, Definitions, Criteria

59

Hence, when −1 < q < 0, Z

1 q −x

x e

Z dx =

k q −x

x e

0

1

Z

xq e−x dx
−1. 0

Z (3) Prove that ∀ q ≤ −1,

1

xq e−x dx(= ∞)

diverges.

0

In (0, 1], the positive function f (x) := xq e−x has singularity at x = 0 (it tends to ∞ as x −→ 0+ ). We compare it with the positive function g(x) := xq by taking the limit lim

x→0+

f (x) = lim e−x = 1. g(x) x→0+

So, there is a constant 0 < k < 1 such that

f (x) 1 > if 0 < x ≤ k, g(x) 2

1 g(x) if 0 < x ≤ k. But then (by the previous Example) 2 Z k Z k 1 q · x dx = ∞, since q ≤ −1. xq e−x dx ≥ 0 0 2

i.e., f (x) >

Hence Z if q ≤ −1,

k

xq e−x dx = ∞.

0

Thus, ∀ q ≤ −1, Z

1

xq e−x dx =

0

Z 0

k

xq e−x dx +

Z

1

xq e−x dx = ∞ + finite = ∞ diverges.

k

We will need the results of this example, when we study the Gamma and Beta functions in Sections 3.10 and 3.11. N Example 1.7.7 We would like to prove that the integral Z ∞ dx ln(x) [ln(x)] 2 converges.

60

Improper Riemann Integrals

One way to do this is to use the Integral Test, since we easily observe the function 1 f (x) = , [ln(x)]ln(x) is positive for x ≥ 2 and decreasing for x ≥ 3. (Also, its limit is zero as x −→ ∞.) So, to prove that this integral converges (is finite) we must prove that the positive series ∞ X 1 ln(n) [ln(n)] n=2 converges. This is done as follows: We use the Cauchy Condensation Theorem, 1.7.3, all the hypotheses of which are satisfied. (Check this.) So, we must prove that the series ∞ ∞ X X 2k 2k = k) ln(2 k [k ln(2)]k ln(2) k=2 [ln (2 )] k=2 converges. We prove that the latter series converges by using the Absolute Root Test. Indeed: s p 2k 2 = lim = 0 < 1. ρ := lim k |ak | = lim k k ln(2) k→∞ k→∞ k→∞ [k ln(2)] [k ln(2)]ln(2) Hence, the last series converges and so the initial series converges, too. Therefore, the given improper integral converges by the Integral Test. Another way to prove that the given integral converges is the following: Using the substitution t = ln(x) ⇐⇒ x = et in the above integral we get Z ∞ Z ∞ t e f (x)dx = dt. t 2 ln(2) t et is positive for t ≥ ln(2) tt and decreasing for t ≥ 3. (Also, its limit is zero as t −→ ∞.) Moreover, the series ∞ X en nn n=1 Then we observe that the function g(t) =

converges. For the convergence of this series we can use, e.g., the Root Test to get r en e lim n n = lim = 0 < 1. n→∞ n n→∞ n

Improper Riemann Integrals, Definitions, Criteria

61

So, the Integral Test applies and the improper integral converges. N Example 1.7.8 With work similar to the previous Example, we can prove that Z ∞ Z ∞ dx dx converges and diverges. ln(x) ln[ln(x)] [ln(x)] 3 3 [ln [ln(x)]] (Work out the details for both proofs.) N Now we consider a nice function y = f (x) with positive and negative values, defined in a Zset A. We say that the improper integral of f (x) over A ⊆ R exists if

f (x) dx is a finite (real) value. However, in such A

a situation, we distinguish the following two cases and definitions: Definition 1.7.2 We say that the improper integral of a nice function f (x) over a set A exists or converges absolutely if Z |f (x)| dx A

is equal to a finite Z non-negative value. Otherwise, |f (x)| dx = ∞ and then we say that the improper inA

tegral of f (x) over the set A diverges absolutely. Definition 1.7.3 We say that the improper integral of a nice function f (x) over a set A converges conditionally if Z f (x) dx A

is equal to a finite real value (and so this improper integral exists), but it diverges absolutely. Now we state the absolute convergence test which claims that absolute convergence implies convergence, but not vice-versa. Theorem 1.7.7 (Absolute Convergence Test) We consider a nice function f (x) defined in a set A. (a) If Z |f (x)| dx

exists

f (x) dx

exists.

A

then

Z A

62

Improper Riemann Integrals

(b) In Case (a) we also have the inequality Z Z f (x) dx ≤ |f (x)| dx. A

A

(c) The converse of this test is not true. Proof (a) The inequality 0 ≤ |f (x)| − f (x) ≤ 2|f (x)| is valid for all x ∈ A. Now we apply the Non-negative Comparison Test, (Theorem 1.7.1), and with a straightforward manipulation we obtain the result. (b) The relation Z Z f (x) dx ≤ |f (x)| dx A

A

is valid, because it is valid for any nice function and any closed interval [p, q] ⊆ A, as we have learnt in a calculus course. (c) In the sequel we shall see several examples that disprove the converse.  Before we present concrete examples using the Absolute Convergence Test and/or other tests, we need to clear out some things about what is a legitimate splitting of integrals into smaller parts. So, begin with the following example. Example 1.7.9 Consider the improper integral Z

∞

Z

N

f (x)dx = lim

N →∞ σ→a+

a

f (x)dx, σ

being improper at both endpoints only. That is, f (x) is nice in the open interval (a, ∞) and on any closed and bounded (finite) subinterval of (a, ∞) its integral is proper. Then, if we pick any two fixed numbers b and c such that a < b < c < ∞, we can always write Z

∞

Z f (x)dx =

a

b

Z f (x)dx +

a

c

Z f (x)dx +

b

∞

f (x)dx. c

Improper Riemann Integrals, Definitions, Criteria

63

In this equality the first and the last summands are improper integrals only at the lower and the upper limit of integration, respectively. The middle summand is a proper integral. The justification of such a splitting of this improper integral goes as follows: In taking the limits in this improper integral we do not lose anything by keeping a < σ < b and c < N < ∞. Also, the following equality is always valid Z N Z c Z b Z N f (x)dx, f (x)dx + f (x)dx + f (x)dx = c

b

σ

σ

because all the integrals of this equality are proper. Therefore, Z ∞ Z N f (x)dx = lim f (x)dx = N →∞ σ→a+

a

"Z

b

lim

b

b

Z

σ

Z f (x)dx + lim

N →∞

Z

c

f (x)dx + a

f (x)dx = c

b b

#

N

c

Z f (x)dx +

Z

Z f (x)dx +

σ

lim

σ→a+

c

Z f (x)dx +

N →∞ σ→a+

σ

Z f (x)dx +

b

N

f (x)dx = c

∞

f (x)dx. c

For the same reasons, if a < a1 < a2 < ... < an < an+1 < ∞, for any n ∈ N, we can write Z

∞

Z f (x)dx =

a

a1

f (x)dx + a

n Z X k=1

ak+1

ak

Z

∞

f (x)dx +

f (x)dx. an+1

N Now we continue with a simple, nevertheless useful, Lemma that tells us when and how we can break an improper integral into denumerable summations of appropriately chosen smaller pieces. As we know from calculus, it is always possible to break a proper integral into summations of countably (finitely or denumerably) many smaller proper integrals. But, even though we can write an improper integral as a summation of finitely many smaller parts, as this was done in the previous Example, not all splittings into denumerable summations of smaller integrals are legitimate. The following Lemma describes conditions under which these denumerable summations are valid. Lemma 1.7.1 (a) Let y = f (x) be a function on the interval [a, c), where −∞ < a < c and c ∈ R or c = ∞. Consider any (strictly)

64

Improper Riemann Integrals

increasing sequence a = a0 < a1 < a2 < ... with lim an = c. Then: n→∞ If Z c

(a1 )

f (x) dx exists and it is equal to a real number L, a

or (a2 )

f (x) ≥ 0, ∀ x ∈ [a, c),

(a3 ) then

f (x) ≤ 0, ∀ x ∈ [a, c),

or c

Z

f (x) dx = a

∞ Z X n=0

an+1

f (x) dx .

an

(b) Let y = f (x) be a function on the interval (c, b], where c < b < ∞ and c ∈ R or c = −∞. Consider any (strictly) decreasing sequence b = b0 > b1 > b2 > ... with lim bn = c. Then: n→∞ If Z b (b1 ) f (x) dx exists and it is equal to a real number L, c or (b2 ) f (x) ≥ 0, ∀ x ∈ (c, b], or (b3 ) f (x) ≤ 0, ∀ x ∈ (c, b], then Z b ∞ Z bn X f (x) dx = f (x) dx . c

bn+1

n=0

Proof We shall prove Case (a) only, for Case (b) is just analogous. Also, Subcase (a3 ) is analogous to Subcase (a2 ) and so it suffices to prove only the Subcases (a1 ) and (a2 ). Z c In Subcase (a1 ), we assume that f (x) dx = L for some real numa

ber L. Then Z

c

Z f (x) dx = lim

a

R→c−

R

f (x) dx = L. a

Since the limit exists, any legitimate limiting process whatsoever gives the number L as value of the limit. Therefore, for any increasing sequence with limit c (an ↑ c as n → ∞) we have Z c Z an f (x) dx = lim f (x) dx = L. a

n→∞

a=a0

Improper Riemann Integrals, Definitions, Criteria

65

But, since we can break any proper integral into a finite sum of successive smaller proper integrals, we get, for n ≥ 1 Z an n−1 X Z ak+1 f (x) dx = f (x) dx. a=a0

ak

k=0

Then putting the last two equations together, we obtain Z an Z c f (x) dx = f (x) dx = lim n→∞

a

lim

n→∞

n−1 X Z ak+1

f (x) dx =

ak

k=0

a=a0 ∞ Z X

an+1

f (x) dx = L.

an

n=0

In Subcase Z c (a2 ), we assume that f (x) ≥ 0, ∀ x ∈ [a, c). If it happens that f (x) dx exists, then we invoke the previous subcase and a

the proof is over. If the integral does not exist, since f (x) ≥ 0 this means that Z Z c

R

f (x) dx = lim

f (x) dx = +∞.

R→c−

a

a

Now, given any R such that a < R < c, since lim an = c, we can n→∞ pick a term an+1 of the sequence such that R ≤ an+1 . Then by the non-negativity of the function, we have Z R Z R Z an+1 f (x) dx ≤ f (x) dx + f (x) dx = a

a

Z

an+1

f (x) dx = a

n Z X k=0

R ak+1

f (x) dx.

ak

Also, for any given term an+1 of the sequence, where n ∈ N0 , we can pick a real R such that a < R ≤ an+1 to obtain again a similar inequality. Z R Since lim− f (x) dx = ∞, a < R ≤ an+1 and f (x) ≥ 0, we get R→c

Z ∞ = lim

R→c−

a

f (x) dx ≤ lim

n→∞

a

Therefore, Z c Z f (x) dx = lim− a

n Z X

R

R→c

a

k=0

ak+1

f (x) dx =

ak

R

f (x) dx =

∞ Z X n=0

∞ Z X n=0

an+1

f (x) dx.

an

an+1

f (x) dx = ∞.

an



66

Improper Riemann Integrals

Examples Example 1.7.10 and Remark: The previous Lemma does not apply in either of the two cases of Z b Z c f (x) dx when these improper integrals do not exist. f (x) dx or a

c

For instance, in Example 1.1.7, we have seen that the integral Z ∞ sin(x) dx does not exist. 0

If we now let an = 2nπ for n = 0, 1, 2, . . . , which satisfies all the requirements of the Lemma, then we get ∞ Z an+1 ∞ Z 2(n+1)π ∞ X X X sin(x) dx = sin(x) dx = 0 = 0, n=0

an

n=0

2nπ

n=0

even though the integral itself does not exist. With an = nπ for n = 0, 1, 2, . . . , we get ∞ Z an+1 ∞ Z (n+1)π X X sin(x) dx = sin(x) dx = n=0

an

n=0 ∞ X

nπ

2(−1)n = does not exist,

n=0

an answer different from the one found before. (For your own practice, find some other sequences (an )n∈N that satisfy the requirements of the Lemma and yield other values for the respective infinite summation in this example.) N Z ∞ sin(x) Example 1.7.11 The improper integral dx is absolutely x2 + 1 0 convergent and therefore convergent. To show this, we notice Z ∞ Z ∞ sin(x) π 1 dx < dx = . x2 + 1 2 x +1 2 0 0 Then we also get, −π < 2

Z 0

∞

sin(x) π dx < . x2 + 1 2 N

Example 1.7.12 Let f (x) =

sin(x) n+1

for nπ ≤ x ≤ (n + 1)π, n = 0, 1, 2, . . . .

Improper Riemann Integrals, Definitions, Criteria

67

With the help of Lemma 1.7.1, Case (a2 ), applied to the non-negative function |f (x)|, we obtain Z ∞ |f (x)| dx = 0 ∞ Z X

(n+1)π

nπ

n=0

  | sin(x)| 1 1 1 dx = 2 1 + + + + ... = ∞. n+1 2 3 4 ∞

Z So, the improper integral

f (x) dx diverges absolutely, and we cannot 0

claim anything about its conditional convergence yet. Since we do not know that this integral converges (exists), we cannot apply Lemma 1.7.1 at this point in order to say that ∞

Z

?

f (x) dx = 0

∞ Z X n=0

(n+1)π

nπ

  1 1 1 sin(x) dx = 2 1 − + − + ... = 2 ln(2). n+1 2 3 4

[The fact that 1−

1 1 1 + − + . . . = ln(2), 2 3 4

at times called Brouncker series13 , follows from the power series ln(1+x) =

∞ X

(−1)n−1

n=1

x2 x3 x4 xn = x− + − +. . . , n 2 3 4

∀ x : −1 < x < 1,

which also converges for x = 1, by Leibniz’s14 alternating series Test.15 Then, we apply Abel’s16 Lemma stated in the footnote of Theorem II 1.2.1. Both Leibniz’s Test and Abel’s Lemma can be found in any good book of mathematical analysis, e.g., Apostol 1974, or Rudin 1976. This series can also be proven in other ways, e.g., one way is shown in Subsection II 1.5.4.] 13 William

Brouncker, English mathematician, 1620–1684. Wilhelm von Leibniz, German mathematician and philosopher, 1646–

14 Gottfried

1716. 15 Leibniz’s alternating series Test states: If (an ), n ∈ N, is a sequence of positive numbers that decreases and has limit zero, ∞ X then the alternating series (−1)n−1 an converges. 16 Niels

n=1

Henrik Abel, Norwegian mathematician, 1802–1829.

68

Improper Riemann Integrals

So, to prove that this integral converges, we proceed as follows: For any R > 0 there is an integer k ≥ 0 such that kπ ≤ R < (k + 1)π. Then, (R −→ ∞) ⇐⇒ (k −→ ∞) (prove this “iff” as an exercise!) and by definition we have Z ∞ Z R def f (x) dx = lim f (x) dx = R→∞

0

"Z

(k+1)π

0

Z f (x) dx −

= lim

R→∞

#

(k+1)π

f (x) dx = R

0

(  )  Z (k+1)π 1 1 1 1 k lim 2 1 − + − + ... + (−1) − f (x) dx . R→∞ 2 3 4 k+1 R But, the two partial limits inside the bracket exist. For the first one, we know   1 1 1 1 lim 2 1 − + − + ... + (−1)k = 2 ln(2). R→∞ 2 3 4 k+1 For the second one, we have "Z lim

R→∞

#

(k+1)π

f (x) dx = 0, R

since we easily observe that Z Z (k+1)π (k+1)π 2 −→ 0, 0< f (x) dx ≤ |f (x)| dx = R k+1 kπ as

k −→ ∞ ⇐⇒ R −→ ∞.

Since these two partial limits exist, we can take their difference to obtain Z ∞ Z R def f (x) dx = lim f (x) dx = 0

R→∞

0

 ∞ X 1 1 1 1 2 1 − + − + ... − 0 = 2 (−1)n−1 = 2 ln(2). 2 3 4 n n=1 

Hence, this improper integral converges conditionally to the number 2 ln(2). N

Improper Riemann Integrals, Definitions, Criteria

69

Example 1.7.13 Let g(x) =

sin(x) (n + 1)2

for nπ ≤ x ≤ (n + 1)π, n = 0, 1, 2, ...

For the absolute convergence, we can apply Case (a2 ) of Lemma 1.7.1 to the non-negative function |g(x)| and use Example 3.6.3 to find ∞

Z

|g(x)| dx = 0

∞ Z X

nπ

n=0

 2 1+

(n+1)π

| sin(x)| dx = (n + 1)2

 1 1 1 π2 π2 + + + ... =2· = . 2 2 2 2 3 4 6 3

(See also Corollary II 1.7.3 of Example II 1.7.23 and Problem II 1.7.52.) Hence, this improper integral converges absolutely, and so it converges. Now, by Case (a3 ) of the Lemma 1.7.1 it is legitimate to say that ∞

Z

g(x) dx = 0

∞ Z X n=0

(n+1)π

nπ

sin(x) dx = (n + 1)2

  1 1 1 π2 π2 2 1 − 2 + 2 − 2 + ... = 2 · = 2 3 4 12 6 (see Problem II 1.7.59). This is the actual finite value of this integral. N The ideas in Example 1.7.12 motivate us to state the following useful Lemma about convergence (existence) and estimation or actual evaluation of improper integrals. Its proof is omitted as analogous to the series of arguments presented in the solution of this example. (You can write it out for practice.) This Lemma can also be used to justify the splitting of improper integrals into an infinite summation of smaller parts, and so it should be viewed together with Lemma 1.7.1. Lemma 1.7.2 (a) Let y = f (x) be a function on an interval [a, c), where −∞ < a < c and c ∈ R or c = ∞. Consider any (strictly) increasing sequence a = a0 < a1 < a2 < ... with lim an = c. n→∞ We assume: n Z an+1 X (a1 ) lim f (x) dx = l, with − ∞ ≤ l ≤ ∞. n→∞

k=0

an

70

Improper Riemann Integrals

(a2 ) For any real number R such that a < R < c and the unique k ∈ N such that ak ≤ R < ak+1 [k depends on R and is unique since the sequence (an ) is strictly increasing] we have: Z ak+1 f (x) dx = 0. lim− R→c

R

Then,

∞ Z X

c

Z

f (x) dx = a

an+1

f (x) dx = l.

an

n=0

(b) Let y = f (x) be a function on an interval (c, b], where c < b < ∞ and c ∈ R or c = −∞. Consider any (strictly) decreasing sequence b = b0 > b1 > b2 > ... with lim bn = c. n→∞ We assume: n Z bn X f (x) dx = l, with − ∞ ≤ l ≤ ∞. (b1 ) lim n→∞

k=0

bn+1

(b2 ) For any real number R such that c < R < b and the unique k ∈ N such that bk+1 < R ≤ bk [k depends on R and is unique since the sequence (bn ) is strictly decreasing] we have: Z lim+

R→c

f (x) dx = 0. bk+1

Then, Z

b

f (x) dx = c

R

∞ Z X n=0

bn

f (x) dx = l.

bn+1

(For the proof, imitate the solutions of the two previous Examples. For a kind of counterexample, see Problem 3.2.20.) Z an+1 Remark 1.7.1 Given that in this Lemma the f (x) dx is asan

sumed to be proper and therefore finite and R is any number in the Z an+1 interval [an , an+1 ), we can only assume that lim f (x) dx = 0 R→c−

R

and not another finite number, since for R close enough to an+1 we can make this partial integral as close to zero as we wish. Remark 1.7.2 Notice that assumption (a2 ) of this Lemma fails in Example 1.1.7. AsZ we have seen in that example and in Example ∞

1.7.10, the integral

sin(x) dx does not exist. 0

Improper Riemann Integrals, Definitions, Criteria

71

We continue with three additional tests which, most of the times, may be superseded by the tests that we have seen so far, in applications. Theorem 1.7.8 (Absolute p-Test) We consider a nice function f : [a, ∞) −→ R, with a > 0. Then we have: p (a) If for some p Z> 1, there is a K > 0 such Z ∞that x |f (x)| ≤ K, ∞ f (x) dx exists as a |f (x)| dx exists and so ∀ x ≥ a, then a

a

real finite value.

(b) If for some p Z≤ 1, there is a K > 0 such that xp |f (x)| ≥ K, ∞ ∀ x ≥ a, then f (x) dx = ∞. a

(c) If forZsome p ≤ 1, there is a K < 0 such that xp f (x) ≤ K, ∀ x ≥ a, ∞

f (x) dx = −∞.

then a

Proof Z ∞ K K , and dx exists, since p p x a>0 x p > 1. Then, the Absolute Convergence Test, Theorem 1.7.7, applies. Z ∞ K K (b) By hypotheses we get f (x) ≥ p > 0, and dx = ∞, since p x a>0 x p > 1. Then, the Non-negative Comparison Test, Theorem 1.7.1, or the Absolute Convergence Test, Theorem 1.7.7, applies. Z ∞ K K (c) By hypotheses we get f (x) ≤ p < 0, and dx = −∞, since p x x a 1 and K < 0. Then, f (x) dx = −∞. (a) By hypotheses we get |f (x)| ≤

a

 Example 1.7.14 Application of the Absolute p-Test, above. (a) Z 1

∞

x2 + 2 (x2 + 1)

2

dx exists.

This follows from the inequality x2 ·

x2 + 2 (x2 + 1)

2

< 1, ∀ x ∈ (1, ∞).

72

Improper Riemann Integrals

(b) ∞

Z

x3 2

(x2 − 1)

2

This follows from the inequality x · (c) Z 3

∞

dx = ∞. x3 2

(x2 − 1)

> 1, ∀ x ∈ (2, ∞).

x dx = −∞. 1 − x2

This follows from the inequality x ·

x < −1, ∀ x ∈ (3, ∞). 1 − x2 N

Theorem 1.7.9 (Absolute Ratio Test. D’ Alembert) We consider Z b a nice function f : [a, ∞) −→ R, with a ∈ R, for which |f (x)|dx a

exists for all b : a < b < ∞.

If

f (x + 1) = L < 1, lim x→∞ f (x)

Z then

∞

f (x) dx

exists.

a

Proof We consider a K such that L < K < 1. Then there exists N ≥ a such that x+1 f (x + 1) |f (x)|dx = ∞. If L = 1, then the test is inconclusive.

73 1, then

a

Example 1.7.15 Z

∞

x−x dx exists.

1

The function x−x is positive in [1, ∞), and " # −x (x + 1)−(x+1) 1 x+1 lim · = lim = x→∞ x→∞ x−x x x+1 "

1 1+ x

= lim

x→∞

−x

# 1 · = e−1 · 0 = 0 < 1. x+1

So, the result follows from the above Absolute Ratio Test. N Theorem 1.7.10 (Root Test. Cauchy) We consider a nice function Z b f : [a, ∞) −→ R, with a ∈ R, for which |f (x)|dx exists for all b : a

a < b < ∞. 1

lim |f (x)| x = R < 1,

If

x→∞

∞

Z then

f (x) dx

exists.

a

Proof We consider a K such that R < K < 1. Then, there exists N ≥ a such that 0 < |f (x)| < K x , ∀ x ≥ N. Z N |f (x)|dx exists by hypothesis, and Since a

Z

∞

N

K x dx =

Z

∞

N

eln(K)x dx = −

KN ln(K)

exists, for

0 < K < 1,

the result follows from the Non-negative Comparison Test, Theorem 1.7.1, or the Absolute Convergence Test, Theorem 1.7.7.  Remark: We can analogously prove that if R > 1, then Z ∞

|f (x)|dx = ∞. If R = 1, then the test is inconclusive. a

74

Improper Riemann Integrals

Example 1.7.16 Z

∞

2

e−x dx exists.

1

The function e

−x2

is positive in [0, ∞), and

lim

x→∞



e−x

2

 x1

= lim e−x = 0 < 1. x→∞

So, the result follows from the above Absolute Root Test. (See also Example 1.7.1 and Section 2.1 for other proofs.) N We conclude this section with the very important and powerful criterion of Cauchy for convergence of improper integrals. We interpret it in the following way: Theorem 1.7.11 (Cauchy Test) Let y = f (x) be a nice function on [a, c), where a ∈ R and c ∈ R with a < c, or c = ∞. Consider the following three statements: Z c (a) f (x) dx converges. a

(b) ∀  > 0, ∃ N ∈ R : a ≤ N < c such that ∀ p ∈ R and ∀ q ∈ R such that N ≤ p, q < c, we have Z q < . f (x) dx p

Z (c) ∀ r ∈ [a, c), the integral

r

f (x) dx exists. a

Then we have: (I) (a) implies (b). (II) If (c) holds, then the converse of (I) is true, i.e., (b) implies (a). Z Remark 1.7.3 We have analogous results for

b

f (x) dx, on (c, b], with c

c < b in R, or c = −∞. (Write down these results explicitly, for practice.) Remark 1.7.4 Hypothesis (c) is needed in Part (II) as seen in the next Example that follows the proof. This hypothesis is valid when f (x) is continuous, or bounded in [a, c), and in some other situations. In most applications, we are interested in using Part (II).

Improper Riemann Integrals, Definitions, Criteria Proof (I) We assume Z Z c f (x) dx = lim− M →c

a

75

M

f (x) dx = L

a

exists as a real finite value L. Then we let Z M F (M ) = f (x)dx, ∀ M ∈ [a, c). a

By our assumption, the function F (M ) is well defined on [a, c) and it is continuous in the variable M . (From calculus, we already know that the definite integral of a nice function is continuous with respect to its upper limit.) Also, lim− F (M ) = L. M →c

 Now, ∀  > 0, we consider > 0 and we use the analytical definition 2 of the existence of a limit to claim that: ∃ N : a ≤ N < c such that ∀ M : N ≤ M < c  the inequality |F (M ) − L| < is true. 2 Then, for any p and q: N ≤ p, q < c we get |F (q) − F (p)| = |F (q) − F (p) + L − L| ≤   |F (q) − L| + |F (p) − L| < + = . 2 2 Since Z |F (q) − F (p)| =

a

q

p

Z f (x) dx − a

Z f (x) dx =

p

q

f (x) dx ,

we obtain the claim: ∀  > 0, ∃ N, a ≤ N < c such that, for any p and q in R : Z q N ≤ p, q < c =⇒ f (x) dx < . p

(II) By hypothesis (c) the function F (r), as defined above in the proof of (I), is well defined on [a, c). Then, the hypotheses of this converse implication are translated as follows: ∀  > 0, ∃ N, a ≤ N < c such that, for any p and q: N ≤ p, q < c =⇒ |F (q) − F (p)| < .

76

Improper Riemann Integrals

By the Cauchy General Criterion for convergence in the real line17 we readily obtain that lim F (M ) exists as a finite real value. M →∞

∞

Z

f (x) dx exists, i.e., it is a finite real value.

Therefore, a



Examples Example 1.7.17 In the Cauchy Test, above, for the converse of (I) in Part (II), hypothesis (c) is necessary. For instance, we let   5, if x = 0,        1 f (x) = , if 0 < x < 1,  x       0, if x ≥ 1. Obviously, f (x) is a nice piecewise continuous function on [0, ∞), and Z ∞ f (x)dx = ln(1) − ln(0+ ) = 0 − (−∞) = ∞, 0

i.e., (a) is false. But (b) is true since for any  > 0 we can pick N ≥ 1. This happens because Condition (c) fails, and so the function F (r), in the above proof, is not well defined for all r ∈ [0, ∞). 17 This Cauchy criterion for convergence in the real line that we have invoked here claims: A sequence of real numbers (xn ) converges to a real number x if and only if it is a Cauchy sequence. This equivalence, by definition, is written in terms of positive ’s as follows:

[ ∃ x ∈ R : ∀  > 0, ∃ N ∈ N : ∀ n ≥ N =⇒ |xn − x| ≤  ] ⇐⇒ [ ∀  > 0, ∃ N ∈ N : ∀ m ∈ N, ∀ n ∈ N : ( m ≥ N, n ≥ N ) =⇒ |xm − xn | ≤  ] . That is,  h

∃ lim xn ∈ R n→∞

i



⇐⇒  lim |xm − xn | = 0  . m→∞ n→∞

This very important criterion is a very powerful tool and it is equivalent to the completeness of the real numbers. It can be found in any book of mathematical analysis or advanced calculus.

Improper Riemann Integrals, Definitions, Criteria

77

Another example is   5,        1 g(x) = ,  x       0,

if x = 0, if

− 1 < x 6= 0 < 1,

if x ≥ 1.

g(x) is a nice piecewise continuous function on [−1, ∞), and Z ∞ g(x)dx = −∞ + ∞ = does not exist, −1

i.e., (a) is false. But again (b) is true since for any  > 0 we can pick N ≥ 1. (To this end study the next three examples.) N Example 1.7.18 In this example, we will prove the following two important results: Z ∞ sin(x) (a) dx exists (converges to a finite real value). x 0 (b) This integral does not converge absolutely and so it converges conditionally. (See also the related Problems 1.8.15, 1.8.16, 3.2.29, 3.2.30, II 1.7.93 and Examples 3.1.8, 3.3.11, II 1.7.35.) sin(x) in (0, ∞) Proof of (a): The continuous function f (x) = x can be continuously extended to x = 0 by letting f (0) = 1, since sin(x) lim = 1. So, the integral (a) is improper only because it is taken x→0 x over an unbounded interval. Using integration by parts, we get that for any 0 < p < q, Z q Z q Z q sin(x) −1 cos p cos q cos(x) dx = d cos(x) = − − dx. x x p q x2 p p p Thus Z

p

q

Z q Z q sin(x) 1 1 | cos(x)| 1 1 1 dx ≤ + + dx ≤ + + dx = 2 2 x p q x p q p p x  q 1 1 −1 2 + + = −→ 0, as p −→ ∞. p q x p p

78

Improper Riemann Integrals

Z q sin(x) 2 Then ∀  > 0, if we pick any q > p > , we get dx < .  x p Z ∞ sin(x) Since here Condition (c) of the Cauchy Test is valid, dx x 0 exists as a finite value, by the Cauchy Test, 1.7.11. Proof of (b): This integral does not converge absolutely. Indeed, |f (x)| ≥ 0, and so by Lemma 1.7.1 we have Z ∞ |f (x)| dx = 0

Z 0

∞

∞ Z (n+1)π X | sin(x)| | sin(x)| dx = dx. x x n=0 nπ

Now for all n = 0, 1, 2, 3, . . . , we have Z (n+1)π | sin(x)| 1 2 dx ≥ | sin(x)| dx = . x (n + 1)π nπ (n + 1)π nπ Z ∞ Z ∞ ∞ X 2 | sin(x)| So, dx ≥ = |f (x)| dx = x (n + 1)π 0 0 n=0   1 1 1 2 1 + + + + ... = ∞. π 2 3 4

Z

(n+1)π

Therefore, Z 0

∞

| sin(x)| dx = ∞. x

(For another proof of the conditional convergence without using the Cauchy Test, 1.7.11, see Problem 1.8.16.) sin(x) Remark 1.7.5 By observing that the function f (x) = is an even x and continuous function over all R, we also obtain that Z ∞ Z ∞ sin(x) sin(x) dx = 2 dx. x x 0 −∞ Therefore, this improper integral over all R converges conditionally, too. Remark 1.7.6 The same results of conditional convergence can be obtained for the improper integrals Z ∞ Z ∞ sin(u) sin (βx) dx = sign(β) · du x u 0 0

Improper Riemann Integrals, Definitions, Criteria and

Z

∞

−∞

sin(βx) dx = sign(β) · x

Z

∞

−∞

79

sin(u) du, u

for any real constant β 6= 0. (If β = 0, the integrals are obviously zero.) The equality is obtained by letting u = βx and by definition   +1, if β > 0, sign(β) =   −1, if β < 0. Remark 1.7.7 Lemma 1.7.2 can also be used to prove convergence, except the Cauchy Test, 1.7.11, is more efficient. N Example 1.7.19 Whereas the Cauchy Test, 1.7.11, applied in the previous Example, it does not apply to the integral Z ∞ cos(x) dx. x 0 This is because Condition (c) does not hold. We have that, ∀ c > 0, Z c cos(x) dx = ∞ x 0 and so, Z 0

∞

cos(x) dx = ∞. x

But the Cauchy Test can apply to prove the conditional convergence of Z ∞ cos(x) dx, ∀ c > 0, x c or the method of Problem 1.8.16. Also, as in the previous Example, this integral does not converge absolutely. (See also Problem 3.2.30.) N Z ∞
2 Example 1.7.20 The Fresnel18 integral sin x dx converges conditionally. 18 Augustin

0

Jean Fresnel, French mathematician and physicist, 1788–1827.

80

Improper Riemann Integrals

(See also the related Problems 1.8.16, 3.13.11, 3.13.12, 3.2.41 and Examples 3.6.2, II 1.7.17.) We have Z ∞ Z R

def 2 sin x dx = lim sin x2 dx. 0 p > 0, using integration by parts, we get lim

u→0

Z

q

Z 2 1 q 1 du √ d [− cos(u)] = sin x dx = sin(u) √ = 2 p2 2 u u p2  q2 Z q2  −1  −1 cos(u) 1 √ cos(u) · d u 2 = + 2 u p2 2 p2   Z 2 −1 cos(q 2 ) cos(p2 ) 1 q cos(u) − du. − 3 2 q p 4 p2 u2 2

p

Z




q2

Therefore, by the properties of absolute value combined with inequalities and integrals and the fact that | cos(u)| ≤ 1, ∀ u ∈ R, we have Z q   Z 2
1 1 1 1 q 1 2 sin x dx ≤ + + du = 2 q p 4 p2 u 23 p =

1 2



1 1 + q p

 −

     q2 1 1 1 1 1 1 1 1 1 √ + − − = . = 2 2 q p 2 q p p u p2

1 Now, as in the previous Example, ∀  > 0 we choose p > to get  1 that ∀ q > p > to guarantee the validity of the inequality  Z q
2 sin x dx < . p

Z

r


sin x2 dx exists, by the Cauchy 0 Z ∞
Test, 1.7.11, [Condition (c) holds], sin x2 dx converges. Since for any r > 0 the integral

0

Improper Riemann Integrals, Definitions, Criteria

81

Now we prove that the integral diverges absolutely. We let x2 = u and use an analysis analogous to the previous Example, to obtain Z ∞ Z ∞ ∞ ∞ X sin(u)
2 1 X 1 sin x2 dx = 1 √ du ≥ 1 √ √ √ = ∞. = 2 0 2 n=1 nπ u π n=1 n 0 (All the steps here are directly justified by the definitions and the fact that we work with a non-negative Z ∞ function.)
sin x2 dx converges but diverges abSo, the Fresnel integral 0

solutely. Therefore, it converges conditionally. With parallel work and Zanalogous adjustments, we also obtain that ∞
the other Fresnel integral cos x2 dx converges conditionally. (See 0

and solve Problem 1.8.3 below.) N Some questions on the conditional convergence of certain improper integrals, such as those addressed in Examples 1.7.18–1.7.20, can be answered without using the Cauchy Test, Theorem 1.7.11, but with the help of the following theorem. Identify these questions and also you may use it in solving Problems 1.8.3, 1.8.15, 1.8.16, etc. Theorem 1.7.12 (Abel’s Test for Convergence of Im. Integrals) Let a ∈ R and suppose f : [a, ∞) −→ R is aZ continuous function, such x f (t) dt ≤ M , ∀ x ≥ 0, that, there is a constant M > 0, such that a and suppose that g : [a, ∞) −→ R is a positive [g(x) > 0] differentiable function, such that, g 0 (x) ≤ 0 [so g(x) is decreasing] and lim g(x) = 0. x→∞ Z x We let h(x) := f (t)dt. Then, a

Z

∞

Z f (x)g(x) dx = −

a

∞

g 0 (x)h(x) dx

exist.

a

Proof By the hypotheses we have h(a) = 0, |h(x)| < M , ∀ x ∈ [a, ∞), (|h| is bounded by M ), g(∞) = 0 and h0 (x) = f (x). Hence, Z

∞

Z ∞ g(x) d[h(x)] = [g(x)h(x)]∞ − g 0 (x)h(x) dx = a a a Z ∞ Z ∞ 0 0 (0 − 0) − g (x)h(x) dx = − g (x)h(x) dx. Z

∞

f (x)g(x) dx = a

a

a

82

Improper Riemann Integrals

Now the existence of these equal integrals follows from −g 0 (x) > 0 and then Z ∞ Z ∞ Z ∞ 0 0 −g 0 (x) dx = [−g (x)] · |h(x)| dx ≤ M |[−g (x)] · h(x)| dx = a

a

a

M [−g(x)]∞ a = M [0 + g(a)] = M g(a) < ∞. Thus the second integral converges absolutely and therefore both integrals exist.  Remarks: (a) In the integral equality of the above proof, the first integral may or may not converge absolutely but it exists and it is equal to the second integral, which converges absolutely. For example, see Problems 1.8.15 (b)-(c) and 1.8.16, Z etc. Z ∞

(b) In the special case where

∞

f (x) dx exist, then h(x) := a

f (t)dt a

is bounded and all hypotheses of this test are satisfied and we can apply it. For example, we can check that the test applies if a = 0 and f (x) sin(x) cos(x) is: sin(x), or cos(x), or . It also applies to , if a > 0 but x x not if a = 0. This Theorem finds convenient applications to the Laplace transform, Chapter 4. (c) From the proof of the Theorem, we seeZ that with relaxed hy∞

g 0 (x)h(x)dx exists,

potheses but the limit g(∞)h(∞) − g(a)h(a) − a

then we also obtain Z ∞ Z f (x)g(x) dx = g(∞)h(∞) − g(a)h(a) − a

∞

g 0 (x)h(x)dx.

a

(d) The Theorem is valid even if f : [a, ∞) −→ R is a piecewise continuous function, with finitely or countably many discontinuities. In such a case h(x) is not differentiable at the discontinuities of f (x), but is continuous. Modify the above proof and give the proof of this Theorem under this relaxed hypothesis.

1.8

Problems

1.8.1 Give examples of functions with discontinuities of the three types that we have stated in the non-standard Definition 1.7.1. √ √ Z ∞ Z ∞ dx dx 2π 3 π 2 1.8.2 (a) Prove: (1) = , (2) = . 1 + x3 9 1 + x4 4 0 0

Improper Riemann Integrals, Definitions, Criteria

83

[Hint: You need partial fractions and the integral rules with the natural logarithm and arc-tangent.] (b) Apply integration by parts to the integrals of the previous part, manipulate what you find and use their values to prove: Z

∞

(1)

dx 2

(1 + x3 )

0

=

√ 4π 3 , 27

∞

Z (2)

dx 2

(1 + x4 )

0

=

√ 3π 2 . 16

Note: This method from (a) to (b), using integration by parts, can be continued successively for computing the same improper integrals of 1 1 and , as long as we the higher powers of the functions 3 1+x 1 + x4 know the results of part (a). [See, e.g., Examples 3.1.6, (b), II 1.7.7, etc.] Similarly, we can apply this method for the powers of functions such 1 1 as , , etc. Write a few examples and generalize. 1 + x2 1 + x5 1.8.3 The Fresnel cosine integral is Z ∞
cos x2 dx. 0

Prove that it exists but diverges absolutely. In the problems (1.8.4–1.8.12) below, check in any possible way the existence or non-existence of the given improper integrals. 1.8.4

Z

∞

(a) 0

1.8.5

∞

Z (a) 0

dx √ , 1 + x3

sin2 (x) dx, x

∞

Z

√

(b) 0

Z

∞

(b) 0

dx . 1 + x4

sin2 (x) dx. x2

1.8.6 Z (a) 0

∞

sin3 (x) dx, x

1.8.7

Z (a) 0

∞

Z

∞

(b) 0

e−x sin(x) dx, x

sin3 (x) dx, x2

Z (c) 0

Z (b) 0

∞

∞

sin3 (x) dx. x3

xdx √ . 1 + x3

84

Improper Riemann Integrals

1.8.8 ∞

Z

sin(ax) dx, ebx − 1

0

where a ∈ R and b > 0 constants.

1.8.9 Z

∞

(a) 2 ∞

Z (c) 1

(b) 2 ∞

Z

sin(x) dx, xp

∞

Z

dx , [ln(x)]p

(d) 1

cos(x) dx, xp

dx , x[ln(x)]p

where p > 0 constant.

1.8.10 ∞

Z

e

−(x−µ)2 σ2

dx,

where µ ∈ R and σ 6= 0 constants.

−∞

1.8.11 Z (a) 1

10

dx √ , x−1

(b) 0

1.8.12 Prove that Z ∞ 1 dx exists, [ln(x)]x 2 1.8.13 Prove: Z ∞ 1 (a) dx = ∞, 2 sin2 (x) 1 + x 0 Z (b)

1

Z

For any α > 6, 0

∞

Z

dx √ , 3 x−1

(c)

1 1

[ln(x)] x

2

Z and 0

x−x dx.

0

∞

Z and

∞

∞

dx = ∞.

1 dx < ∞. 1 + x3 sin2 (x)

x dx is positive finite. 1 + xα sin2 (x)

(Problem II 1.8.7 provides the sharp answers to these questions!) [Hint: (The limit comparison test, Theorem 1.7.6, does not apply.) Now:

Improper Riemann Integrals, Definitions, Criteria

85

1 (a) Consider the positive function f (x) = on [0, ∞) 1 + x2 sin2 (x) Z nπ+ π2 and let an = f (x) dx, for n = 1, 2, 3, . . . . Then, ∀ n ∈ N, nπ− π 2  1 π = an > 0, f (nπ) = 1, f 0 (nπ) = 0, f nπ ±
, and π 2 2 1 + nπ ± 2 h π π πi . Now, |f 0 (x)| ≤ 1 + + nπ < 2nπ for every x ∈ nπ − , nπ + 2 2  2  1 ∀ n ∈ N, the isosceles triangle with vertices the points nπ ± ,0 2nπ 1 and (nπ, 1) has base length , height 1, and its two equal sides have nπ slopes ± 2nπ. Argue why this triangle lies completely under the graph of the function f (x) and therefore, ∞

Z

f (x) dx ≥ 0

∞ X

an ≥

n=1

∞ X 1 = ∞. 2nπ n=1

x (b) Consider the positive function g(x) = on [0, ∞) α 1 + x sin2 (x) Z nπ+ π2 g(x) dx, and n = 1, 2, . . . . with α > 6 and let bn = nπ− π 2

sin2 (x) = 1, x→nπ (x − nπ)2 2 π 1 there exists  such that 0 <  < and sin2 (x) > (x − nπ)2 , for all 2 2 x ∈ (nπ − , nπ + ). So, for any β > 0 there is an N ∈ N such that    2 1 1 1 1 1 2 2 ≤  and sin (x) ≥ sin > = n−4−2β for all 2+β 2+β n2+β n 2 n 2    S π 1 1 π x ∈ nπ − , nπ − 2+β nπ + 2+β , nπ + and for all n ≥ N . n n h 2 2  π α π πi Also, xα ≥ nπ − . > nα , for all x ∈ nπ − , nπ + 2 2 2 1 So, for all n ≥ N , we have 1 + xα sin2 (x) > xα sin2 (x) > nα−4−2β 2 and therefore, 

Since sin2 nπ ±

Z

π

= 1, sin2 (nπ) = 0, and lim

1 n2+β

nπ−

Z

nπ− π 2 −α+4+2β

Z

nπ−

2n

nπ− π 2

nπ+ π 2

g(x) dx +

g(x) dx < nπ+

1 n2+β

x dx + 2n

1 n2+β

−α+4+2β

Z

nπ+ π 2

x dx = nπ+

2π 2 n−α+5+2β − 4πn−α+3+β .

1 n2+β

86

Improper Riemann Integrals

For the remaining middle part of the integral, we have Z

nπ+

nπ−

1 n−2−β

Z

nπ+

g(x) dx
6 and we pick 0 < β < 2 n=1 1.8.14 Why the integral Z

1

ln(x) sin 0

  1 dx x

is improper? Prove that it converges absolutely. 1 [Hint: Use x = and any method and test that applies.] u 1.8.15 Prove: (a) The following integrals do not converge absolutely. Z ∞ Z ∞ cos(x) sin(x) dx and dx x + 1 x+1 0 0 (b) The following integrals converge absolutely. Z ∞ Z ∞ sin(x) cos(x) dx and dx 2 (x + 1) (x + 1)2 0 0 (c) The following integrals converge conditionally. Z ∞ Z ∞ sin(x) cos(x) dx and dx x + 1 x+1 0 0 [Hint: Use integration by parts and (b) to prove (c), or apply Abel’s Test, Theorem 1.7.12.] 1.8.16 Combine integration by parts with some information provided in Examples 1.7.18 and 1.7.20 to prove the existence (conditional convergence) of the following three integrals without using the Cauchy Test, 1.7.11. Z 1 Z ∞ Z ∞ sin(x) sin(x) sin(x) dx = dx + dx, (a) x x x 0 1 0

Z (b)

Improper Riemann Integrals, Definitions, Criteria Z ∞ ∞

sin x2 dx, and (c) cos x2 dx

0

87

0

[Hint: Split the last two integrals about the number 1. The parts of the integrals on [0, 1] are proper. For the parts of the integrals on [1, ∞), use x2 = u and as we suggest in the hint of the previous Problem, use integration by parts and notice that the new integrals you get converge absolutely, or apply Abel’s Test, Theorem 1.7.12. (See also Examples 3.1.8 and 3.6.2, and Problems 3.2.25 and 3.2.26.)] 1.8.17 Prove that the four integrals converge only conditionally. Z ∞ Z ∞ (a) sin (ex ) dx, (b) cos (ex ) dx, (use u = ex ) 0 0 Z ∞ Z ∞ cos(x) sin(x) √ √ dx, (d) dx. (c) 2 x +1 x2 + 1 0 0 1.8.18 (a) Prove that Z 0

∞

hπ 2

i − arctan(x) dx = +∞.

(b) If a > 0 constant, prove that the following integral is positive finite. Z ∞ π 2 − arctan(x) dx. x a (c) If a = 0, prove that the integral in (b) is +∞. 1

  1 1 sin dx is improper at x = 0 x x 0 and it converges conditionally (not absolutely). (See also Problem 3.2.22.) Z

1.8.19 Prove that the integral

1.8.20 Prove that ∀ m, n ∈ N the integral Z ∞ (m − 1)! n! 1 1

xn (x + 1)−m−n−1 dx = = m+n = m+n . (m + n)! m m 0 m n [Hint: Prove that the integral exists and then use integration by parts n to prove the recursive formula Im,n = Im,n−1 . m+n Also, recognize it as a case of the Beta function, Section 3.11 and see Problem 1.8.25.]

88

Improper Riemann Integrals

1.8.21 Prove that for any n ≥ 0 integer the following two integrals exist: Z ∞ Z ∞ n −x2 (a) x e dx, (b) xn e−|x| dx. −∞

−∞

If the integer n is odd, their values are zero. Why? (See and compare Problem 3.13.6.) 1.8.22 (a) Prove (by elementary method) that r √ n+1 n n ∀ n ∈ N, < n! ≤ . 2 2 (b) Use (a) to prove that if n ∈ N, then √ n lim n! = ∞. n→∞

[Compare with Problem 3.13.66, (c).] 1.8.23 (a) If (an ) with n ∈ N is a sequence of positive numbers and the limit √ an+1 lim exists, then prove that the limit lim n an also exists, n→∞ an n→∞ and the two limits are equal. (See also Problem II 1.2.4.) (b) Use (a) to prove that √ n n!  ∞  1 lim = = . n→∞ n ∞ e 1.8.24 Consider positive integers α1 , α2 , . . . , αn whose sum is m ≥ 2, a1 , a2 , . . . , an different one another real or complex numbers, P (z) a polynomial of degree ≤ m − 2. Let in the partial faction decomposition of P (z) , (z − a1 )α1 (z − a2 )α2 . . . (z − an )αn the coefficient of the fraction Prove that

n X k=1

Ak = 0.

1 be Ak , for k = 1, 2, . . . , n. z − ak

Improper Riemann Integrals, Definitions, Criteria

89

1.8.25 (a) For integers m and n such that 0 ≤ m < n − 1, prove that Z ∞ Z ∞ n−m−2 xm x dx = dx = n (1 + x) (1 + x)n 0 0 Z ∞ Z ∞ (u − 1)n−m−2 (u − 1)m du = du = un un 1 1 m   n−m−2 X X n − m − 2 (−1)k m (−1)k = . k n−m+k−1 m+k+1 k k=0

k=0

(b) For real numbers α and β such that −1 < α < β − 1, prove that the integral Z ∞ Z ∞ β−α−2 xα u Iβ := dx = du β (1 + x) (1 + u)β 0 0 exists, and for any γ ≥ β, we have the recursive formula   α+1 Iγ+1 = 1 − Iγ . γ [See also Examples 3.1.6, (b), II 1.7.7 and Problems 1.8.20, 3.2.47, 3.7.18, 3.13.64, and properties (B, 5) and (B, 8) of the Beta function. Then practice by computing some integrals for some real numbers α and β and γ that satisfy the above condition.]

1.9

Three Important Notes on Chapter 1

(I) By now, the definitions, theorems, examples, and problems presented in this chapter make it clear that there is an analogy between the improper integrals and the series of numbers. Besides the Integral Test that for certain types of improper integrals exposes an immediate connection of the two, we have seen many analogous results. When a series converges absolutely, then it can be summed in any way without altering the final answer. Similarly, when an improper integral converges absolutely the limiting process that defines and evaluates it plays no significant role for the final answer.

90

Improper Riemann Integrals

If a series converges conditionally (not absolutely), then by the Riemann rearrangement Theorem on series (see bibliography, e.g., Rudin 1976, 75–78), we can find a way to sum it up in order to achieve any answer we wish. E.g., the series   ∞ X (−1)n−1 1 1 (−1)n−1 = lim 1 − + − . . . + = ln(2) n→∞ n 2 3 n n=1 converges conditionally, and as the stated limit of the initial partial sums gives the answer ln(2). Otherwise, we can find ways to sum it up and achieve as limit any number, or ±∞, or even the limit does not exist. Also, notice that ∞ X n= odd ≥1

(−1)n−1 = ∞, n

But, we do not say that

∞ X

and

n= even ≥2

(−1)n−1 = −∞. n

∞ X (−1)n−1 = −∞ + ∞, which is meaningless. n n=1

Instead, we have   ∞ X (−1)n−1 (−1)n−1 1 1 = lim 1 − + − . . . + = ln(2). n→∞ n 2 3 n n=1 This phenomenon is also true for the improper integrals that converge conditionally (not absolutely). E.g., the improper integral Z 0

∞

sin(x) dx = lim M →∞ x

Z 0

M

sin(x) dx, x

as we have seen, converges conditionally (not absolutely). As the stated π limit, the value of this improper integral is , as we shall see in the next 2 chapter, Example 3.1.8. If now for n = 0, 1, 2, . . . , we consider the sequence   Z (n+1)π > 0, if n is even ≥ 0, sin(x) dx = an =  x nπ  < 0, if n is odd ≥ 1, then

lim an = 0

n→∞

and

∞ X n=0

an = lim (a0 + a1 + a2 + . . . + an ) = n→∞

π . 2

But again, we can find ways to sum up this sequence in order to achieve as limit any number, or ±∞, or even the limit does not exist.

Improper Riemann Integrals, Definitions, Criteria

91

We also have ∞ X

an = ∞

and

n= even ≥0

Z But, if we say 0

∞

∞ X

an = −∞.

n= odd ≥1

sin(x) dx = ∞ − ∞, renders the integral meaningless. x

However, according to Definition 1.1.2 of the improper integral, we have Z ∞ Z M sin(x) π sin(x) dx = lim dx = , M →∞ x x 2 0 0 as we prove in Example 3.1.8. So, in the definition of the above improper integral, the stated particular limit process is absolutely significant. Similar things can be interpreted with any improper integral that converges conditionally. In another situation, we have seen that certain improper integrals defined as double limits of two independent limiting processes do not exist, but the particular limiting process of the Cauchy principal value gives a real number as an answer. (II) Beyond the Fundamental Theorem of Integral Calculus, 1.1.1, we also have the following theorem concerning the integrals of functions. Theorem Z1.9.1 Let f : R → R be a real Riemann integrable function, |f (x)|dx < ∞. We pick any point a ∈ R and define the

such that function

R

Z F (x) :=

x

f (x) dx,

∀ x ∈ R.

a

Then: (a) The function F (x) is continuous. (b) If also f (x) is bounded and so there is M ≥ 0 such that |f (x)| ≤ M , then F (x) is Lipschitz19 continuous with Lipschitz constant M . That is (by the definition of Lipschitz continuity), ∀ u ∈ R and ∀ v ∈ R, it holds |F (u) − F (v)| ≤ M |u − v|. 19 Ruldolf

Lipschitz, German mathematician, 1832-1903.

92

Improper Riemann Integrals

(The proof of this theorem is easy.) (III) Final Note and Example: In a course of mathematical analysis, we prove that the four series ∞ X sin(n) , n n=1

∞ X (−1)n sin(n) , n n=1

∞ X cos(n) , n n=1

∞ X (−1)n cos(n) n n=1

converge conditionally (not absolutely). [E.g., see Rudin 1976, page 71, Theorem II 1.44.20 See also Problem II 1.2.8, (c) and hint.] Also, the following two series ∞ X (−1)n | sin(n)| , n n=1

∞ X (−1)n | cos(n)| n n=1

converge conditionally (not absolutely). The proof of conditional convergence is hard and depends on Number Theory and continuous fractions. [See Kumchev 2013. This question, however, appeared in Lang 1983, pp. 195, exercise 3 (b), to be answered by the students!] The proof of absolute divergence of all of the six series above is easy. We observe that, for all n ∈ N, | sin(n)| < 1 and | cos(n)| < 1 and so sin2 (n) < | sin(n)| and

cos2 (n) < | cos(n)|.

Then, for all n ∈ N, we have: 0
0, such that ∀ u ∈ D and ∀ v ∈ D, |u − v| < δ =⇒ |f (u) − f (v)| < . That is, the choice of δ > 0 depends only on the `a-priori choice of  > 0 (not on the points u ∈ D and v ∈ D), i.e., δ = δ(). The simple continuity on D is defined point-wise and so for the simple continuity at a point w ∈ D, the positive number δ depends on both `a-priori choices of  > 0 and w ∈ D, i.e., δ := δ(, w) > 0. We notice that the uniform continuity depends on both the function f and its domain D and it implies the simple continuity. (Simple continuity does not necessarily imply uniform continuity.) Also, it is obvious from the definition that if f is uniformly continuous on D, then it is uniformly continuous on any subset of D. We also say that a function f : R −→ R is H¨ older continuous of order α > 0 (uniformly) on (the whole) R, if ∃ M ≥ 0 constant : ∀ u ∈ R and ∀ v ∈ R, |f (u) − f (v)| ≤ M |u − v|α . In particular, if α = 1, we say that f is Lipschitz continuous21 (uniformly) on R. 21 There are several variations of these definitions of H¨ older or Lipschitz continuities, which depend on the validity of the above definition at some points only, or locally in a domain, or globally in a whole domain, and on the fact that the domain is bounded

94

Improper Riemann Integrals We can prove the following:

(a) If α > 1, then f is identically constant. (b) If f is differentiable and its derivative is (uniformly) bounded on R, then f is Lipschitz continuous. (c) If f (x) is defined by

f (x) =

 0,          1

if x ≤ 0, , if 0 < x ≤

ln(x)       −1    , ln(2)

if x >

1 , 2

1 , 2

then f (x) is uniformly continuous over the whole R, but not H¨older continuous of any order. [Hint: For the non H¨ older continuity consider any α > 0, u = x, v = 0 and argue by contradiction on the inequality condition of the definition near v = 0.] In many instances and problems, we encounter the importance of the real functions which are uniformly continuous on their domains. Here, we would like to make a compendium on the uniformly continuous real functions. So, we consider real functions f : D −→ R with domain a set D ⊆ R and we can provide proofs or counterexamples to the following facts: 1. A Lipschitz continuous function is uniformly continuous. 2. If a function is uniformly continuous on D, then so is its absolute value and any multiple of it by a constant. 3. If two functions are uniformly continuous on D, then their sum and difference are uniformly continuous. or unbounded. There are many properties of these local or global continuities, which may vary drastically depending on considering the respected functions on bounded or unbounded domains. (In the above problem, for the sake of brevity, we have considered as domain the whole unbounded R, omitting all the other cases.)

Improper Riemann Integrals, Definitions, Criteria

95

4. If two functions are uniformly continuous on D, then their maximum and minimum22 are uniformly continuous. 5. The product and the quotient of two uniformly continuous functions may not be uniformly continuous. 6. The composition of two uniformly continuous functions is uniformly continuous. 7. A continuous function with domain D, a closed and bounded set, is uniformly continuous. 8. A uniformly continuous function defined on an open interval (a, b), with ∞ < a < b < ∞, can be extended continuously to the endpoints a and b. 9. The uniform limit of a sequence of uniformly continuous functions is uniformly continuous. 10. The point-wise limit of a sequence of uniformly continuous functions may or may not be uniformly continuous or even continuous. 11. A continuous function defined in (a, b), where −∞ ≤ a < b ≤ ∞ such that the limits lim+ f (x) and lim− f (x) exist, is uniformly x→a

x→b

continuous. 12. A uniformly continuous function with domain D a bounded set is bounded. 13. A uniformly continuous function with domain D an unbounded set may or may not be bounded. 14. A H¨older continuous function is uniformly continuous, but not viceversa. 15. An absolutely integrable and uniformly continuous function defined in R has limits lim f (x) = 0. x→±∞

16. Now, verify or give counterexamples to the statements of the following table and hint: 22 To prove this, we can use the preceding facts and the known formulae for the maximum and minimum of two functions f and g

max(f, g)(x) := max[f (x), g(x)] =

1 (f + g + |f − g|) (x) 2

min(f, g)(x) := min[f (x), g(x)] =

1 (f + g − |f − g|) (x). 2

and

96

Improper Riemann Integrals Continuous function f : D→R D⊆R

D = (a, b) with −∞ < a < b < ∞

D = [a, ∞) with a ∈ R

D=R

f bounded

May/may not be unif. cont.

May/may not May/may not be unif. cont. be unif. cont.

f unbounded

Not uniformly continuous

May/may not May/may not be unif. cont. be unif. cont.

f 0 bounded

Uniformly continuous

f 0 unbounded

May/may not be unif. cont.

May/may not May/may not be unif. cont. be unif. cont.

f bounded and f 0 unbounded

May/may not be uniformly continuous

May/may not May/may not be uniformly be uniformly continuous continuous

f unbounded and f 0 unbounded

Not uniformly continuous

May/may not May/may not be uniformly be uniformly continuous continuous

Uniformly continuous

Uniformly continuous

{If the domain D is of the type [a, b), (a, b], (−∞, b], we can figure out the answers to the same questions with the help of this table.} [Hint: Examine the following table and check the examples provide and give the proofs asked.

Improper Riemann Integrals, Definitions, Criteria Continuous function f : D→R D⊆R

f bounded

D = (a, b) with −∞ < a < b 0. 1 Leonhard

Euler, Swiss mathematician, 1707–1783. Denis Poisson, French mathematician, 1781–1840. 3 Johann Carl Friedrich Gauß, German mathematician, 1777–1855. 2 Sim´ eon

DOI: 10.1201/9781003433477-2

99

100

Improper Riemann Integrals y

1.0 0.8 0.6 0.4 0.2 -2

O

-1

1

-0.2

2

x

2

FIGURE 2.1: Function y = e−x

It is usually more convenient to switch to polar coordinates when we work with circular discs centered at the origin. So, x = r cos(θ),

x2 + y 2 = r2 ,

y = r sin(θ),

dxdy = rdrdθ

and D(0, a) = { (r, θ) | 0 ≤ r ≤ a, 0 ≤ θ ≤ 2π}. Hence the integral in polar coordinates r, θ is Z Z Z 2π Z a 2 −(x2 +y 2 ) e−r r dr dθ = e dxdy = 0

0

D(0,a)

"

−e−r 2π 2

2

#a

  2 = π 1 − e−a .

0

(b) Take the limit as a → ∞ to find Z Z   2 2 2 e−(x +y ) dxdy = lim π 1 − e−a = π(1 − 0) = π. a→∞

R2 2

2

[Since the function e−(x +y ) is continuous and positive in R2 , advanced integration theory proves that it is integrable and for any non-negative integrable function any legitimate limit process yields the unique non-negative real or +∞ value of its integral. See Section 3.6 condition I.] (c) Now we view the integral Z Z R2

2

e−(x

+y 2 )

dxdy,

Calculus Techniques

101

as the limit of integrals over the rectangles Ra = [−a, a] × [−a, a], as 0 < a → ∞, i.e., Z Z Z aZ a 2 2 2 2 e−x e−y dxdy = π. e−(x +y ) dxdy = π, or lim lim a→∞

a→∞

−a

Ra

Then

a

Z

e

lim

a→∞

−x2

Z

a

dx ·

−a

e

−y 2

−a

 dy

= π.

−a

(d) In Example 1.7.1, we have proved that Z a Z ∞ 2 −x2 lim e dx = e−x dx exists. a→∞

−a

−∞

So, the last equation can be rewritten as     Z a Z a 2 2 lim e−x dx · lim e−y dy = π. a→∞

a→∞

−a

−a

These two limits are the same, and so 

a

Z lim

a→∞

Z

e

−x2

2 Z dx =

−a

∞

∞

e

−x2

2 dx = π.

−∞ 2

e−x dx > 0 as an integral of a positive function,

Finally, since −∞

we can take square roots of both sides of the last equality to get the result Z ∞ Z ∞ √ √ 2 2 e−x dx = π, which implies, ∀ c ∈ R, e−(x−c) dx = π. −∞

−∞ 2

Remark: Since f (x) = e−x is an even function over R, we have: √ R∞ R0 2 2 (a) 0 e−x dx = −∞ e−x dx = 2π . (b)

R∞ c

2

e−(x−c) dx =

Rc −∞

2

e−(x−c) dx =

√ π 2 ,

∀ c ∈ R.

Important Note: If a function of two variables is non-negative or nonpositive (i.e., does not change sign), then its double integral over a domain can be manipulated in any way and iterated in any order without affecting the final result. Otherwise, we would need the integral of its absolute value over its domain to exist or some other conditions that we will study later in Section 3.6. In the above example, the function

102

Improper Riemann Integrals 2

2

e−(x +y ) in R2 is positive. Therefore, our manipulations and iterations do not alter the existence Z and the uniqueness of the final answer. x

2

e−t dt cannot be found in closed form, we de-

Since the integral 0

fine, by means of the integral computed here, and use the following two standard functions in theory and application. (1) The error function erf(x) Z x 2 2 e−t dt. (2.2) ∀ x ∈ R, erf(x) = √ π 0 Notice that the error function is odd, i.e., erf(−x) = −erf(x), increasing, −1 < erf(x) < 1,

erf(0) = 0,

lim erf(x) = 1, and lim erf(x) = −1.

x→∞

x→−∞

This function is very important to application and has been tabulated. (2) The complementary error function erfc(x) Z x Z ∞ 2 2 2 2 ∀ x ∈ R, erfc(x) = 1 − erf(x) = 1 − √ e−t dt = √ e−t dt. π 0 π x Notice that erfc(x) = 1 − erf(x) is decreasing, 2 > erfc(x) > 0, erfc(0) = 1, lim erfc(x) = 0, and x→∞

erfc(−x) = 2 − erfc(x), lim erfc(x) = 2.

x→−∞

For another useful integral representation of either function, see and adjust Example 3.1.15 and Problem 2.3.31. In this context, we also encounter the scaled complementary error function, defined by 2 erfcx(x) := e(x ) erfc(x),

which sometimes is more convenient for numerical computations. Example 2.1.1 For any b ∈ R and s > 0, we have Z √sb Z b √ 2 dv v= s u −su2 e−v √ = e du = s r0 r 0 √
1 π √
 1 π erf sb = 1 − erfc sb . 2 s 2 s Also notice, by Hˆ opital’s rule, we find !  r    Z b Z b √
1 π 0 −0u2 lim erf sb = = ... = b = e du = 1 du . s→0 2 s 0 0 0 N

Calculus Techniques

2.2

103

Applications

Application 1: In probability and statistics, any real continuous random variable X takes values x in a set A, where A ⊆ R has nonempty interior. To such a random variable, we associate a probability density function y = f (x), i.e., a function by which we find the probabilities Z b

P r(a ≤ x ≤ b) =

f (x)dx, a

for any real constants a and b with a ≤ b. Notice that P r(a < x < b) = P r(a < x ≤ b) = P r(a ≤ x < b) = P r(a ≤ x ≤ b). We set f (x) = 0 for every x ∈ R−A. Then, without loss of generality, the function f : R −→ R satisfies the following three properties: 1. y = f (x) is piecewise continuous. 2. f (x) ≥ 0, ∀ x ∈ R. Z ∞ 3. f (x) dx = 1. −∞

Now we define the rth moment about the origin of X to be Z ∞ µ0r = xr f (x) dx, −∞

for r = 0, 1, 2, 3, . . . . Depending on f (x) and the value of r, µ0r may or may not exist. Notice that µ00 = 1. In applications, we are especially interested in the following three computations: (a) The expected value or the mean value or simply the mean or average value of the continuous random variable X associated to the probability density function f (x), is denoted and defined by: Z ∞ µ := E(X) := µ01 = xf (x) dx. −∞

Depending on f (x), µ := E(X) may or may not exist.

104

Improper Riemann Integrals

(b) The variance of X is defined and denoted by and equal to: Z ∞ var(X) := V (X) := σ 2 := (x − µ)2 f (x) dx = −∞ Z ∞
2 2 x − 2xµ + µ f (x) dx = µ02 − 2µµ + µ2 = µ02 − µ2 . −∞

Depending on f (x), var(X) := σ 2 may or may not exist. The value p σ = var(X) is called standard deviation from the mean of the continuous random variable X. (c) The moment-generating function of X denoted and defined by: Z

∞

MX (t) =

etx f (x) dx.

−∞

Notice that MX (0) = 1. Depending on f (x), MX (t) may or may not exist in the whole R or in some intervals of R or may exist for t = 0, only. (See also Problem 3.2.34.) Now, for any real constants µ and σ > 0, we define the function n(x; µ, σ) =

1 x−µ 2 1 √ e− 2 ( σ ) σ 2π

for

− ∞ < x < ∞.

We can prove that this function qualifies as a probability density function. It obviously satisfies properties (1.) and (2.) above, and we must prove property (3.). x−µ We use the change of variables u = √ and the result of Integral 2σ (2.1) to get: Z ∞ Z ∞ 1 x−µ 2 1 n(x; µ, σ)dx = √ e− 2 ( σ ) dx = σ 2π −∞ −∞ Z ∞ Z ∞ √ 2 2 1 1 1 √ √ e−u 2 σ du = √ e−u du = √ π = 1. π −∞ π σ 2π −∞ If a real random variable X = x has probability density function the function n(x; µ, σ), we say that X has the normal distribution with mean value µ and variance σ 2 . When µ = 0 and σ = 1, then 1 2 1 n(x; 0, 1) = √ e− 2 x 2π

for

−∞ 0 and sign(β) = −1 if β < 0. For β = 0, the integral is trivially zero.) Here, we are going to evaluate this important integral Z ∞ sin (βx) dx x 0 where β 6= 0 is a real constant, which is called Dirichlet sine integral. To apply the Main Theorem, 3.1.1, we need some absolute convergence and limit process. To this end, for any β ∈ R fixed, we consider this integral to be a part of the following more general integral Z ∞ sin (βx) F (α) = e−αx dx, ∀ α ≥ 0. x 0 The parameter α ∈ I := [0, ∞) of this integral is considered to be the variable of the function F (α). If α = 0, we notice that the integral Z ∞ sin (βx) dx exists, F (0) = x 0 as we have already proved in Example 1.7.18 and its Remark 1.7.2. If α > 0, we shall prove that the integral F (α) exists because it converges absolutely. We can show this as follows:

132

Improper Riemann Integrals

sin(βx) is continuously defined at x x = 0 by assigning the value f (0) = β. This follows by L’ Hˆopital’s rule, since sin(βx) β cos(βx) β lim = lim = = β. x→0 x→0 x 1 1 Next, at ∞ we get sin(βx) lim = 0. x→∞ x Then, by a simple modification of the Extreme Value Theorem for Continuous Functions (study it one more time and make the modification in this situation), we get that there exists a finite constant M such that: sin(βx) < ∞. 0 < M = maximum 0≤x 0, the integral Z ∞ sin(βx) F (α) = e−αx dx x 0 converges absolutely, and therefore it converges. For α = 0, as we explained at the beginning, the integral Z ∞ sin(βx) F (0) = dx x 0 converges conditionally. In the sequel, we will firstly show that for any given β 6= 0, the realvalued function F (α) is continuous for all 0 ≤ α < ∞. Notice here that the parameter interval is I = [0, ∞) including α = 0. (When β = 0 then F (α) is obviously continuous, since in this case F (α) = 0, constant for all α.) Secondly, for any given β 6= 0 fixed, we will evaluate F (α) for any α > 0. Then we will compute the limit of F (α) as α → 0+ , and so by the Continuity Part Z ∞of the Main Theorem, 3.1.1, we will find the sin (βx) value of the integral dx, which exists (as we already knew). x 0 (Note: This kind of method is used quite often with improper integrals. We imbed the given integral into a more general one which we can

Real Analysis Techniques

133

evaluate, and then we manipulate the value of the more general integral to find the value of the given particular integral.) To prove the continuity of F (α) in α, with α ∈ I = [0, ∞), we must apply Part (I) of the Main Theorem, 3.1.1. We therefore need to find an appropriate function g(x), as in the Theorem. This is not so immediate in this example because we include the value α = 0, and so we cannot find a good g(x) that works throughout the whole I = [0, ∞). Therefore, we are going to work as follows: For any given β 6= 0 fixed and 0 ≤ α < ∞, we first write Z 1 Z ∞ sin (βx) sin (βx) F (α) = e−αx dx + e−αx dx. x x 0 1 That is, we write the integral F (α) as a sum of these two smaller parts. Then we need to prove that each of these parts is itself a continuous function of α. In doing so, we may use the Continuity Part of the Main Theorem, 3.1.1, applied to some or all of the smaller parts of this integral. Hence, the part Z 1 sin (βx) dx (*) e−αx x 0 is a continuous function of α, for all 0 ≤ α < ∞. This is so because we can apply the Continuity Part of the Main Theorem, 3.1.1, if, for example, we pick g(x) = M,

where

0 < M = maximum 0≤x≤1

for which

| sin (βx)| < ∞, x

1

Z

g(x)dx = M < ∞. 0

Notice, of course, that ∀ α ≥ 0 constant, 0 < e−αx ≤ 1, ∀ x ∈ [0, ∞) and M is a positive finite constant, guaranteed by the Extreme Value Theorem for Continuous Functions on a closed and bounded interval [a, b]. Next, the second partial integral of F (α) Z ∞ sin (βx) dx e−αx x 1 must be first transformed in the following way: We perform integration by parts after choosing u=

1 x

and

dv = e−αx sin(βx)dx.

134

Improper Riemann Integrals

The indefinite integration of dv up to a constant, which we take here to be zero, gives: v(x) =

−e−αx [α sin(βx) + β cos(βx)] . α2 + β 2 α

β and sin(φ) = p to implicitly de2 + α + β2 fine the function φ(α) (remember β 6= 0 is fixed). By adding to φ multiples of 2π whenever we go across such multiples, the function φ(α) can be defined to be continuous in α. So, we can rewrite We let cos(φ) = p

α2

β2

v(x) = v(x; α) =

−e−αx sin[βx + φ(α)] p α2 + β 2

and v(x) = v(x; α) is continuous in α ≥ 0, since β = 6 0. Now performing integration by parts, we obtain Z ∞ Z ∞ −αx e−α sin(β + φ) e sin(βx + φ) −αx sin (βx) p dx = p e − dx. 2 2 2 x α +β x α2 + β 2 1 1 We observe that the function (**)

e−α sin[β + φ(α)] p α2 + β 2

is continuous in α, for all 0 ≤ α < ∞, because β 6= 0. Next, the integral Z ∞ −αx e sin(βx + φ) p (***) dx 2 x α2 + β 2 1 is continuous in α for 0 ≤ α < ∞, because we can apply the Continu1 . ity Part (I) of the Main Theorem, 3.1.1, if we choose g(x) = |β|x2 Finally, F (α) is continuous in α, for 0 ≤ α < ∞, since it is the sum of the three continuous functions stated in (*), (**), (***) above. Now we are going to evaluate the integral F (α) for any α > 0. To do this, we look at the integral as a function of the parameter β. That is, we let Z ∞ sin (βx) H(β) = e−αx dx with β ∈ R. x 0 As we proved earlier, H(β) exists ∀ β ∈ R and ∀ α ≥ 0. (We trivially observe H(0) = 0, ∀ α ≥ 0.)

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By the Differentiability Part (II) of the Main Theorem, 3.1.1, d we can compute the derivative [H(β)] by differentiating under the dβ integral sign because:     sin (βx) d sin (βx) d e−αx = e−αx = e−αx cos(βx), dβ x dβ x and so for fixed given α > 0, regardless of β ∈ R, this derivative is absolutely bounded by the positive function g(x) = e−αx , whose integral Z ∞ 1 e−αx dx = is finite. Therefore, the Differentiability Part of the α 0 Main Theorem, 3.1.1, applies to obtain: Z ∞ α d [H(β)] = e−αx cos(βx) dx = 2 . dβ α + β2 0 Since H(0) = 0, we find that H(β) satisfies the initial value-problem  d α    dβ [H(β)] = α2 + β 2 ,   

H(0) = 0.

The solution of this initial value-problem is found easily to be   Z β α β H(β) = du = arctan . 2 + u2 α α 0 So, we have obtained the important result that for all α > 0 and β ∈ R constants, we have   Z ∞ β sin (βx) e−αx dx = arctan . x α 0 Hence, by continuity with respect to α in [0, ∞), we obtain Z ∞ sin(βx) F (0) = lim F (α) = lim dx, e−αx + + x α→0 α→0 0 or, we find that the evaluation of the Dirichlet sine  π  ,   2       Z ∞  sin(βx) β dx = lim+ arctan = 0,  x α α→0 0       − π , 2

integral is if β > 0, if β = 0, if β < 0.

136

Improper Riemann Integrals Z ∞ ∞ sin(x) sin(2x) π [For example, dx = dx = , etc. See also x x 2 0 0 Problem 3.7.1, for another method of proving this important result!] sin(βx) Also, since the function is even in R, we have: x   π, if β > 0,      Z ∞  sin(βx) dx = 0, if β = 0,  x −∞      −π, if β < 0. Z

N Example 3.1.9 In the course of computation in the previous Example, we have shown the following useful integral result that we sometimes find in integral tables:   Z ∞ β sin (βx) dx = arctan . ∀ α > 0 and β ∈ R constants, e−αx x α 0 As we have proved above that this integral is differentiable in β, for β ∈ R, so we can prove that it is differentiable in α, for α > 0. Taking the derivatives of both sides, we find the known integral: Z ∞ β ∀ α > 0 and β ∈ R constants, e−αx sin (βx) dx = 2 . α + β2 0 N Example 3.1.10 In advanced problems and computations (e.g., see Applications 1 and 4 of Subsection II 1.7.6), many times we need to write the function of absolute value in integral form. Thus: Z 2β ∞ sin(βx) For any − ∞ < β < ∞, we have |β| = dx. π 0 x N Example 3.1.11 The function Z Z 1 ∞ sin(βx) 2 ∞ sin(βx) F (β) = dx = dx, π −∞ x π 0 x is defined for every β ∈ R, is discontinuous at β = 0 and its range is the three element set {−1, 0, 1}, if β < 0, β = 0, β > 0, respectively. This function can be used as a discontinuous multiplier or factor. [See also Problems 3.2.23, (b), II 1.7.106.] N

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Example 3.1.12 Let α ∈ R. Then,   Z ∞ Z ∞ Z ∞ sin2 (αx) sin2 (|α|x) −1 2 dx = dx = sin (|α|x) d = x2 x2 x 0 0 0  ∞ Z ∞ − sin2 (|α|x) 2|α| sin(|α|x) cos(|α|x) + dx = x x 0 0 Z ∞ Z ∞ sin(2|α|x) π 1 − cos2 (αx) dx. 0 + |α| dx = |α| · = x 2 x2 0 0 (See also Problem 3.9.27.) N Example 3.1.13 Let α ∈ R. Then, Z ∞ Z ∞ 1 − cos(|α|x) 1 − cos(αx) dx = dx = 2 x x2 0 0
Z ∞ Z ∞ 2 sin2 |α| x2 π sin2 (|α|u) x=2u dx = du = |α| . x2 u2 2 0 0 (See also Problem 3.9.27.)

N

Example 3.1.14 In Example 3.1.8, we have used the Technique of Differentiability of the Main Theorem, 3.1.1. We continue with another important example illustrating this technique. Note that the final result obtained here is also obtained in Problem 3.5.16 by real analysis, and in Example II 1.7.18 and Problem II 1.7.32, (a), by complex analysis. (Also, compare the result of this example with Problem 3.2.43.) In this example, we are going to evaluate the following integral of Laplace3 stated here as: Z ∞ 2 I(β) = e−αx cos(βx) dx, ∀ a > 0 and ∀ β ∈ R. 0

We notice that for any α > 0 constant and any given β ∈ R constant, the integral converges absolutely, since by Problem 2.3.11 r Z ∞ Z ∞ 1 π −αx2 −αx2 cos(βx) dx ≤ e dx = < ∞. e 2 α 0 0 This also proves that I(β) is a continuous function for all β ∈ R. (Here, we are interested in the finiteness of this integral regardless of β and not in its exact value. The exact value proves finiteness, but most of the times we can prove finiteness without knowing the exact value.) 3 Pierre-Simon

Laplace, French mathematician, 1749–1827.

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Next, we show that I(β) is differentiable. We have: i 2 d h −αx2 e cos(βx) = −xe−αx sin(βx). dβ Then for any real constant β 2 2 −xe−αx sin(βx) ≤ xe−αx and for any α > 0 if we let u = αx2 , we find Z ∞ Z ∞ 2 du 1  −u ∞ 1 1 xe−αx dx = e−u = −e ·1= < ∞. = 0 2α 2α 2α 2α 0 0 So, we can apply Part(II) of the Main Theorem, 3.1.1, to get ! Z ∞ Z ∞ 2 d e−αx −αx2 [I(β)] = = e (−x) sin(βx) dx = sin(βx) d dβ 2α 0 0  ∞ Z ∞ 2 β β −αx2 sin(βx) e − e−αx cos(βx) dx = − I(β). 2α 2α 0 2α 0 Therefore, I(β) satisfies the homogeneous first-order ordinary differential equation β d [I(β)] = − I(β). dβ 2α By Problem 2.3.11, I(β) also satisfies the initial condition r Z ∞ Z ∞ 2 2 1 1 π I(0) = . e−αx dx = √ e−u du = 2 α α 0 0 The solution of the initial value-problem  d β   [I(β)] = − I(β),   2α  dβ     

I(0) =

1 2

r

π , α

dI β =− dβ when I 2α I(β) 6= 0 and then integrating. We finally find: r Z ∞ 1 π −β2 −αx2 ∀ α > 0, and ∀ β ∈ R, I(β) = e cos(βx) dx = e 4α . 2 α 0

is obtained easily by separating the variables

(Fill in the details of the integrations.)

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2

Note: Since e−αx cos(βx) is an even function in (−∞, ∞), we also have Z ∞ Z ∞ 2 2 e−αx cos(βx) dx = e−αx cos(βx) dx = 2 0 −∞ r Z 0 2 π −β2 −αx e cos(βx) dx = 2 e 4α . α −∞ (Compare this result with Example II 1.7.18 and Problems II 1.7.32, (a), and 3.2.43.) N Example 3.1.15 In this example, we prove that for any a ≥ 0 and k ≥ 0 constants but not both zero   √ Z Z 2 1 ∞ e−kx sin(a x) a 2 ∞ e−ku sin(au) √ dx = du = erf π 0 x π 0 u 2 k and thus obtain an integral representation of the error function, defined by (2.2). We easily see that the two integrals exist and are equal by means of the change of variables x = u2 . We observe that for k = 0, by letting x = u2 √ Z Z 1 ∞ sin(a x) 2 ∞ sin(au) 2π dx = du = = 1 = erf(∞). π 0 x π 0 u π2 [Also for k = ∞, we find that both integrals are 0 = erf(0).] For 0 < k < ∞ constant, the two integrals converge √ absolutely and sin(a x) ≤ √a , and then are continuous in a. Indeed, for all x ∈ R x x in any open finite interval (b, c) containing a we have: √ Z ∞ −kx Z ∞ −kx e sin(a x) ae √ dx < dx ≤ x x 0 0     Z 1 Z ∞ 1 e−k e−k −kx √ dx + a a e dx = a 2 + 0, b > 0, and c > 0,

Z

∞

(b) Prove: ∀ a > 0,

0

sin(ax) dx. x+1

∞

sin(ax) dx converges condixb c tionally. Also, find the values of b for which the integral converges absolutely.

3.2.30 Z

∞

cos(x) dx converges conditionally. x+a 0 Z ∞ cos(ax) dx converges con(b) Prove: ∀ a > 0, b > 0, and c > 0, xb c ditionally. Also, find the values of b for which the integral converges absolutely. Z ∞ cos(x) (c) Prove: dx = +∞. x 0 Z ∞ Z ∞ cos(x) cos(ax) (d) Prove: ∀ a > 0, dx = dx. x + a x+1 0 0 (See also Example 1.7.19.) (a) Prove: ∀ a > 0,

3.2.31 (a) Consider parameter α ∈ R and the function defined by  2 −(x− α x) ,  if x ∈ R − {0}, e fα (x) =   0, if x = 0, Show that: if α 6= 0, then fα (x) is continuous in x ∈ R, but if α = 0, then f0 (x) is continuous at x = 0 if we set f0 (0) = 1. Show also that fα (x) is even (about 0) in x ∈ R, for any α ∈ R. (b) For any α ∈ R, we let Z ∞ Z α 2 I(α) = e−(x− x ) dx = 0

0

2

e−(x− x ) dx. α

−∞

Show that I(α) exists and is continuous for all α ∈ R.

158

Improper Riemann Integrals

(c) Show that I(α) is differentiable ∀ α ∈ R − {0}. Z ∞ a α −(u− αu )2 (d) For any α > 0, use x = to show that I(α) = e du. 2 u 0 u d [I(α)] = 0. dα √ π . (f) Show that if α ≥ 0, then I(α) = 2

(e) Show that if α > 0, then

(g) Prove that if α ∈ R, then ∞

Z

e−(x

2

+α2 x−2 )

Z

√

0

2

e−(x

dx =

+α2 x−2 )

dx =

−∞

0

π −2|α| e . 2

(h) Prove that for any α ∈ R and β ∈ R − {0} Z

∞

e

−(β 2 x2 +α2 x−2 )

Z

0

dx =

0

−(β 2 x2 +α2 x−2 )

e −∞

√ π −2|αβ| e . dx = 2|β|

(If β = 0, the integral is equal to ∞.) (i) If α < 0, prove I 0 (α) = 4I(α) and then I(α) =

√ π 4α e −→ 0. α→ −∞ 2

3.2.32 Imitate Example 3.1.17 to prove the following Frullani integral: Z ∞ −x e − e−tx dx. ln(t) = x 0 (See also Example 3.8.1.) 3.2.33 For s ∈ [0, ∞), consider the function Z ∞ −sx e F (s) = dx. 2+1 x 0 π and lim F (s) = 0. s→∞ 2 π (b) Show that 0 < F (s) ≤ , ∀ s ≥ 0, the convergence of the integral 2 is absolute, and F (s) is decreasing. (a) Prove that F (0) =

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(c) Show that F (s) is continuous in 0 ≤ s < ∞. (d) Show that if s > 0, then ∞

Z

e−sx arctan(x) dx

F (s) = s 0

and the convergence is absolute. (e) Show that at s = 0+ the ∞

Z

e−sx arctan(x) dx

F (s) = s · 0

is of the form 0 · ∞ and  Z lim+ s · s→0

0

∞

e

−sx

 π arctan(x) dx = . 2

(f) Show that for s > 0, F (s) is twice continuously differentiable. [Hint: For any s > 0, consider a p such that 0 < p < s and use the Main Theorem, 3.1.1, on the interval [p, ∞).] (g) Prove that lim+ F 0 (s) = −∞ and lim+ F 00 (s) = ∞. s→0 s→0 Z ∞ Z ∞ x x2 [Hint: Observe that dx = ∞ and dx = ∞.] x2 + 1 x2 + 1 0 0 (h) Prove that for s > 0, the function F (s) satisfies the ordinary dif1 ferential equation F 00 (s) + F (s) = . s (i) Prove that lim+ F 00 (s) = ∞ by using (a), (c) and (h) only. s→0

(j) Use the method of variation of parameters to find that the general solution of the ordinary differential equation in (h) is Z s Z ∞ sin(t) cos(t) F (s) = c1 cos(s)+c2 sin(s)−cos(s) dt−sin(s) dt, t t 0 s where c1 and c2 are arbitrary real constants.

160

Improper Riemann Integrals The two integrals Z s sin(t) Si(s) := dt, t 0

Z and

∞

Ci(s) := s

cos(t) dt t

are very important and cannot be found in closed form. They define two new functions in [0, ∞), called integral sine and integral cosine, respectively. (See also Example 1.1.21 and Problem II 1.2.37.) π and c2 = 0. That is, the final answer 2 for the given integral with parameter F (s) is Z s Z ∞ sin(t) cos(t) π dt − sin(s) dt. F (s) = cos(s) − cos(s) 2 t t 0 s

(k) Use (a) to prove that c1 =

(l) Prove that for s ≥ 0, the function F (s) found in (k) can also be written as Z ∞ Z ∞ sin(t) cos(t) F (s) = cos(s) dt − sin(s) dt = t t s s Z ∞ Z ∞ Z ∞ sin(t − s) sin(u) sin(su) dt = du = du. t u + s u+1 s 0 0 (Notice that the convergence of these integrals is only conditional and we cannot differentiate under the integral sign in the last three.) (m) For s > 0, justify why it is legitimate to take the derivative of F (s) and with the help of the first equality in (l) show Z ∞ −sx Z
s ∞ −sx xe F 0 (s) = − dx = − e ln x2 + 1 dx = 2 x +1 2 0 0 Z ∞ Z ∞ sin(t) cos(t) dt − cos(s) dt = − sin(s) t t s s Z ∞ Z ∞ Z ∞ cos(t − s) cos(u) cos(su) − dt = − du = − du. (cont.) t u + s u+1 s 0 0 (Notice that the convergence of the first two integrals is absolute, but of the others only conditional and we cannot differentiate under the integral sign in the last three.) (See also Problems 3.2.29, 3.2.30 and II 1.7.91.)

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3.2.34 Show that the functions  2    π (1 + x2 ) ,

1 f (x) = , for x ∈ R, and g(x) =  π (1 + x2 )   0,

if x > 0, if x ≤ 0,

are probability density functions, but the mean of f (x) does not exist (except as principal value in which case it is zero), the mean of g(x) is ∞ and the moment generating functions of both do not exist (except for t = 0, only). 3.2.35 Let

Z

∞

F (t) =

te−tx dx,

for

0 ≤ t < ∞.

0

Show: (a) F (0) = 0 and F (t) = 1, ∀ t > 0. (b) F is continuous at every t > 0. (c) F is discontinuous at t = 0. In fact, lim+ F (t) 6= F ( lim+ t) = F (0). t→0

t→0

(d) Explain why the Continuity Part of the Main Theorem, 3.1.1, does not apply. 3.2.36 Prove that for all constants B ≥ 0, h ≥ 0 and H > 0, the integral Z Hπ p 1 + 4H 2 h2 + B 2 + 2B sin(2Hs) [1 + B sin(2Hs)] ds 1 + B 2 + 2B sin(2Hs) 0 has a positive value. Then examine separately its continuity when one of the constants approaches zero while the other two are fixed. 3.2.37 Z (a) Prove that for integers n ≥ m ≥ 2, the integral converges absolutely.

0

∞

sinn (x) dx xm

(b) Now for any integer n ≥ 1, explain what happens with respect to (conditional and/or absolute) of the integral Z ∞the convergence sinn (x) dx. x 0 Z ∞ sin(x) π [Hint: E.g., we have seen that dx = , but this x 2 0

162

Improper Riemann Integrals Z ∞ sin2 (x) integral diverges absolutely. Also, dx = ∞ and x 0 Z ∞ 4 sin (x) dx = ∞, and so these integrals of positive integrands x 0 diverge, etc. ]

(c) For any a 6= 0 constant, prove Z ∞ 1 − cos(ax) dx = ∞. x 0 (See also Examples 3.8.8, 3.8.10, 3.10.7 and 3.10.8 and Problems 3.2.28, 3.2.38, 3.9.17, 3.13.12, 3.13.13 and II 1.7.105.) 3.2.38 Compute precisely each of the following integrals: Z ∞ Z ∞ sin3 (x) sin3 (x) dx, and dx with k = 1, 3. (a) k x xk −∞ 0 (For k = 2, see Problems 3.9.17 and 3.9.19.) Z (b) 0

∞

sin4 (x) dx, xl

Z

∞

and −∞

sin4 (x) dx xl

with l = 1, 2, 4.

(For l = 3, see Problem 3.9.17. See also Example 3.10.8 and Problem 3.13.13.) [Hint: Prove and use trigonometric identities, such as: (1) sin3 (x) =

3 −1 sin(3x) + sin(x), 4 4

(2) sin4 (x) = sin2 (x)[1 − cos2 (x)], (3) sin(x) cos(x) =

1 sin(2x). 2

E.g.: to derive the first of these trigonometric identities expand the sin(3x) = sin(x + 2x) = ..., etc.

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Use appropriate integration by parts as many times as necessary, results previously obtained and the previous Problem. Finally, find Z ∞ Z ∞ sin3 (x) sin3 (x) 3π π dx = dx = , , 3 x 4 x 8 0 0 Z ∞ Z ∞ sin4 (x) π sin4 (x) dx = , dx = ∞, 2 x x 4 0 0 Z ∞ 4 sin (x) π dx = . ] 4 x 3 0 (c) Use a half angle formula to prove Z ∞ Z ∞ [1 − cos(x)]2 [1 − cos(x)]2 dx = 2 dx = x4 x4 0 −∞ Z 0 [1 − cos(x)]2 π 2 dx = . 4 x 3 −∞ (See also Examples 3.10.7, 3.10.8 and Problems 3.2.28, 3.13.12 and II 1.7.105.) 3.2.39 Use any results established in the text and in the problems and appropriate trigonometric identities to prove: R∞ (a) −∞ sin(ax)x2sin(bx) dx = π min{a, b}, (b)

R∞ −∞

sin2 (ax) sin2 (bx) dx x4

=

π 2

min{a, b},

where without loss of generality a ≥ 0 and b ≥ 0 are real constants. [For (a), see also Example II 1.7.38.] [Hint: In (a) you may begin with integration by parts.] 3.2.40 Obviously sin4 (x) sin2 (x)[1 − cos2 (x)] = = x4 x4 sin2 (x) − sin2 (x) cos2 (x) sin2 (x) sin2 (2x) = −4 . 4 4 x x (2x)4 0≤

164

Improper Riemann Integrals

So, Z

∞

sin4 (x) dx = x4

Z

∞

0< 0

0

0

∞

∞



0

2

sin (x) dx − 2 x4

 sin2 (x) sin2 (2x) − 4 dx = x4 (2x)4

Z 0

∞

sin2 (2x) d(2x) = (2x)4

Z ∞ sin2 (u) sin2 (x) dx − 2 du = x4 u4 0 0 Z Z ∞ ∞ sin2 (x) sin2 (x) sin2 (x) dx − 2 dx = − dx < 0, x4 x4 x4 0 0 Z

Z

Z

∞

wow and woe! Find where the error has occurred and explain why. 3.2.41 (a) Prove that for p > 1, the following integrals converge conditionally: Z ∞ Z ∞ sin(xp ) dx and cos(xp ) dx. 0

0

(b) Analyze what happens when p ≤ 1. (For each of the two integrals, examine the following cases: p = 1, 0 < p < 1, p = 0 and p < 0 separately.) (See also Example 3.10.8 and Problems 3.13.12 and 3.13.13.) [Hint: Let u = xp and work as in Example 1.7.20 or as in Problem 1.8.16.] 3.2.42 Justify why for any a ∈ R and any b 6= 0 constants, Z ∞ sin(ax) dx = 0. 2 + b2 x −∞ 3.2.43 (a) Prove that for α > 0 and β ∈ R constants, the integral Z ∞ 2 I(β) = e−αx sin(βx) dx 0

converges absolutely, and so it exists. (b) Imitate the work done in Example 3.1.14 to find the initial valueproblem that this integral satisfies.

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(c) Solve this initial value-problem to find 1 − β2 I(β) = e 4α 2α

Z

β

ρ2

e 4α dρ. 0

(We cannot put this result in closed form. Compare with Problem II 1.7.33.) (d) Find the exact numeric value of Z ∞ 2 I(β) = e−αx sin(βx) dx, −∞

Justify your answer. (Observe the integral and think first. Do not compute.) 3.2.44 (a) If α > 0, β 6= 0 and c ≥ 2 constants, prove: Z ∞ sin(βx) I(β) = e−αx dx = sign(β) · ∞. xc 0 Z ∞ 2 sin(βx) J(β) = e−αx dx = sign(β) · ∞. xc 0 (b) If c < 2, prove that these integrals are finite. 3.2.45 (a) If α > 0, β 6= 0 and c ≥ 1 constants, prove: Z ∞ cos(βx) dx = ∞. I(β) = e−αx xc 0 Z ∞ 2 cos(βx) J(β) = e−αx dx = ∞. xc 0 (b) If c < 1, prove that these integrals are finite. 3.2.46 For r ∈ R, consider the following Poisson-Dini5 integral Z π   I(r) = ln r2 − 2r cos(x) + 1 dx. 0

5 Ulisse

Dini, Italian mathematician, 1845-1918.

166

Improper Riemann Integrals

Prove: (a) I(0) = 0 and I(1) = 0. (b) I(r) is an even, i.e., [I(−r) = I(r)], and continuous function. (c) Z

2π

  ln r2 ± 2r cos(x) + 1 dx = 2I(r)

0

(d) I(r) =

  0,

if |r| ≤ 1,

  2π ln(|r|), if |r| > 1. (See also Problem 2.3.19, Subsection II 1.5.4 and Example II 1.7.50.) [Hint: Notice that for any r ∈ R, (1 − |r|)2 ≤ r2 − 2r cos(x) + 1 ≤ (1 + |r|)2 and I(0) = 0. Use this to prove that I(r) −→ 0, as r −→ 0 and so I(r) is continuous at r = 0. Now, prove that
2I(r) = I(−r) + I(r) = . . . = I r2 . So, for all n ∈ N, we have I(r) =

1  2n  I r . 2n

Then, if |r| < 1, I(r) = 0. For I(±1) = 0, use Problem 2.3.19, (d) or (f ), or Example II 1.7.50. For |r| > 1, notice that   1 1 2 2 r − 2r cos(x) + 1 = r 1 − 2 cos(x) + 2 , r r with

1 < 1 and derive the answer. [So, I(r) is continuous at r = ±1.] |r|

Different method: Use the differentiability part of the Main Thed orem, 3.1.1, to compute the derivative of the integral I(r). In the dr   x computation, you may use the half angle substitution u = tan . 2

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For instance, check and adjust the following Poisson-Dini integral: If −π < x < π, then use rationalizing substitutions to show that Z

2r − 2 cos(x) 1 dx = 2 r − 2r cos(x) + 1 r

   x  1+r + C, x − 2 arctan tan 1−r 2

where C is the constant of integration. Then find I(r) itself with the help of an initial value, e.g., I(0) = 0, I(1) = 0, etc.] 3.2.47 Prove that for all m ≥ 0 and n ≥ 2 integers such that 0 ≤ m < n − 1, we have: Z ∞ (n − m − 1)π xm 2 dx = 2 n (1 + x ) n sin[ (m+1)π ] 0 n Generalize to higher powers in the denominator! [See also Examples 3.1.6, (b), II 1.7.7, Problems 1.8.25, 3.7.18, and Properties (B, 5) and (B, 8) of the Beta function.] [Hint: See Problem 1.8.2, (2), and follow the method suggested there.] 3.2.48 Justify the application of the differentiability part of the Main Theorem, 3.1.1, to the Euler integral Z ∞ p−1 t π dt = , ∀ 0 < p < 1, 1+t sin (pπ) 0 to prove that Z ∞ p−1 t ln(t) dt = −π 2 cot(pπ) csc(pπ), 1 + t 0

∀ 0 < p < 1.

(See also Examples II 1.7.8, II 1.7.47, II 1.7.49 and Problems 3.13.63, II 1.7.142, II 1.7.145.) 3.2.49 Prove that

Z lim

n→∞

0

b

π  ,   2      sin(nx) dx = 0,  x        −π , 2

if b > 0, if b = 0, if b < 0.

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Improper Riemann Integrals

3.2.50 For a ≥ 0, consider the integral Z

∞

f (a) := −∞

2

2

ea x dx. x2 + 1

Show that we can take the derivative of f (a) and find that √ f 0 (a) = −2 π + 2af (a). Solve this differential equation and find 2

2

f (a) = π ea [1 − erf(a)] = π ea erfc(a). Then show that Z ∞ −x2 Z ∞ −x2 Z 0 2 e e e−x dx = 2 dx = 2 dx = 2 2 1 + x2 −∞ 1 + x 0 −∞ 1 + x π e [1 − erf(1)] = π e erfc(1), and with b > 0, Z

∞

−∞

Z

0

2 −∞

3.3

2

2

2

e−a x dx = 2 b2 + x2

Z 0

∞

2

2

e−a x dx = b2 + x2

2

π 2 2 π 2 2 e−a x dx = ea b [1 − erf(ab)] = ea b erfc(ab). 2 2 b +x b b

Commuting Limits and Integrals

In this section, we give the definitions of point-wise and uniform convergence of sequences and series of functions and of improper integrals. Then, we state the most important theorems concerning the most applicable sufficient conditions for commuting limits and integrals. The section is related to the continuity part of the Main Theorem, 3.1.1, of this chapter. We do not give the proofs. These can be found in practically all books of mathematical and/or real analysis. The interested reader who has not learnt these theorems yet is recommended to study them in a good book on these subjects.

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In a calculus course, we have learnt that we can integrate power series Rd term by term. This means that we can commute the integral c with the   P∞ Pk infinite summation n=0 = lim n=0 , or as we say, we can switch 0≤k→∞

the order of integration and the limit process of the infinite summation. ∞ X Hence, if the power series f (x) = an xn , taken with center a = 0 n=0

without loss of generality, converges in the open interval (−r, r) ⊆ R, where r > 0 or r = ∞, and [c, d] ⊂ (−r, r) then ! Z d Z d X ∞ ∞ Z d X n an xn dx = an x dx = f (x) dx = c

c ∞ X n=0

 an

n=0

 n+1 d

x n+1

= c

n=0 ∞ X

 an

n=0

(See also Theorem II 1.2.1.)Z In this situation, the integral

c

dn+1 − cn+1 n+1

 .

r

f (x) dx could be an improper Rie-

−r

mann integral at either or both endpoints. When this is the case, the integral is treated by means of the limiting processes we have seen so far. Notation: In what follows, we use the following notation: N = {1, 2, 3, 4, ... }, the set of natural numbers, and N0 = {0, 1, 2, 3, 4, ... } = N ∪ {0}. In the important remark immediately following Example 1.1.21, we have indicated that, whereas this commuting is always legitimate with integrals of power series and limits of integration in the open interval of convergence of the power series, it is not valid in every situation with limits of sequences or series of functions, even if the limits of integration are within the domain of definition of all functions involved. Serious mistakes may occur if such a commutation is performed while not valid! For instance: Example 3.3.1 The functions 2

fn (x) = nxe−nx ,

∀ n∈N

are all continuous at every x ∈ R and therefore Riemann integrable over any interval [a, b] ⊂ R.

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Improper Riemann Integrals

Now, for x = 0, we have that fn (0) = 0 for every n ∈ N, and for any x 6= 0, we find that d nx dn (nx) = lim = 2 n→∞ n→∞ d (enx2 ) n→∞ enx dn 1 1 x = 0. lim 2 = lim 2 = n→∞ xenx n→∞ x2 enx ±∞

lim fn (x) = lim

Therefore, for every x ∈ R, this sequence of functions converges to the continuous functions f (x) = 0, i.e., lim fn (x) = f (x) = 0, ∀ x ∈ R.

n→∞

Now we evaluate the proper Riemann integrals of all fn (x) Z

1

"

1

Z fn (x) dx =

nxe

0

−nx2

0

e−nx dx = − 2

2

#1 = 0

1 − e−n 2

and of f (x) = 0 1

Z

Z f (x) dx =

0

Hence Z and Z

1

1 1 − e−n = n→∞ 2 2

0

1

Z f (x) dx =

0

So, we see that Z lim n→∞

0

0 dx = 0. 0

fn (x) dx = lim

lim

n→∞

1

1

Z

1

lim fn (x) dx =

0 n→∞

1

fn (x) dx =

1 6= 0 = 2

0 dx = 0. 0

Z

1

lim fn (x) dx.

0 n→∞

Therefore, this is an example in which we cannot commute (switch) the order of limit and integration. Otherwise, the mistake would be imminent. We could have used the interval A = [0, ∞) instead of [0, 1] to deal with improper Riemann integrals, getting again Z ∞ Z ∞ 1 lim fn (x) dx = 6= 0 = lim fn (x) dx. n→∞ 0 n→∞ 2 0

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But, on the interval B = [1, ∞), we get Z ∞ fn (x) dx = lim n→∞ 1 " # ∞   2 −e−nx 0 − (−e−n ) lim = lim = n→∞ n→∞ 2 2 1 Z ∞ Z ∞ e−n f (x) dx = 0 dx = lim =0= n→∞ 2 1 Z ∞ 1 lim fn (x) dx. n→∞

1

Hence, with interval B = [1, ∞), the commutation of limit and integral Z ∞ Z ∞ lim fn (x) dx = lim fn (x) dx n→∞

1

1

n→∞

is valid. N This example shows that we cannot commute (switch) the order of limit and integral in general, something that we can do freely with power series as long as we stay inside their intervals of convergence. We also see that the legitimacy of this commutation could depend on the interval of integration. Thus, in this part of this section, we expose conditions under which this commuting is legitimate. The continuity part of the Main Theorem, 3.1.1, is implicitly related to this material. But first, we must begin with two definitions that follow: Definition 3.3.1 Let ∅ = 6 A ⊆ R and fn : A → R, ∀ n ∈ N be a sequence of real functions on A. Suppose there is a function f : A → R such that ∀ x ∈ A, lim fn (x) = f (x). Then we say that the sequence n→∞

of real functions (fn )n∈N converges point-wise to the function f in the set A, or the function f is the point-wise limit of the sequence of functions (fn )n∈N in the set A. pw

pw

We write fn (x) −→ f (x) in A, as n → ∞, or lim fn (x) = f (x) in A, n→∞

or simply fn (x) −→ f (x) in A, as n → ∞, or lim fn = f in A. n→∞

pw

The condition lim fn (x) = f (x) in A, can be equivalently expressed n→∞

by the x − (ε − N )−condition: Given any x ∈ A : ∀  > 0, ∃ N := N (, x) ∈ N : ∀ n ∈ N, [n ≥ N =⇒ |fn (x) − f (x)| < ] .

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Improper Riemann Integrals

Example 3.3.2 We let fn : (−1, 1] → R, fn (x) = xn , ∀ n ∈ N and

f : (−1, 1] → R, f (x) =

  0, if − 1 < x < 1,   1, if x = 1.

We readily see that the sequence of functions (fn )N ∈N converges point-wise to the function f in the set A = (−1, 1]. We also observe that all the functions (fn )N ∈N are continuous on (−1, 1], whereas the function f is not, since it has a jump discontinuity at x = 1. So, the point-wise limit of continuous functions may or may not be a continuous function. But, this f , with only one discontinuity at x = 1, is Riemann integrable on (−1, 1]. N Remark: There are more advanced examples in which a sequence of continuous functions has point-wise limit a function which is not even Riemann integrable, let alone continuous. We are not going to explore these examples here. [See Problem 1.3.9 Part II, (5).] Definition 3.3.2 Let ∅ 6= A ⊆ R and fn : A → R, ∀ n ∈ N be a sequence of real functions. Suppose there is a function f : A → R such that ∀  > 0, ∃ N := N () ∈ N : ∀ n ∈ N [(n ≥ N and x ∈ A) =⇒ |fn (x) − f (x)| < ] . Then we say that the sequence of real functions (fn )n∈N converges uniformly to the function f in the set A, or the function f is the uniform limit of the sequence of functions (fn )n∈N in the set A. un

We write lim fn (x) = f (x) on A, or lim fn = f uniformly on A, or un

n→∞

n→∞

fn −→ f on A, as n −→ ∞. Remark: (a) In the point-wise convergence, the convergence lim fn (x) = f (x) n→∞ is checked ∀ x ∈ A point by point as a convergence of a sequence

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of real numbers. The N ∈ N depends on both the ´a-priori chosen  > 0 and the individual x ∈ A. But, in the uniform convergence the number N ∈ N in the definition depends only on the ´a-priori chosen  > 0 and not on the individual x ∈ A. That is, for any given  > 0, this  > 0 is the same for all x ∈ A. Therefore, the uniform convergence depends not only on the functions of the sequence but also on their common domain ∅ 6= A ⊆ R. (b) If B ⊂ A and the convergence is uniform in A, then it is automatically uniform in B. But, it can happen that the convergence is uniform in B, even though it is not uniform in A. We will see such examples in the sequel. (c) We observe that the uniform convergence implies the point-wise convergence, that is ∀ x ∈ A, lim fn (x) = f (x), but not vice n→∞ versa, as we show in some examples. Now, Definition 3.3.2 implies the following Corollary: Corollary 3.3.1 The condition ∀  > 0, ∃ N := N () ∈ N : ∀ n ∈ N [(n ≥ N and x ∈ A) =⇒ |fn (x) − f (x)| < ] in Definition 3.3.2 of uniform convergence, is equivalent to   lim max |fn (x) − f (x)| = 0. n→∞

x∈A

In many situations, this equivalent condition is very convenient in proving uniform convergence. We also get the negative result that: If this limit is not zero, then the convergence is not uniform. Example 3.3.3 In the previous Example with fn : (−1, 1] → R, fn (x) = xn , ∀ n ∈ N and f : (−1, 1] → R, f (x) =

  0, if − 1 < x < 1,   1, if x = 1,

the convergence is not uniform but only point-wise. To see this, by means of the definition, √ we can pick any 0 <  < 1 . Then for any given n ∈ N if x satisfies n  < |x| < 1, f (x) = 0 and √ |fn (x) − f (x)| = |xn − 0| = |xn | > , ∀ x : n  < |x| < 1.

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Improper Riemann Integrals

However, the convergence is uniform on any interval [a, b] ⊂ (−1, 1). To see this, we let k = max{|a|, |b|} ∈ (−1, 1), and we notice that 0 ≤ k < 1 and f (x) = 0 and ∀ n ∈ N, max |fn (x) − f (x)| = max |fn (x)| = k n . x∈[a,b]

x∈[a,b]

Since 0 ≤ k < 1, lim k n = 0. So, by the previous Corollary, we n→∞

get

un

lim fn (x) = f (x) ≡ 0,

n→∞

on

[a, b] ⊂ (−1, 1). N

Example 3.3.4 We let sin(nx) √ , ∀x∈R n and f : R → R, f (x) = 0, ∀ x ∈ R.

fn : R → R, fn (x) =

Then (fn )n∈N is a sequence of functions that converges uniformly to the function f in R. Indeed,      sin(nx) = − 0 lim max {|fn (x) − f (x)|} = lim max √ n→∞ x∈R n→∞ x∈R n    sin(nx) ±1 √ = lim √1 = 0. lim max √ = lim n→∞ x∈R n→∞ n n→∞ n n N Example 3.3.5 In Example 3.3.1, with the sequence of functions 2 fn (x) = nxe−nx for n ∈ N and x ∈ A = [0, ∞) ⊂ R, we had seen that this sequence converges point-wise to the continuous functions f (x) = 0 for every x ∈ A = [0, ∞). Let us now check if this pointwise convergence is or is not   uniform in A = [0, ∞) by evaluating the lim

n→∞

max {|fn (x) − f (x)|} . x∈A

To find this maximum, we use the derivative of each fn (x): 2 fn0 (x) = (n − 2n2 x)e−nx , ∀ n ∈ N. This derivative is zero at 1 x= ∈ A only. Then ∀ n ∈ N the maximum of the function fn (x) on 2n A = [0, ∞) occurs at this point because ∀ n ∈ N, fn (x) ≥ 0, fn (0) = 0 and lim fn (x) = 0. Therefore, x→∞

    lim max|fn (x) − f (x)| = lim max|fn (x) − 0| = n→∞ x∈A n→∞ x∈A     1 1 1 1 lim fn = lim = 0 = = 6 0. 1 n→∞ n→∞ 2n 2e 2 2e 4n

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Since this limit is not zero, the convergence is not uniform in A = [0, ∞). But, the convergence is uniform in B = [1, ∞). This is so because for each n ∈ N, fn (x) is decreasing in [1, ∞) and     lim max|fn (x) − f (x)| = lim max|fn (x) − 0| = n→∞ x∈B n→∞ x∈B   n lim max|fn (x)| = lim |fn (1)| = lim = 0. n→∞ x∈B n→∞ n→∞ en N A theorem which at times conveniently proves the uniform convergence of sequences of real continuous functions defined on a closed and bounded interval is the following. Theorem 3.3.1 (Dini’s Theorem for uniform convergence) We consider a sequence of real continuous functions fn : [a, b] −→ R, where a < b are real numbers, for all n ∈ N0 , which satisfies the following two conditions: (a) For all x ∈ [a, b] and n ∈ N0 , fn (x) ≥ fn+1 (x). (Decreasing sequences of functions.) (b) fn (x) converges point-wise to a continuous function f (x) on [a, b]. Then fn converges uniformly to f on [a, b] (not just point-wise), as n → ∞. (For the proof of this Theorem, see, e.g., see Apostol 1974, exercise 9.9, 248, or Rudin 1976, Theorem 7.13, 150, or other books in analysis.) Example 3.3.6 Consider the decreasing sequence of the continuous 1 functions fn (x) := , n ∈ N on the interval (0, 1]. This converges 1 + nx to the continuous f (x) = 0 on (0, 1] point-wise, as n → ∞ (easy). This convergence on (0, 1] is not uniform and the interval (0, 1] is not closed. {We can easily see that the convergence is not uniform by extending the problem to [0, 1]. Then, fn (0) = 1, ∀ n ∈ N and so f (0) = 1 6= 0 and so the limit function f (x) is discontinuous at x = 0.} Now, on the interval [0, ∞) consider the decreasing sequence of the x continuous functions fn (x) := , n ∈ N. This converges to the continn uous f (x) = 0 on [0, ∞) point-wise, as n → ∞, (easy). But the convergence is not uniform, as, e.g., can be seen from sup fn (x) = ∞ 6= 0 = f (0), ∀ n ∈ N. 0≤x 0 and N ∈ N, as we have seen in the Definitions 3.3.1 and 3.3.2. Also,

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in the case of improper integrals, we can use other types of domains of integration and not just A = [a, ∞) ⊂ R. (Practice by translating these four definitions by using the  > 0, N ∈ N conditions and also by writing these definitions in the other cases of improper integrals.) Z ∞ t dx, for t ∈ [0, ∞). Notice Example 3.3.7 Consider (tx + 1)2 0 that for any 0 < M < ∞, and for any t ∈ [0, ∞), we have   0, if t = 0  Z M   t pw dx = −→ 2 1   (tx + 1) 0 1 − , if t > 0 tM + 1   0, if t = 0 −→ F (t) := , as M −→ ∞.   1, if t > 0 Now, for 0 < M < ∞ and 0 < t < ∞, we get Z M 1 t dx , = F (t) − 2 (tx + 1) tM +1 0 which, for any fixed 0 < M < ∞, approaches 1, as t −→ 0+ . Therefore, for any 0 <  < 1 there is no N = N () > 0 such that Z M t dx F (t) − < , ∀ M ≥ N. 2 (tx + 1) 0 Hence, the point-wise convergence of this improper integral is not uniform. Notice that the limit function F (t) is not continuous at t = 0 ∈ [0, ∞). N For uniform convergence of series of functions, the following criteria are very convenient. Theorem 3.3.2 (Cauchy Test for uniform convergence of s & s) (I) For sequences of functions: We consider A ⊆ R and a sequence of functions fn : A −→ R, for all n ∈ N0 . The two following claims are equivalent: (a) ∀  > 0, ∃ N ∈ N0 such that ( ∀ n > m ≥ N in N0 ) =⇒ [ |fn (x) − fm (x)| < , ∀ x ∈ A.] (b) The sequence of functions fn (x), n ∈ N0 , converges uniformly on A to some function f : A −→ R as n −→ ∞. (See Definition 3.3.2.)

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Improper Riemann Integrals

(II) For series of functions: The test (I) is adjusted to the series of a sequence of functions as n X follows: Let Sn (x) := fk (x), n ∈ N0 , be the sequence of the initial k=0

partial sums of the sequence of functions fn (x), n ∈ N0 , in (I). The two following claims are equivalent: (a) ∀  > 0, that (∀ n > m ≥ N in N0 )   ∃ N ∈ N0 such P n =⇒ |Sn (x) − Sm (x)| = k=m+1 fk (x) < , ∀ x ∈ A. (b) The series of functions

∞ X

fn (x) converges uniformly on A to

n=0

some function F : A −→ R. (See Definition 3.3.3.) Theorem 3.3.3 (Weierstraß M-Test for uniform conv. of ser.’s) We consider A ⊆ R and a sequences of functions fn : A −→ R, for all n ∈ N0 , that satisfies the following condition: ∀ n ∈ N0 , ∀ x ∈ A,

∃

Mn ≥ 0 : ∞ X and Mn < ∞.

constant

|fn (x)| ≤ Mn

n=0

Then, the series of functions

∞ X

fn (x) converges uniformly on A to

n=0

some function f : A −→ R. (See Definition 3.3.3.) Theorem 3.3.4 (Abs. Ratio Test for un. converg. of s.’s of f.’s) We consider A ⊆ R and a sequences of functions fn : A −→ R, for all n ∈ N0 , that satisfies the following condition: There exist constant 0 < q < 1 and M ∈ N such that for (at least) one n ≥ M , the function fn is bounded on A, and fn+1 (x) fn (x) ≤ q < 1, ∀ n ≥ M and ∀ x ∈ A. Then, the series of functions

∞ X

fn (x) converges absolutely and uni-

n=0

formly on A to some function f : A −→ R. (See Definition 3.3.3.) Remark 3.3.1 Under the conditions of this Theorem, if fN is bounded on A for some N ∈ N, then all functions fn with n ≥ N are also bounded on A.

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Remark 3.3.2 by the Absolute Ratio Test for series of numbers (see, Theorem 1.7.5, or better the one stated in terms of the limit superior, Apostol 1974, Theorem 8.14, p. 193, Rudin 1976, Theorem 3.34, p. 66) the series of functions converges point-wise absolutely, even without the condition on the boundedness of one of the functions. We need the condition of boundedness for the uniform convergence, as we explain in Example 3.3.9 below. The proofs of these three Theorems can be found in advanced calculus or mathematical analysis books. E.g., for the latter see Titchmarsh 1939, p. 4. Note: As we have seen in several examples the limit function of an infinite sequence or series of functions can be computed explicitly but a lot of times cannot. Example 3.3.8 Pick any number −1 < p < 1 and define fn : [−π, π] −→ R : fn (θ) = p n cos(nθ), ∀ n ∈ N0 . Then with Mn = |p|n , we have: ∀ n ∈ N0 and ∀ θ ∈ [−π, π], |fn (θ)| ≤ Mn

and

∞ X

Mn =

∞ X

|p|n =

n=0

n=0

1 < ∞. 1 − |p|

Therefore, by the Weierstraß M-Test the series ∞ X n=0

fn (θ) =

∞ X

pn cos(nθ)

n=0

converges uniformly to some function on [−π, π]. (In this way, we obtain the same result if fn (θ) = p n sin(nθ). See also Problem II 1.5.42. You may be able to solve it now. Otherwise, solve it when you study Section II 1.5.) N Example 3.3.9 The boundedness of at least one function fn for n ≥ M in the Absolute Ratio Test 3.3.4 is necessary. ex Consider fn (x) = n−1 , with n ∈ N, and with x ∈ A = [0, ∞). Then, 2 fn+1 (x) 1 = < 1, ∀ x ∈ A. ∀ n ≥ 1, fn (x) 2

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Improper Riemann Integrals

The corresponding series of these functions converges point-wise to F (x) :=

∞ X

fn (x) =

n=1

∞ X ex = 2ex , n−1 2 n=1

∀ x ∈ A.

Also, ∀ N ≥ 1, the partial sum is N X

  N X ex 1 SN (x) = fn (x) = = 2 − N −1 ex , n−1 2 2 n=1 n=1

∀ x ∈ A.

Then, ∀ N ≥ 1, the difference in absolute value   x 1 ex |F (x) − SN (x)| = 2e − 2 − N −1 ex = N −1 2 2 is not bounded on A = [0, ∞), and therefore the convergence of the series is not uniform on A = [0, ∞). We notice that no function fn (x) is bounded on A = [0, ∞). Note: Remember that unbounded functions may converge uniformly. 1 E.g., for any n ∈ N, let fn (x) = x + , ∀ x ∈ R. Then, n un. fn (x) −→ f (x) := x, as n −→ ∞, in R. N For the uniform convergence of improper integrals, the first two criteria above can be stated as follows. Theorem 3.3.5 (Cauchy Test for unif. converg. of impr. int.’s) Let f (x, t) be a real function “nice enough” in x ∈ [a, ∞), with a ∈ R, and t ∈ I ⊆ R and such that: Z x f (u, t) du exists, ∀ x > a and ∀ t ∈ I. a

The two following claims are equivalent: (a)

∀>0

∀ x2 > x1 ≥ N ()

in

∃ N () > 0

(independent of t) such that Z x2 =⇒ f (u, t)du ≤ , ∀ t ∈ I.

[a, ∞)

x1

Z (b)

∞

f (x, t) dx a

converges uniformly on I to some function F (t). (See Definition 3.3.4.)

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Theorem 3.3.6 [Weierstraß M(x)-Test - un. con. of im. in.’s] Let f (x, t) be a real function “nice enough” in x ∈ [a, ∞), with a ∈ R, and t ∈ I ⊆ R, that satisfies the following two conditions: Rx (a) a f (u, t) du exists, ∀ x > a and ∀ t ∈ I (b) ∃ function M (x) ≥ 0 on [a, ∞), such that R∞ |f (x, t)| ≤ M (x) ∀ x ∈ [a, ∞) and ∀ t ∈ I, and a M (x)dx < ∞. Then

Z

∞

f (x, t) dx a

converges uniformly on I to some function F (t). (See Definition 3.3.4.) The proofs of these two Theorems can be found in advanced calculus or mathematical analysis books. Example 3.3.10 We have seen in Problem 1.6.13 [see also Problem II 1.7.58, Example II 1.7.25, Corollary II 1.7.5, (C)] that   does not exist, if t = 0,  Z ∞  e−tx sin(x) dx =  0   1 , if t > 0. t2 + 1 Due to the discontinuity of the limit function at t = 0, the convergence is not uniform on the interval [0, ∞). [See also Theorem 3.1.1, (I), in the sequel.] But, the convergence is uniform if t ∈ [c, ∞), with c > 0. This follows by Theorem 3.3.6, if we consider M (x) = e−cx . We see that M (x) > 0, e−tx sin(x) ≤ M (x) ∀ t ∈ [c, ∞) and ∀ x ≥ 0, Z ∞ Z ∞ 1 and M (x) dx = e−cx dx = < ∞, c 0 0 and so, the Theorem applies to prove the claim. N We also observe that any theorem concerning sequences and/or series of functions finds a corresponding interpretation in the context of improper integrals. For instance, there is a theorem of Dirichlet for converges of sequences and/or series, which can find an analogous interpretation in the context of improper integrals. Find these theorems in various books of mathematical analysis and/or advanced calculus and study them. Here, we state and prove the Theorem of Dirichlet in the context of improper integrals.

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Improper Riemann Integrals

Theorem 3.3.7 (Dirichlet Test for unif. conver. of impr. int.’s) Let f (x, t) be a real function “nice enough” in x ∈ [a, ∞), with a ∈ R, and t ∈ I ⊆ R, which is also decreasing with respect to x. We suppose, there exists function g(x) on [a, ∞) that satisfies: (a) g(x) ≥ 0, (b) lim g(x) = 0, (c) |f (x, t)| ≤ g(x), for all x ∈ [a, ∞) and x→∞ all t ∈ I. We also suppose, there exists a differentiable function h(x) on [a, ∞) such that |h(x)| ≤ M , bounded on [a, ∞) by some constant M > 0, and Z b f (x, t)h0 (x) dx exists, ∀ b > a and ∀ t ∈ I. a

Then

Z

∞

f (x, t)h0 (x) dx

a

converges uniformly on I to some function F (t). (See Definition 3.3.4.) Proof Using integration by parts and for any q ≥ p ≥ a, we obtain Z q Z q q 0 f (x, t)h (x) dx = [f (x, t)h(x)]p − h(x)dx f (x, t) = p

p

Z f (q, t) · h(q) − f (p, t) · h(p) −

q

h(x)dx f (x, t). p

Then, Z

p

q

Z f (x, t)h (x) dx ≤ |f (q, t)|·|h(q)|+|f (p, t)|·|h(p)+ 0

p

q

h(x)dx f (x, t) .

Since, |h(x)| ≤ M and f (x, t) is decreasing with respect to x, we get Z q Z q Z q ≤ |h(x)|[−d f (x, t)] ≤ −M dx f (x, t) dx = h(x)d f (x, t) x x p

p

p

M f (p, t) − M f (q, t). By the hypotheses |h(x)| ≤ M , g(x) ≥ 0 and |f (x, t)| ≤ g(x), and the last two inequalities, we get Z q 0 f (x, t)h (x) dx ≤ M [g(q) + 2g(p)]. p

But also, lim g(x) = 0. Therefore, ∀  > 0, ∃ N ≥ a such that x→∞  |g(x)| = g(x) < , (M > 0), ∀ x ≥ N . Thus, we obtain 3M Z q     0 0. Therevalue 1. Also, x x fore, the integral exists and defines a function F (t) for t > 0. For proving that the given integral converges uniformly, for t ≥ 0, we apply the Dirichlet Test, Theorem 3.3.7. We observe that −tx −tx sin(x) e ≤ 1 , for all t ≥ 0 and so we consider: f (x, t) = e x x x 1 that bounds f (x, t), and which is decreasing in x if t ≥ 0, g(x) = x h(x) = cos(x) whose absolute value is bounded by M = 1. Hence, the convergence is uniform on [0, ∞), t = 0 included. So F (t) is defined and continuous on [0, ∞). Now,   ∂ −tx sin(x) e = −e−tx sin(x). ∂t x But then, as we have computed in Problem 1.6.13, we have Z ∞ 1 . e−tx sin(x) dx = 2 t +1 0 The convergence of this integral is uniform on [a, ∞), for any a > 0, by the Cauchy Test, 3.3.5, for instance. (Check this easily!). Then, we have: ∀ t > 0,   Z ∞ Z ∞ ∂ 1 −tx sin(x) G(t) := . e dx = − e−tx sin(x) dx = − 2 ∂t x t +1 0 0 So, by the previous Theorem, 3.3.8, we must have Z 1 dt = − arctan(t) + C. F (t) = − 2 t +1 Now, we must determine the constant C. For this purpose, we use lim F (t) = lim [− arctan(t) + C] = −

t→∞

t→∞

π + C. 2

But, for t > 0 Z

∞

lim |F (t)| ≤ lim

t→∞

t→∞

e 0

Thus, lim |F (t)| = 0 t→∞

−tx

Z ∞ sin(x) 1 dx ≤ lim e−tx dx = lim = 0. x t→∞ 0 t→∞ t

and so

lim F (t) = 0.

t→∞

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π π and F (t) = − arctan(t), as was expected. This also 2 Z ∞2 sin(x) π π holds at t = 0 as: F (0) = dx = − 0 = , by the uniform x 2 2 0 convergence. Finally: C =

Remark: For t = 0, this Z ∞ result proves in a remarkable and quicker sin(x) π dx = , as a result of the uniform way the established fact x 2 0 convergence and continuity. (See Examples 3.1.8 and II 1.7.35.) N We now continue with some other important and powerful theorems and examples. un

Theorem 3.3.9 Let lim fn = f in a set A: ∅ 6= A ⊆ R. Then we have n→∞ the following three results: (a) If all fn ’s are continuous in A, then f is continuous in A and therefore all these functions are Riemann integrable. (b) If all fn ’s are Riemann integrable in A, then f is Riemann integrable in A (not necessarily continuous). (c) In either Case (a) or (b), if A is bounded we have: Z  Z Z lim fn (x) dx = lim [fn (x)] dx = f (x) dx. n→∞

A

A n→∞

A

Remark 3.3.3 The boundedness of the set A is necessary in (c). For 1 example, we consider ∀ n ∈ N, fn (x) = , ∀ x ∈ R (constant funcn tions). This sequence of functions converges uniformly to the function f (x) = 0 in R. Now, computing the respective improper integrals, we find: Z Z ∀ n ∈ N, fn (x) = ∞, and f (x) = 0, that is, equality fails. R

R

This hypothesis may be removed if we assume that there is a Riemann integrable real function g(x) dominating the functions fnZ(x) on the set A. That is, 0 ≤ |fn (x)| ≤ g(x), ∀ n ∈ N and ∀ x ∈ A and

g(x) dx < ∞. A

Also, if (c) is valid the convergence does not have to be uniform. See Example 3.3.12 below. Remark 3.3.4 As we have seen in Part II of Problem 1.3.9, Item

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Improper Riemann Integrals

(4.), the point-wise limit of a sequence of Riemann integrable functions may not be a Riemann integrable function necessarily. But, as we see here, the uniform limit is! By Part II of Problem 1.3.9, Item (5.), even the point-wise limit of a sequence of continuous functions may not be a Riemann integrable function. Corollary 3.3.2 Suppose (fn ), n ∈ N is a sequence of Riemann integrable functions on an interval [a, b), where ∞ < a < b < ∞, such that fn −→ f , as n −→ ∞, point-wise in [a, b), and ∀ r, a < r < b, fn −→ f as n −→ ∞ uniformly in [a, r]. Then f (x) is Riemann integrable and # Z "Z Z b

b

n→∞

b

lim [fn (x)] dx =

fn (x) dx =

lim

a n→∞

a

f (x) dx. a

[At b the integral(s) involved may or may not be generalized Riemann integral(s).] 6 pw

Corollary 3.3.3 Let lim fn = f in a set A: ∅ 6= A ⊆ R. Then we have n→∞ the following three results: (a) If all fn ’s are continuous in A but f is not continuous in A, then the convergence is not uniform (only point-wise). (b) If all fn ’s are Riemann integrable in A but f is not Riemann integrable in A, then the convergence is not uniform (only point-wise). (c) If A is bounded and Z  Z lim fn (x) dx 6= n→∞

Z lim [fn (x)] dx =

A n→∞

A

f (x) dx, A

pw

then the convergence lim fn = f in the set A is not uniform (only n→∞

point-wise). Remark: The term by term integration of a power series, as claimed in the beginning of this section, follows from Weierstraß M-Test, 3.3.3, and Theorem 3.3.9 or Corollary 3.3.2, etc. We apply these n X Theorems to Sn (x) := ck (x − a)k , n ∈ N0 , which, as we can check, k=0

satisfies all the required hypotheses, etc. See Problem 3.5.25. 6 See

also, Rudin 1976, 138, exercise 7.

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Example 3.3.12 In Examples 3.3.2 and 3.3.3, we had fn : (−1, 1] → R, fn (x) = xn , ∀ n ∈ N,

f : (−1, 1] → R, f (x) =

  0, if − 1 < x < 1,   1, if x = 1,

pw

and lim fn (x) = f (x) in A = (−1, 1] ⊂ R. n→∞

Since all the functions (fn )n∈N are continuous on (−1, 1] whereas the function f is not (it has a jump discontinuity at x = 1), we conclude that the point-wise convergence in the Examples 3.3.2 and 3.3.3 is not uniform over (−1, 1]. But, for n ∈ N, Z 1 Z 1 1 − (−1)n+1 fn (x) dx = −→ f (x) dx = 0, as n −→ ∞. n+1 −1 −1 So, the converse of Theorem 3.3.9, (c) fails. N Example 3.3.13 In Example 3.3.1, we had the sequence of the con2 tinuous functions fn (x) = nxe−nx , n ∈ N, on the interval [0, 1]. We saw that: pw lim fn (x) = f (x) = 0, ∀ x ∈ A = [0, 1] ⊂ R n→∞

and Z lim

n→∞

0

1

Z 1 1 6 0= fn (x) dx = = lim fn (x) dx = 2 0 n→∞ Z 1  Z 1 f (x) dx = 0 dx. 0

0

Therefore, the point-wise convergence in Example 3.3.1 is not uniform. N Next, we state three big theorems of advanced real analysis. We will adjust them to the level of this text; therefore, we do not state them in the greatest possible generality, and we do not include their proofs. But, they can easily and efficiently be used in applications. All that someone needs to do is to check the stated hypotheses, which are fairly straightforward. So, at this level, we can at least learn how to use them as efficient and powerful tools.

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Improper Riemann Integrals

In these theorems, we give sufficient conditions under which we can switch the order between limit and integral. We consider sequences of Riemann integrable functions, and we require their limit functions to be Riemann integrable, too. We do so because, as we have seen in Part II of project Problem 1.3.9, Items (3.) and (4.), the point-wise limit of a sequence of Riemann integrable functions is not always a Riemann integrable function. In advanced real analysis, we use the class of the Lebesgue integrable functions which contains the class of Riemann integrable functions (but not all generalized Riemann integrable functions. See Problem 3.2.20.). Now, it is always the case that the point-wise limit of a sequence of Lebesgue integrable functions is a Lebesgue integrable function. Therefore, such a requirement on the point-wise limit function, in the general interpretation of these Theorems within the Lebesgue theory of integration, is not necessary because it follows as a result from the already stated hypotheses. Theorem 3.3.10 (Lebesgue Monotone Convergence Theorem) We consider a sequence of real functions (fn )n∈N in a set ∅ 6= A ⊆ R that satisfies the following three conditions: (a) f1 (x) ≤ f2 (x) ≤ f3 (x) ≤ f4 (x) ≤ ... ≤ ∞, ∀ x ∈ A (increasing sequence of real functions in A). pw

(b) lim fn (x) = f (x) in A. n→∞

(c) All fn ’s are Riemann integrable in A and also f is Riemann integrable in A (possibly in the generalized Riemann sense). Then, under these conditions, we have the following: R R R (1) ∀ n ∈ N, f (x) dx ≤ A fn+1 (x) dx ≤ A f (x) dx, A n R R R (2) lim A fn (x) dx = A lim fn (x) dx = A f (x) dx. n→∞

n→∞

Remark 3.3.5 The condition that the limit function f is Riemann integrable in A is necessary, as easily seen by project Problem 1.3.9, Part II, Items 4 and 5, where A = [0, 1]. (In this context, study also bibliography, Thompson 2010.) Remark 3.3.6 If we have a series of real functions

∞ X

pw

fn (x) = f (x),

n=1

with x ∈ A ⊆ R, for which the sequences of the initial partial sums Sn (x) = f1 (x) + f2 (x) + . . . + fn (x),

n ∈ N,

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satisfies the conditions of the above Theorem, then we can switch the integration and the summation #  Z "X Z ∞ Z ∞ X fn (x) dx = fn (x) dx = f (x) dx. n=1

A

A

A

n=1

Theorem 3.3.11 (Lebesgue Dominated Convergence Theorem) We consider a sequence of real functions (fn ), n ∈ N in a set A, ∅ 6= A ⊆ R, that satisfies the following three conditions: pw

(a) lim fn (x) = f (x) in A. n→∞

(b) All fn ’s are Riemann integrable in A and also f is Riemann integrable in A (possibly in the generalized Riemann sense). (c) There exists a Riemann integrable function (possibly in the generalized Riemann sense) g : A → [0, ∞] (in this context, called the dominating function) such that Z g(x) dx < ∞ and ∀ n ∈ N, |fn (x)| ≤ g(x), ∀ x ∈ A. A

Then, under these conditions, we have the following: (1) |f (x)| ≤ g(x), ∀ x ∈ A, R (2) lim A |fn (x) − f (x)| dx = 0, n→∞

(3) lim

R

n→∞ A

fn (x) dx =

R

lim f (x) dx A n→∞ n

=

R A

f (x) dx.

Remark 3.3.7 As with Theorem 3.3.10 so in this Theorem, the condition that the limit function f is Riemann integrable in A is necessary, as easily seen again by project Problem 1.3.9, Part II, Items 4 and 5. The dominating function is g(x) ≡ 1 in A = [0, 1]. (In this context, study also bibliography, De Silva 2010.) Remark 3.3.8 If we have a series of real functions

∞ X

pw

fn (x) = f (x),

n=1

with x ∈ A ⊆ R, for which the sequences of the initial partial sums Sn (x) = f1 (x) + f2 (x) + . . . + fn (x), n ∈ N, satisfies the conditions of the above Theorem, then we can switch the integration and the summation #  Z "X Z ∞ Z ∞ X fn (x) dx = fn (x) dx = f (x) dx. n=1

A

A

n=1

A

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Improper Riemann Integrals

Remark 3.3.9 The Continuity Part of the Main Theorem, 3.1.1, is essentially the Lebesgue Dominated Convergence Theorem, 3.3.11, where the parameter t plays the role of the index n ∈ N. Example 3.3.14 For all n ∈ N, find Z ∞ x n −2x 1+ e lim dx. n→∞ 0 n  x n −2x We have that the functions fn (x) = 1 + e , with n ∈ N, n satisfy the following conditions: (a) f1 (x) ≤ f2 (x) ≤ f3 (x) ≤ f4 (x) ≤ ... ≤ ∞, ∀x ∈ A = [0, ∞), x n is nonbecause for any x ≥ 0 the sequence an := 1 + n decreasing. (Prove this as an exercise!) (b) We know that (prove it one more time)  x n lim an = lim 1 + = ex . n→∞ n→∞ n Then we get pw

 x n −2x e = ex e−2x = e−x . 1+ n→∞ n

lim fn (x) = lim

n→∞

(c) All the functions fn ’s and the limit function f (x) = e−x are continuous in A = [0, ∞) and therefore Riemann integrable (in the generalized sense). Then by the Lebesgue Monotone Convergence Theorem, 3.3.10, we get Z ∞ Z ∞  x n −2x x n −2x lim 1+ e dx = e dx = lim 1 + n→∞ 0 n→∞ n n 0 Z ∞  ∞ e−x dx = −e−x 0 = 0 − (−1) = 1. 0

(See also Problem 3.5.3.) N Example 3.3.15 In the previous Example, we can use the Lebesgue Dominated Convergence Theorem, 3.3.11, instead of the Lebesgue Monotone Convergence Theorem, 3.3.10, with dominating function g(x) = e−x . (Check that all the conditions of the two Theorems are satisfied.) N

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Example 3.3.16 Working as in the previous two examples, we can achieve the following result: Z n x n −2x 1+ lim e dx = 1. n→∞ 0 n Here, we consider the sequence of Riemann integrable functions defined by  x n −2x   e , if 0 ≤ x ≤ n, 1 +  n fn : R −→ R : R 3 x −→ fn (x) =    0, if n < x < ∞, for all n = 1, 2, 3, 4, . . . . Then we take Z ∞ lim fn (x) dx. n→∞

0

Now we can use either the Lebesgue Monotone Convergence Theorem, 3.3.10, or the Lebesgue Dominated Convergence Theorem, 3.3.11, to switch limit and integral signs and obtain the result, as in the previous two examples. (Fill in the details. See also Problem 3.5.3.) N Example 3.3.17 Suppose that for all n ∈ N, we have Riemann integrable functions satisfying: x ∈ [a, b], 0 ≤ fn (x) ≤ c and

pw

lim fn (x) = f (x) = 0.

n→∞

Then we can apply the Lebesgue Dominated Convergence Theorem, 3.3.11, with dominating function g(x) = c ≥ 0 for all x ∈ [a, b], Z b Z b since g(x) dx = c dx = c(b − a) < ∞. Hence, in such a situation, we get

a

a

Z lim

n→∞

b

Z fn (x)dx =

a

b

Z lim fn (x)dx =

a n→∞

b

0 dx = 0. a

[See also Problem 3.13.33, (b) and solve it.] N We also state the Lebesgue Criterion for Riemann integrable functions in R. But first we need: Definition 3.3.5 A set A subset of R (A ⊂ R) has Lebesgue measure zero if for every  > 0 there are open intervals whose union contains A and the sum of their lengths is less than .

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Improper Riemann Integrals

Example 3.3.18 Every countable A ⊂ R has measure zero. Suppose A = {a1 , a2 , a3 , a4 , . . .} and consider any  > 0. Then we consider the sequence of open intervals     In = an − n+2 , an + n+2 , n ∈ N. 2 2 Obviously, A⊂

∞ [

In

n=1

and for the sum of the lengths we have ∞ X

  = < . n+1 2 2 n=1 (There are also uncountable subsets of R with Lebesgue measure zero. A classical example is the Kantor7 set. Have a look at it in the bibliography.) N Now we state: Theorem 3.3.12 (Lebesgue Criterion for Riemann Integrability) Let a < b be real numbers and suppose f : [a, b] −→ R is bounded, on the closed and bounded real interval [a, b]. Then f is Riemann integrable on [a, b] if and only if the set of discontinuity points of f has Lebesgue measure zero. Corollary 3.3.4 Let f ◦g be a composition of two real bounded functions defined on an interval [a, b]. Then, if one function is continuous and the other is Riemann integrable, the composition is Riemann integrable. Remark: The Corollary fails if both functions f and g are Riemann integrable. See project Problem 1.3.9, Part II. In such a case, f ± g, f f · g are Riemann integrable and , away from the zeroes of g. g Example 3.3.19 The Dirichlet function on the closed interval [0, 1]   if x = rational in [0, 1], 1, y = f (x) = χ[0,1]∩Q (x) =   0, if x = irrational in [0, 1], 7 Georg Ferdinand Ludwig Philipp Cantor or Kantor, German mathematician, 1845–1918.

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(see also Problem 1.3.9) is bounded and discontinuous at every point of [0, 1]. Since the Lebesgue measure of [0, 1] is the length of this interval, i.e., 1 > 0, this function is not Riemann integrable. This result can also be checked elementarily, for the upper and lower sums of this function have constant difference 1. N Example 3.3.20 The Riemann Dirichlet function  p 1  if x = rational in reduced representation in [0, 1]  q, q y = h(x) =    0, if x = irrational in [0, 1] (see also Problem 1.3.9) is bounded and discontinuous exactly at the rational points of the closed interval [0, 1]. Since the rational numbers are countable, the Lebesgue measure of [0, 1] ∩ Q is 0, and so this function is Riemann integrable. Its Riemann integral is zero. N We also state the following inequality Theorem, which is a restricted version of the general Fatou8 Theorem in integration theory. In the Lebesgue theory of integration, this is a key result and is stated in the most general terms, such as liminfimum instead of limit here, Lebesgue integrable function, etc. So, we modify it to our context and level. Theorem 3.3.13 (Fatou Theorem) We consider a sequence of Riemann integrable non-negative real functions (fn ≥ 0), n ∈ N, defined on pw a set A : ∅ 6= A ⊆ R, and let f (x) := lim fn (x), ∀ x ∈ A. n→∞

We assume that the point-wise limit function f (x)(≥ 0) is a Riemann Z integrable function and lim

n→∞

fn (x) dx exists in [0, +∞]. A

Then the following inequality holds Z Z Z lim fn (x) dx = f (x)dx ≤ lim A n→∞

A

n→∞

fn (x) dx.

A

Remark: As with Theorems 3.3.10 and 3.3.11, the condition that in this setting the limit function f is Riemann integrable in A is necessary, as easily seen again by project Problem 1.3.9, Part II, Items 4 and 5. 8 Pierre

Fatou, French mathematician, 1878–1929.

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Improper Riemann Integrals

Example 3.3.21 Strict inequality may occur in Fatou Theorem.   if n − 1 ≤ x ≤ n, 1, ∀ n ∈ N we let fn (x) =   0, otherwise. Then we have:

pw

lim fn (x) = 0,

n→∞

Z

∞

Z

∞

∀ n ∈ N and

fn (x)dx = 1,

f (x)dx = 0. −∞

−∞

Hence Z ∞

∞

Z

−∞ n→∞

∞

Z

lim fn (x) dx =

f (x)dx = 0 < lim

fn (x) dx = 1.

n→∞

−∞

−∞

N Example 3.3.22 Another example with a strict inequality in Fatou Theorem is the following:  1   n, if 0 ≤ x ≤ ,   n ∀ n ∈ N we let fn (x) =   1  0, if < x ≤ 2. n Then we have:   0,

if

  ∞,

if x = 0,

∀ n ∈ N and

lim +

pw

lim fn (x) =

n→∞

and so Z

0 < x ≤ 2,

2

2

Z fn (x)dx = 1,

0

Hence Z 2

Z lim fn (x) dx =

0 n→∞

f (x)dx = 0.

0 0 or r = ∞, and u ∈ (−r, r) then ! ∞ ∞ X X d d d n [f (x)] = a x = (an xn ) = n dx x=u dx x=u n=0 dx x=u n=0 ∞ X

∞ ∞ X X an nxn−1 x=u = an nun−1 = nan un−1 .

n=0

n=0

n=1

(See also Theorem II 1.2.1.) At the endpoints −r and r, anything can happen. Sometimes the appropriate side-derivatives exist. Whereas this commuting is always legitimate with derivatives of power series at any point in the interval of convergence, it is not valid in every situation with limits of sequences or series of functions and derivatives taken at points within the domain of definition of all functions involved. Serious mistakes may occur if such a commutation is performed while not valid! For instance: Example 3.4.1 In Example 3.3.4 we saw that the sequence of functions sin(nx) , ∀x∈R fn : R → R, fn (x) = √ n converges uniformly to the function f : R → R, f (x) = 0, ∀ x ∈ R. Let us now examine the derivatives:   √ d d sin(nx) √ = n cos(nx), ∀ x ∈ R [fn (x)] = dx dx n and

d d f (x) = (0) = 0, ∀ x ∈ R. dx dx

We see that   d d sin(nx) √ lim [f (x)] = lim = n n→∞ dx x=0 n→∞ dx x=0 n √ √ √ lim n cos(nx)|x=0 = lim n cos(0) = lim n · 1 = ∞,

n→∞

n→∞

but

n→∞

d [f (x)] = 0. dx x=0

Therefore, we see that lim

d

n→∞ dx x=0

[fn (x)] = ∞ 6= 0 =

d [f (x)] . dx x=0

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Improper Riemann Integrals

Hence, in this example, even though the sequence of the differentiable functions (fn )n∈N converges to a differentiable function f uniformly in R, we cannot switch the order of limit and differentiation. N By this example, we see that we need some stronger conditions to guarantee the legitimacy of commuting (switching) the order of a limit process and taking the derivative (differentiating). The most-seen general theorem in the literature, or slight variations of it, is the following: Theorem 3.4.1 Suppose that a sequence of real functions (fn )n∈N on an interval (a, b) ⊆ R satisfies the following conditions: (a) All functions fn ’s are differentiable, i.e., ∀ n ∈ N and ∀ x ∈ (a, b), fn0 (x) exists. (b) There exists a point x0 ∈ (a, b) such that the sequence of real numbers fn (x0 ) converges. (We require convergence at at least one point of the domain.) (c) The sequence of the derivative functions fn0 (x) converges uniformly to some function g(x) on (a, b), as n → ∞. I.e., ∃ g(x) real function on (a, b), such that un

lim fn0 (x) = g(x) in (a, b).

n→∞

Then: The sequence of fn (x)’s converges uniformly to a differentiable function f (x) on (a, b), as n → ∞, that is, (I)

un

lim fn (x) = f (x) in (a, b),

n→∞

and it holds, (II)

h

i0 lim fn (x) = lim fn0 (x),

n→∞

n→∞

or

f 0 (x) = g(x).

Remark: If a power series ∞ X

bk (x − c)k

k=0

has radius of convergence R > 0, then the radius of convergence of the power series ∞ X

k bk (x − c)k−1

k=0

√ k+1 n k = 1, or lim = 1. See k→∞ k k→∞ also Section II 1.2 and Problem II 1.3.1.)

is also R. (Prove this by noticing lim

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203

Then, the sequences of the initial partial sums of the two above power series n n X X Sn (x) = bk (x − c)k and S 0n (x) = k bk (x − c)k−1 , k=0

k=1

respectively, satisfy all of the conditions (and even more) of the above Theorem in their common open interval of convergence. So, we apply the above Theorem in their common open interval of convergence to prove that the derivative of a power series is equal to the power series of the derivatives of its terms, at any x ∈ (c − R, c + R). See Problem 3.5.25. In general, the above Theorem, 3.4.1, can be reformulated in terms of series of functions and not just power series as follows: Theorem 3.4.2 Suppose that a sequence of real functions (fn )n∈N on an interval (a, b) ⊆ R satisfies the following conditions: (a) All functions fn ’s are differentiable, i.e., ∀ n ∈ N and ∀ x ∈ (a, b), fn0 (x) exists. (b) There exists a point x0 ∈ (a, b) such that the series of numbers ∞ X fn (x0 ) converges. (We require convergence at at least one point n=1

of the domain.) (c) The series of the derivative functions

∞ X

fn0 (x) converges uni-

n=1

formly to some function g(x) on (a, b). I.e., there exists a real function g(x) on (a, b), such that ∞ X

un

f 0n (x) = g(x) in (a, b).

n=1

Then: The series

∞ X

fn (x) converges uniformly to a differentiable

n=1

function f (x) on (a, b), that is, (I)

∞ X

un

fn (x) = f (x) in (a, b),

n=1

and " (II)

∞ X n=1

#0 fn (x)

=

∞ X n=1

fn0 (x),

or

f 0 (x) = g(x).

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Improper Riemann Integrals

Example 3.4.2 Condition (b) is necessary. For instance, if we let ∀ n ∈ N fn (x) = n, ∀ x ∈ R, then the conditions (a) and (c) are satisfied as fn0 (x) = 0, ∀ x ∈ R and ∀ n ∈ N. So, as n → ∞ the sequence of the derivative functions converges uniformly in R to the differentiable function g(x) = 0, ∀ x ∈ R. But lim fn (x) does not exist at any point x ∈ R,

n→∞

and therefore we cannot claim any of the conclusions of Theorem 3.4.1. N Example 3.4.3 Condition (c) is necessary. For instance, as we saw before the sequence of functions fn : R → R, fn (x) =

sin(nx) √ , ∀ x ∈ R, n

converges uniformly to the function f : R → R, f (x) = 0, ∀ x ∈ R, and we could not switch the order of taking limit and derivative. We observe that the derivative functions   √ d d sin(nx) √ [fn (x)] = = n cos(nx), ∀ x ∈ R dx dx n do not converge even point-wise to any differentiable function, let alone uniformly. For example, at the points kπ, k ∈ Z, we have   d d sin(nx) √ lim [f (x)] = lim = n n→∞ dx x=kπ n→∞ dx x=kπ n √ √ lim n cos(nx)|x=kπ = lim n cos(nkπ) = n→∞ n→∞   +∞, if k is even,   does not exist, if k is odd. N Example 3.4.4 Let us examine the situation with the functions fn : (−1, 1) → R, fn (x) = xn , ∀ n ∈ N and their point-wise limit f : (−1, 1) → R, f (x) = 0. We have ∀ n ∈ N,

d fn (x) = nxn−1 , −1 < x < 1. dx

Real Analysis Techniques

205

We can easily prove pw d [fn (x)] := g(x) = 0, −1 < x < 1. n→∞ dx

lim

The convergence here is only point-wise, not uniform. (Prove this as a calculus exercise! See Problem 3.5.1.) Obviously, d d [f (x)] = (0) = 0, −1 < x < 1. dx dx So, here we do not have uniform convergence of the sequence of the derivatives of the functions, and so we have lost the uniform converge of the sequence of the functions (fn )n∈N on (−1, 1), otherwise claimed by Theorem 3.4.1. [See also Problem 3.5.12, (a)-(g).] N

3.5

Problems

3.5.1 Prove that for n ∈ N and x ∈ (−1, 1),


pw lim nxn−1 = 0

n→∞

and the convergence is not uniform over (−1, 1). Is the convergence uniform on a closed interval [a, b] ⊂ (−1, 1)? Prove your answer! 3.5.2 Prove that

Z 0

∞

x2

1 π dx = . 2 +t 2t

and the convergence is uniform when t ∈ [a, ∞), for any a > 0. Is the convergence uniform on the interval [0, ∞)? Prove your answer! 3.5.3 Prove the following two limits:
n x R∞ (a) lim 0 1 − nx e 2 dx = 2. n→∞

(b) lim

Rn

n→∞ 0

1−


x n n

x

e 2 dx = 2.

[Hint: Imitate the ideas of Examples 3.3.14 and 3.3.16.]

206

Improper Riemann Integrals

3.5.4 Consider a continuous function f : [0, 1] −→ R and n ∈ N. Prove Z 1 f (xn ) dx = f (0). lim n→∞

0

3.5.5 Let fn (x) = nx(1 − x2 )n , for 0 ≤ x ≤ 1, and n = 1, 2, 3, . . . . (a) Find lim fn (x) point-wise. n→∞

(b) Is the convergence uniform? (c) Can you commute limit and integrals? (d) Can you commute limit and derivatives? Justify your answers. Give answers to these questions if we let −1 ≤ x ≤ 1. (Notice that all functions fn (x) are odd in −1 ≤ x ≤ 1 and so their integrals are zero.) 3.5.6 For any a > −1, we let  X ∞  ∞ X 1 1 x−a fa (x) = − = . n+1+a n+1+x (n + 1 + a)(n + x + a) n=0 n=0 (a) Prove that fa (x) is well defined in (−1, ∞) and ∞ X d 1 [fa (x)] = > 0, dx (n + 1 + x)2 n=0

∀ x > −1,

and so fa (x) is increasing in (−1, ∞). (b) Investigate further the definition of fa (x) for a ∈ R, and the d [fa (x)] for x ∈ R. dx (See also Problem 3.5.32.) 3.5.7 (a) For any α > 0 and any 0 ≤ x < 1, by the geometric series, we have ∞ X

(−1)n xαn =

n=0

1 . 1 + xα

(b) Show that for any 0 ≤ a < b < 1, the convergence in (a) is uniform on the closed interval [a, b].

Real Analysis Techniques (c) Use Corollary 3.3.2 to prove that Z ∞ Z 1 X (−1)n xαn dx = n=0

and so

207

∞ 1X

(−1)n xαn dx

0 n=0

0

Z 1 ∞ 1 X (−1)n 1 dx < 1. < = α 2 n=0 1 + αn 0 1+x

[See also Problem II 1.7.71, (d).] (d) Prove the result in (c) by using the Lebesgue Dominated Convergence Theorem, 3.3.11, or Example 3.3.17 on the closed interval [0, 1]. (Why can’t we use the Lebesgue Monotone Convergence Theorem, 3.3.10 ?) 1 1 , , 1, 2, 3, 4 to obtain, respectively, 3 2   ∞ ∞ X X (−1)n (−1)n 1 = 3 ln(2) − , (II) = 2[1 − ln(2)], 1 + n3 2 1 + n2 n=0 n=0

(e) Let α = (I)

(III)

∞ X (−1)n = ln(2), 1+n n=0

(V) (VI)

(IV)

∞ X (−1)n π = , 1 + 2n 4 n=0

" √ # ∞ X (−1)n 1 π 3 = ln(2) + , 1 + 3n 3 3 n=0 √ h  ∞ i X √  (−1)n 2 = ln 3 + 2 + π . 1 + 4n 8 n=0

(f) Use partial fractions to prove √ ∞ ∞ X X 3π (−1)n 2 (−1)n = . (II) = ln(2). (I) (n + 1)(3n + 1) 6 (3n − 2)(3n − 1) 3 n=1 n=0 (g) Compute the four sums (1) (3)

∞ X (−1)n , −1 + n n=2 ∞ X (−1)n , −1 + 3n n=0

(2) (4)

∞ X (−1)n , −1 + 2n n=0 ∞ X (−1)n . −1 + 4n n=0

208

Improper Riemann Integrals

(h) Prove that for β < 0 Z 0< 0

1

∞ ∞ X X 1 (−1)n 1 (−1)n−1 dx = = < . β 1+x 1 − β(n + 1) n=1 1 − βn 2 n=0

[Note: For positive series (and so not alternating, as they are here), especially if they are not telescopic, see the methods used in Problems 3.5.28–3.5.34 and their hints.] [Remark: Notice that the computations and results in this Problem could also be carried out with the help of exponentials, since Z ∞ 1 = e−(1+αn)x dx 1 + αn 0 and Z 0

1

1 dx = 1 + xα

Z 0

∞

e−x dx, 1 + e−αx

etc., but this way seems to be less convenient. If you like try it out.] 3.5.8 Imitate Example 3.3.23 to evaluate, as series, the integrals Z 1 Z 1 √ Z 1 − x ln(x) dx, cos(x) ln(x) dx, sin(x) ln(x) dx. 0 0 1−x 0 3.5.9 Consider any α > 0 and any 0 ≤ a < b < 1. (a) Prove that on the closed interval [a, b], the convergence in the geo∞ X 1 , (0 ≤ a ≤ x ≤ b < 1), is uniform. metric series xαn = 1 − xα n=0 Z (b) Under the hypotheses prove: a

1

! ∞ X 1 1 dx = ∞ = . 1 − xα 1 + αn n=0

(c) Prove that on the closed interval [0, 1] the convergence of the geometric series in (a) is not dominated but only monotone increasing. 3.5.10 Justify fully the following results: # Z 1 "X ∞ 1 (a) dx = 1. (n + x)2 0 n=1

Real Analysis Techniques

209

∞ ∞ X X xn 1 π2 dx = = − 1. 2 2 n n (n + 1) 6 0 n=1 n=1 # # Z π "X Z π "X ∞ ∞ sin(2nx) sin(nx) cos(nx) (c) dx = dx = 0. n2 2n2 −π n=1 −π n=1

Z

!

1

(b)

3.5.11 Z (a) Prove

lim

n→∞

0

1

xn (2x + 1) dx = 0. x2 + x + 1

(b) Then prove lim

n→∞

1

 Z n·

 xn−1 ln(x2 + x + 1) dx = ln(3).

0

3.5.12 For all n ∈ N, consider the real functions fn (x) =

nx2 , nx2 + 1

x ∈ R.

(a) Prove that these functions are non-negative, even and ∀ x ∈ R ( 1, if x 6= 0, pw lim fn (x) = f (x) = n→∞ 0, if x = 0. (b) Prove that the convergence in (a) is not uniform on the whole R. (c) Prove that in any set Aε = (−∞, −ε] ∪ [ε, ∞) ⊂ R, where ε > 0, the convergence in (a) is uniform. (d) Compute the derivatives fn0 (x) for all n ∈ N (simplify). (e) Prove that ∀ x ∈ R,

pw

lim fn0 (x) = 0.

n→∞

(f) Prove that the convergence in (e) is not uniform. ( i0 h 0, if x 6= 0, 0 0 (g) Prove lim fn (x) 6= lim fn (x) = f (x) = n→∞ n→∞ DN E, if x = 0. Z ∞ Z ∞ (h) Prove that ∀ n ∈ N, fn (x)dx = ∞ and f (x)dx = ∞. −∞

−∞

3.5.13 Consider a sequence of real numbers (an ) with n ∈ N0 such that ∞ X an ≥ 0 (non-negative) for all n ∈ N0 and an = s ∈ R. n=0

210

Improper Riemann Integrals ∞ X

tn converges absolutely n! n=0 and the convergence is uniform on every bounded interval of R. Z ∞ (b) Use Theorem 3.3.10 and the fact that tn e−t dt = n! to prove (a) Prove that for every t ∈ R, f (t) :=

an

0

that

Z

∞

f (t) e−t dt = s.

0

3.5.14 Justify fully the following result: For any r > 2 real and for any x ∈ R, it holds ! ∞ ∞ X cos(nx) d X sin(nx) = . dx n=1 nr nr−1 n=1 3.5.15 (a) Justify fully: ∀ x ∈ (0, 1), "∞ # n ∞  X d X  2n x2 −1 2n ln 1 + x . = dx n=0 1 + x2n n=0 (b) Observe that, ∀ x ∈ (0, 1), k X

n

ln(1 + x2 ) =

n=0

" ln(1 − x) +

k X



n

ln 1 + x2



#

n+1

− ln(1 − x) = ln

n=0

So,

∞ X

   n ln 1 + x2 = ln

n=0

1 1−x

1 − x2 1−x

! .

 ,

∀ x ∈ (0, 1).

The above two equations are also true for x = 0. (c) Prove that ∀ x ∈ (0, 1), n ∞ X 2n x2 −1 1 = , 2n 1 + x 1 − x n=0

which is also true for x = 0. (So this sum is also equal to the geometric series for 0 ≤ x < 1.)

Real Analysis Techniques

211

If we replace x by |x|, we find n ∞ ∞ X X 2n |x|2 −1 1 = = |x|n , ∀ x ∈ (−1, 1). n 2 1 + x 1 − |x| n=0 n=0

3.5.16 The integral of Laplace found in Examples 3.1.14, II 1.7.18 and Problem II 1.7.32, (a), can also be found by switching summation and integration. For instance, use the power series cos(2bx) =

∞ X

(−1)n

n=0

(2bx)2n (2n)!

and justify the switching of summation and integration to obtain √ Z ∞ π −b2 −x2 e . e cos(2bx) dx = 2 0

3.5.17 (a) Prove, if a ≥ 0, Z I(a) := 0

∞

  ln (ax)2 + 1 dx = π ln(1 + a). x2 + 1

[Hint: Show that I(a) converges uniformly on [0, b] for any b > 0. d π Use partial fractions to prove that [I(a)] = . Then, use da 1+a Theorem 3.3.8 and I(0) = 0 (obvious) to obtain the result. (See also Problem II 1.7.138 for another method using complex analysis.)]  Z ∞  ln (ax)2 + b2 dx. (b) Now, if a ≥ 0, b ≥ 0 and c > 0, find x2 + c2 0 (c) Use this result to prove Problem 2.3.19, (e)-(f ). See also Problem 2.3.24. 3.5.18 (a) Justify each equality of the following integral representations of the Euler-Mascheroni constant γ.

212

Improper Riemann Integrals (See also Problem 2.3.32 and its footnote. At some steps you need to perform an appropriate u-substitution. Some ideas in Examples 3.3.14 and 3.3.16 may help at some equalities.) Z ∞ − e−x ln(x) dx = 0 Z ∞ − e−x ln(x) dx = 0 1

Z

−e

−x

Z

∞

− e−x ln(x) dx =

ln(x) dx +

0

1 1

Z

−x

ln(x) de 0

Z =

Z

∞

+

ln(x) de−x =

1

1


ln(x) d e−x − 1 +

0

Z

∞

ln(x) de−x =

1 ∞ −x

1

Z e 1 − e−x dx + dx = x x 1 0 Z 1  Z n 1 − (1 − nx )n (1 − nx )n dx − dx = lim n→∞ x x 1 # "Z0 1 Z 1 n 1 − (1 − t)n (1 − t)n dt − dt = lim n→∞ 1 t t 0 n "Z 1  # Z 1 n 1 − (1 − t)n 1 1 − (1 − t)n lim dt + − dt = n→∞ 1 t t t 0 n "Z # Z 1 1 1 1 − (1 − t)n lim dt − dt = n→∞ 1 t t 0 n Z 1   1 1 − tn lim dt + ln = n→∞ 1−t n 0 Z 1  
1 lim tn−1 + tn−2 + ... + t + 1 dt + ln = n→∞ n 0   1 1 1 lim + + ... + + 1 − ln(n) := γ, n→∞ n n−1 2 Z

by the definition of γ. [E.g., see footnote of Problem 2.3.32, (d).] (b) Now let u = e−x and achieve the integral Z 1 ln[− ln(u)]du = −γ. 0

Real Analysis Techniques

213

3.5.19 (a) See the previous Problem and show the useful result Z ∞ − e−x ln(x) dx = 0

Z ∞ −x 1 − e−x e dx − dx = x x 0 1 Z 1 −1 Z 1 −1 e t 1 − e−t − e t dt − dt = dt = t t 0 0 Z ∞ −1 1 − e−s − e s ds = γ. s 1

Z Z 0

1

1 − e−t t

1

Hence

∞

Z 0

(b) Prove Z

∞

0

1 − e−v − e v 

−1 v

1 − e−s 1+s

dv = 2γ.



ds = γ. s

(See also Problem 2.3.32.) 3.5.20 With the help of the previous Problem prove that if a > 0 and b > 0 constants, then ∞

a

b

e−u − e−u a−b du = γ · . u ab 0 [Hint: Let t = ua and change the integral into    b Z 1 Z ∞ −t ab −t a 1 − e 1 e −γ − − dt + dt . a t t 0 1 Z

b

Then, let s = t a and get the result. (Notice that this integral is not Frullani. Why?)] 3.5.21 Justify 1+

Z 1 Z n 1 1 − tn dy 1 1 + + . . . + − ln(n) = dt − = 2 3 n 1−t y 0 1 Z nh Z n  y n i dy dy 1− 1− − = n y y 0 1 Z 1h Z n  y n i dy y n dy 1− 1− − 1− . n y n y 0 1

214

Improper Riemann Integrals

Now check that it is legitimate to take limits, as n −→ ∞, to find an integral representation of the Euler-Mascheroni constant10 , as Z 1 Z ∞
dy −y dy 1−e γ= e−y . − y y 0 1 (The first integral cannot be separated about the −.) 3.5.22 (a) For a ∈ R, b > 0 and −1 ≤ c ≤ 1 use the geometric series, the methods of this Section, justify the switching of summation with integration and Problem 1.6.13 to prove the result obtained with complex parameters in Example II 1.7.25 (and Corollary II 1.7.5). # Z ∞ Z ∞ "X ∞ sin(ax) n −b(n+1)x dx = (−c) e sin(ax) dx = ebx + c 0 0 n=0 ∞ a X (−c)n−1 a(−c)n =
. a2 + b2 (n + 1)2 b2 n=1 n2 + a 2 n=0 b ∞ X

(b) Divide the equation in (a) by a and take the limit as a → 0 to obtain that for any b > 0 and −1 ≤ c ≤ 1, Z ∞ ∞ x 1 X (−c)n−1 dx = 2 . ebx + c b n=1 n2 0 (c) If b > 0 and −b < a < b, use Problem 1.6.17 and justify the integral Z Z 0 Z ∞ 1 ∞ sinh(ax) sinh(ax) sinh(ax) dx = dx = dx = 2 −∞ sinh(bx) sinh(bx) sinh(bx) −∞ 0 " # Z ∞ X ∞ ∞ X 2a 2 sinh(ax) e−b(2n+1)x dx = . 2 (2n + 1)2 − a2 b 0 n=0 n=0 (d) Divide the equation in (c) by a, take the limit as a → 0 and use the result of Example 3.6.3 to obtain that for any b > 0, Z Z ∞ Z 0 1 ∞ x x x 1 π2 dx = dx = dx = 2 . 2 −∞ sinh(bx) sinh(bx) b 4 0 −∞ sinh(bx) 10 Using

complex contour integration we can also prove Z 1 Z ∞ dy dy γ= [1 − cos(y)] − cos(y) . y y 0 1 (The first integral cannot be separated about the −.) See Problem II 1.7.38

Real Analysis Techniques

215

(See also Problems II 1.7.43, II 1.7.77 and II 1.7.78.) (e) Differentiate the equations in (a) and (c) with respect to a to obtain two additional equalities. 3.5.23 Use appropriate power series (and possibly the previous Problem) to evaluate, as series, the integrals Z ∞ Z ∞ cos(x) cos2 (x) dx, dx. 2ex ± 1 2ex ± 1 0 0 3.5.24 (Phragm´en.11 ) (a) For s, t, and τ real numbers, prove

lim

∞ X (−1)k−1

s→∞

k=1

k!

eks(t−τ )

  1,       h i  1 −es(t−τ ) = lim 1 − e = 1− , s→∞  e       0,

if t > τ, if t = τ, if t < τ.

(b) For any B > 0 constant and any continuous function g : [0, B] −→ R, use (a) and justify the switchings of limits and integrals involved to prove lim

s→+∞

Z ∞ X (−1)k−1 k=1

k!

0

B

e

ks(t−τ )

Z

t

g(τ ) dτ, ∀ t ∈ [0, B].

g(τ ) dτ = 0

(c) If B > 0 and there exists a constant M ≥ 0 such that for the function g in (b), it holds Z B nt e g(t) dt ≤ M, ∀ n ∈ N, 0 e.i., all the exponential moments of g are uniformly bounded over the closed interval [0, B], prove that g(t) ≡ 0 on [0, B]. 11 Lars

Edvard Phragm´ en, Swedish mathematician, 1863–1937.

216

Improper Riemann Integrals

(d) If b > 1 and g : [1, b] −→ R is continuous and there is a constant K ≥ 0 such that Z b tn g(t) dt ≤ K, ∀ n ∈ N, 1 e.i., all the moments of g are uniformly bounded over the closed interval [1, b], prove that g(t) ≡ 0 on [1, b]. (Show that this is not true in [0, 1].) 3.5.25 Use the appropriate theorems and results of this section to prove that to integrate or differentiate a power series in its open interval of convergence, we can integrate or differentiate it term by term. (See also Theorem II 1.2.1 and its Remark.) [Hint: Without loss of generality, you may assume that the center of the power series is c = 0 and so the open interval of convergence is (−R, R), where R > 0 is the radius of convergence. Use Remark after Corollary 3.3.3 and Theorem 3.4.1 and its Remark.] 3.5.26 The real Riemann Zeta function is defined by the series ζ(x) =

∞ X 1 . nx n=1

Prove: (a) This series diverges (= ∞) for any x ≤ 1. (b) This series converges point-wise for all x > 1. (c) This series converges uniformly on any interval [a, ∞) with a > 1. (d) The convergence is not uniform on the open interval (1, ∞). Note that this is not a power series. Replacing x with the complex z, we obtain the complex Riemann Zeta function for all z with Re(z) > 1. This function is very important in real and complex analysis, analytic number theory, in the special functions, etc. See Problem II 1.5.18. 3.5.27 Prove that the series uniformly in R.

∞ X

x converges absolutely and n (1 + nx2 ) n=1

Real Analysis Techniques

217 √

∞ X

1 3 π 3 ln(3) = + . (n + 1)(3n + 1) 12 4 n=0 (For alternating series see Problem 3.5.7.)

3.5.28 If n ∈ N0 prove

[Hint: Use the appropriate results and theorems to complete and justify the following steps. (For a different method use Problem 3.5.32.) 3 1 1 2 = − 2 . Then, First notice (n + 1)(3n + 1) 3n + 1 n + 1 Z 1 Z 1 1 1 ∀ n ∈ N0 , xn dx and x3n dx. = = n+1 3n + 1 0 0 Following this way, we observe that the given infinite sum becomes of the form ∞ − ∞, even though it exists. (See also Problem 3.5.9.) So continuing in this way, we must use limits of partial sums as follows: ∞ X

1 = (n + 1)(3n + 1) n=0  N N  3 1 X X 1 2 2 = lim − = lim N →∞ (n + 1)(3n + 1) N →∞ n=0 3n + 1 n + 1 n=0  Z N  Z 1 X 3 1 1 n lim x3n dx − x dx = N →∞ 2 0 2 0 n=0 ! Z N 1 1X n x dx − x dx = = lim N →∞ 2 0 n=0 0 n=0 ( "Z #)
Z 1 1 3 1 − x3(N +1) 1 1 − xN +1 lim dx − dx = N →∞ 2 0 1 − x3 1−x 0 "Z   # 1 (x2 + x − 2) + 3x3(N +1) − xN +1 (x2 + x + 1) 1 lim dx = N →∞ 2 (x − 1)(x2 + x + 1) 0 Z 1 1 x+2 dx+ 2 0 x2 + x + 1 "Z
# 1 N +1 3x2N +1 + 3x2N + . . . + 3x2 + 2x + 1 x 1 + lim dx = N →∞ 2 x2 + x + 1 0 √ Z 1 3π 1 1 xN +1 (2x + 1) ln(3) + + lim dx+ N →∞ 2 0 4 12 x2 + x + 1
Z 3 1 xN +3 x2N + x2N −1 + . . . + x + 1 lim dx. N →∞ 2 0 x2 + x + 1 3 2

Z

N 1X

3n

218

Improper Riemann Integrals

By the Lebesgue Dominated Convergence Theorem, 3.3.11, we have that Z 1 1 xN +1 (2x + 1) lim dx = 0. N →∞ 2 0 x2 + x + 1 Therefore, we must prove
Z 1 3 1 xN +3 x2N + x2N −1 + . . . + x + 1 dx = ln(3) lim 2 N →∞ 2 0 x +x+1 2 or Z lim

N →∞

0

1


xN +3 x2N + x2N −1 + . . . + x + 1 ln(3) dx = . 2 x +x+1 3

Performing the division of the expression (1 + x + x2 ) + (x3 + x4 + x5 ) + . . . + (x2N −2 + x2N −1 + x2N ) (written in increasing order of powers starting with 1 and grouped three by three terms), by 1+x+x2 and disregarding the remainder (because, by the Lebesgue Dominated Convergence Theorem, 3.3.11, the limit of the corresponding integral of this remainder multiplied by xN +3 is 3k +1 zero), or equivalently, we may only consider 2N = 3k+2, i.e., N = 2 with k even positive integer, so that this remainder is zero, we find

xN +3 x2N + x2N −1 + . . . + x + 1 = xN +3 1 + x3 + x6 + . . . + x3k = x2 + x + 1
3k 3k 3k 3k x 2 +4 1 + x3 + x6 + . . . + x3k = x 2 +4 + x 2 +7 + . . . + x 2 +3k+4 . The integral of this expression on [0, 1] is 1 1 1 + 3k + . . . + 3k = +5 2 +8 2 + 3k + 5 2 2 2 + + ... + . 3k + 10 3k + 16 3k + 6k + 10 3k 2

2 (This is a sum of k + 1 positive terms and it is bounded below by and 9 2 above by for all k.) 3 Hence, we must prove that for k ∈ N the sequence Ak :=

k 2 2 2 2 1X 1 + + ... = · 10 3k + 10 3k + 16 3k + 6k + 10 3 k i=0 1 + 3k +

converges to

1 ln(3), as k −→ ∞. This can be done as follows: 3

2i k

Real Analysis Techniques If for all k ∈ N, we let fk (x) =

1 1+

10 3k

+ 2x

lim fk (x) := f (x) =

k→∞

219

, then for x ≥ 0 we obtain

1 1 + 2x

and the convergence is monotonically increasing. Notice also that Ak is the left Riemann sum of the function fk on 1 the interval [0, 1] with ∆x = . Then (by the Lebesgue Monotone k Convergence Theorem, 3.3.10, see also the next Problem), we get lim Ak =

k→∞

2 · lim 3 k→∞

Z

1

fk (x) dx = 0

2 · 3

Z 0

1

1 dx = 1 + 2x

   1 2 ln(1 + 2x) 2 ln(3) ln(3) · = · −0 = .] 3 2 3 2 3 0 3.5.29 Prove the following limits which are useful in problems similar to the previous Problem. (a) If n, k ∈ N such that b ∈ R, then: lim

k→∞

n = r ≥ 1 fixed ratio, a ∈ R − {0} and k ! n=rk X 1 1 = ln(r). ai + b a i=k

r (b) If n ∈ N, a ∈ R − {0}, b ∈ R and r ∈ R, such that ≥ 0, then: a ! ! n n−1 X X 1 1  r 1 = lim = ln 1 + . lim n→∞ n→∞ an + b + ri an + b + ri r a i=1 i=0 [For r = 0 the answer is

1 , which agrees with L’ Hˆopital’s rule.] a

3.5.30 (a) [Another proof of the previous Problem (a).] Suppose that n n, k ∈ N, such that = r ≥ 1 fixed ratio. Knowing that for k "m # X1 m ∈ N, lim − ln(m) = γ (Euler-Mascheroni constant), m→∞ i i=1 prove: ! n=rk X 1 (1) lim = ln(r). k→∞ i i=k

220

Improper Riemann Integrals (2) If a ∈ R − {0} and b ∈ R, then lim

k→∞

n=rk X i=k

1 ai + b

! =

1 ln(r). a

(b) Now, if m ≥ 1, p ≥ 1 and q ≥ 1 integers, prove that     1 1 1 1 1 1 1 lim 1 + + + + ... + − + + ... + m→∞ 3 5 7 2mp − 1 2 4 2mq   1 p = ln(2) + ln . 2 q Recognised this as a rearrangement of the alternating series ∞ X (−1)n−1 = ln(2). n n=1 3.5.31 Prove that   1,      k  X 1 F (a) := lim = ln(2),  k→∞ k + ia  i=1      0,

if a < 1, if a = 1, if a > 1.

3.5.32 (a) For all p > 0 and q > 0 prove Z 0

1

 X ∞  ∞ X xq−1 − xp−1 1 1 p−q dx = − . = 1−x n + q n + p (n + p)(n + q) n=0 n=0

So, for all p > 0 and q > 0, this integral exists and it is equal to the given series. (For p = q all sides are equal to zero. If we let p = 0 and q = 1, then we get −∞ = −∞, and so on with other combinations of the two exponents.) The sum has only finitely many nonzero terms for any positive integers p 6= q ∈ N. Find these terms and the sum in such a case. (b) Compute the integral explicitly for q = 1 fixed (x0 = 1) and 1 4 3 p= , , and thus evaluate explicitly the corresponding series. 4 3 2 However, for almost all other p’s, we cannot reduce it beyond the given series.

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(c) Evaluate the sum of Problem 3.5.28 by using this integral with 1 p = and q = 1. 3 (d) For any k ∈ N prove that  lim

q→1

dk dq k

(−1)k k

1

xq−1 − 1 dx 1−x 0 ∞ X 1

Z

n=0

(n + 1)k+1

Z 1 [ ln(x)]k [ ln(1 − x)]k dx = dx = 1−x x 0 0 ∞ X 1 = (−1)k k = (−1)k kζ(k + 1). k+1 n n=1



Z

1

=

{See also Example 3.3.23 and Problems 2.3.17, 3.5.6 [and 3.5.26 for the zeta function ζ(x)].} 3.5.33 (a) Find the following six sums. (1)

(3)

(4)

∞ X

1 , n(n + 2) n=1

(2)

∞ X

1 , n(n + 1)(n + 2) n=1

∞ X

1 . 2 (n + 2) (n + 1) n=0 ∞ X

∞ X 1 1

{= 4[1 − ln(2)]}. = 1 n n+ 2 (n + 1) n + 32 n=1 n=0 ∞ X

1 (2n + 1)(3n + 1) ( "√ (5) n=0 ∞  #) 1 1X 1 27 3π

= = + ln . 1 1 6 n=0 n + 2 n + 3 2 3 16 (6)

∞ X

∞ n+2 n+2 1X
. = 1 2 2 (2n + 1)(n + 1) 2 n + 2 (n + 1) n=0 n=0

[Hint: Introduce partial fractions. You may use the method of Problem 3.5.28 or compute the corresponding integral of the previous Problem. Also Z 1 q−1 Z 1 Z 1 x − xp−1 1 − xp−1 1 − xq−1 dx = dx − dx. 1−x 1−x 1−x 0 0 0 In (3) and (6), you also need the Euler’s sum, Example 3.6.3. Notice also Problem 1.8.24.]

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Improper Riemann Integrals

(b) Compare sum (6) with Problem II 1.7.72 and its footnote and derive some byproducts. 3.5.34 Compute the four sums  ∞  X 2 1 1 (1) − + , n + 12 n + 13 n + 41 n=0 (2)

∞  X n=0

(3)

(4)

1 n+

1 2

2 + n+

1 3

3 − n+

 1 4

,

∞ X

n+1 , (2n + 1)(3n + 1)(4n + 1) n=0 ∞ X

n+1 . (2n − 1)(3n − 1)(4n − 1) n=0 [Hint: Use the hint of the previous Problem.]

3.5.35 Prove the following four results Z 1 xk (a) If a > 0, lim dx = 0. k→∞ 0 a + xk Z 1
k 1 (b) lim kx2 1 − x3 dx = . k→∞ 0 3 Z q sin(kx) (c) If q > p > 0, lim dx = 0. k→∞ p x Z p sin(kx) π (d) If p > 0, lim dx = . k→∞ 0 x 2 [Hint: In (c) use integration by parts first. In (d) use the main result of Example 3.1.8 and (c).] 3.5.36 Z (a) Use Riemann sums for the integral 0

that lim

k→∞

k X i=1

k k 2 + i2

1

1 π to prove dx = 1 + x2 4

! =

π . 4

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(b) Use Riemann sums for the integral Z 0

1

√

√ 1 dx = ln(1 + 2) 2 1+x

to prove that k X

lim

k→∞

i=1

1 √ k 2 + i2

! = ln(1 +

3.5.37 Prove that for any n ∈ N  Z n x − [[x]]   ln(n) + 1 − dx,   n x2  1 X 1 = Z n  kr  k=1 [[x]] 1    + r dx, r+1 nr−1 x 1

√

2).

if r = 1,

if r 6= 1,

where [[x]] is the integer part of the real number x, which by definition is the greatest integer number less or equal to x. [Hint: [1, n) = [1, 2) ∪ [2, 3) ∪ . . . ∪ [n − 1, n) and [1, n] = [1, n) ∪ {n}.] 3.5.38 Consider any four positive numbers a > 0, b > 0, c > 0, and r > 0. (a) Justify: Z Z ∞ cos(ac) b ∞ cos(ax) sin(ax) dx = ... = − − dx −→ 0, xb acb a c xb+1 c as c −→ ∞. (b) Prove: Z ∞ ∞ X 1 sin(nx) dx r n c xb n=1 converges absolutely. (c) Prove: " lim

c→∞

# Z ∞ ∞ X 1 sin(nx) dx = 0. nr c xb n=1

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Improper Riemann Integrals

3.6

Double Integral Technique

As we have seen in Section 2.1, the double integrals can be used to evaluate improper integrals. Here we investigate this technique much further. It consists of imbedding a given improper integral into an appropriate double integral. Then we evaluate the double integral, usually by switching the order of integration. (Therefore, we could name this technique as: “Technique of Switching the Order of Integration in Double Integrals.”) From the result we find about the double integral, we can now find the value of the originally given improper integral. In a course of multi-variable calculus, we see that if a function of two variables z = f (x, y) is defined on the closed rectangle R = [a, b] × [c, d], where a < b and c < d real numbers, and it is integrable and bounded (e.g., when it is continuous, but not only continuous), then we have # # Z Z Z b "Z d Z d "Z b f (x, y) dx dy. f (x, y) dA = f (x, y) dy dx = a

R=[a,b]×[c,d]

c

c

a

But, if z = f (x, y) is not bounded in the closed rectangle R, (and therefore not continuous in R) and its double integrals – one for each order of integration – exist, it is not automatic that they are equal. Two often-encountered examples (counterexamples) are the one in Problem 3.7.7 and the following: We consider R = [0, 1] × [0, 1] and  x−y  if (x, y) ∈ (0, 1] × (0, 1],   (x + y)3 , f (x, y) =    0, if (x, y) = (0, 0). Then Z 0

Also, Z 1 Z 0

0

x

1

  Z 1 Z 1 x−y 2x − (x + y) dy dx = dy dx = 3 (x + y)3 0 0 (x + y) 0   Z 1 Z 1  2x 1 1 − dy dx = ... = , 3 2 (x + y) (x + y) 2 0 0  Z 1 Z 1 x−y −1 whereas dx dy = ... = . 3 2 0 0 (x + y)

Z

1

 x−y dy dx = ∞, (x + y)3

Z

1

Z

and 0

0

y

 x−y dx dy = −∞, (x + y)3

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and Z x−y dA = (x + y)3

Z Z

1

0

R=[0,1]×[0,1]

1

Z 0

x−y (x + y)3 dxdy = ∞ + ∞ = ∞.

(Carry out the details to convince yourselves. See Problem 3.7.6.) On this example, we notice the following five things: 1. 2. 3.

lim

f (x, y) = 0.

0 0, we have Z ∞ Z ∞  π −(c+v2 )u sin(u) du dv < √ < ∞. e 2 c 0 0 Therefore, by the Tonelli Conditions II or III, we are allowed to switch the order of integration in the double integral (3.2) above (without the absolute value). Thus, using Problem 1.6.13, we obtain Z ∞ sin(u) (3.3) e−cu √ du = u 0  Z ∞ Z ∞ Z ∞ 2 2 dv 2 √ e−(c+v )u sin(u) du dv = √ 2. π 0 π 1 + (c + v 2 ) 0 0 Since ∀ c ≥ 0,

1

1 0< 2 ≤ 1 + v4 2 1 + (c + v )

Z and 0

∞

√ dv π 2 = , 1 + v4 4

we also conclude that the integral Z ∞ dv 2 is continuous as a function of c ≥ 0. 1 + (c + v 2 ) 0 Then, in equality (3.3), we let c → 0+ to get √ √ Z ∞ Z ∞ sin(u) 2 dv 2π 2 π 2 √ du = √ = . =√ · 4 1 + v 4 2 u π π 0 0 This finally verifies that the Fresnel integral is equal to √ √ Z ∞ Z
1 ∞ sin(u) 1 2π 2π 2 √ du = · sin x dx = = . 2 0 2 2 4 u 0 Remark 3.6.1 Similar work shows √ Z ∞
2π 2 cos x dx = . 4 0

Real Analysis Techniques 233

Remark 3.6.2 Note that since sin x2 and cos x2 are even functions in R, then: √ Z ∞ Z ∞

2π . (a) sin x2 dx = 2 sin x2 dx = 2 −∞ 0 Z

∞

cos x

(b)

2




√

∞

Z

cos x

dx = 2

−∞

2




dx =

0

2π . 2

Also for any real number a 6= 0, by means of u-substitution, we have: Z ∞ Z ∞ r

π . (a) sin ax2 dx = 2 sin ax2 dx = sign(a) 2|a| −∞ 0 Z

∞

(b)


cos ax2 dx = 2

Z

−∞

∞


cos ax2 dx =

0

r

π . 2|a|

[See also Problem 3.7.16, (a).] Remark 3.6.3 We observe that, using Problem 1.6.13, for β > 0 and c > 0 constants, the equality (3.3) is written more generally Z ∞ Z ∞
2 sin(βu) 2 e−cx sin βx2 dx = du = e−cu √ u 0 0 Z ∞ β dv 2 √ . π 0 β 2 + (c + v 2 )2 (See Problem II 1.7.19 for the final answer in closed form.) Remark 3.6.4 The integrals defined by Z x Z
S(x) := sin t2 dt, and C(x) := 0

x


cos t2 dt

0

are called Fresnel sine integral and Fresnel cosine integral, respecπ tively. Some authors like to use t2 instead of just t2 in the argument. 2 So, check what definition the book you study uses. N Example 3.6.3 In this example we compute Euler’s sum, or ζ(2) (see Problem 3.5.26) ∞ X 1 π2 ζ(2) = = , 2 n 6 n=1

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Improper Riemann Integrals

which we have already used in a few instances, (e.g., see Example 3.3.23, etc.), by means of double integration and basic analysis. (We also computed this sum later in Example II 1.7.23, Corollary II 1.7.3 and Problem II 1.7.52, using methods of complex analysis.) To this end, we examine the double integral Z 1Z 1 1 I := dxdy 1 − xy 0 0 in two ways. Way 1: We notice that for 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1, by the geometric series, we have ∞ X 1 . (xy)n = 1 − xy n=0 At x = y = 1 both sides are +∞. The convergence is uniform on any closed rectangle [0, a] × [0, b], where 0 < a, b < 1. (Prove! E.g., by an analogous Weirstraß M-Test, 3.3.3, and/or by other ways.) By a result analogous to Corollary 3.3.2, we can switch integration and summation to obtain Z 1Z 1 1 I := dxdy = 0 0 1 − xy Z 1Z 1X ∞ Z 1Z 1 ∞ X n (xy)n dxdy. (xy) dxdy = 0

0 n=0

n=0

0

0

[Switching integration and summation here, can also be justified by the non-negativity of (xy)n , ∀ n ∈ N, for 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1, and invoking Remark 2 that follows the Lebesgue Monotone Convergence Theorem (Theorem 3.3.10, if we adjust it to R2 ). Or, we can invoke the Beppo-Levi Theorem 3.3.14 (Theorem 3.3.14, if we adjusted to R2 ).] Therefore, we have I :=

∞ Z X n=0

0

1

Z

1 n

(xy) dxdy = 0

∞ Z X n=0

1 n

Z

x dx ·

0

1 n

y dy

 =

0

∞ X 1 1 = . 2 (n + 1) n2 n=1 n=0 ∞ X

Way 2: We change the variables by letting x = s − t and y = s + t, or y−x y+x and t = . This change of variables changes 2 2 the square S := [0, 1] × [0, 1] of the (x-y)-integration onto the square T equivalently s =

Real Analysis Techniques

235

formed by the lines s−t = 0,s+t =0,s−t =  1, and s+t = 1. Therefore, 1 1 1 1 T has vertices (0, 0), (1, 0), , , , − , in the s-t-plane. (Draw 2 2 2 2 the figures for S and T .) Now, as we know from calculus, ∂ [x(s, t), y(s, t)] ds dt = |1 · 1 − (−1) · 1| ds dt = 2 ds dt. dx dy = ∂(s, t) So, Z Z

1 dx dy = 1 − xy

I= S

Z Z

2 ds dt. 1 − s2 + t2

T

Since T is symmetrical about the s-axis and we can write Z 12 Z I=2 0

s

0

2 dt 1 − s2 + t2



Z

1

Z

ds + 2 1 2

0

2 is even in t, 1 − s2 + t2

1−s

2 dt 1 − s2 + t2

 ds.

Using the known integrals with arc-tangent, we find 1 2

  1 s √ I=4 arctan √ ds + 1 − s2 1 − s2 0   Z 1 1 1−s √ 4 arctan √ ds. 2 1 1−s 1 − s2 2   s in the first integral above Now, by letting u = arctan √ 1−rs2    1−s 1−s and v = arctan √ = arctan in the second, we find 1+s 1 − s2 1 1 du = √ ds, −2dv = √ ds, and 2 1−s 1 − s2 Z 0 Z π6 u du + 4 (−2v) dv = I=4 Z

π

0

4  π 2 2

6

 −0+0−

 6  π 2 −8  π 2 π2 =6 = . 2 6 6 6

Finally, by the two results in way 1 and way 2, we obtain Z

1

Z

I := 0

0

1

∞ X 1 1 π2 dx dy = = . 2 1 − xy n 6 n=1

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Improper Riemann Integrals

This is a mathematical analysis proof of Euler’s sum ζ(2) =

∞ X 1 π2 = . 2 n 6 n=1

π2 , see Apostol 1974, [For a nice different elementary proof of ζ(2) = 6 216-217, exercise 8.46. Also, in exercise 8.47, there is an elementary proof π4 of ζ(4) = . This uses some of the results of Problem II 1.1.46, (II).] 90 Remark: Similarly, for any a ≥ 0, we have the integral Z 1Z 1X Z 1Z 1 ∞ (xy)a (xy)n+a dxdy = dxdy = I(a) := 0 0 n=0 0 0 1 − xy  X Z 1 ∞ Z 1Z 1 ∞ Z 1 ∞ X X 1 (xy)n+a dxdy = xn+a dx y n+a dy = . (n + a)2 0 0 0 n=0 0 n=0 n=1 Then, for any k ≥ 0 integer, we get Z 1Z 1 ∞ X dk I(a) (xy)a lnk (xy) 1 k := dxdy = (−1) (k + 1)! . dak 1 − xy (n + a)k+2 0 0 n=1 (Justify the differentiation under the integral sign.) Then, with a = 0, we find that for any k ≥ 0 integer, we have Z 1Z 1 k ∞ X ln (xy) 1 dxdy = (−1)k (k + 1)! = (−1)k (k + 1)!ζ(k + 2). k+2 1 − xy n 0 0 n=1 (For more examples and applications see Aksoy and Khamsi 2010 and Nahin 2015.) N Example 3.6.4 In Problems II 1.7.63 and II 1.7.64, we ask to show Z 1 Z 1 ln(u) −π 2 ln(1 − x) −π 2 I2 = du = ⇐⇒ I2 = dx = . 6 x 6 0 1−u 0 By the previous Example, we have Z 1Z 1 1 π2 dxdy = . 6 0 0 1 − xy We let v = 1 − xy, dv = −xdy, etc., to perform the y-integration and obtain Z 1 π2 − ln(1 − x) dx = . x 6 0

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237

Thus, we have obtained the second integral above. If in this integral, we now let x = 1 − u, we obtain the first. Remark: We can also obtain this result by using the power series − ln(1 − x) =

∞ X xn , n n=1

− 1 < x < 1.

for

Thus, we find ∞ ∞ X − ln(1 − x) 1 X xn xn−1 = = , x x n=1 n n n=1

for

− 1 < x < 1.

At x = 0, both sides are equal to 1. (The first side is obtained by L’ Hˆopital’s rule.) The convergence is uniform on any interval [0, r], with 0 < r < 1. So, by Corollary 3.3.2 (or by the Remark that follows Theorem 3.3.10, or by Theorem 3.3.14), we get Z 1X Z 1 ∞ ∞ Z 1 n−1 ∞ X X xn−1 1 x − ln(1 − x) dx = dx = dx = . x n n n2 0 n=1 0 n=1 0 n=1 Therefore, by Euler’s sum computed in the previous Example, we obtain Z 1 Z 1 ∞ X 1 − ln(1 − x) − ln(u) π2 u=1−x ζ(2) = dx = du = . = x 1−u n2 6 0 0 n=1 N Example 3.6.5 For −1 ≤ x ≤ 1, we define the functions Li1 (x) = − ln(1 − x) =

∞ X xn , n n=1

[where Li1 (1− ) = +∞] and Z x Z 1 ∞ X − ln(1 − t) xn − ln(1 − xt) dt = dt = . Li2 (x) = t t n2 0 0 n=1 The function Li2 (x) is called dilogarithm function. By the previπ2 ous Example, or by the two above integrals, we have Li2 (1) = and 6 Li2 (0) = 0. Now, for 0 ≤ x ≤ 1, we let f (x) := Li2 (x) + Li2 (1 − x)

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Improper Riemann Integrals

Then, we get f 0 (x) =

− ln(1 − x) ln(x) + = [− ln(x) ln(1 − x)]0 . x 1−x

So, f (x) = C − ln(x) ln(1 − x), where C is a constant.

π2 , and lim− [ ln(x) ln(1 − x) ] = 0. (This 6 x→1 limit is of the type 0 · ∞ and can be found by L’ Hˆopital’s rule. With x = 0+ , we obtain the same results.) Thus, we find the useful formula For x = 1, we have f (1) =

Li2 (x) + Li2 (1 − x) =

π2 − ln(x) ln(1 − x), 6

0 ≤ x ≤ 1,

or, if 0 ≤ x ≤ 1, Z x Z 1−x − ln(1 − t) − ln(1 − t) π2 dt + dt = − ln(x) ln(1 − x). t t 6 0 0 This formula is called Landen’s14 formula. 1 If we let x = in this formula and simplify, we obtain the integral 2 and the sum   Z 1 ∞ X 2 − ln(1 − t) 1 1 π2 ln2 (2) Li2 = dt = = − . 2 t 2 n n2 12 2 0 n=1 So, by the previous Example, we also get Z 1 − ln(1 − t) π2 ln2 (2) dt = + . 1 t 12 2 2 (Compare and use these results with: Example 3.3.23 and Problems II 1.7.62, II 1.7.63 and II 1.7.64.) Note: For k ≥ 2 integer, we define the so-called Polylogarithmic function by the recursive definition Z x ∞ X Lik−1 (t) xn Lik (x) = dt = , for − 1 ≤ x ≤ 1. t nk 0 n=1 Also, if the real variable x is replaced by the complex variable z these special functions can be studied in the complex domain to yield many interesting results. They have many applications in mathematics, science and technology. (See also Problem 3.7.9.) N 14 Landen

John, English amateur mathematician, 1719–1790.

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239

Application In the theory of curves in R2 , the natural equation of a curve C is its curvature κ given as a function of the arc-length parameter. I.e., κ = f (s), where s is the arc-length parameter of C. y

2π 4a

-

2π 4a

x O

2π 4a

-

2π 4a

FIGURE 3.4: Clothoid or Spiral of Cornu spiral Then, up to the rigid motions (isometries) of the plane, the curve is given by Z s Z t  Z t   Z s C(s) = cos f (u) du dt , sin f (u) du dt ?

??

?

??

with the frame {T, N } of the tangent and normal unit vectors positively oriented. So, the curve with κ = 2a2 s, where a > 0 constant, i.e., the curvature is a positive multiple of s, without loss of generality is given by Z s  Z s

C(s) = cos a2 t2 dt , sin a2 t2 dt . 0

0

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Improper Riemann Integrals

Such a curve is called clothoid or spiral of Cornu15 (see Figure 3.4). Cornu used it in optics for the calculation of diffraction of light in some diffraction problems, and Fresnel for lens designing. It was also known to Leonhard Euler and Jakob Bernoulli. This curve is also very important in designing and constructing railways and exits and entrances of highways. We observe: (1) C(0) = 0. (2) The curve is symmetrical about the origin since C(−s) = −C(s). (Check this!) (3) Using v = at and the values of the Fresnel integrals, found in Example 3.6.2, we find that √ ! √ √ ! √ 2π 2π 2π 2π , and lim C(s) = − ,− . lim C(s) = s→−∞ s→∞ 4a 4a 4a 4a

3.7

Problems

3.7.1 (a) Prove that for every x > 0,

1 = x

Z

∞

e−xt dt.

0

Z ∞  Z N sin(x) dx = sin(x) e−xt dt dx, x 0 0 0 for any 0 < N < ∞. Justify the switching of the order of integration and the resulting limit, to derive the Dirichet sine integral Z N Z ∞ sin(x) π sin(x) dx = lim dx = , N →∞ 0 x x 2 0 Z

N

(b) From (a) we get

(a result obtained in Example 3.1.8, by a different method). (c) Now, use the result in (b) to obtain the final general result of Example 3.1.8. (d) For any k ≥ 1 integer, justify the differentiation of the equation in Z ∞ 1 1 −xt k−1 (a) k times to find k = e t dt. x (k − 1)! 0 15 Marie

Alfred Cornu, French mathematician, 1841–1902.

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241

[See also Problem 3.13.6, (b) and compare.] Z 1 1 (e) Whereas du = ∞, prove that 0 u 2 Z ∞ 1 1 − e−t dt = du dv = t 0 u+v 0 0 Z ∞ t2−2 1 ds = 2 ds = 2 ln(2). (s + 1)(s + 2) (s + 1)(s + 2) 0 1

Z

∞

Z =2 0

Z

1

[Hint: Use (a) and change the double integral to a triple one. Switch order of integration and use (d) and carry our the resulting computations.] (f) Imitate (e) and prove and evaluate Z 0

Z 3 0

∞

1

Z

1

Z

0

0

1

1 du dv dw = u+v+w

s3−2 ds = 3 (s + 1)(s + 2)(s + 3)

0



0

∞

Z

∞

Z

1 − e−t t

3 dt =

s ds. (s + 1)(s + 2)(s + 3)

Generalize! Z 3.7.2 Prove completely that

∞


cos x2 dx =

0

√ 2π . 4

3.7.3 Use the power series of sin(x) and cos(x) to express the following two integrals as power series: Z x Z x

2 S(x) := sin u du and C(x) := cos u2 du. 0

3.7.4 Evaluate the integrals Z ∞ cos(x) √ dx x 0

0

Z and 0

∞

sin(x) √ dx. x

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Improper Riemann Integrals

3.7.5 (a) By looking at the double improper integral Z ∞Z ∞
sin x2 + y 2 dxdy 0

0

as the limit of double proper integrals over the rectangles [0, a] × [0, b] as a → ∞ and b → ∞, use the values of the Fresnel integrals π to find that its value obtained in this way is . 4 (b) What happens if you try to evaluate this integral by using polar coordinates? (c) Can you explain the discrepancy between the results of (a) and (b)? 3.7.6 (a) Justify the results of the initial example of this section:   Z 1 Z 1 Z 1 Z 1 x−y 2x − (x + y) dy dx = dy dx = 3 (x + y)3 0 0 (x + y) 0 0 Z 0

1

1

Z 0

whereas Z 0

(b) Prove Z 1 Z 0

0

x



1

  1 1 2x − dy dx = ... = 3 2 (x + y) (x + y) 2

Z 0

1

 −1 x−y dx dy = ... = . (x + y)3 2

  Z 1 Z y x−y x−y dy dx = ∞ and dx dy = −∞. 3 (x + y)3 0 0 (x + y)

x−y assumes (x + y)3 values near 0, equal to 0, unbounded positive and unbounded negative.

(c) Show that near (0, 0) the function f (x, y) =

3.7.7 Let

 2 x − y2    2 , if (x, y) 6= (0, 0).  2 (x + y 2 ) f (x, y) =     0, if (x, y) = (0, 0).

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243

(a) Show that near (0, 0), f (x, y) assumes values near 0, equal to 0, unbounded positive and unbounded negative.
(b) Write x2 − y 2 = 2x2 − x2 + y 2 and then break the fraction to prove that  Z 1 Z 1 π f (x, y) dx dy = − . 4 0 0 (c) Similarly, prove Z

1

Z

0

0

1

 π f (x, y) dy dx = . 4

(d) Also, prove 1

Z 0

Z

x

 Z f (x, y) dy dx = ∞ and

0

1

0

Z

y

 f (x, y) dx dy = −∞.

0

3.7.8 Define the functions  1   q , if 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1,    y + x − 1 2 f (x, y) =     0, otherwise, and g(x, y) =

 1   , if  q   x − 1

0 ≤ x ≤ 1 and 0 ≤ y ≤ 1,

2

    0,

otherwise.

(a) Show that 0 ≤ f (x, y) ≤ g(x, y). Z Z √ (b) Show that g(x, y)dxdy = 2 2 and so R2

Z Z 0< R2

√ f (x, y)dxdy < 2 2

(is positive finite).

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Improper Riemann Integrals

(c) Let Z v(x) :=

f (x, y)dy. R

Show that 0 ≤ v(x) < ∞ for any x 6=

1 and v 2

  1 = +∞. 2

(d) Pick any 0 < x0 < 1, and answer the same three questions for the function  1     y + p|x − x | y , if 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1, 0 f (x, y) =     0, otherwise. (e) Provide another such example on your own. (See Remark after Condition IV.) 3.7.9 Let Z 1Z I := 0

1

0

(a) Prove:

Z 0

1

1 dxdydz and J := 1 − xyz I=

R1R1 0

0

− ln(1−xy) xy

1

Z

dxdy =

0

Z 0

P∞

1

Z

1

0

1 n=1 n3

1 dxdydz. 1 + xyz = ζ(3).

[Hint: See and imitate Example 3.6.4.] (b) Generalize statement (a). (c) Prove 3I = 4J. [Hint: Use x = u2 , y = v 2 , z = w2 and partial fractions.] 3.7.10 Consider a > 0 and b ∈ R constants. Use a Tonelli condition to justify the change of order of integration in  Z ∞ Z 1 e−ay sin(2bxy) dx dy 0

0

and then assume b 6= 0 and prove that  2  Z ∞ Z ∞ 2 1 a + 4b2 −ay sin (by) −ay 1 − cos(2by) e dy = e dy = ln . 2y y 4 a2 0 0

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Prove that this formula is also correct for b = 0, as 0 = 0, and for a = 0 and b 6= 0, as ∞ = ∞. (See also and compare with Problem 3.9.23.) 3.7.11 Consider a ≥ 0 and b ∈ R constants. Then: (a) Without computing, explain why for every 0 < N < ∞  Z "Z Z Z N

1

N

1

e−axy sin(by) dx dy =

0

0

#

e−axy sin(by) dy dx.

0

0

(b) Now, perform the inner integration in both double integrals in (a) and write the obtained equation. (c) Take limit as N → ∞ of the equal expressions found in (b), and prove: For a ≥ 0 and b ∈ R constants, we have   Z ∞ a sin(by) π b I(α) := e−ay dy = − arctan = arctan . y 2 b a 0 [Hint: In (c), commuting the limit and the integral in the second side of the equation obtained in (b), you need to use either Part (I) of the Main Theorem, 3.1.1, or Definition 3.3.2 and Theorem 3.3.9. ] (d) The integral I(α) found in (c) can also be found as follows: (1) Use the Main Theorem, 3.1.1, and Problem 1.6.13 to find d [I(α)] . dα (2) What is I(0) and why? (3) Now find I(α). (Note: This was also found by a different method inside the computation of Example 3.1.8 and stated in Example 3.1.9. It is also reported in many integral tables. See also and compare with Problem 3.9.23.) 3.7.12 (a) For α ≥ 0 and β ∈ R constants, prove  2    Z ∞ α 4β sin2 (βu) 2β − ln J(α) := e−αu du = β arctan + 1 . u2 α 4 α2 0

246

Improper Riemann Integrals [Hint: Use appropriate integration by parts and the results in Problems 3.7.10 and 3.7.11. For α = 0, as usual, you may use a continuity argument.]

(b) The integral J(α) found above can also be found as follows: (1) Use the Main Theorem, 3.1.1, and Problem 3.7.10 to find d [J(α)] . dα (2) Find J(0). (See Problem 3.2.16). (3) Now find J(α). 3.7.13 Change the integral16 Z ∞ Z ∞
x2 − y 2 2 e−(x−y) sin2 x2 + y 2 2 dy dx (x2 + y 2 ) x=0 y=x to polar coordinates and prove that it is equal to 1 [4 arctan(2) − ln(5) − 2π] ≈ −0.2165018. 16 3.7.14 For k and l real constants, prove: Z

∞

e−x

(a) If l ≥ k + 1, 0

Z

∞

(b) If l ≥ 1,

e−x

0

Z 3.7.15 For l = 0, 1, 2 and 3, find

sink (x) dx = +∞. xl

cosk (x) dx = +∞. xl

∞

e−x

0

sin3 (x) dx. xl

−1 3 [Hint: sin (x) = sin(3x) + sin(x), etc.] 4 4 3

3.7.16 (a) For any real numbers a 6= 0, b and c, use the known identity  2 b 4ac − b2 2 ax + bx + c = a x + + , the Fresnel integrals, and 2a 4a 16 American Mathematical Monthly, Problem 11650, Vol. 119, Number 6, June-July 2012.

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247

appropriate trigonometric formulae, to prove Z ∞
sin ax2 + bx + c dx = −∞      r π 4ac − b2 4ac − b2 sign(a) cos + sin 2|a| 4a 4a and Z

∞


cos ax2 + bx + c dx = −∞      r 4ac − b2 4ac − b2 π cos − sign(a) sin . 2|a| 4a 4a (b) Use (a) and the appropriate trigonometric formulae, to compute the integrals Z ∞
I1 = sin −2x2 cos(4x + 6) dx, −∞ Z ∞
I2 = sin 3x2 sin(−2x + 1) dx, −∞ Z ∞
I3 = cos x2 cos(−4x + 1) dx. −∞

3.7.17 (a) Prove Z 2 0

1 3

− ln(1 − t) dt + t

2 3

Z 1 3

− ln(1 − t) π2 dt = + ln(3) ln t 6

  2 . 3

(b) Find Z 0

3 5

− ln(1 − t) dt + t

Z 0

2 5

− ln(1 − t) dt. t

3.7.18 We have proven, in Example 3.1.6, (b) (and Example II 1.7.7), that if m and n are integers such that 0 ≤ m < n − 1, then Z ∞ Z ∞ n−m−2 xm x π
. I1 := dx = dx = n n x +1 x +1 n sin m+1 0 0 n π

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Improper Riemann Integrals

Now for any integer k ≥ 2, prove that Z ∞ Z ∞ kn−m−2 xm x Ik := dx = dx = n + 1)k n + 1)k (x (x 0 0  k−1 Y π m+1
· 1 − , nj n · sin m+1 n π j=1 and also prove that Ik −→ 0, as k −→ ∞. [See also Problems 1.8.25, 3.2.47 and properties (B, 5) and (B, 8) of the Beta function.] [Hint: See Problem 1.8.2, (2), and follow the method suggested there.]

3.8

Frullani Integrals

A special category of improper integrals are the Frullani17 or CauchyFrullani integrals. They have the general type Z ∞ f (bx) − f (ax) I(a, b) = dx, where 0 < a, b < ∞, x 0 with f : (0, ∞) −→ R is a “nice” function (according to the nonstandard definition, 1.7.1,) not necessarily continuous but satisfies the following condition which we assume throughout this section: Z µ (z) f (x) dx exists, ∀ 0 < λ < µ < ∞. λ

For functions continuous in (0, ∞), this condition is obviously fulfilled. Also, if f (x) is defined and continuous at x = 0, we could have taken 0 ≤ λ < µ < ∞. (See also Problems 3.9.13 and 3.9.14 for examples of “nice” functions that do not satisfy this condition. Another 1 simpler such example is f (x) = , with 0 ≤ x 6= 2. Check it! We can x−2 also check that the results proven below do not yield the correct results for this function.) 17 Giuliano

Frullani, Italian mathematician, 1795–1834.

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249

We observe that: (1) I(a, a) = 0 for any 0 < a < ∞. (2) I(a, b) = −I(b, a) for any 0 < a, b < ∞. (3) For I(a, b) to exist when 0 < a 6= b < ∞, the limits lim+ [f (bx) − x→0

f (ax)] and lim [f (bx) − f (ax)] must either be zero or oscillate x→∞ about zero. These two limit conditions are necessary but not sufficient, as we will see in examples and problems that follow. Otherwise, i.e., if either condition or both is/are not valid, then I(a, b) is either ±∞ or does not exist (oscillates depending on the two limiting processes). We can prove these assertions, by applying the Limit Comparison Test, Theorem 1.7.6, to the func1 f (bx) − f (ax) and h(x) = and using the facts: tions g(x) := x x Z r Z ∞ 1 1 dx = ∞ and dx = ∞, ∀ 0 < r, c < ∞. x 0 x c Z ∞ f (x) dx = r (= a finite value), then (4) If x 0 ∀ a>0

and ∀ b > 0,

I(a, b) = 0.

This follows from the fact that for any s > 0 constant, by making the substitution x = su, we find Z ∞ Z ∞ Z ∞ f (x) f (su) f (su) r= dx = s du = du, ∀ s > 0. So, x su u 0 0 Z ∞0 Z ∞ Z ∞ f (bx) − f (ax) f (bx) f (ax) dx = dx − dx = r − r = 0. x x x 0 0 0 For instance, see Examples 3.8.3 and 3.8.7 of this section. Remarks: Z ∞ f (x) dx = r (= a finite value) implies that the (a) The condition x 0 two limits lim+ f (x) and lim f (x) must either be zero or oscillate x→0

x→∞

about zero. (Justify!) (b) Z If a = 6 b, for I(a, b) 6= 0, it is necessary that the integral ∞ f (x) dx either is ±∞ or oscillates. (Justify!) If it is ±∞, then x 0 we obtain the indeterminate form ±∞ ∓ ∞, whose value we must determine.

250

Improper Riemann Integrals

In the sequel, we want to compute I(a, b) and state the important general results and the hypotheses under which are valid. To this end, we begin with the following.18

Preliminary Computation Consider any 0 < a, b < ∞. Using the standing assumption (z) for f (x), we can write

Z lim

h→∞ →0+



h

I(a, b) = "Z # Z h h f (bx) − f (ax) f (bx) f (ax) dx = lim dx − dx . h→∞ x x x   →0+

Performing the substitutions t = bx and t = ax in the first and second integral, respectively, and using the standing assumption (z) for f (x), we find # "Z Z ah bh f (t) f (t) dt − dt = I(a, b) = lim h→∞ t t a b + →0 "Z # Z bh Z b Z ah ah f (t) f (t) f (t) f (t) = lim dt + dt − dt − dt = h→∞ t t t t b ah a b + →0 "Z # Z b bh f (t) f (t) lim dt − dt . h→∞ t t ah a →0+

[For this splitting of the integral to be legitimate, we need the condition (z) and then the two extreme integrals cancel each other.] If the individual limits exist as real numbers, or are ±∞, we can write Z bh Z b f (t) f (t) I(a, b) = lim dt − lim+ dt. (3.4) h→∞ ah t t →0 a Here, the case ∞ − ∞ means that the I(a, b) does not exist (oscillates). For giving more practical results, we will deal with these two limits under certain conditions. We assume that f (x) is continuous in (0, δ) for some δ > 0 and in (µ, ∞), for some µ > 0. In the first limit, we use the substitution t = eu and consider h large enough so that ah > µ and bh > µ to invoke the Mean Value Theorem for integrals. Then we find   Z bh Z ln(bh) f (t) b u c dt = f (e ) du = [ln(bh) − ln(ah)] f (e ) = ln f (ζ), t a ah ln(ah) 18 In the exposition of this material, we follow bibliography: Ostrowski 1949 and Agnew 1951, with some additions and modifications.

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251

where c is between ln(ah) and ln(bh) or ζ := ec is between ah and bh. If h → ∞, then ζ → ∞. We let f (∞) := lim f (x) ∈ R ∪ {±∞} or x→∞

does not exist (oscillates). Then, we have: Z

bh

lim

h→∞

ah

f (t) dt = ln t

    b b · lim f (ζ) = ln · f (∞). ζ→∞ a a

Now using the substitution t = e−u in the second limit, with analogous work, we find   Z b f (t) b lim+ dt = ln · f (0), t a →0 a where f (0) := lim f (x) ∈ R ∪ {±∞} or does not exist (oscillates). x→0+

This preliminary computation along with the stipulated hypotheses proves the following general Theorem, which is more general than the classical results on this subject, as we shall see in Example 3.8.2 and its generalization: Theorem 3.8.1 Let f : (0, ∞) −→ R be a nice function that satisfies (z) and 0 < a, b < ∞. Assume: (1) f (x) is continuous in (0, δ) for some δ > 0 and in (µ, ∞) for some µ > 0. (2) lim+ f (x) = f (0) ∈ [−∞, ∞]. x→0

(3) lim f (x) = f (∞) ∈ [−∞, ∞]. x→∞

Then Z I(a, b) = 0

∞

f (bx) − f (ax) dx = [f (∞) − f (0)] ln x

  b a

which, depending on the value of f (∞) − f (0), may be a real number, or ±∞, or does not exist. (See the next Remark.) Remark 3.8.1 In general, an improper integral that takes final answer ∞ − ∞ does not exist. If under the hypotheses of the above Theorem one of the limits f (0) and f (∞) is finite and the other oscillates, or both oscillate, then I(a, b) oscillates and so does not exist.

252

Improper Riemann Integrals

Remark 3.8.2 A byproduct of the above proof [see equation (3.4)] is: For any f : (0, ∞) −→ R nice function that satisfies condition (z), any 0 < a, b < ∞, and any 0 < , h < ∞, we have Z 

h

f (ax) − f (by) dx = x

Z

b

a

f (t) dt − t

Z

bh

ah

f (t) dt. t

Letting b = t and a = 1 in the above Theorem, in general, we have the following: Corollary 3.8.1 Under the conditions of the Theorem and if f (∞) − f (0) 6= 0, or ±∞, or ∞ − ∞, or does not oscillate, we obtain the following integral formula for ln(t): Z ∞ 1 f (tx) − f (x) ln(t) = dx = f (∞) − f (0) 0 x Z ∞ 1 f (x) − f (tx) dx. f (0) − f (∞) 0 x Remark: Integral expressions of ln(t) are useful in applications to special integrals and functions. Obviously, besides b = t and a = 1, other combinations can give the result of the Corollary.

Examples In the examples that follow here and throughout this section and also in the problems that follow, make sure that the functions involved satisfy the prerequisite condition (z), when necessary. Otherwise, indicate the opposite or the counterexample. Example 3.8.1 For 0 < a, b < ∞,   Z ∞ −bx a e b − e−ax dx = − ln = ln x a b 0 since f (x) = e−x satisfies the conditions of Theorem 3.8.1, f (0) = 1 and f (∞) = 0. [The ´ a-priori existence of this integral can be worked out as in Example 3.1.17. See also Application (d), after Example 4.1.8.] From this example, we have the integral expression of ln(t) Z ∞ −x e − e−tx ln(t) = dx. x 0

Real Analysis Techniques

253

(This can also be found by the method of Example 3.1.17. See also Problem 3.2.32.) Also, if we put t = e−x , we find Z 1 a−1 Z 1 a−1 a t − tb−1 t −1 , and ln(a) = dt = ln dt, ∀ a > 0. ln(t) b ln(t) 0 0 (The second integral can also be found by the differentiation method of Section 3.1, as in Example 3.1.17. Solve it with this method as an exercise.) N Example 3.8.2 We easily prove that for any t > 0, Z

∞

Z arctan(tx)dx = ∞,

0

and 0

∞

arctan(tx) dx = ∞. x

Then, for 0 < a, b < ∞, the integral Z ∞ arctan(bx) − arctan(ax) dx, x 0 broken about the − sign of the numerator, becomes ∞ − ∞. But as proven in Example 3.1.17, this integral exists as a finite value. Now, since f (x) = arctan(x) satisfies the conditions of Theorem π 3.8.1 and f (∞) = and f (0) = 0, by the same Theorem, we find that 2 the value of this integral is   Z ∞ b arctan(bx) − arctan(ax) π dx = ln . x 2 a 0 Similarly, for 0 < a, b < ∞ and r > 0, we get Z ∞  π r  b  arctanr (bx) − arctanr (ax) dx = ln . x 2 a 0 Now, letting a −→ 0+ and keeping b > 0 fixed we find Z ∞  π r  b  arctanr (bx) dx = ln = ∞, x 2 0+ 0 which is a´-priori true, ∀ r ∈ R. From the first integral above, we find the integral expression of ln(t) Z 2 ∞ arctan(tx) − arctan(x) ln(t) = dx, π 0 x and similar result from the second integral.

254

Improper Riemann Integrals

(See also Example 3.1.17 and Problem 3.9.24.) Alternative method: Any example similar to this (and the previous one) may also be carried out in the following elementary way, independent of the knowledge of the theory of the Frullani integrals. In this Example we can also work in the following way: Z ∞ Z ∞ n o arctan(bx) − arctan(ax) 1 u=b [arctan(ux)]u=a dx = dx = x x 0 0 # ) Z ∞ (Z b Z ∞ "Z b 1 d 1 x du dx = [arctan(ux)] du dx = x x a 1 + (xu)2 0 a du 0 #  Z ∞ "Z b Z b Z ∞ 1 1 du dx = dx du = 2 1 + (xu)2 0 a 1 + (xu) a 0   Z b Z b 1 π b 1π x=∞ {[arctan(ux)]x=0 } du = du = ln . 2 a a u a u 2 (The switching of the order of integration in the double integral is justified by positivity condition I in Section 3.6, for instance.) (Solve the previous Example by applying this method. Also, notice that with Examples 3.8.7 and 3.8.8, this method does not work.) N Example 3.8.3 We easily prove that: (1) The function

f (x) :=

arctan2 (x) x2 +1

≥ 0,

∀ x ∈ R,

is a positive even function. (2) Both integrals Z ∞ Z f (x) dx and 0

0

(3) lim f (x) = f (0) = 0, x→ 0

∞

f (x) dx exist, as positive (finite) values, x and

(4)

lim f (x) = f (∞) = 0.

x→∞

So, as we have seen at the beginning of this section, result (4), for any 0 < a, b < ∞, Z ∞ Z ∞ Z ∞ f (x) f (bx) f (ax) dx = dx = dx, x x x 0 0 0 is a positive (finite) value. So, Z 0

∞

f (bx) − f (ax) dx = x

Z 0

2 ∞ arctan2 (bx) (bx) +1

− x

arctan2 (ax) (ax)2 +1

dx = 0,

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255

a result that agrees with Theorem 3.8.1, since f (0) = 0 and f (∞) = 0. Now, we may think that if we let b = 1 and a −→ 0+ , we will find the value of Z ∞ arctan2 (x) dx. x (x2 + 1) 0 But, on the one hand Z lim Z lim+ a→0+

a→0

2 (x) ∞ arctan x2 +1

0 ∞

0

−

arctan2 (ax) (ax)2 +1

dx = x f (x) − f (ax) dx = lim+ 0 = 0, x a→0

and on the other hand Z

∞

0

Z 0

arctan2 (x) dx = x (x2 + 1) 2 (x) ∞ arctan x2 +1

− x

Z



∞

lim 

0

a→0+

arctan2 (0x) (0x)2 +1

arctan2 (x) x2 +1

−

arctan2 (ax) (ax)2 +1

x 

dx = (0 − 0) ln

1 0+

  dx =

 = 0 · ∞.

This is an indeterminate form, that does not give the value of Z ∞ arctan2 (x) dx. x (x2 + 1) 0 Does this violate Theorem 3.8.1? The answer is no, because in the Theorem we assume 0 < a, b < ∞, whereas here a = 0. The limit process yields the undeterminate form 0 · ∞ whose value must be determined. So, there is no violation of Theorem 3.8.1. The value of this integral must then be found by a different method. Using the indicated substitution, integration by parts, we find Z ∞ Z π2 arctan2 (x) u=arctan(x) dx = u2 cot(u) du = (integration by parts) x (x2 + 1) 0 0 Z π2 Z π2   u= π π 2 −v −2 u ln[sin(u)] du = −2 − v ln[cos(v)] dv. 2 0 0 By Example II 1.5.5 and Problems 2.3.19, (d), and II 1.7.68, (a), we finally find Z ∞ ∞ ∞ X arctan2 (x) π2 1 π2 7X 1 dx = ln(2) − ln(2) − = . x (x2 + 1) 4 (2k − 1)3 4 8 n=1 n3 0 k=1

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Improper Riemann Integrals

[Compare with Problems 2.3.19 and 2.3.21. Also, notice that Z ∞ arctan2 (x) dx = 0, 2 −∞ x (x + 1) since the integrand is odd.] N Generalization of the Alternative Method to the Frullani Integrals The alternative method used in Example 3.8.2 was the classical method of dealing with the Frullani integrals. This method uses the derivative of f (x), especially when it is continuous, and therefore is less general than the method we have exhibited on the basis of the Preliminary Computation in which we use the mean values of f (x) and thus we only need the continuity or, depending on the case, the piecewise continuity of f (x). In general, we can evaluate Frullani integrals by using this alternative method, whenever we are able to perform the following two steps: 1. Apply the Fundamental Theorem of Calculus, 1.1.1, to the derivative of f (x) [e.g., when f 0 (x) is continuous, etc.]. 2. Switch the order of integration in the double integral that follows. We also observe that under these two conditions, we can apply this method to evaluate the more general integrals: Z

L

g(x)[f (bx) − f (ax)]dx, K

where K and L are in [−∞, ∞], a and b are in R, and the functions f and g possess all the necessary properties in order to apply the above two steps in the respected intervals of computation. (See also Problem 3.9.29.) This integral, when possible to be computed in this way, is more general than the Frullani integral where K = 0, L = ∞, 0 < a, b < ∞, 1 and g(x) is the specific function . In the sequel, on the other hand, x we will study some Frullani integrals that are not amenable to the two conditions stipulated above. (E.g., Examples 3.8.7, 3.8.8, etc.) Under the above conditions, (1) and (2), following the steps analogous to those in the alternative method of Example 3.8.2, we

Real Analysis Techniques find Z L

Z

b

(Z

L

g(x)[f (bx) − f (ax)]dx = K

a

K



d[f (u)] |u=tx du

257

)



· g(x) · x dx dt,

(3.5) d[f (u)] d[f (u)] |u=tx , we first compute the derivative and {Note: In du du then we replace u by tx}. Remark: Equation 3.5 may fail to give the correct result if one of the two conditions, (1) and (2) above, fails. So, we must be careful when we apply it. See Remark of Example 3.8.7 in which condition (2) fails. We can apply Equation 3.5 to some questions in Problems 3.9.23–3.9.27 in conjunction with some established results. For example: Example 3.8.4 Let β ≥ 0 and a, b be real constants and consider e−βx . Then, we easily check the functions f (x) = cos(x) and g(x) = x that conditions (1) and (2) hold. [(1) is immediate. For (2) use the Tonelli conditions, Section 3.6.] So, by equation (3.5), we have  Z ∞ Z b Z ∞ −βx cos(bx) − cos(ax) −βx dx = − sin(tx) e dx dt. e x 0 a 0 Then by the result in Problem 1.6.13, we get Z b Z ∞ t cos(bx) − cos(ax) dx = − dt = e−βx 2 x β + t2 0  2a 2
b 1 1 β +a − ln β 2 + t2 a = ln . 2 2 β 2 + b2 [Do also Problem 3.9.23, (b) and check Problems 3.9.23– 3.9.27.] N Example 3.8.5 For 0 < a, b < ∞

Z 0

∞

ebx − eax x

  +∞,         b dx = (∞ − 1) ln = 0,  a      −∞,

a result that can also be proven elementarily.

if b > a, if b = a if b < a, N

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Improper Riemann Integrals

Example 3.8.6 For 0 < a 6= b < ∞ and f (x) = ∞ ebx bx

Z 0

− x

eax ax

dx = (∞ − ∞) ln

ex on (0, ∞), we get x

  b = does not exist, a

a result that can also be proven elementarily.

N

We now continue with further results followed from the preliminary computation. If we investigate the preliminary computation further, we will see that under certain additional conditions we can evaluate I(a, b) even if one of the limits lim f (x) and lim f (x) does not exist due to x→0+

x→∞

oscillation and the other limit exists or is ±∞. In fact, let us assume: 1. f (x) is continuous in (0, δ) for some δ > 0. 2. lim+ f (x) ∈ [−∞, ∞] or oscillates. x→0

Z

∞

3. c

f (x) dx is finite for some c > 0. x

[We assume nothing about lim f (x).] x→∞

As in the preliminary computation, hypothesis (1) implies Z

b

lim

→0+

a

f (t) dt = ln t

  b f (0). a

Also, using the Cauchy Test, 1.7.11, hypothesis (2) implies Z bh f (t) dt = 0. lim h→∞ ah t Therefore,   Z ∞ f (bx) − f (ax) b I(a, b) = dx = −f (0) ln . x a 0 [If the limit f (0) oscillates, the integral does not exist.] Similarly, we may assume: 1. f (x) is continuous in (µ, ∞) for some µ > 0. 2. lim f (x) ∈ [−∞, ∞] or oscillates. x→∞

Z 3. 0

r

f (x) dx is finite for some r > 0. x

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[We assume nothing about lim+ f (x).] x→0

Then again Z lim+

→0

b

a

f (t) dt = 0 t

[by the Cauchy Test 1.7.11 and its Remark 1], and we find:   Z ∞ f (bx) − f (ax) b dx = f (∞) ln . x a 0 [If the limit f (∞) oscillates, the integral does not exist.] So, we have proven the following: Theorem 3.8.2 Let f : (0, ∞) −→ R be a nice function that satisfies (z) and 0 < a, b < ∞. (I) If f (x) is continuous in (0, δ) for some δ > 0, lim+ f (x) ∈ x→0 Z ∞ f (x) dx is finite for some c > 0, then [−∞, ∞] or oscillates, and x c   Z ∞ b f (bx) − f (ax) dx = −f (0) ln . x a 0 [If the limit f (0) = lim f (x) oscillates, the integral does not exist.] x→0+

(II) If f (x) is continuous in (µ, ∞) for some µ > 0, lim f (x) ∈ x→∞ Z r f (x) [−∞, ∞] or oscillates, and dx is finite for some r > 0, then x 0   Z ∞ f (bx) − f (ax) b dx = f (∞) ln . x a 0 [If the limit f (∞) = lim f (x) oscillates, the integral does not exist.] x→∞

Letting b = t and a = 1, we have the following corollary: Corollary 3.8.2 Under the conditions of the two cases of the Theorem and if the limits f (∞) and f (0) are not 0, or ±∞, and do not oscillate, we obtain the following integral formulae for ln(t), respectively: Z ∞ Z ∞ f (tx) − f (x) 1 f (x) − f (tx) 1 (I) ln(t) = dx = dx. f (∞) 0 x −f (∞) 0 x

(II) ln(t) =

1 −f (0)

Z 0

∞

f (tx) − f (x) 1 dx = x f (0)

Z 0

∞

f (x) − f (tx) dx. x

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Improper Riemann Integrals

Examples Example 3.8.719 For 0 < a, b < ∞, Z ∞ sin(bx) − sin(ax) dx = 0, x 0 since f (0) = sin(0) = 0 and by Example 3.1.8, f (x) = sin(x) satisfies the conditions of Theorem 3.8.2, (I). This result was already evident in view of the general result in Example 3.1.8, since in this case, we can split the integral and we get Z 0

∞

sin(bx) − sin(ax) dx = x

∞

Z 0

sin(bx) − x

Z 0

∞

sin(ax) π π dx = − = 0. x 2 2

Remark: This result cannot be found by using Equation 3.5 because the change of the order of integration is not allowed. Indeed, we have # Z ∞ Z ∞ "Z b sin(bx) − sin(ax) dx = cos(tx) dt dx, x 0 0 a but we cannot switch the order of integration, since in  Z b Z ∞ cos(tx) dx dt, a

0

the integral Z

∞

cos(tx) dx 0

does not exist. (We can easily check that the Tonelli conditions, Section 3.6, do not hold. Also, notice that the method of Example 3.1.17 does not work for similar reason.) Next, by Theorem 3.8.2, (II), we also find

Z ∞ 1 1 − sin ax sin bx dx = 0. x 0     1 1 = sin(0) = 0, but lim sin does not exist (see Here lim sin x→∞ x x x→0+ Problem 3.9.8). Instead, by Example 3.1.8, we have the condition
Z ∞ Z ∞ sin x1 u= 1 π sin(u) dx =x du = , x u 2 0 0 and so the Theorem applies.

N

19 For more examples on the Frullani integrals, see bibliography: Albano, Amdeberhan, Beyerstedt and Moll 2010 and Gradshteyn and Ryzhik 2007.

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Example 3.8.8 (a) For 0 < a, b < ∞, Z ∞ a cos(bx) − cos(ax) dx = ln x b 0 since f (0) = cos(0) = 1 and by Problem 3.2.30, (b), f (x) = cos(x) satisfies the conditions of Theorem 3.8.2, (I). a (If here a < 0 or b < 0, then the result is written as ln .) b In this example, the splitting of integral about the minus “−” is not legitimate because then it becomes ∞ − ∞. Remark: This integral and the integral in the previous Example cannot be computed by the other method described in Example 3.8.2. (Check this to see where the method fails. Also, check that the method of Example 3.1.17 does not work.) We observe that if we keep a > 0 fixed and we let b −→ 0+ , we find Z ∞  a  1 − cos(ax) dx = ln + = ln(∞) = ∞, x 0 0 a result that can be shown earlier and without any knowledge of the Frullani integrals [see Problem 3.2.37, (c)]. This agrees with Problem 3.2.28, (c). (b) From this example, we have the integral expression of ln(t) Z ∞ cos(x) − cos(tx) ln(t) = dx. x 0 (See also Problem 3.9.3.) (c) Again, by Theorem 3.8.2, (II) we also find

  Z ∞ 1 1 − cos ax cos bx b dx = ln . x a 0     1 1 = cos(0) = 1, but lim+ cos does not exist Here lim cos x→∞ x x x→0 (see Problem 3.9.8). Instead, by Problem 3.2.30, (b), we have the condition
Z r Z ∞ cos x1 u= 1 cos(u) ∀ r > 0, dx =x du converges conditionally, 1 x u 0 r

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Improper Riemann Integrals

and so the theorem applies. (d) Since for all real numbers a and b we have sin(ax) sin(bx) =

1 {cos[(a − b)x] − cos[(a + b)x]} = 2

1 {cos(|a − b|x) − cos(|a + b|x)} , 2 we find that if a 6= 0 or b 6= 0, then   Z ∞ 1 |a + b| sin(ax) sin(bx) dx = ln . x 2 |a − b| 0 If a = 0 or b = 0, then this integral is obviously zero. If a = b 6= 0 the integral is ∞, known also from Problem 3.2.28, (c), etc. N We continue with an interesting and useful theorem and two corollaries in which we do not impose any concrete assumptions on the limits lim+ f (x) and lim f (x), but we require a particular mean value to exist. x→0

x→∞

Before we state and prove the Theorem, we must define this mean value and make an important remark. We consider f : (0, ∞) −→ R a nice function that, as always in this section, satisfies condition (z). For any τ > 0, we define the mean value Z 1 x f (t) dt. Mτ (f ) := lim τ 0. Then for any σ > 0 (either 0 < ρ ≤ σ or 0 < σ ≤ ρ) the mean value Mσ (f ) exists and Mσ (f ) = Mρ (f ). Therefore, if the mean value exists for some ρ > 0, we can drop the index and simply write M (f ) for a function f that satisfies (z). This observation follows immediately from the fact that by hypothZ σ

esis (z) the integral

f (t) dt is finite, and so ρ

1 lim {ρ,σ} 0. x 0 We fix a ρ > 0 and assume that the mean value Z 1 x f (t) dt exists (as finite value). M (f ) := lim ρ 0 we have   Z ∞ f (bx) − f (ax) b dx = M (f ) ln . x a 0 Z δ f (x) dx exists for Proof Since f (x) is a nice function such that x 0 some δ > 0, then (by the Cauchy Test, 1.7.11, and its Remark 1) for a, b > 0 Z b f (t) lim+ dt = 0. t →0 a Z x Since ρ > 0, by (z) we have f (t) dt exists ∀ x ≥ ρ > 0, and so ρ

we can define 1 F (x) := x

Z

x

f (t) dt,

∀ x ≥ ρ > 0.

ρ

Then, under the given hypotheses, we conclude that F (x) is continuous for all x ≥ ρ, [xF (x)]0 = f (x) at the points of continuity of f (x) in [ρ, ∞), and lim F (x) = M (f ). ρ 0 such that ah > ρ and bh > ρ. Applying integration by parts, we get Z bh Z bh Z bh f (t) 1 F (t) dt = d[tF (t)] = F (bh) − F (ah) + dt. t t t ah ah ah Using the substitution t = eu and the Mean Value Theorem for integrals, we have that there is ζ = ec is between ah and bh such that Z bh Z ln(bh) f (t) dt = F (bh) − F (ah) + F (eu ) du = t ah ln(ah)   b . F (bh) − F (ah) + F (ζ) ln a

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Improper Riemann Integrals

If h → ∞, then ah → ∞, bh → ∞, and ζ → ∞. Then, we get   Z bh f (t) b lim dt = M (f ) ln , since M (f ) exists. h→∞ ah t a Finally, by equation (3.4) in the preliminary computation, we get Z ∞ f (bx) − f (ax) I(a, b) := dx = x 0   Z b Z bh f (t) f (t) b lim dt − lim+ dt = M (f ) ln . h→∞ ah t t a →0 a  Corollary 3.8.3 Under the hypotheses of Theorem 3.8.3, we get: (1) If lim f (x) = f (∞) is a real number, then M (f ) = f (∞) [see x→∞

Problem 3.9.9, (a)] and so   Z ∞ b f (bx) − f (ax) dx = f (∞) ln . x a 0 [So, Theorem 3.8.3 generalizes Theorem 3.8.2, (II).] Z p (2) If f (x) is periodic with period p > 0 and f (x) dx exists, then 0

∀ u ≥ 0,

1 M (f ) = p

Z 0

p

1 f (x) dx = p

Z

u+p

f (x) dx. u

[See Problem 1.3.8, Item (8.), where ρ = 0, and adjust its proof to the more general definition of M (f ), as defined in the previous Theorem with any fixed ρ ≥ 0.] Then, for any u ≥ 0,
Z p
Z u+p Z ∞ ln ab ln ab f (bx) − f (ax) dx = f (x) dx = f (x) dx. x p p 0 0 u (3) For b = t and a = 1, we get the corresponding integral representations of ln(t) (as before). Z δ f (x) If in Theorem 3.8.3 we replace the hypothesis “ dx exists x 0 for some δ > 0” by “f (x) is continuous in (0, δ] for some δ > 0 and lim f (x) = f (0) ∈ [−∞, ∞]”, then by adjusting the proof of the Theox→0+

rem and using the preliminary computation, we obtain:

Real Analysis Techniques

265

Corollary 3.8.4 Suppose f : (0, ∞) −→ R is a nice function that satisfies condition (z) (see beginning of the section), is continuous in (0, δ] for some δ > 0 and lim f (x) = f (0) ∈ [−∞, ∞] or oscillates. x→0+

We fix a ρ > 0 and assume that the mean value Z 1 x f (t) dt exists (as finite value). M (f ) := lim x→∞ x ρ Then for any a, b > 0 we have   Z ∞ f (bx) − f (ax) b dx = [M (f ) − f (0)] ln . x a 0 [Under the conditions of this Corollary, if the limit f (0) oscillates, the integral does not exist.    b If f (0) = ±∞, then the integral is ∓∞ · sign ln .] a [For b = t and a = 1, we get the corresponding integral representations of ln(t) (as before).] Note: If in the previous Corollary M (f ) = f (0) = ±∞, and a 6= b, then the integral does not exist due to oscillation. Also, we have:    b (1) If M (f ) = ±∞ and f (0) = ∓∞, the integral is ±∞·sign ln . a    b (2) If M (f ) = ∓∞ and f (0) = ±∞, the integral is ∓∞·sign ln . a (3) Keep in mind, in integral situations, ±∞·0 = 0. See also Problem 3.9.9. Example 3.8.9 In Problem 2.3.19, (f ), we have proved Z π ln | cos(x)|dx = −π ln(2). 0

h πi [notice that The function f (x) = ln | cos(x)| is continuous on 0, 4 f (x) f (0) = 0] and lim+ = 0 (use L’ Hˆ opital’s rule). So, x x→0 Z π4 f (x) dx exists. x 0 Since f (x) is periodic with period π [by Theorem 3.8.3 and Part (2) of Corollary 3.8.3, or Corollary 3.8.4,], we get M (f ) =

1 [−π ln(2)] = − ln(2). π

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Improper Riemann Integrals

[See also Problem 1.3.8, Item (8.).] Then, ∀ a > 0 and b > 0 we obtain Z ∞ cos(bx) 1 dx = ln cos(ax) x 0   a b . − ln(2) ln = ln(2) ln a b [Notice: f (x) is not continuous in (0, ∞) and lim f (x) does not exx→∞

ist due to oscillation, but f (0) = 0, the mean value M (f ) exists and therefore the above integral exist. Compare with Problem 3.9.11.] N Example 3.8.10 In Problems 3.2.28, (c) and 3.2.37, (b) hint, we prove that Z ∞ sin2 (x) dx = ∞. x 0 We can prove this here, by using Corollary 3.8.4, as follows. We have that cos2 (0) = 1,

sin2 (x) = 1 − cos2 (x) = cos2 (0 · x) − cos2 (1 · x),

and cos2 (x) is periodic with period π. Then, by Corollary 3.8.3, (2), we get   M cos2 (x) = Z 1 π cos2 (x) dx = π 0 Z 1 π 1 + cos(2x) 1 π 1 dx = · = . π 0 2 π 2 2 Then by Corollary 3.8.4, Z ∞

sin2 (x) dx = x 0 Z ∞ cos2 (0 · x) − cos2 (1 · x) dx = x  0   + 1 0 1 − 1 ln = − · (−∞) = ∞. 2 1 2 Z c cos2 (x) [Notice here that, for any c > 0, dx = ∞.] x 0

N

Real Analysis Techniques

3.9

267

Problems

3.9.1 (a) For real numbers 0 < a, b < ∞, prove:     Z ∞ arctan x − arctan x   b a dx = π ln a . x 2 b 0 (b) Use (a) to write the corresponding integral expression of ln(t). 3.9.2 For real numbers 0 < a, b < ∞, prove: Z 0

∞

1 1 − a 2 1 + (bx) 1 + (ax)2 dx = ln . x b

From this, write the corresponding integral expression of ln(t). 3.9.3 Establish an a ´-priori existence of Z ∞ cos(x) − cos(tx) dx[= ln(t)], x 0 without using the result in Example 3.8.8. Does this integral converge absolutely? 3.9.4 If a, b, c are real constants, give answers to Z ∞ cos(ax + c) − cos(bx + c) dx x 0 and

Z 0

∞

sin(ax + c) − sin(bx + c) dx x

for the various combinations of values (positive, negative or zero) of a, b and c. 3.9.5 Consider any four real numbers a, b, c and d. Prove Z b dx Z d bt e − ecx e − eat dx = dt. x t a c

268

Improper Riemann Integrals

3.9.6 1 x and then adjust it to equation (3.5) and use Problem 2.3.11. I.e, prove: If a ≥ 0 and b ≥ 0 constants, Z ∞  √ √ −a −b e x2 − e x2 dx = πb − πa.

(a) Solve Problem 2.3.15, (c) by first making the substitution u =

0

(b) Now, if p ≥ 0 and q ≥ 0 constants, compute Z ∞ h −1 −1 i e (px)2 − e (qx)2 dx. 0

(You may treat the cases p = 0 and/or q = 0 separately.) 3.9.7 (a) For real numbers 0 < a, b < ∞, prove:       Z ∞ 1 1 b b sin − a sin dx = ln . bx ax a 0 (b) What is the corresponding integral expression of ln(t)? (c) Also prove:   +∞, if b > a > 0,           Z ∞  1 1 − a cos dx = 0, b cos if b = a > 0  bx ax 0      −∞, if 0 < b < a.

3.9.8 (a) If a function f (x) is continuous in an interval (0, δ) for some δ > 0, Z δ f (x) dx exists and lim+ f (x) exists, then prove lim+ f (x) = 0. x x→0 x→0 0 (b) If a function Z ∞f (x) is continuous in an interval (r, ∞) for some f (x) r > 0, dx exists and lim f (x) exists, then prove x→∞ x r lim f (x) = 0. x→∞

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269

3.9.9 (a) Consider a nice function f : (0, ∞) −→ R such that for some ρ ≥ 0 Z A f (x)dx exists ∀ A > ρ and lim f (x) = f (∞) ∈ [−∞, ∞]. x→∞

ρ

Prove M (f ) := lim

x→∞

1 x

Z

x

f (t) dt = f (∞). ρ

(b) Consider Z ∞ a nice absolutely integrable function f : (0, ∞) −→ R, i.e., |f (x)|dx < ∞. Prove M (f ) = 0. 0

(c) Give examples of non-periodic and not absolutely integrable functions f : (0, ∞) −→ R for which M (f ) is either a finite number, or ±∞, or oscillates. (You may take ρ = 0.) 1 with 0 < x < ∞. Use Problem 1.8.13, 1 + sin2 (x) (a), part (a) of this problem, and Corollary 3.8.4 to prove that for any 0 < a, b < ∞    Z ∞ f (ax) − f (bx) b dx = ∞ · sign ln . x a 0

(d) Let f (x) =

x2

x with 0 < x < ∞. Use Problem 1.8.13, 1 + sin2 (x) (b), part (b) of this problem, and Corollary 3.8.4 to prove that for any 0 < a, b < ∞ Z ∞ g(ax) − g(bx) dx = 0. x 0

(e) Let g(x) =

x7

123 with 0 < x < ∞. Use Problem 1.8.13, 1 + x7 sin2 (x) (b), part (b) of this problem, and Corollary 3.8.4 to prove that for any 0 < a, b < ∞   Z ∞ a b h(ax) − h(bx) dx = −123 ln = 123 ln . x a b 0

(f) Let h(x) =

3.9.10 For 0 < a, b < ∞, write and evaluate the Frullani integrals ex I(a, b) of the following functions: f (x) = x , g(x) = xp e−x , e +1 u(x) = sin(xp ), v(x) = cos(xp ), where p ≥ 1. Justify your answers.

270

Improper Riemann Integrals

3.9.11 (a) In Problem 2.3.19, (f ), we have proved Z π ln | sin(x)|dx = −π ln(2). 0

However, show that for 0 < a, b < ∞   +∞,      Z ∞  sin(bx) 1 dx = 0, ln  sin(ax) x 0      −∞,

if b > a > 0, if a = b > 0, if

0 < b < a.

(b) Use Problem 2.3.22 to prove the same result for ln | tan(x)| in the place of ln | sin(x)|. (Compare with Example 3.8.9.) 3.9.12 For 0 < a, b < ∞, show   Z ∞ 2 b | sin(bx)| − | sin(ax)| dx = ln x π a 0 and

Z 0

∞

| cos(bx)| − | cos(ax)| dx = x



   2 b − 1 ln . π a

(See also Problem 3.9.14.) 3.9.13 Show that the functions tan(x) and | tan(x)| do not satisfy the condition (z) at the beginning of this section. Use the result in Problem 3.13.19 to prove that for any 0 < α < 1, | tan(x)|α satisfies (z) and for any a, b > 0 we get   Z ∞ | tan(bx)|α − | tan(ax)|α 1 b
J(α) := dx = ln . απ x a cos 2 0 Notice that ´ a-priori J(0) = 0. Then explain why the above formula for J(α) with 0 < α < 1, has discontinuity at α = 0 when a 6= b. Investigate the integral obtained by replacing tangent with cotangent for the same and/or new range of α. 3.9.14 Show that the functions sec(x) and | sec(x)| do not satisfy the condition (z) at the beginning of this section.

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271

Use the result in Problem 3.13.20 to prove that for any α < 1, | sec(x)|α satisfies (z) and for any a, b > 0 we get " #  
Z ∞ | sec(bx)|α − | sec(ax)|α 1 Γ 1−α b 2
J(α) := − 1 ln dx = √ · . 2−α x a π Γ 2 0 Show ∞

Z 0

   | sec(bx)| − | sec(ax)| b dx = sign ln · ∞, x a

which is the same as J(1) in the above formula. Replace secant with cosecant and investigate the new integral. 3.9.15 Consider the function f (x) =

sin(x) on the interval (0, ∞). x

(a) Prove that this function satisfies all of the conditions of both Theorems 3.8.1 and 3.8.2 for x ∈ (0, ∞). (b) For any real numbers 0 < a, b < ∞, write and compute the Frullani integral of this function. 3.9.16 (a) For real numbers 0 < a, b < ∞, prove: Z

2 ∞ sin (bx) bx

0

− x

sin2 (ax) ax

dx = 0.

(b) Without using the result in (a), show directly that this integral is proper over any interval [0, ε], with ε > 0. 3.9.17 (a) Prove that Z 0

∞

sin3 (x) 3 dx = ln(3) x2 4

∞

Z and 0

sin4 (x) dx = ln(2). x3

(b) Then evaluate the following four integrals: Z

0

−∞

sin3 (x) dx, x2

Z

∞

−∞

sin3 (x) dx, x2

272 Z

0

−∞

Improper Riemann Integrals Z ∞ sin4 (x) sin4 (x) dx, dx. x3 x3 −∞

(Compare with Problems 3.2.38 and 3.9.19. See also Example 3.10.8 and Problems 3.13.13, II 1.7.105.) [Hint: Use integration by parts, trigonometric identities and Example 3.8.8 and possibly equation (3.5).] 3.9.18 For i = 1, 2, 3, . . . , k, we consider any positive numbers ai > 0 k P Ai = 0. and any real numbers Ai that satisfy the condition i=1

(a) Let y = f (x) be a function that satisfies the conditions of the Frullani integrals of Theorem 3.8.1. Then prove Z

∞

Pk

i=1

0

k X Ai f (ai x) dx = [f (∞) − f (0)] · Ai ln(ai ). x i=1

(b) Use (a) to show Z ∞ A1 arctan(a1 x) + A2 arctan(a2 x) + ... + Ak arctan(ak x) x 0 π dx = [A1 ln(a1 ) + A2 ln(a2 ) + . . . + Ak ln(ak )] . 2 (c) Let y = f (x) be a function that satisfies the conditions of the Frullani integrals of Theorem 3.8.2. Then prove Z 0

∞

Pk

i=1

k X Ai f (ai x) Ai ln(ai ). dx = −f (0) · x i=1

(d) Use (c) to show Z ∞ A1 cos(a1 x) + A2 cos(a2 x) + A3 cos(a3 x) + ... + Ak cos(ak x) x 0 dx = −A1 ln(a1 ) − A2 ln(a2 ) − A3 ln(a3 ) − . . . − Ak ln(ak ). (J. Wolstenholme20 ) 20 Joseph Wolstenholme, English mathematician, 1829–1891. He computed the above integral using contour integration in complex analysis.

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273

(e) If, in the integral in (d), we replace cosine by sine, prove that the sin(x) answer is zero, by (c). But, if we replace cosine by f (x) = , x then the answer is the same, by (a). 3.9.19 (a) Use the trigonometric formula in Problem 3.2.28, (a), the relation in (h), and the previous Problem, (c), to prove the for n ∈ N, Z ∞ sin2n+1 (x) dx = x2 0   n (−1)n+1 X k 2n + 1 (−1) [2(n − k) + 1] · ln[2(n − k) + 1] = k 22n k=0   n−1 (−1)n+1 X k 2n + 1 (−1) [2(n − k) + 1] · ln[2(n − k) + 1]. 22n k k=0

(b) Now, compute the integral when 2n + 1 = 5. (c) Also, for n ∈ N and a ∈ R find Z 0

∞

sin2n+1 (ax) dx. x2 Z

[See also Problem 3.9.17. For the integrals 0

∀ n ∈ N0 , see Problem 3.2.28, (f ).]

∞

sin2n (x) dx, x2

∞

sinN (x) dx, where xL 0 N and L are integers such that 1 ≤ L ≤ N , are given as rational fractions of π or as combinations of logarithms with rational coefficients. A general process for evaluating these integrals is given in the following three steps: Z

3.9.20 (I) In general, the answers to the integrals

1. Write the power sinN (x) in term of sines or cosines of degree one, by using the formulae in (a), (b) and (e) of Problem 3.2.28. 2. Apply integration by parts L − 1 times to reduce the denominator xL to just x.

274

Improper Riemann Integrals

3. (1) If N − L ≥ 0 is even, then the final integral has sines in the numerator and use the results of Example 3.1.8. The integral is a rational fraction of π. (2) If N − L ≥ 1 is odd, then the final integral has cosines (and a constant) in the numerator and use the result of Problem 3.9.18, (d), above. The integral is a rational linear combination of logarithms. Now follow this process to prove: R∞ 5 45 (a) 0 sinx4(x) dx = 125 96 ln(5) − 32 ln(3). R∞ 6 27 (b) 0 sinx5(x) dx = 16 ln(3) − ln(4). R ∞ sin6 (x) (c) 0 dx = π8 . x4 Evaluate: R∞ 5 R ∞ sin6 (x) (d) 0 sinx3(x) dx, and (e) dx. x3 0 (f) Write some integrals of this kind on your own and evaluate them. [With sines on the square in the numerator Problem 3.2.28, (f ) could also be used, but is not a short cut. For L = 1 or L = 2 see Problems 3.2.28, (a) and (f ) and the previous Problem 3.9.19, (a) for general formulae. For L = N see the general formula in Problem II 1.7.105 to avoid L−1 = N −1 integration by parts.] Z ∞ sinN (x) [For 0 ≤ N < L integers, dx = ∞, due to the xL 0 degree of the unboundedness of the integrand at x = 0.] (II) Summarizing the above, derive the following general formulae for the integral Z ∞ sinN (x) dx, xL 0 where N and L belong to Z. Case N − L = even ≥ 0. The integral is a rational fraction of π. Subcases: (1) 1 ≤ L = 2l + 1 ≤ N = 2n + 1. (Both numbers are positive odd.) (2) 2 ≤ L = 2l ≤ N = 2n. (Both numbers are positive even ≥ 2.)

Real Analysis Techniques Then Z 0

∞

275

  n sinN (x) (−1)n+l π X k N dx = (−1) (N − 2k)L−1 . xL 2N (L − 1)! k k=0

[In subcase (2) the maximum index in the summation may be obviously replaced by n − 1, as the last summand obtained for k = n is zero.] Case N − L = odd ≥ 1. The integral is a rational combination of logarithms. Subcase: (3) 2 ≤ L = 2l ≤ N = 2n + 1. Then   Z ∞ n−1 X (−1)n+l sinN (x) k N dx = N −1 (−1) (N − 2k)L−1 ln(N − 2k). xL 2 (L − 1)! k 0 k=0

Subcase: (4) 3 ≤ L = 2l + 1 ≤ N = 2n. Then   Z ∞ n−1 (−1)n+l+1 X sinN (x) k N (−1) (N − 2k)L−1 ln(N − 2k). dx = N −1 k xL 2 (L − 1)! 0 k=0

(III) In all other cases of combinations of N and L in Z, the integral does not exist (is infinity or oscillates). 3.9.21 State the necessary conditions and give the proofs so that for any  > 0 and for all 0 < a, b < ∞   Z ∞ f (bx) − f (ax) b (1) dx = [f (∞) − f ()] ln , x a  or

Z (2) 

∞

f (bx) − f (ax) dx = −f () ln x

  b . a

3.9.22 (a) Use the previous Problem to prove that for any  > 0 and any 0 < a, b < ∞, we have Z 

∞

cos2 (bx) cos2 (ax) − cos2 ()  a  bx ax dx = ln . x  b

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Improper Riemann Integrals

(b) What happens when  → 0+ ? (c) Without using this result, show directly that this integral is improper over any interval [0, ε], with ε > 0 and its value is +∞ when a > b. 3.9.23 For β ≥ 0 and a, b real constants, show:  2 2 R∞ β +b 1 (a) I(β) := 0 e−βx cos(ax)−cos(bx) dx = ln x 2 β 2 +a2 . (See also Example 3.8.4.) If we replace cos(ax) − cos(bx) with cos(ax) + cos(bx), then prove that the new integral is equal to ∞. R∞ (b) J(β) := 0 e−βx sin(ax)±sin(bx) dx =  x a b arctan β ± arctan β . [Hint: These integrals are not Frullani type integrals (why?). Use equation (3.5). Or use Problems 3.7.10 and 3.7.11 and/or the result in Example 3.1.9, after you justify the splitting of the integral J(β).] (c) The integral I(β) found in (a) can also be found by the following steps: (1) Use the Main Theorem, 3.1.1, and Problem 1.6.15 d to find [I(β)]. (2) What is I(0) and why? (3) Now find I(β). dβ (d) The integral J(β) found in (b) can also be found by the following steps: (1) Use the Main Theorem, 3.1.1, and Problem 1.6.13 d to find [J(β)]. (2) What is J(0) and why? (3) Now find J(β). dβ (e) Knowing I(β) and J(β) ´ a-priori, explain how you can use them to find the integrals in Examples 3.8.7 and 3.8.8. (See also hint below.) That is, to prove that: For all 0 < a, b < ∞, we have
R∞ (1) 0 cos(bx)−cos(ax) dx = ln ab x and R∞ (2) 0 sin(bx)−sin(ax) dx = 0. x [Hint: You must justify why you can let β = 0 by using a continuity argument, i.e., you can switch limit and integral.]

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3.9.24 (a) Prove that Z Z ∞ 1 ∞ arctan2 (x) arctan2 (x) dx = dx = ... = 2 −∞ x2 x2 0 Z ∞ arctan(1 · x) − arctan(0 · x) 2 dx = ... = π ln(2). x(x2 + 1) 0 [Hint: In the first part “...” use integration by parts. In the second part “...” use equation 3.5 and partial fractions. See also Problem II 1.7.143.] (b) Use (a) and prove Z Z ∞ 1 ∞ arctan3 (x) arctan3 (x) 3 π3 dx = dx = π ln(2) − . 3 3 2 −∞ x x 2 16 0 (c) For p > 0 and q > 0, find the cases for which the integral Z ∞ arctanp (x) dx is finite or infinite. xq 0 (See also Example 3.8.2.) 3.9.25 For β ≥ 0 and a, b real constants, find: R∞ R∞ (a) 0 e−βx cos(ax)xcos(bx) dx, (b) 0 e−βx sin(ax)xsin(bx) dx, R∞ R∞ (d) 0 e−βx sin(ax)xcos(bx) dx. [For β = (c) 0 e−βx cos(ax)xsin(bx) dx, 0, see also Problem 3.2.23, (a), and Example 3.8.8, (d).] (e) Now, replace the x in the denominators by x2 and compute the new four integrals. [Hint: Use equation (3.5). Or use appropriate trigonometric identities to split the trigonometric products and Problem 3.9.23, (a) and (b).] 3.9.26 For β ≥ 0 and a, b real constants, use Problem 3.9.23, (a) and (b), to find the values of the following four integrals: R∞ 2 2 (bx) (a) 0 e−βx cos (ax)−cos dx, x (b)

R∞ 0

e−βx

sin2 (ax)−sin2 (bx) x

dx,

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Improper Riemann Integrals

(c)

R∞

(d)

R∞

0 0

e−βx e−βx

cos3 (ax)−cos3 (bx) x 3

3

sin (ax)−sin (bx) x

dx, dx.

(e) Now, replace the x in the denominators by x2 and compute the new four integrals. [Hint: The first two integrals can be found easier by using the double angle trigonometric formulae or using Problems 3.7.10 and 3.7.11. For the other two integrals, use the trigonometric formulae that reduce the cubic powers to first powers, etc. Or use equation (3.5).] 3.9.27 For a, b real constants, prove: Z 0 Z ∞ cos(bx) − cos(ax) cos(bx) − cos(ax) dx = 2 dx = (a) 2 x x2 −∞ −∞ Z ∞ cos(bx) − cos(ax) 2 dx = (|a| − |b|)π. x2 0 Z

∞

(b) 0

∞

Z (c)

−∞

Z 0 sin(bx) − sin(ax) sin(bx) − sin(ax) dx = − dx = 2 x x2 −∞   +∞, if b > a,       = 0, if b = a,       −∞, if b < a.

cos2 (bx) − cos2 (ax) dx = 2 x2 ∞

Z 2 0

Z

∞

(d) −∞

Z 0

∞

−∞

cos2 (bx) − cos2 (ax) dx = x2

cos2 (bx) − cos2 (ax) dx = (|a| − |b|)π. x2

sin2 (bx) − sin2 (ax) dx = 2 x2

2

0

Z

Z

0

−∞

sin2 (bx) − sin2 (ax) dx = x2

sin2 (bx) − sin2 (ax) dx = (|b| − |a|)π. x2

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[Hint: Use equation (3.5) and the result in Example 3.1.8. Or use Examples 3.1.12 and 3.1.13, or come up with a different method.] 3.9.28 Justify the following steps and fill in the missing details to prove the following Ramanujan’s21 integral for any a > 0, constant. Z ∞  x −x R := e−ae + e−ae − 1 dx = 0

1 2

Z

∞



e

−aex

+e

−ae−x

−∞

1 2

Z

∞

Z −a  1 ∞ e−au + e u − 1 − 1 dx = du = 2 0 u

  −a   −1 −1 (e−au − e−u ) + e u − e u + e−u + e u − 1 u

0

du.

Now, use Problems 3.5.19, (3.5.18, and 2.3.32) and Example 3.8.1 to show that R = − ln(a) − γ.

3.9.29 Show that the integral of Problem 3.5.20 is equal to Z ∞  au bu e−e − e−e du, −∞

where a > 0 and b > 0 constants, but cannot be evaluated by using the formula 3.5, in this section. (Switching the order of integration does not hold here.)

3.10

The Real Gamma Functions

In this section, we study the real Gamma22 function and its fundamental properties. The complex Gamma function and the great number of its properties and applications are studied in advanced complex analysis. This chapter of complex analysis is outside the scope of this text. 21 Srinivasa 22 Gamma,

Ramanujan, Indian mathematician, 1887–1920. γ, Γ, the third letter of the Greek alphabet.

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Improper Riemann Integrals

This special function is very important in Mathematics, Statistics, Engineering and Science. It was first defined and used by Euler. The Gamma function is defined as the improper integral of a real parameter p Z ∞ Γ(p) = xp−1 e−x dx. 0

This is also called Euler’s integral of the second kind. We notice that the integrand is always non-negative, and this integral is obviously improper since the interval of integration, [0, ∞), is unbounded. When 0 < p < 1, the integral is improper for one more reason: The integrand becomes +∞ at x = 0+ . (In complex analysis, the real p is replaced with the complex variable z = x + iy, and so we must rename the dummy variable with a letter other than x.) In Example 1.7.6, we have established the convergence of this integral for all p > 0 and its divergence (= ∞) for all p ≤ 0. So, the study of the real Gamma function begins with examining the properties of this integral for p ∈ (0, ∞). We are also going to provide some preliminary estimates which, besides reproving the existence of Γ(p) for all 0 < p < ∞, allow us to also establish lim+ Γ(p) = ∞, and lim Γ(p) = ∞, p→∞

p→0

facts that can be established elementarily by the material exposed so far. (Prove these as exercise!)

Preliminary Estimates In proving various facts about the Gamma function, it is convenient to write Z ∞ Z ∞ Z 1 1 p−1 −x p 1 dx + xp x dx. (3.6) Γ(p) = x e dx = x x xe xe 1 0 0 With the help of this relation, we will derive some preliminary estimates for the Gamma function. These estimates are useful in proving some results and solving some problems. (I) We estimate the first part of the integral in (3.6), namely: Z

1

x 0

p−1 −x

e

Z dx = 0

1

xp

1 dx for all xex

p > 0.

This integral is improper when 0 < p < 1 and is proper when p ≥ 1.

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We observe the following inequality: ∀ 0 < p < ∞ and ∀ 0 ≤ x ≤ 1, xp−1 e−1 ≤ xp−1 e−x ≤ xp−1 . Therefore, 1 e

Z

1

x

p−1

1

Z dx
0 is very close to zero (small swing), then we can approximately consider k ' 0, and the period is given by the known much simpler but approximate law s l T ' 2π . g

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Improper Riemann Integrals

The special integral π 2

Z

dψ

K(k) :=

q

0

,

1 − k 2 sin2 (ψ)

where 0 < k < 1, that appears in this physical law, is a complete elliptic integral of the first kind in its Legendre form. With the help of the Gamma function, we will evaluate this integral π 1 π = √ < 1. We have when θ0 = and so 0 < k = sin 2 4 2  K

By letting sin(φ) =

1 √ 2



Z = 0

√ 2 sin

π 2

dφ q 1−

1 2

.

sin2 (φ)

  θ , this integral changes to (work out 2

this substitution carefully) 1 √ 2

Z

π 2

0

dθ

1 p =√ 2 cos(θ)

Z

π 2

cos

−1 2

(θ) dθ.

0

By the properties (B, 6) and (B, 7) of the Beta and Gamma functions and the result of Example 3.11.5, this is equal to √   π Γ( 41 ) Γ 2( 1 ) 1 1 1 Γ( 21 ) Γ( 14 ) 1 1 1 √ √ √ · ·B , = = = √4 . 1 3 1 2 4 4 π 2 2 2 2 Γ( 2 + 4 ) 2 2 Γ( 4 ) Therefore,  K

1 √ 2



Z

π 2

= 0


Γ 2 14 q = √ . 4 π 1 − 12 sin2 (φ) dφ

So, the precise period of a pendulum with swing angle θ0 = s T =

l · Γ2 gπ

π is 2

  1 . 4

π For swing angles 0 < θ0 < , the integral in the formula of the period 2 of the pendulum is given by an elliptic integral of the first kind which cannot be reduced to the Gamma function. See also Problem 3.13.44.

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The derivation of the above formula for the actual period is as follows: We look at the provided Figure, 3.6, and we use Newton’s law “mass × acceleration = force.” Since distance = arc-length = l·θ = l·θ(t) [θ = θ(t) is in radians] and tangential force = −mg sin(θ) in this circular d2 θ motion, we get acceleration = l 2 and then dt ml

d2 θ = −mg sin(θ), dt2

and so

d2 θ g = − sin(θ). 2 dt l

To integrate this second-order ordinary differential equation, we first dθ 31 multiply both sides by , i.e., dt "  # 2 d2 θ dθ g dθ 1 d dθ g d[cos(θ)] · = − sin(θ) · , and so = . 2 dt dt l dt 2 dt dt l dt We integrate once and obtain 1 2



dθ dt

2 =

g cos(θ) + c. l

We assume that the pendulum is released at time t = 0 from the dθ = 0 (zero velocity at the start), position θ = θ0 . When t = 0, dt t=0 and so g c = − cos(θ0 ). l Therefore, r dθ 2g p =± · cos(θ) − cos(θ0 ). dt l dθ When the pendulum goes from θ = θ0 to θ = 0, is negative dt because θ is decreasing and the time needed is one quarter of the period. So, s Z 0 T dθ l p =− · . 4 2g θ0 cos(θ) − cos(θ0 ) Then, using the half angle trigonometric formulae, we find s Z θ0 l dθ q T =2 ·

. g 0 sin2 θ0 − sin2 θ 2

31 Another

method is to use the transformation

2

dθ = φ(θ). dt

320

Improper Riemann Integrals

Finally, we use the change of variables     θ θ0 sin = sin sin(ψ) 2 2 to find

s T =4 

where k = sin

l · g

Z 0

π 2

dψ q , 1 − k 2 sin2 (ψ)

 θ0 . (See also Problem 3.13.47.) 2

Application 2: In advanced geometry, analysis, partial differential equations (especially elliptic), potential theory, etc., we need to know the volume and the surface area of balls in the Euclidean spaces. In Rn , n = 1, 2, 3, . . . , the n-dimensional closed ball with center −c = (c , c , . . . , c ) ∈ Rn and radius R > 0 is the set given by a point → 1 2 n ( ) n X → − → − n 2 2 B ( c , R) = x = (x , x , . . . , x ) ∈ R | (x − c ) ≤ R . n

1

2

n

i

i

i=1

−c , R), has dimension The surface or this ball, denoted here by Sn or Sn (→ −c and n − 1 and is called the (n − 1)-dimensional sphere with center → radius R. This is the subset of Rn analytically given by ( ) n X → − → − 2 2 n S ( c , R) = x = (x , x , . . . , x ) ∈ R (x − c ) = R . | n

1

2

n

i

i

i=1

With the help of the Gamma function, we will prove n

n −c , R)] = 2R π 2
Volume [Bn (→ nΓ n2

and

n

n−1 2 π −c , R)] = 2R
. Area [Sn (→ Γ n2

For n = 1, 2, 3, we can check that these formulae give the correct −c = (c ) is just a real number c answers. When n = 1, the center → 1 on the real line. So, B1 (c, R) is the closed interval [c − R, c + R] ⊂ R. Its length, 2R, is viewed as its volume and its two point set boundary, {c − R, c + R}, as its surface. The formulae for n = 1 give √ 1 2Rπ 2 2R π
= √ Volume [B1 (c, R)] = = 2R π 1Γ 21

Real Analysis Techniques and

321

√ 1 2R 0 π 2 2 π
= √ = 2. Area [S1 (c, R)] = π Γ 21

−c = (c , c ) When n = 2, we deal with the usual circle with center → 1 2 2 and radius R. Its area, πR , is viewed as its volume and the length of its circumference, 2πR, as its surface. The formulae for n = 2 give 2

2 2 −c , R)] = 2R π
2 = R π = πR2 Volume [B2 (→ 2 1 2Γ 2

and

2

1 −c , R)] = 2R π
2 = 2Rπ = 2πR. Area [S2 (→ 1 Γ 22 −c = (c , c , c ) When n = 3, we get the usual ball in R3 with center → 1 2 3 3π 3 and radius R. Its volume is R , and the area of its boundary surface, 4 −c and radius R in R3 , is 4πR2 . which is the usual sphere with center → The formulae for n = 3 give 3

3

3

3 3 3 −c , R)] = 2R π
2 = 2R π 2
= 4R√π 2 = 4π R3 , Volume [B3 (→ 3 3 π 3Γ 32 3 21 Γ 12

and 3

3

3

3−1 2 2 2 π −c , R)] = 2R
π 2 = 2R π 2
= 4R √ Area [S3 (→ = 4πR2 . 3 1 1 π Γ 2 Γ 2 2

So, we need to prove the formulae for n ≥ 4. We will prove the volume − −c = → formula first. Without loss of generality, we assume → 0 the origin n of R . Then, as we know, Z h → − i Volume Bn 0 , R = 1 dx1 dx2 . . . dxn . → − Bn ( 0 ,R) → −  We use the parametrization of Bn 0 , R in Rn , which is the extension of spherical coordinates as we know them in R3 . These coordinates are 0 ≤ ρ ≤ R,

0 ≤ θ ≤ 2π and 0 ≤ φj ≤ π for j = 1, 2, . . . , n − 2,

322

Improper Riemann Integrals

and satisfy x1 = ρ cos(φ1 ), x2 = ρ sin(φ1 ) cos(φ2 ), x3 = ρ sin(φ1 ) sin(φ2 ) cos(φ3 ), ... xn−1 = ρ sin(φ1 ) . . . sin(φn−2 ) cos(θ), xn = ρ sin(φ1 ) sin(φ2 ) . . . sin(φn−2 ) sin(θ). The Jacobian32 determinant of this change of variables is ∂(x1 , x2 , x3 , . . . , xn−1 , xn ) = ∂(ρ, φ1 , φ2 , . . . , φn−2 , θ) ρn−1 sinn−2 (φ1 ) sinn−3 (φ2 ) . . . sin2 (φn−3 ) sin(φn−2 ), as it is verified either inductively or by writing it explicitly and pulling the common factors out of the determinant. (Here the Jacobian matrix is in Hessenberg33 form,34 and so its determinant is computed fairly easily.) Since this Jacobian is positive, the volume is h − i Z −c , R)] = Vol B → 1 dx1 dx2 . . . dxn = Vol [Bn (→ 0 ,R = n → − Bn ( 0 ,R) Z RZ π Z π Z 2π ... ρn−1 sinn−2 (φ1 ) . . . sin(φn−2 )dθdφ1 . . . dφn−2 dρ = 0 0 0 0 Z π Z π Z π 2πRn n−2 n−3 · sin (φ)dφ · sin (φ)dφ . . . sin(φ)dφ = n 0 0 0




Γ 12 Γ 12 Γ n−2 Γ(1)Γ 12 2πRn Γ n−1 2 2


· · ... n Γ n2 Γ n−1 Γ 23 2 (by using the second integral of Example 3.11.9. See also Problem 3.13.48 foranother way of computing this.).  √ 1 Since Γ = π, after simplifying we get 2
n n−2 1 n 2R n π 2 −c , R)] = 2πR · Γ 2

Volume [Bn (→ = . n Γ n2 nΓ n2 32 Carl Gustav Jacob Jacobi or Carolus Gustavus Iacobus Iacobi, German mathematician, 1804–1851. 33 Karl Adolf Hessenberg, German mathematician and engineer, 1904–1959. 34 A square matrix is in Hessenberg form if all entries (i, j) with j ≥ i + 2, or (i, j) with 1 ≤ j ≤ i − 2, are zero.

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(See also Problems 3.13.48 and 3.13.53 for other methods of deriving this formula.) We observe that Z → −  → − Vol [Bn ( c , R)] = Vol[Bn 0 , R ] = 1 dx1 dx2 . . . dxn = → − Bn ( 0 ,R) Z π Z 2π Z RZ π ρn−1 sinn−2 (φ1 ) . . . sin(φn−2 )dθdφ1 . . . dφn−2 dρ = ... 0

0

0

0

Z = 0

R

Z → − Sn ( 0 ,R)

where dS is the surface element of Sn Therefore,

dS dρ,

→ −  0 ,R .

d −c , R)]} = {Volume [Bn (→ dR "Z Z # Z R d −c , R)], dS dρ = dS = Area[Sn (→ → − − dR 0 Sn (→ 0 ,R) Sn ( 0 ,R) and so

n

n−1 −c , R)] = 2R
π 2 . Area [Sn (→ Γ n2

n [The area in Rn may be viewedhh as (n − 1)-dimensional   n  volume in R .] n ii We now observe that since −1 !≤Γ , we get the rather 2 2 counterintuitive results # " n n 2 2R π −c , R)]} = lim
= 0, and (a) lim {Volume [Bn (→ n→∞ n→∞ nΓ n2 " # n 2R n−1 π 2 → −
(b) lim {Area[Sn ( c , R)]} = lim = 0. n→∞ n→∞ Γ n2

(See also Problem 3.13.50.) A Corollary of this application is the computation of the volume of an n-dimensional ellipsoid with semi-axes ai > 0 for i = 1, 2, . . . , n ( ) n X x2i → − En = x = (x1 , x2 , . . . , xi ) | ≤1 . a2 i=1 i We make the change of variable xi ui = ⇐⇒ xi = ai ui , ai

i = 1, 2, . . . , n,

324

Improper Riemann Integrals

and we use the volume of the unit ball found above. Then we find n

2 a1 a2 . . . an π 2
Volume [En ] = . nΓ n2 (The surface area of such an ellipsoid is a much harder matter.) Application 3: Referring to definitions in Application 1 of Section 2.2, the function  1 x α−1 − β  e , for x > 0,   β α Γ(α) x g(x : α, β) =    0, for x ≤ 0, where α > 0 and β > 0 constants, is a probability density function. Indeed, it is piecewise continuous and greater than or equal to zero everywhere. We only need to show that its integral over R is equal to 1. x We let u = , and we have: β Z ∞ Z ∞ x 1 g(x : α, β)dx = xα−1 e− β dx = α Γ(α) β −∞ 0 Z ∞ Z ∞ 1 1 Γ(α) α−1 α−1 −u β u e β du = uα−1 e−u du = = 1. β α Γ(α) 0 Γ(α) 0 Γ(α) If a random variable X = x > 0 has density function g(x : α, β), we say that it has the gamma distribution. We can compute its expected value, variance and moment-generating function. For all r = 0, 1, 2, 3, . . . , we have that the moment of order r about the origin is Z ∞ x 1 (x=βu) 0 µr = xα−1 e− β dx = xr α β Γ(α) Z 0∞ βr β r Γ(α + r) uα+r−1 e−u du = . Γ(α) 0 Γ(α) So, the expected value is E(X) = µ = µ01 =

βΓ(α + 1) βαΓ(α) = = αβ. Γ(α) Γ(α)

Similarly, we find µ02 =

β 2 Γ(α + 2) β 2 (α + 1)αΓ(α) = = α(α + 1)β 2 . Γ(α) Γ(α)

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Therefore, the variance is var(X) = σ 2 = µ02 − (µ01 )2 = α(α + 1)β 2 − (αβ)2 = αβ 2 . The moment-generating function is Z ∞ Z ∞ x(1−βt) x 1 1 etx α MX (t) = xα−1 e− β dx. xα−1 e− β dx = α β Γ(α) β Γ(α) 0 0 We use the substitution v =

x(1 − βt) to obtain β

Z ∞ 1 β α−1 v α−1 (1 − βt)−α+1 e−v β(1 − βt)−1 dv = β α Γ(α) 0 Z (1 − βt)−α β α ∞ α−1 −v 1 (1 − βt)−α β α Γ(α) = . v e dv = α β Γ(α) β α Γ(α) (1 − βt)α 0   1 This exists for all t ∈ R − . β Application 4: In physics, physical and/or quantum chemistry, study of gasses, thermodynamic, etc., improper integrals appear in various laws and computations. For example, finding the average kinetic and potential energy of the H (hydrogen) atom at an excited state uses repeatedly the integral Z ∞ n! xn e−ax dx = n+1 , a 0 MX (t) =

where n = 0, 1, 2, 3, 4, . . . , integer and a > 0 constant. This is a standard integral rule of the Gamma function. See property (Γ, 11) and its extension in Problem 3.13.6 for more generality. In this application, we skip the derivations of formulae and the meanings of various physics constants and terms, and we concentrate on the mathematics of the integrals involved. We consider a particular example of an excited state of the H (hydrogen) atom for which the average value of the kinetic energy is given by the following improper integral: Z 2π Z π 1 ~2 Average Kinetic Energy = − · × dθ sin(φ)dφ 2me 32πa30 0    0  Z ∞ ρ ρ ρ 1 d d ρ 2− e− 2a0 2 ρ2 2− e− 2a0 ρ2 dρ = a0 ρ dρ dρ a0 0 ρ  Z ∞ ~2 1 ρ e− a0 2 2 · · 2π · 2 × 2 − 3 ρ (16a0 − 10a0 ρ + ρ )ρ dρ. 2me 32πa30 a 4 a 0 0 0

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Improper Riemann Integrals

We break the integral into four smaller integrals. This is legitimate because each of these four integrals exists. Then we simplify and get  Z ∞ ρ ~2 9 1 ρ2 e− a0 dρ− Average Kinetic Energy = − · 3× 2 2me 8a0 a0 0 8 a0

Z

∞

0

ρ

ρe− a0 dρ −

3 a30

Z

∞

ρ

ρ3 e− a0 dρ +

0

1 4a40

∞

Z

 ρ ρ4 e− a0 dρ .

0

We use the standard Gamma integral we wrote above, and after direct computation we find Average Kinetic Energy =

~2 . 8me a20

The average value of the potential energy of this atom is given by the following integral: Z 2π Z π e2 1 Average Potential Energy = − × · dθ sin(φ)dφ · 4π0 32πa30 0 0    Z ∞ ρ 1 ρ ρ e2 1 ρ 2− · 2π · 2× e− 2a0 e− 2a0 ρ2 dρ = − · 2− a0 ρ a0 4π0 32πa30 0   Z ∞ Z ∞ Z ∞ ρ ρ ρ 4 1 4 ρe− a0 dρ − ρ2 e− a0 dρ + 2 ρ3 e− a0 dρ = a0 0 a0 0 0 e2 1 e2 − (2a20 ) = − . 3 4π0 8a0 16π0 a0 Application 5: We can use the Beta and Gamma functions to evaluate the sum of certain series of numbers or power series. (I) We will evaluate the double series of positive terms ∞ X ∞ X

n2 . (n + m)! n=1 m=1 We have n!(m − 1)! Γ(n + 1)Γ(m) = = B(n + 1, m) = Γ(n + m + 1) (m + n)!

Z

1

xn (1 − x)m−1 dx.

0

So, ∞ X ∞ X

Z 1 ∞ X ∞ X n2 n2 = xn (1 − x)m−1 dx. (n + m)! n!(m − 1)! 0 n=1 m=1 n=1 m=1

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We can switch summations and integral by, e.g., the Monotone Convergence Theorem, 3.3.10, since we have summations of nonnegative functions. Hence, Z 1X ∞ X ∞ ∞ X ∞ X n2 n2 = xn (1 − x)m−1 dx = (n + m)! n!(m − 1)! 0 n=1 m=1 n=1 m=1 # " ∞ # Z 1 "X ∞ X (1 − x)m−1 nxn · dx. (n − 1)! (m − 1)! 0 n=1 m=1 Now, we observe that the second sum is ∞ ∞ X X (1 − x)m−1 (1 − x)m = = e1−x . (m − 1)! m! m=1 m=0

For the first sum we work as follows: ∞ X

∞ ∞ X X nxn (n + 1)xn+1 (n + 1)xn = =x = (n − 1)! n=0 n! n! n=1 n=0 ∞ ∞ ∞ X X X nxn xn xn−1 2 +x =x + xex = x n! n! (n − 1)! n=0 n=0 n=1

x2

∞ X xn + xex = x2 ex + xex = (x + 1)xex . n! n=0

[Note: Form this, with x = 1 and x = −1, we obtain the byproducts ∞ X

n = 2e and (n − 1)! n=1

∞ X (−1)n n = 0, (n − 1)! n=1

etc.

The second sum here implies ∞ X k=1

∞

X 2l − 1 2k = .] (2k − 1)! (2l − 2)! l=1

Finally, the given series is ∞ X ∞ X

n2 = (n + m)! n=1 m=1

Z

1 2

x 1−x

(x + x)e e 0

x3 x2 dx = e + 3 2 

1 = 0

5e . 6

(II) Now we will prove that for every real number x, we have:   ∞ X ∞ X n · m n+m x 1 x = x2 e x + . (n + m)! 6 2 n=1 m=1

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Improper Riemann Integrals

Again, we follow steps and results analogous to the ones in (I) above and we get ∞ X ∞ X

0

n · m n+m x = (n + m)! n=1 m=1 Z 1 ∞ X ∞ X nxn mxm tn (1 − t)m−1 dt = n!(m − 1)! 0 n=1 m=1 "∞ #" ∞ # X (xt)n−1 X m[x(1 − t)]m t dt = 1 − t n=1 (n − 1)! m=1 (m − 1)!

Z

1

1

Z x

  t etx x2 (1 − t)2 + x(1 − t) ex(1−t) dt = 0 1−t     Z 1 x 1 1 [xt(1 − t) + t] dt = x2 ex xB(2, 2) + = x2 e x + . x2 e x 2 6 2 0 x

If we let x = 1, we find ∞ X ∞ X

n·m 2e = , (n + m)! 3 n=1 m=1 and if we let x = −3, we find ∞ X ∞ X

n·m (−3)n+m = 0. (n + m)! n=1 m=1 (See also Problem 3.13.60 and Furdui 2013, for more applications.) Application 6: The so-called Dirichlet Gamma Integrals are ndimensional integrals of the type presented below. We here examine the cases n = 3 and n = 2. The formulae evolve analogously for all n ∈ N, (n ≥ 4). We consider W to be the solid bounded by the coordinate planes in the first octant of R3 (i.e., x ≥ 0, y ≥ 0, and z ≥ 0) and the surface given implicitly by the equation  x p  y q  z r + + = 1, a b c where a > 0, b > 0, c > 0, p > 0, q > 0, and r > 0 are positive constants. Then for positive constants α > 0, β > 0, and γ > 0, we have    
β γ α Z Z Z α β γ Γ p Γ q Γ r a b c α−1 β−1 γ−1  . D3 := x y z dxdydz = pqr Γ 1 + α + β + γ W

p

q

r

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The proof of this formula is obtained by the substitutions  x p  y q  z r = u, = v and = w. a b c 1 1 1 Then x = au p , y = bv q and z = cw r , 1 1 1 1 1 1 dx = a u p −1 du, dy = b v q −1 dv and dz = c w r −1 dw, p q r and the solid W in the x-y-z-space is transformed to the pyramid V bounded by the coordinate planes in the first octant of R3 (i.e., all coordinates are non-negative) and the plane u+v+w = 1, in the u-v-w-space. Performing this change of variables and simplifying, we thus obtain Z Z Z β γ α aα bβ cγ u p −1 v q −1 w r −1 dudvdw = D3 = pqr V Z   Z 1−u Z 1−u−v α aα bβ cγ 1 −1 β −1 γr −1 p q dw dv du = u v w pqr 0 0 0  Z Z 1−u β γ α aα bβ cγ r 1 u p −1 v q −1 (1 − u − v) r dv du. pqr γ 0 0 Now, for the integral with respect to v, we perform the change of variables v = (1 − u)t and we find Z 1−u β γ v q −1 (1 − u − v) r dv = 0   Z 1 γ γ β β β γ β γ , +1 . (1 − u) q + r t q −1 (1 − t) r dt = (1 − u) q + r B q r 0 Therefore, by the definition of the Beta function and property (B, 7), we obtain  
Z β γ 1 Γ α β γ β γ α q Γ r +1 a b c r   u p −1 (1 − u) q + r du = D3 = pqr γ Γ β + γ + 1 0 q r      
β γ β γ α Γ Γ + 1 Γ Γ + + 1 α β γ q r p q r a b c r     = β γ α pqr γ Γ β + γ + 1 Γ + + + 1 q r p q r    
β γ α aα bβ cγ Γ p Γ q Γ r  . pqr Γ α + β + γ + 1 p

q

r

(See also Problems 3.13.46, 3.13.52 and 3.13.53.)

330

Improper Riemann Integrals

Note 1: If α = 1, β = 1 and γ = 1, we find the volume of W .    
1 1 1 abc Γ p Γ q Γ r .  Volume(W ) = pqr Γ 1 + 1 + 1 + 1 p q r Note 2: We have analogous formulae for Dn in all dimensions n = 1, 2, 3, . . . . For example, in two variables (n = 2), we consider S to be the region bounded by the coordinate axes in first quadrant of R2 (x ≥ 0 and y ≥ 0,) and the curve given implicitly by the equation  x p  y q + = 1, a b where a > 0, b > 0, p > 0 and q > 0, are positive constants. Then for positive constants α > 0 and β > 0, we have     β α Z Z α β Γ p Γ q a b α−1 β−1  . D2 := x y dxdy = pq Γ 1 + α + β p

S

q

If α = 1 and β = 1, we find the area of S.     1 1 Γ p Γ q ab  . Area(S) = pq Γ 1 + 1 + 1 p

q

(Now, write the formula of Dn in any dimension n = 1, 2, 3, . . . . The case n = 1 is straightforward and basic and give its geometric meaning.) Note 3: With the help of the Dirichlet integrals we can evaluate integrals of linear combinations of products of powers of n variables, n = 1, 2, 3, . . . , with exponents > −1, over the domain in Rn analogous to W ⊂ R3 above. (E.g., see Problem 3.13.54.) Application 7: Here we examine three integrals that lead to conclusions on the convergence or divergence of certain positive series. First, we have Z 0

π

Z π 1 1 du = 2 1 − cos(u) 0 2 sin

Z
du = u 2

0

π 2

π

csc2 (θ) dθ = [− cot(θ)]02 = +∞.

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By using the geometric series, Examples 3.11.9, 3.11.10, 3.11.11 or Problem 3.13.25, etc., and the fact that the integrals of the odd powers of the cosine over [0, π] are zero, we get Z π Z πX ∞ ∞ Z π X 1 du = cosn (u)du = cosn (u)du = 1 − cos(u) 0 0 n=0 n=0 0
∞ Z π ∞ 2k X X 2k k = +∞. cos (u)du = π 22k 0 k=0

k=0

{The Lebesgue Monotone Convergence Theorem,h3.3.10, can jusπi tify the switching of the integral with the sum over 0, and then 2 adjust the other half. Or use the uniform convergence over any close interval [a, b] ⊂ (0, π).} Therefore, the final series equals infinity, i.e.,
∞ 2k X k = +∞. 22k k=0

Second, we would like to compute the integral Z πZ π dudv . 0 0 1 − cos(u) cos(v) We use the half-angle trigonometric substitutions u v x = tan and y = tan . 2 2 So, we have: du =

2dx 2dy 1 − x2 1 − y2 , dv = and cos(u) = , cos(v) = . 2 2 2 1+x 1+y 1+x 1 + y2

After substituting and simplifying, we find Z πZ π Z ∞Z ∞ dudv dxdy =2 . x2 + y 2 0 0 1 − cos(u) cos(v) 0 0 In polar coordinates x = r cos(θ) and y = r sin(θ), the last integral becomes Z ∞ Z π2 Z ∞ Z π rdrdθ dr 2 π ∞ π 2 =2 dθ = 2 · [ln(r)]0 · = 2 · ∞ · = ∞. 2 r r 2 2 0 0 0 0 Hence, Z 0

π

Z 0

π

dudv = ∞. 1 − cos(u) cos(v)

332

Improper Riemann Integrals

Using the geometric series and the integrals of the powers of cosine as before, we get # 2 Z π Z π "X ∞ ∞ Z π X n n n cos (u) cos (v) dudv = cos (x) dx = 0

0

n=0

0

" 2 ∞ π X 2k 2 cos (x) dx = π

2k k 22k

n=0 ∞ Z X 0

k=0

k=0


#2 = ∞.

Therefore, we find ∞ X k=0

Third, the integral Z πZ 0

π

0

Z 0

π

"


#2

2k k 22k

= ∞.

dudvdw 1 − cos(u) cos(v) cos(w)

is treated along the same lines as the previous one, with the same trigonometric substitutions and using spherical coordinates in this case. It con  1 4 1 . (Its interesting history and verges and is computed to be Γ 4 4 computation are exposed in Nahin 2015, 206–212, along with two other difficult, interesting and important integrals.) So, as before, by the geometric series and the integrals of powers of cosine, we have "
#3
Z πZ πZ π ∞ 2k X Γ4 14 dudvdw 3 k = =π . 4 22k 0 0 0 1 − cos(u) cos(v) cos(w) k=0

Hence, and in view of property (B, 8), we finally find ∞ X k=0

"


#3

2k k 22k


Γ4 14 π = = 4 3
= 1.393203929... . 4π 3 Γ 4

[The divergence of the first two series and the convergence of the third one can easily be checked by the pertinent criteria we find in advanced calculus and mathematical analysis. One such strong criterion is the Dini-Kummer35 criterion for the convergence or divergence of positive series.36 See, e.g., Knopp 1990, 310–311. 35 Ernst

Eduard Kummer, German mathematician, 1810–1893. convenient interpretation of the Dini -Kummer criterion for the convergence of positive series (which includes three statements) is the following: 36 One

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333

Also, the convergence of the latter series implies the convergence of all series "
#q ∞ 2k X k < ∞, with q ≥ 3, real. 22k k=0

It takes a lot of work to figure out their exact values, even when q is integer, which can involve hypergeometric forms. See also Problem 3.13.66, (f ).]

3.13

Problems

3.13.1 In our lists of the properties of the Gamma and the Beta functions, the proofs of some claims were left as exercises or incomplete with or without hints. Identify all these claims and provide their full proofs. 3.13.2 Prove that the improper integral that defines the Gamma function in (0, ∞) converges uniformly. (See Definition 3.3.4 and the criteria that follow it.) “Dini -Kummer Criterion for Convergence of Positive Series: Consider (an ) and (bn ), with n ∈ N, two sequences of positive numbers and define the sequences cn = bn − bn+1

an+1 an bn − an+1 bn+1 = , an an

n ∈ N,

and dn = bn

an an+1

− bn+1 =

an bn − an+1 bn+1 an+1

 = cn ·

an an+1

 ,

n ∈ N. Then:

(1) If there is σ > 0 and N ∈ N such that cn > σ for all n ≥ N , (or dn > σ for all ∞ X n ≥ N , whichever is more convenient), then the series an converges (to a positive n=1

number). (2) If there is N ∈ N such that cn ≤ 0 for all n ≥ N , (or dn ≤ 0 for all n ≥ N , ∞ X 1 whichever is more convenient), and the series diverges (= ∞), then the series b n=1 n ∞ X an diverges (= ∞). n=1

(3) The converses of both (1) and (2) are also true.” Note: This criterion is very powerful and combined with the harmonic series and the comparison criterion for positive series can yield most of the well known criteria for convergence or divergence of positive series as corollaries.

334

Improper Riemann Integrals

3.13.3 Prove that: (a)   1,       lim p α Γ(p) = 0,  0 1, if α < 1.

In particular, Γ(pb) 1 1 and lim + = , b b 0 0,

if α ≤ 0, if α > 0.

(c) ∀ p ≥ 8, Γ(p) > ep . (d) ∀ α ∈ R,

lim

0 0 fixed, lim

Γ(p) = 0. (We may also switch p and q.) + q)

p→∞ Γ(p

[Hint: Prove this directly when q = n ∈ N. Otherwise, for any q > 0, appeal to properties (B, 11) or (B, 12). Also, see project Problems 3.13.65, Item (6.), and 3.13.66, (a).] (h) For r ∈ R and ∀ 0 < a < ∞,   Z ∞ ∞, Γ(x) dx =  Γ(x + r) a  finite > 0,

if 0 ≤ r ≤ 1, if r > 1.

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Examine what happens with this integral when r < 0. [Hint: See properties (B, 11) and (B, 12) or Problems 3.13.65, Item (6.) and 3.13.66, (a).]   dn Γ(p) = (−1)n n! . (i) For n = 0, 1, 2, 3, . . . , lim + pn+1 · dpn 0 0 fixed,

lim B(p, q) = Γ(0+ ) = +∞.

0 0, let I(r) = B(r, r) = Γ(2r) (1) lim I(r) = ∞.

(k)

lim

(p, q)→(0+ , 0+ )

r→0

(2) lim I(r) = 0. r→∞

(3) ∀ p, q > 0, ∃ r > 0 such that B(p, q) = B(r, r). 3.13.4 Prove that: (a) If p ≥ 1 and q > 1 or if p > 1 and q ≥ 1, then Γ(p)Γ(q) < Γ(p + q). (b) If p = q = 1, then Γ(p)Γ(q) = Γ(p + q). (c) If 0 < p ≤ 1 and 0 < q < 1 or if 0 < p < 1 and 0 < q ≤ 1, then Γ(p)Γ(q) > Γ(p + q). (d) If p and q satisfy other combinations of inequalities, then Γ(p)Γ(q) and Γ(p + q) may satisfy different inequalities from those above. In fact, using properties (B, 11) and (B, 12) or Problems 3.13.65, Item (6.) and 3.13.66, (a), prove that for any 0 < p < 1 there is q > 1 such that any of the following three relations can be achieved: (1) Γ(p)Γ(q) > Γ(p + q), (2) Γ(p)Γ(q) < Γ(p + q), (3) Γ(p)Γ(q) = Γ(p + q). (Obviously, we have analogous conclusions for a given 0 < q < 1 or q ≥ 1, etc.) (e) Besides p = q = 1, there are infinitely many p > 0 and q > 0 such that B(p, q) = 1.

336

Improper Riemann Integrals [Hint: Use the relation between the Beta and Gamma functions and the definition of the Beta function.]

3.13.5 (a) Use the change of variables x = − ln(u) to prove that for p > 0 Z ∞ Z 1 Z 1 Γ(p) = xp−1 e−x dx = [− ln(u)]p−1 du = | ln(u)|p−1 du. 0

0

0

(b) Prove: Z 1   − 12 Z 1 Z 1 √ 1 1 1 p p (1) dx = dx = dx = π. ln x − ln(x) | ln(x)| 0 0 0 √ Z 1p Z 1    12 Z 1p π 1 − ln(x) dx = | ln(x)| dx = . (2) dx = ln x 2 0 0 0 Z 1    1 (3) ln ln dx = − γ. x 0 [See also Problems 2.3.17, 3.5.18, 3.13.8, 3.13.67, II 1.7.64, footnote of Problem 2.3.32 and properties (Γ, 5) and (Γ, 9).] (c) Let x = eu in the definition of the Gamma function to prove Z ∞ u ∀ p > 0, Γ(p) = epu−e du. −∞

3.13.6 [Extension of property (Γ, 11).] (a) For α > −1, β > 0 and c > 0 constants, use u = βxc to prove that   Z ∞ α+1 1 α −βxc . x e dx = α+1 Γ c cβ c 0 (b) Use (a) to prove ∀ p > 0,

Γ(p) = 2

1−p

Z

∞

1

2

u2p−1 e− 2 u du,

0

Z and for

c = 1,

we get

∞

xα e−βx dx =

0

[See also Problem 3.7.1 (d) and compare.]

Γ(α + 1) . β α+1

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(c) Many (not all) integrals in Problems 2.3.1-2.3.14 and 1.8.21 can be evaluated by means of the formulae in (a) and (b).√ Identify Z ∞ 2 π x2 e−x dx = those integrals and evaluate them. E.g., , etc. 4 0 [See also Problem 3.7.1, (a) and (d).] (d) In (a), under the same conditions, take the derivative with respect to α to find       Z ∞ c α+1 α+1 1 0 Γ − Γ ln(β) . ln(x)xα e−βx dx = α+1 c c c2 β c 0 3.13.7 Use (Γ, 11) or the previous Problem to prove the following integral representations of the real Zeta function. ∀ p > 1, we have: Z ∞ p−1 Z ∞ p−1 ∞ X 1 1 1 x ln (u) ζ(p) := dx = du = = p x−1 n Γ(p) e Γ(p) (u − 1)u 0 1 n=1 Z 1 Z 1 1 1 [− ln(v)]p−1 [− ln(1 − t)]p−1 dv = dt. Γ(p) 0 1−v Γ(p) 0 t (Notice that for 0 ≤ p ≤ 1, all of the above equalities are of the type ∞ = ∞. See also Example II 1.7.25 and the results in its Corollary, II 1.7.5 and Example II 1.7.26.) So we have: Z ∞ p−1 Z 1 x [− ln(1 − t)]p−1 ∀ p > 1, ζ(p) · Γ(p) = dx = . . . = dt. ex − 1 t 0 0 3.13.8 (a) Prove that for n = 0, 1, 2, 3, ... integers and α > −1 constant Z 1 (−1)n n! xα [ln(x)]n dx = . (α + 1)n+1 0 (See also Problems 2.3.17 and 3.13.5.) [Hint: Let x = e−u and use the Gamma function as in Problem Z 1 1 3.13.6, (a) or (b). Or, use xα dx = and justify that you α+1 0

338

Improper Riemann Integrals can take derivatives of both sides with respect to α, as explained in Section 3.1.]

(b) For a ∈ R and b > 0 constants, use the exponential power series and the result in (a) to prove Z

1

b x(ax ) dx =

0

∞ X

(−a)n . (b n + 1)n+1 n=0

(Johann Bernoulli’s37 integral.) Notice that for b = 0 this result is true only if −1 < a < 1. Investigate this integral when b = 0 and a ∈ R. 3.13.9 Imitate Example 3.10.7 and provide the proof of Result 3.8 Z ∞ cos(x) π dx = ∀ 0 < p < 1. pπ , p x 2Γ(p) cos( 0 2 ) 3.13.10 Use the results of Example 3.10.8 and compute the integrals Z ∞ Z ∞ √ I1 = x2 sin(x4 ) dx, I2 = x cos(x3 ) dx, 0 0 Z ∞ Z ∞ −1 −5 2 2 I3 = x sin(x ) dx, I4 = x−2 cos(x−3 ) dx. 0

0

3.13.11 With the help of the Gamma function, compute the values of Z ∞ Z ∞ both Fresnel integrals sin(x2 ) dx and cos(x2 ) dx. 0

0

(See also Examples 3.6.2 and II 1.7.17.) 3.13.12 For p ∈ R, investigate the convergence of the two integrals Z ∞ Z ∞ sin(xp ) dx and cos(xp ) dx. 0

0

Compute those integrals that converge by means of the Gamma function. (See also Problem 3.2.41.) 37 Johann

Bernoulli, Swiss mathematician, 1667–1748.

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3.13.13 Problems 3.2.22, 3.9.17 and some of the questions in Problems 3.2.37–3.2.41 can be answered by the two results contained in Example 3.10.8. Identify those questions and give the answers with the help of these two general results. 3.13.14 Prove directly, or by induction, or by using property (B, 8) or the byproduct of the duplication formula in (B, 10), that: (a) For n = 0, 1, 2, 3, ... , we have   √ √ 1 (2n)! π (2n)! π Γ n+ = = . 2 22n n! 4n n! This expression for n = 1, 2, 3, ... simplifies to   1 1 · 3 · 5 · · · (2n − 1) √ π. Γ n+ = 2 2n (b) Use the definition of Γ(p) for p < 0 (have a look at it if you have forgotten it) to prove that for m = 1, 2, 3, 4, ...   √ 1 (−1)m 2m (−1)m 22m m! √ Γ −m + = π= π. 2 1 · 3 · 5 · · · (2m − 1) (2m)! 3.13.15 For any m ≥ 0 and n ≥ 0 integers, use the Beta and Gamma functions and the Binomial Theorem to prove that B(m + 1, n + 1) = Z = 0

1

xm (1 − x)n dx =

n X k=0

m

X (−1)l (−1)k [nCk] = [mCl] = m+k+1 n+l+1 l=0

m! n! 1 1 = = , (m + n + 1)! (m + n + 1) [(m + n)Cm] (m + n + 1) [(m + n)Cn]   n n! where nCk = = are the combination numbers. k!(n − k)! k 3.13.16 Let x3 = 8u to prove that Z 2 p 16π 3 x 8 − x3 dx = √ . 9 3 0 3.13.17 Let f (x) be a continuous probability density for −∞ < x < ∞.

340

Improper Riemann Integrals

(See Definition in Application 1, Section 2.2). Prove that for any n ∈ N and r = 1, 2, . . . , n n−r r−1 Z ∞ Z y n! f (x)dx g(y) = f (x)dx f (y) (r − 1)!(n − r)! −∞ y is also a probability density. (This density is called the density of the rth order statistics.) Z ∞ Z y f (x) dx = 1 − u, u(−∞) = 0, f (x) dx. Then, [Hint: Let u = −∞

y

u(∞) = 1 and use property (B, 7).] 3.13.18 Evaluate: (a) B(2.3¯3 . . . , 3.6¯ 6 . . .). R ∞ x3.5 (b) 0 (1+x)5 dx, and R ∞ x3 (c) 0 (1+x)5.2 dx, and

R∞ 0

x1.5 (1+x)5

R∞ 0

dx.

x5 (1+x)8.7

dx.

[Hint: See property (B, 8) and Example 3.11.6.] 3.13.19 For all −1 < α < 1, prove the two equalities Rπ Rπ (a) 0 | tan(x)|α dx = 2 02 tanα (x) dx = cos παπ . (2) π Rπ R (b) 0 | cot(x)|α dx = 2 02 cotα (x) dx = cos παπ . (2) Show that the formulae give the right answer (∞) even when α = ±1. 1 1 Evaluate the integrals for α = and α = . 2 3 3.13.20 For all α < 1, prove the two equalities Rπ Rπ √ Γ( 1−α 2 ) (a) 0 | sec(x)|α dx = 2 02 secα (x) dx = π · Γ 2−α . ( 2 ) (b)

Rπ 0

| csc(x)|α dx = 2

R

π 2

0

cscα (x) dx =

√

π·

Γ( 1−α 2 ) Γ( 2−α 2 )

.

Show that the formulae give the right answer (∞) even when α = 1. 1 1 Evaluate the integrals for α = and α = . 2 3 3.13.21 Using the properties of the Beta and Gamma functions and v 2 = tan(x) (or Example 3.11.15), prove √ Z ∞ du π 2 = . 1 + u4 4 0

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[See also Examples 3.1.6, (b), 3.6.2, 3.11.15, II 1.7.7 and II 1.7.8.] 3.13.22 Use property (Γ, 8) of the Gamma function and property (B, 7) of the Beta function to prove the recursive properties (B, 3) and (B, 4), of the Beta function. 3.13.23 Use x4 = b4 u to prove that if b > 0, then Z

b

0

 1 2 Γ( 4 ) dx √ √ , = 4b 2π b4 − x4

Z

b

and −b

 1 2 Γ( 4 ) dx √ √ . = 2b 2π b4 − x4

(See also Example 3.11.7.) 3.13.24 Use x2 = b2 tan(θ) to prove that if b > 0, then Z 0

∞

 1 2 Γ( 4 ) dx √ √ , = 4 4 4b π b +x

Z

∞

and −∞

 1 2 Γ( 4 ) dx √ √ . = 4 4 2b π b +x

3.13.25 (a) Using the Beta and Gamma functions, prove the following result which we frequently see in many calculus books and integral tables. For the integers p = 0, 1, 2, 3, . . . , we have: Z 0

                 

π 2

sinp (φ) dφ =

Z

π 2

cosp (φ) dφ =

0

π , 2

if p = 0,

1,

if p = 1,


p p 1 · 3 · 5 · · · (p − 1) π p! π π  2  · =  p
2 · = p · ,   p 2 · 4 · 6 · · · p 2 2 2 2  2 · 2 !        2  p−1 2 · 4 · 6 · · · (p − 1)  2p−1 ·[( p−1 2 )!]   = = 2 p−1 ,  1 · 3 · 5···p p! p ·( p−1 )

if p = even ≥ 2,

if p = odd ≥ 3.

2

(b) Prove this result by means of calculus and induction. [See also Problems 3.2.28, (g), II 1.8.12, II 1.8.17 and Examples 3.11.10, II 1.8.4, II 1.8.5, II 1.8.6.]

342

Improper Riemann Integrals

3.13.26 h(a) Integrate the trigonometric identity of Problem 3.2.28, πi (a) over 0, and use the previous Problem to prove that for n ∈ N0 , 2
n 2n+1 X 2 · 4 · 6... 2n 2 4n (n!)2 k k (−1) = (−1)n 22n = (−1)n . 2(n − k) + 1 1 · 3 · 5... (2n + 1) (2n + 1)! k=0

(b) Do the same thing with the trigonometric identities of Problem 3.2.28, (b) and (e). 3.13.27 Use the power series of sine and cosine only (not of e−x ), to evaluate as series of numbers the integrals Z

∞

Z


e−x cos x2 dx

∞

√ e−x sin( x) dx.

Z

1

and

0

0

3.13.28 Compute the following integrals 1

Z

4

5

t (1 − t) dt

(a) Z

0 1

p 4 t(1 − t) dt Z ∞ −t e 5 dt t4 0

(b) 0

(c)

Z

x10 e−x dx

0 ∞

(e) 0

Z

Z

1

(f) 0

0 1

p 5 x3 (1 − x)8 dx.

and 0

∞

Z and

e−8t 1

(5t) 4

0

∞

Z (d)

x4 (1 − x)97 dx.

and

∞

Z

5

dt. 15

x 2 e− 22 x dx.

and 0

1 2

x dx (1 + x)2

Z 0

13

x2 dx (1 + x)15

∞

and 1

Z and 0

−1

x2 dx. (1 + x)3

x3 + x10 dx. (1 + x)15

3.13.29 As we do in Example 3.10.9, compute the integrals ∞

Z (a)

5 3

0

Z (b) 0

sin4 (x)

∞

∞

Z dx

x sin7 (x) dx x4.25

and 0

Z and 0

∞

cos5 (x)

dx, 1 x3 cos7 (x) dx. x0.75

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3.13.30 Prove √ ∞ ∞ X X (n + 1)(n!)2 4 3π n+1
=2+ (a) = . 2n (2n)! 27 n n=0 n=0 (b)

∞ ∞ X X 3π (n + 1)2n (n!)2 (n + 1)2n
=5+ = . 2n (2n)! 2 n n=0 n=0

[Hint: Use the method of Application 5 of this Section.] 3.13.31 (a) Prove that for any p > 0, 1 2

Z

xp−1 (1 − x)p−1 dx = 2

B(p, p) = 2

Z

1

xp−1 (1 − x)p−1 dx = 1 2

0

Γ2 (p) . Γ(2p)

(b) Use this to derive the duplication formula of the Gamma function found in (B, 10).
(c) Use the duplication formula to derive the formula for Γ 2k p , for k ∈ N. 3.13.32 Prove that for p > 0, q > 0 and t ∈ R Z

1

ext xp−1 (1 − x)q−1 dx =

0

∞ X Γ(n + p) Γ(q) n t . n! Γ(n + p + q) n=0

3.13.33 (a) If a > −1, b > −1 and c > 0, let u = xc to prove that Z

1

0

1 x (1 − x ) dx = B c Z

(b) Prove lim

n→∞

c b

a

1



 a+1 ,b + 1 . c

n

(1 − xn ) dx = 1. (Use the Gamma function or

0

the Lebesque Dominated Convergence Theorem, 3.3.11.) Z 1r 2048 x √ (c) Show dx = = 2.955267.... 4 693 1 − x 0 (d) Check the Chebyshev’s Conditions for the integral in (c) and compute it in that way. (See also Example 3.11.1.) 3.13.34 Prove the following six Chebyshev integrals Z −3 1 + x2 (1) x3 (1 + 2x2 ) 2 dx = √ + C. 2 1 + 2x2

344

Improper Riemann Integrals ! √ Z 4 p  dx 1 x−4 + 1 + 1 1 4 √ (2) = ln √ x−4 + 1 + C. − arctan 4 4 4 2 1 + x4 x−4 + 1 − 1 √
Z 2x2 − 1 1 + x2 dx √ + C. (3) = 3x2 x4 1 + x2 √     Z dx 1 3 (u − 1)2 2u + 1 √ √ (4) = ln 2 + arctan + C, 10 u +u+1 5 3 x 3 1 + x5   p 3 with u = 1 + x5 . Z dx −4 − 3x2 (5) 5 = 2 + C. x2 (2 + x3 ) 3 8x (2 + x3 ) 3 Z  −3  23 dx 4 + 1 (6) = −2 x + C. √ p √ 3 4 x3 1 + x3 (See also Example 3.11.1.) 3.13.35 (a) If a > 0, p > 0 and q > 0, prove Z a1 1 xp−1 (1 − ax)q−1 dx = p B(p, q). a 0 1 (b) Replace a by and simplify the new formula. a (c) What happens when 0 < a −→ 0+ or 0 < a −→ +∞ ? [See also Problems 3.13.41 below, and 4.2.19 (h) and (i).] 3.13.36 (a) If p > 0 and q > 0, prove that Z x y p−1 (x − y)q−1 dy = B(p, q)xp+q−1 . 0

(See also Problem 4.2.30.) (b) By (a) we get  Z ∞ Z x Z p−1 q−1 y (x − y) dy e−x dx = B(p, q) 0

0

∞

xp+q−1 e−x dx.

0

Change the order of integration, make a change of variables and prove property (B, 7), Γ(p)Γ(b) = B(p, q)Γ(p, q). 3.13.37 The real hypergeometric function with real parameters a, b and c and real variable x38 is the function defined by the power series ∞ X [a]n · [b]n n F (a, b; c; x) = x , where [c]n · n! n=0 38 If

a, b and c and the variable is the complex z, the same formula defines the complex hypergeometric function.

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345

c is a not a negative integer (why?) and [r]n is the Pochhammer symbol, or the symbol of the rising or ascending factorial of r, defined earlier. (a) Prove that F (a, b; c; x) has radius of convergence R ≥ 1. (b) Prove that if a > 0 and |x| < 1, then F (a, b, b; x) = (1 − x)−a , and so R = 1. [Hint: Use the formula for the binomial series. See Problem 3.13.59.] (c) Find the conditions on a, b and c so that R > 1. (d) Prove that if a > 0, c > b > 0 and |x| < 1, then Z 1 1 ua−1 (1 − u)c−b (1 − ux)−a du. F (a, b; c; x) = B(b, c − b) 0 (See also the second half of Appendix II 1.8.3.) 3.13.38 By letting x = e−t , prove that for all p > 0 and q > 0 Z ∞
q−1 B(p, q) = e−p t 1 − e−t dt 0

and then show

∞

Z

e−3t 1 − e−t


5

dt =

0

1 . 168

3.13.39 By letting x = u2 , prove that for all p > 0 and q > 0 Z 1
q−1 B(p, q) = 2 u2p−1 1 − u2 du 0

and then show Z 0

1

u9 1 − u2


8

du =

1 . 12870

3.13.40 (a) Let p = q := s in the definition of the Beta function and use the transformation u = (1 − 2x)2 to prove that for all s > 0,   1 1−2s B(s, s) = 2 B ,s . 2 (b) Use (a) to derive the duplication formula of the Gamma function that was derived in (B, 10).

346

Improper Riemann Integrals

3.13.41 If b > a, p > −1 and q > −1, use u = Z

b

(x − a)p (b − x)q dx = (b − a)p+q+1

a

1 (x − a) to prove b−a

Γ(p + 1) Γ(q + 1) . Γ(p + q + 2)

[See also Problems 3.13.35 and 4.2.19 (h) and (i).] 3.13.42 (a) Prove that for all a and b real such that a > b > 0, Z ∞ cosh(2bx) Γ(a + b) Γ(a − b) dx = 4a−1 B(a + b, a − b) = 4a−1 . 2a Γ(2a) cosh (x) 0 [Hint: Change to exponentials, simplify, use u = e−2x and (B, 5, IV).] (b) So, for any a > 0 Z ∞ Z sech2a (x) dx =

∞

1 [Γ(a)]2 dx = 4a−1 . 2a Γ(2a) cosh (x) 0 0 Z ∞ 1 (c) For n = 1, 2, 3, 4, 5, 6, use (b) to evaluate dx. coshn (x) 0 (d) Make the substitution y = rx and use (a) to obtain the general b integral for r > 0, b > 0 and a > r Z ∞ cosh(2bx) dx = cosh2a (rx) 0

  b b 4a−1 Γ a + rb Γ a − rb 4a−1 B a+ , a− = . r r r r Γ(2a) Z ∞ Z ∞ 1 (e) Find as a series the integral dx = sech(x2 ) dx. cosh (x2 ) 0 0 3.13.43 Prove that for all a and b real such that a > b > −1,

  Z ∞ Γ a−b sinhb (x) 1 b+1 a−b 1 Γ b+1 2 2
dx = B , = . cosha (x) 2 2 2 2 Γ a+1 0 2 [Hint: cosh2 (x) − sinh2 (x) = 1. Let sinh(x) =

√ w and use (B, 5) (I ).]

3.13.44 Look at Application 1 of this Section and find the period of the pendulum attached to a solid thin rod, if the swing angle θ0 = π, i.e., we let it swing from the upper vertical position.

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347

3.13.45 (a) Verify the triplication formula of the Gamma function,39     1 33p− 2 1 2 ∀ p > 0, Γ(3p) = Γ(p) Γ p + Γ p+ . 2π 3 3 [Hint: Possible method is to work analogously to Problem 3.13.40 for deriving the duplication formula, that was first derived in (B, 10). Another method is to use the Gauß multiplication formula of the Gamma function or the product representation of the reciprocal of the Gamma function in the footnote below. Also, we may work as follows: ∀ n ∈ N, we have: Γ(3n) = (3n − 1)! =

n−1 1 Y (3n)! = (3k + 1)(3k + 2)(3k + 3) = 3n 3n k=0

1 3n

n−1 Y

n−1 Y

n−1 Y

k=0

k=0

k=0

(3k + 1) ·

(3k + 2) ·

(3k + 3) =

  n−1 n−1 n−1 Y Y Y 1 2 1 n n ·3 k+ ·3 k+ · 3n (k + 1) = 3n 3 3 k=0 k=0 k=0

2 Γ n − 31 + 1 3n−1 Γ n − 3
+ 1
3 · · · (n − 1)! = Γ 13 Γ 23 √     3 1 2 3n−1 · Γ(n) · Γ n + ·Γ n+ = 3 · 2π 3 3     1 33n− 2 1 2 Γ(n) Γ n + Γ n+ . 2π 3 3 We may be able to extend this argument to any p > 1, as we do in 39 The duplication and triplication formulae are subcases of the Gauß multiplication formula of the Gamma function. This states: ∀ p > 0 and ∀ n ∈ N, 1       nnp− 2 1 2 n−1 Γ(np) = Γ(p) Γ p + Γ p + . . . Γ p + . n−1 n n n (2π) 2

For a proof, see Problem 3.13.65, Item (13.). Another proof can be easily achieved by using the product representation of the reciprocal of the Gamma function ∞ h Y z  −z i 1 = zeγz 1+ e n Γ(z) n n=1 which is valid for every real and complex number z. [γ is the Euler-Mascheroni constant. See also Problem 3.13.65, Item (8.).] The proof of this important formula for the complex variable z is presented in complex analysis.

348

Improper Riemann Integrals

(B, 8). Then, if the formula is true for p > 1, prove that it is also true for 0 < p < 1.     1 2 3p− 21 Another method is to write 3 Γ(p) Γ p + Γ p+ as a 3 3 triple integral and show that it is equal to 2πΓ(3p). (Similarly, the duplication formula and property (B, 8) could be proven by an analogous double integrals.)] [See and compare with duplication property (B, 10) of the Gamma function and Problems 3.13.40 and 3.13.65, Item (13.).] 1 Γ(3p) = . (b) Use this formula and (B, 8) to prove: lim 3 p→0+ Γ(p) [See and compare with duplication property (B, 10) and Problems 3.13.3, (a), 3.13.65, Item (14.).] (c) Prove that √   1 2 π2 3 B , = 3 3 3 and √   1 2 2πΓ(3p) 3 πΓ(3p) ∀ p > 0, B p + , p + = 3p− 1 . = 3p 3 3 3 p Γ(p)Γ(2p) 2 Γ(p)Γ(2p + 1) 3 Now use the duplication formula to replace the Γ(2p) and simplify. Also, find the limit of the last fraction as p 7−→ 0+ . 3.13.46 (a) Prove that for all p > 0, q > 0, r > 0, and s > 0, as we have Z

1

xp−1 (1 − x)q−1 dx =

0

Γ(p) Γ(q) , Γ(p + q)

in property (B, 7), we also have Z

and Z

1

1

Z

0

0

Z

1−x

1−x

 Γ(p) Γ(q) Γ(r) xp−1 y q−1 (1 − x − y)r−1 dy dx = Γ(p + q + r)

Z

1−x−y

x 0

0

p−1 q−1 r−1

y

z

(1 − x − y − z)

s−1





dz dy dx =

0

Γ(p) Γ(q) Γ(r) Γ(s) . Γ(p + q + r + s) (b) Generalize this formula to all dimensions n = 1, 2, 3, . . . . (See also Application 6.)

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349

3.13.47 In Application 1, we have seen the elliptic integral Z

π 2

dψ

K(k) :=

q

0

,

where

0 < k < 1.

1 − k 2 sin2 (ψ)

Prove: (a) This integral is an analytic function of k. π and the integral is continuous at k = 0. 2

(b) K(0) =

(c) lim K(k) := K(1) = ∞. What is the physical meaning of this k→1−

result for the mathematical pendulum as examined in Application 1. 3.13.48 For any n = 1, 2, 3, . . . , develop the integral in the equality Z ∞ n 2 n π2 = e−x dx (see an extension in the next problem) −∞

as a multiple integral over Rn in the spherical coordinates that we have used in Application 2 [in a way analogous to the way of obtaining the Integral (2.1) in Section 2.1] to obtain that π

n 2

1 n = Γ 2 2

Z

π

π

Z

2π

Z

sinn−2 (φ1 ) . . . sin(φn−2 )dθdφ1 . . . dφn−2 ,

... 0

0

0

and so the area of the (n − 1)-dimensional unit sphere is n

2π 2
. Area [Sn (1)] = Γ n2 [Note: We can use this result to derive the formulae in Application 2.] 3.13.49 Suppose that the constant real is symmet matrix An×n = [aij ] → − → − − → − T → ric (a = a ) and positive definite x A x > 0, if x 6= 0 . Prove ij

ji

that Z

∞

Z

∞

Pn

e−(

... Z

−∞ ∞

Z

−∞ ∞

... −∞

−∞

i=1

→ − T A→ − x

e− x

aii x2i +2

− d→ x =

Pn

1≤i 0, height h > 0 and axis of symmetry the x1 -non-negative axis, is the set   q h − Cn = → x = (x1 , x2 , . . . , xn ) | x22 + x23 + . . . + x2n ≤ x1 ≤ 1 . R Prove that 2hRn−1 π Volume [Cn ] = n(n − 1)Γ

n−1 2

n−1 2


.

(Note: When n = 2, the respective cone is an isosceles triangle with base 2R and height h, and this formula gives its area.) 3.13.52 In this problem, use the Dirichlet gamma integrals presented in Application 6. (a) Prove that the volume of the ellipsoid  x 2 a

+

 y 2 b

+

 z 2 c

=1

4π abc. 3 (If a = b = c = R, we find the volume of a sphere with radius R, 4π 3 V = R .) 3 (b) If the mass density of the ellipsoid in (a) is given by the even function ρ(x, y, z) = x2 y 2 z 2 + 1, prove that its total mass is with semi-axes a > 0, b > 0 and c > 0, is

M=

V =

4π 4π (abc)3 + abc. 945 3

[Hint: In (a) and (b) do not forget to multiply by 8, given all the symmetries involved.] (c) Find the volume of the solid enclosed (separately) by each of the two surfaces r r r |x| 3 |y| 4 |z| + + = 1, 2 3 4 and  3  4  5 |x| |y| |z| + + = 1. 2 3 4  x 2  y 2 (d) Prove that the area of the ellipse + = 1, with semia b axes a > 0 and b > 0, is A = πab.

352

Improper Riemann Integrals

(e) Find the area of the astroid of Figure 1.3, in Application 4 of Section 1.2. 3.13.53 Write the formula for the n-dimensional Dirichlet gamma integrals Dn (see Application 6) and use it to find the volumes of the (n − 1)-dimensional sphere and ellipsoid in Rn , that they were found in Application 2 by a calculus method. 3.13.54 Refer to W of Application 6 and compute Z Z Z   p p 2 + 3x − 4yz + 5 xy 3 z + 6 3 x2 yz 4 dxdydz. W

3.13.55 (a) For any n ≥ 2 integer, prove the general Euler’s formula n−1 Y

  Y   n k k = Γ = n n k=1 k=1         r 1 2 n−1 n (2π)n−1 Γ ·Γ · ... · Γ Γ . = n n n n n Γ

[See also Problem 3.13.65, Item (13.).] [Hint: Use the result in Problem II 1.2.34, (c) and (B, 8). ] (b) Use (a) and the hint below to prove Raabe’s40 integral Z 1 √  1 2π . ln[Γ(x)]dx = ln(2π) = ln 2 0 [Hint: Take the logarithm of both sides in (a), form a Riemann sum and take its limit.] (c) Use property (Γ, 8) to prove that Z 2 √  2π = −0.0810615 . . . , ln[Γ(x)] dx = −1 + ln 1

and ∀ p ≥ 2, Z p+1 Z ln[Γ(x)]dx = p ln(p) − (p − 1) ln(p − 1) − 1 + p

ln[Γ(x)]dx.

p−1

Z Now show 40 Joseph

p

0≤p→∞

p+1

ln[Γ(x)]dx = ∞.

lim

p

Ludwig Raabe, Ukrainian-Swiss mathematician, 1801–1859.

Real Analysis Techniques

353

(d) Use (c) to prove that for any n ∈ N = {1, 2, 3, . . . , } Z n+1 √  ln[Γ(x)]dx = n ln(n) − n + ln 2π , n

which is also true for n = 0+ , by (b). Then compute

Z

3

ln[Γ(x)] dx. 2

(e) In general prove that for s ≥ 0, Z s+1 Z 1 √ ln[Γ(x)] dx = ln[Γ(x + s)] dx = s · ln(s) − s + ln( 2π). s

0

(f) Use (d) to prove that for every n = 1, 2, 3, . . . , Z

n

ln[Γ(x)] dx = 0+

n−1 √ −(n − 1)n X + k ln(k) + n ln( 2π), 2 k=0

where in the summation, we use the convention 0 · ln(0) = 0. [See also Problem 3.13.65, Item (15.).] 3.13.56 Use the results of Example 3.10.7 and property (B, 8) to prove that for any 0 < m < 1 Z ∞  mπ  cos(x) (a) dx = Γ(m) cos , x1−m 2 0 Z ∞  mπ  sin(x) (b) dx = Γ(m) sin . x1−m 2 0 (See also Problem II 1.7.20.) 3.13.57 The t-Student41 statistical sampling distribution with ν degrees of freedom, where ν is a positive integer, has probability density given by
  −(ν+1) 2 Γ ν+1 t2 2
f (t) := f (t; ν) := √ , 1 + ν πν Γ ν2

for − ∞ < t < ∞.

Prove the following: (a) f (t) is a positive even function and Z ∞ 0 µ0 := f (t) dt = 1. −∞ 41 Student is the pseudonym of William Sealy Gosset, English statistician who discovered the important t-Student distribution for small statistical samples, 1876–1937.

354

Improper Riemann Integrals

(b) For 1 ≤ n ≤ ν − 1 odd integer Z ∞ µ0n := tn f (t) dt = 0. −∞

(c) For n ≥ ν odd integer ∞

Z

tn f (t) dt = 0.

P.V. −∞

(d) For n ≥ ν odd integer Z ∞ µ0n := tn f (t) dt = does not exist. −∞

(e) For n ≥ ν even integer Z µ0n :=

∞

tn f (t) dt = ∞.

−∞

(f) For 2 ≤ n ≤ ν − 1 even integer, the integral Z ∞ µ0n := tn f (t) dt −∞

exists. Then, evaluate its value. 3ν 2 ν , if ν > 4, then µ04 = , etc. E.g., if ν > 2, then µ02 = ν−2 (ν − 2)(ν − 4) [Hint: Making appropriate u-substitutions may be necessary in some of the above questions. E.g., use the fact that f (t) is an even function and √ 1 t2 let t = ν · tan(u), or = 1 + , etc. Also, you may need to use the u ν properties of the Beta and Gamma functions.] (g) Refer to the definitions in Application 1 of Section 2.2 and justify why for the t-distribution µn = µ0n , for all n, and σ 2 = µ02 . (h) Prove t2 1 lim f (t; ν) = √ e− 2 := n(x; 0, 1), ν→∞ 2π the standard normal probability density.

[Hint: Use property (B, 12) and the limit for an exponential.]

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3.13.58 The F-statistical sampling distribution42 with ν1 and ν2 degrees of freedom (or parameters), where ν1 and ν2 are positive integers, has probability density given by g(f ) := g (f ; ν1 , ν2 ) =

and

Γ Γ

ν1 2

ν1 +ν2
2

Γ







ν2 2

ν1 ν2

 ν21 f

ν1 2

−1

 −(ν12+ν2 )  ν1 , 1+ f ν2 if 0 < f < ∞, if − ∞ < f ≤ 0.

g(f ) := g (f ; ν1 , ν2 ) = 0,

(Notice here that the letter f is used as the variable of the function g.) (a) Prove g(f ) is non-negative and Z ∞ g(f ) df = 1. −∞

(b) For ν2 > 2 integer, prove Z ∞ µ01 := f g(f ) df = −∞

ν2 . ν2 − 2

(c) Find the positive integers ν2 for which the integral Z ∞ µ02 := f 2 g(f ) df −∞

exists and evaluate it. Also prove that if ν2 > 4, then σ 2 = µ02 − (µ01 )2 =

2ν22 (ν1 + ν2 − 2) . ν1 (ν2 − 2)2 (ν2 − 4)

1 ν1 ν2 1 − u = 1 + f ⇐⇒ f = and use the u ν2 ν1 u properties of the Beta and Gamma functions.] [Hint: The transformation

(d) Find lim g (f ; ν1 , ν2 ), for ν1 given fixed. Is this limit a probabilν2 →∞

ity density? 3.13.59 In a second-semester calculus course, we see Newton’s binomial series, which is very useful to applications. That is: For any real 42 The F-distribution is very important for small statistical samples. It was discovered independently by Sir Ronald Aylmer Fisher, famous English geneticist and statistician, 1890–1962, and George Waddel Snedecor, American statistician, 1881– 1974.

356

Improper Riemann Integrals

number p as the exponent and for all −1 < x < 1, we have the power series expansion ∞ X p (p − 1) . . . (p − n + 1) n (1 + x) = 1 + x . n! n=1 p

(Review this very important power series from a calculus book.) (a) Prove that this expansion agrees with the Binomial Theorem when p ≥ 0 integer. (b) Prove that if p < 0 real and −1 < x < 1, then we can also write (1 − x)p = 1 +

∞ X Γ(n − p) n x . n! Γ(−p) n=1

Why in this formula have we considered negative real exponents only and not every real exponent? 3.13.60 Prove that for any real numbers a and b we have  b ae − bea   + 1, if a 6= b, ∞ ∞  n m XX a b b−a = (n + m)!   n=1 m=1  if a = b. (a − 1)ea + 1, If we let

a = b = 1,

we find

∞ X ∞ X

1 = 1. (n + m)! n=1 m=1

(See also Application 5.) 3.13.61 Prove

Z 0

π

Z 0

π

dudv = ∞. 2 − cos(u) − cos(v)

3.13.62 For any p > 0 prove the following series expansion of the Gamma function Z 1 Z ∞ ∞ ∞ X X (−1)n Γ(p) = xp−1 e−x dx + xp−1 e−x dx = + cn pn , n!(n + p) 0 1 n=0 n=0 Z Z ∞ 1 ∞ −1 −x n 1 n+1 with cn = x e ln (x) dx = e−x ln (x) dx = n! 1 (n + 1)! 1 Z e Z ∞ 1 1 e−x lnn+1 (x) dx + e−x lnn+1 (x) dx. (n + 1)! 1 (n + 1)! e [See also properties (Γ, 2), (Γ, 3) and (Γ, 8) and Problems II

Real Analysis Techniques

357

1.2.38 and II 1.6.22 in which p can be a complex number.] [Hint: In the first integral substitute the power series of e−x and justify the switching of integral and sum. The second integral is convergent for all p ∈ R and then use the Taylor series expansion about p = 0. For cn √ use integration by parts and show lim n cn = 0.] n→∞

3.13.63 Use Property (B, 8) to prove that for 0 < p < 1, we have d d B(p, 1 − p) = [Γ(p)Γ(1 − p)] = dp dp     Z 1 Z 1 x 1−x xp−1 (1 − x)−p ln x−p (1 − x)p−1 ln dx = dx = 1−x x 0 0 −π 2 cos(pπ) = −π 2 csc(pπ) cot(pπ). sin2 (pπ) (See also Examples II 1.7.8, II 1.7.47, II 1.7.49 and Problems 3.2.48, II 1.7.142, II 1.7.145.) 3.13.64 If 0 ≤ m and n are integers such that n > m + 1, use the property (B, 5) to prove that m   X m (−1)k m!(n − m − 2)! = = k n−m+k−1 (n − 1)! k=0 n−m−2 X n − m − 2 (−1)k . k m+k+1 k=0

[See also Problem 1.8.25, (a).] 3.13.65 Project on the Gamma function and some inequalities for which you may need to consult the bibliography. Let f : A −→ R and g : A −→ R be piecewise continuous integrable functions over an appropriate set A ⊆ R. When we study the integrals of such functions, in a more complete theory, for every number 1 ≤ s < ∞, we define: Z  1s s kf ks := |f (x)| dx the s-norm of f . A

For the special case s = ∞, we define kf k∞ := Maximum|f (x)| the ∞-norm of f . x∈A

(Similar definitions are for g. These definitions can be generalized in the larger class of measurable functions, in a more general setting.)

358

Improper Riemann Integrals

1. Prove that for any two real numbers and/or infinity s and t such that 1 ≤ s, t ≤ ∞ and 1 1 + = 1, s t [equivalently s+t = st, and/or (s−1)(t−1) = 1, etc.], and any two non-negative numbers a ≥ 0 and b ≥ 0, we have the inequality: ab ≤

as bt + . s t

Equality holds if and only if as = b t . Then, for functions f and g as above and assuming that Z f (x) · g(x) dx exists, prove the inequalities: A

Z Z |f (x) · g(x)| dx = (a) f (x) · g(x) dx ≤ A A Z Z Z 1 1 s |f (x)| · |g(x)| dx ≤ |f (x)| dx + |g(x)|t dx, s A t A A or using the norm notation kf · gk1 ≤

1 1 kf kss + kgktt . s t

Equality holds if and only if |f |s = |g|t . Z Z f (x) · g(x) dx ≤ |f (x) · g(x)| dx = (b) A A Z  1s Z  1t Z s t |f (x)| · |g(x)| dx ≤ |f (x)| dx · |g(x)| dx , A

A

A

or using the norm notation kf · gk1 ≤ kf ks · kgkt .

(3.10)

Equality holds if and only if there are real numbers α and β such that |α| + |β| > 0 (i.e., not both α and β are 0) and α|f |s = β|g| t . (So, α · β ≥ 0, that is, α and β have the same sign.)

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Inequality (b) is usually called H¨ older’s43 inequality in the mathematical literature. However, it was also discovered independently by Viktor Yakovlevich Bunyakovsky44 a few years earlier. Therefore, many times it is called Bunyakovsky-H¨ older’s inequality. The exponents s and t as above are called conjugate exponents. Notice that s = 1 is conjugate to t = ∞ and vice-versa. In this case, the proof of the inequality is easier than in the other cases. Also, the case of equal conjugate exponents s = t = 2 is a very important one in theory and applications. In this case, the inequality is also called Cauchy-Schwarz45 inequality and generalizes the homonymous inequality in Linear Algebra with inner products. We can use the inequality 3.10 for establishing absolute convergence of integrals. For, if we prove that the right side is finite (kf ks · kgkt < ∞), then we obtain the absolute convergence (and therefore the convergence) of the integral in the left side, Z Z |f (x) · g(x)| dx = |f (x)| · |g(x)| dx < ∞. A

A

2. Use the Cauchy-Schwarz inequality, Problem 2.3.17, and the facts that if x ≥ 1 then 0 ≤ ln(x) ≤ x − 1 and if 0 ≤ x ≤ 1 then 1 1 ≤ e−x ≤ 1 to prove that for any p > and any n = 1, 2, 3, ... e 2 integer, we have (n) Γ (p) < Z ∞ p−1 −x x e [ln(x)]n dx = Z ∞ 0 −x p−1 −x x e 2 · e 2 [ln(x)]n dx < . . . < 0

 1 1 Γ(2n + 1) 2 [Γ(2p − 1)] 2 (2n)! + . e 1 So, here we prove that: For p > the integral that represents 2 th the n derivative of the Gamma function, for all integers n ≥ 1, converges absolutely, and so it exists. 43 Otto

Ludwig H¨ older, German mathematician, 1859–1937. Yakovlevich Bunyakovsky, Russian mathematician, 1804–1889. 45 Hermann Amandus Schwarz, German mathematician, 1843–1921. 44 Viktor

360

Improper Riemann Integrals

3. We have already said the integral that represents the nth derivative of the Gamma function, for all integers n ≥ 1, exists for all p > 0 1 and not just for p > that we examined in the previous item. 2 Prove the absolute convergence of this integral for any 0 < p < 1 1 (or even any 0 < p) by picking an integer k ≥ 2 such that < p, k k and use H¨ older’s inequality with conjugate exponents s = k−1 and t = k to derive an inequality similar to the previous one by completing (n) Γ (p) < Z ∞ p−1 −x x e [ln(x)]n dx = Z ∞ 0 −x p−1 −x x e 2 · e 2 [ln(x)]n dx < . . . < 0

Z 0

∞

Z kp−1  k−1 k −kx k−1 −1 2(k−1) e · x dx

0

∞

−kx  k1 2 nk [ln(x)] dx . . . e

and then use Problems 3.13.6 and 2.3.17, etc. (The choice of k to be an integer is not necessary. We can find analogous upper bounds for any k > 0 such that kp > 1.) 4. Use the Cauchy-Schwarz inequality appropriately to prove that for any n = 1, 2, 3, ... integer, the derivatives of the Gamma function satisfy the strict inequality h i2 Γ (n) (p) < Γ(p) · Γ (2n) (p). 5. Use the previous result to prove that the ln [Γ(p)] is strictly convex by showing that its second derivative is positive. I.e., prove: h 0 i2 00 Γ (p) · Γ(p) − Γ (p) 00 [ln [Γ(p)]] = > 0. Γ(p)2 [So, here we prove that not only the Gamma function itself is convex, as we have seen in (Γ, 4), but also the logarithm of the Gamma function is convex. (See also Problem 3.13.67.)] 6. Prove that for any p ∈ R fixed lim

0< q → ∞

Γ(p + q) = 1. Γ(q) · q p

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[See property (B, 12). Compare with Problem 3.13.66, (a).] [Hint: Either appeal to property (B, 12) or provide a different proof as follows: First, prove this directly when p ≥ 0 integer by using property (Γ, 8). Then, prove that the given fraction is eventually (i.e., for large enough q’s) increasing in p, when p > 0. (You may do this directly or prove that the derivative of the fraction with respect to p is eventually positive.) Then, for any non-integer 0 < p, if [[p]] is the integer part of p, we have that [[p]] < p < [[p]] + 1 and use the Squeeze Lemma from calculus. Finally, reduce the case p < 0 to the case p > 0.] Then for any p > 0 fixed, prove that (a) (b)

lim

0< q → ∞

lim

Γ(q) = 0+ , Γ(q + p)

0< q → ∞

q p · B(p, q) = p

lim

0< q → ∞

or lim

0< q → ∞

Z

lim

0< q → ∞

qp ·

Z

Γ(q + p) = +∞, Γ(q)

1

xp−1 (1 − x)q−1 dx =

0

1

q ·

0< q → ∞

lim

x

q−1

(1 − x)p−1 dx =

0 p

q ·

Γ(p)Γ(q) = Γ(p). Γ(p + q)

[See also and compare with properties (B, 11) and (B, 12). Also, combine (b) with Problem 3.13.33, (a) to obtain other limits.] 7. Prove the Gauß formula for the Gamma function. ∀ p > 0 and ∀ n ∈ N, we have n! np = n→ ∞ p (p + 1)(p + 2) . . . (p + n) n! np−1 lim = n→ ∞ p (p + 1)(p + 2) . . . (p + n − 1) (n − 1)! np lim . n→ ∞ p (p + 1)(p + 2) . . . (p + n − 1)

Γ(p) = lim

[Hint: Use (B, 7) on B(p, n + 1) and also apply integration by parts to the integral of the Beta function until the exponent n is eliminated. Then use the results of the previous two items.]

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Improper Riemann Integrals

8. Write the reciprocal of the fraction of the previous item as  p  p  p  −p ln(n) p 1+ 1+ ... 1 + e = 1  2  n     −p p −p p −p p p 1+ 1+ e e 2 ... 1 + e n egn p , 1 2 n 1 1 1 + + . . . + − ln(n) (−→ γ, as n −→ ∞). 2 3 n Now take limit, as n −→ ∞, to obtain the where gn = 1 +

∞  Y p  −p 1 1+ en . = p eγp Γ(p) n n=1

Taking the reciprocal of this result, we express the Gamma function as an infinite product. 9. Use the previous result to obtain  ∞  Y 1 1 p2 2 · = −p 1− 2 . Γ(p) Γ(−p) n n=1 Use the result of Problem II 1.7.83, (c) (the sine function as an infinite product), to show the result of (B, 8) π = Γ(p)(−p)Γ(−p) = Γ(p)Γ(1 − p). sin(πp) 10. Two characterizations of the Gamma function. (I) Let f : (0, ∞) −→ (0, ∞) be a function that satisfies the following three properties: (a) f (1) = 1. (b) f (p + 1) = p f (p). (c)

ln[f (p)] is convex.

Prove that f (p) = Γ(p). (I.e., these properties characterize the Gamma function.) [Hint: Prove that such a function f (p) is equal to lim

n→ ∞

n! np p (p + 1)(p + 2) . . . (p + n)

and therefore is unique. Then it is equal to the Gamma function, since the Gamma function satisfies these three properties! (See also Rudin 1976, Theorem 8.19, 193.)]

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(II) Let f : R−{0, −1, −2, . . . } −→ R be a function that satisfies the following three properties: (a) f (1) = 1. (b) f (p + 1) = p f (p). f (p + n) (c) lim p = 1. n→∞ n f (n) Prove that f (p) = Γ(p). (I.e., these properties characterize the Gamma function.) 11. Show that there is no function f : [0, ∞) −→ R (f is defined at x = 0), satisfying:

(a) f (1) = 1

and

(b) f (p + 1) = p f (p).

Construct or give an example of a function f : (0, ∞) −→ R [i.e., f is defined in the open interval (0, ∞)] that satisfies these two properties, (a) and (b), but is not the Gamma function. 12. Given n ∈ N, for all q > 0, define the function 1

g(q) =

nq− 2 (2π)

Prove

Γ

n−1 2

q n

 Γ

q+1 n

     q+2 q+n−1 Γ ...Γ . n n

g(q) = Γ(q).

[Hint: Show that g(q) satisfies the three characteristic conditions (a), (b) and (c) of the Gamma function, stated in (8.) above. You also need the result of Problem 3.13.55, (a).] 13. Use the result in (10.) to give a proof to the Gauß multiplication formula of the Gamma function. Prove: ∀ n ∈ N and ∀ p > 0, 1

Γ(np) =

nnp− 2 (2π)

n−1 2

      1 2 n−1 Γ(p) Γ p + Γ p+ ...Γ p + . n n n

1 , we get back the result in Problem 3.13.55, n (a). See also and compare with (B, 10) and Problem 3.13.45.]

[Notice: With p =

14. Prove: (a)

      1 2 n−1 lim Γ 1 + Γ 1+ ...Γ 1 + = 0. n→∞ n n n

[Hint: Use the above result in (10.), with q = n, or in (11.), with

364

Improper Riemann Integrals p = 1. Prove that ! this limit goes to zero as does the expression √ n 1 1 2π √ ·√ · , when n −→ ∞.] e n 2π (b) ∀ n ∈ N,

lim+

p→0

Γ(np) 1 = . Γ(p) n

[Hint: Use the result in Item (10.) or Item (11.) above and Problem 3.13.55, (a), or easier use Problem 3.13.3, (a).] ∀ α > 0,

(c)

lim

p→0+

1 Γ(αp) = . Γ(p) α

k [Hint: Using (b), prove this for α = > 0 rational, with k ∈ N l and l ∈ N, and then use the density of the rationals in the reals and the continuity of the Gamma function, or easier use Problem 3.13.3, (a).] 15. Prove that ∀ p > 0,    √  Z p+1 Γ(np) 1 ln np− 1 lim = − ln 2π + ln[Γ(x)] dx. n→∞ n 2 n p [Hint: Use (10.) above and the argument in the hint of Problem 3.13.55, (b).] Now prove    Γ(np) 1 ln np− 1 =∞= lim 0≤p→∞ n→∞ n 2 n     1 Γ(np) lim lim ln np− 1 n→∞ 0≤p→∞ n 2 n 

lim

and   Γ(np) 1 ln np− 1 lim lim = 0 6= 0≤p→0 n→∞ n 2 n     1 Γ(np) ln np− 1 lim . ∞ = lim n→∞ 0≤p→0 n 2 n 



16. Prove       3 n ∞ ∞ Γ3 n Γ X X p p 1 1   = lim  2    = lim  3n 3n 1≤p→∞ p 3 1≤p→∞ 3p n=1 Γ 1 + p n=1 nΓ p     √ ∞ 2 n ∞ Γ X p 2π 3 X 1 =     lim  = ζ(3). 3n 2 1≤p→∞ 3p n=1 n 3 p Γ n + 1 Γ n + 2 n3 n=1 p 3 p 3 

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365

(If, in this equality, we switch the order of limit and the infinite summation, the equality is immediate. But in this order, use Problem 3.7.9, the Dirichlet integral D3 of Application 6 and the triplication formula in Problem 3.13.45.) 17. In a common real analysis course, the material concerning H¨ older’s inequality is usually combined with the following: For any 1 ≤ s ≤ ∞, prove that the real number kf ks , defined above, satisfies the following four properties of what we call to be the properties that define a norm function: (a) kf ks ≥ 0 (positivity). (b) kf ks = 0 ⇐⇒ f ≡ 0 (non-degeneracy). (c) ∀ c ∈ R, kc · f ks = |c| · kf ks (semi-linearity). (d) kf + gks ≤ kf ks + kgks (triangle inequality). The real number kf ks is called the Ls -norm of the function f . In this context, the triangle inequality [in (d)] is also called Minkowski’s46 inequality. I.e., ∀ s : 1 ≤ s ≤ ∞, we have Z

s

|f (x) + g(x)| dx

 1s

 1s Z  1s s |f (x)| dx + |g(x)| dx .

Z

s

≤ A

A

A

This inequality holds as equality if and only if there are real numbers α and β such that α · β ≥ 0 (i.e., the two numbers have the same sign), |α| + |β| > 0 (i.e., not both α and β are 0) and αf = βg. When s = 1 or s = ∞, the proof of Minkowski’s inequality is easy. (Write the inequalities explicitly and argue about their proofs.) To prove Minkowski’s inequality when 1 < s < ∞, we need H¨older’s inequality. We can use Minkowski’s inequality for establishing absolute convergence of integrals. For, if we prove that the right side is finite, then we obtain the convergence (finiteness) of the integral in the left side. For example, prove: Z 0 46 Hermann

1



1 1 √ + √ 3 5 x x

 32

" 2 3

dx < 2 +



10 7

Minkowski, German mathematician, 1864–1909.

 23 # 32 .

366

Improper Riemann Integrals

18. We conclude this project with the following two pertinent inequalities. Let f : A −→ R and g : A −→ R be piecewise continuous integrable functions over an appropriate set A ⊆ R. Consider any 1 1 s such that 0 < s < 1 and then let t such that + = 1, i.e., s t s < 0. Then we have: t= s−1 (1) ∀ 0 < s < 1 and

1 1 + =1: s t

kf · gk1 ≥ kf ks · kgkt ,

1 1 + = 1, we have s t Z Z |f (x) · g(x)| dx = |f (x)| · |g(x)| dx ≥

i.e., ∀ 0 < s < 1 and

A

A

 1t  1s Z |g(x)|t dx . |f (x)|s dx ·

Z

A

A

(2) ∀ 0 < s < 1 :

kf + gks ≥ kf ks + kgks ,

i.e., ∀ 0 < s < 1, we have Z

 1s Z  1s Z  1s s s |f (x) + g(x)| dx ≥ |f (x)| dx + |g(x)| dx . s

A

A

A

(E.g., see Hardy, Littlewood & P´ olya, 1988, p. 139, or, De Barra G., 1974, exercise 27, pp. 132, 256.) 19. For i = 1, 2, 3, . . . , n, with 3 ≤ n ∈ N, consider piecewise continuous integrable functions fi : A −→ R over an appropriate set n X 1 = 1. A ⊆ R and real numbers pi ≥ 1 or pi = ∞ such that p i=1 i Use H¨older’s inequality in item 1 of this project and induction to prove the general H¨ older’s inequality kf1 · f2 · . . . · fn k1 ≤ kf1 kp1 · kf2 kp2 · . . . · kfn kpn . 20. For i = 1, 2, 3, . . . , n, with n ∈ N, consider piecewise continuous integrable functions fi : A −→ R over an appropriate set A ⊆ R. Use induction and H¨ older’s inequality (in item 1 of this project) to prove the inequality

1 1

|f1 · f2 · . . . · fn | n ≤ (kf1 k1 · kf2 k1 · . . . · kfn k1 ) n . 1

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367

3.13.66 Project on Stirling’s Formula. Find and consult appropriate bibliography and provide a detailed proof of the general Stirling’s formula √ Γ(x + 1) x Γ(x) lim  x x √ = lim  x x √ = 1, 0< x → ∞ 0< x → ∞ 2πx 2π e e or, for x > 0 r

2π  x x [1 + R(x)], x e where R(x) is a remainder that satisfies the asymptotic relation Γ(x) =

|R(x)| ≤ C ·

1 , x

as x −→ ∞,

for some constant C > 0. Use this to prove: (a) If p is any real constant, then Γ(p + q) = 1. 0< q → ∞ Γ(q) · q p lim

[See also property (B, 12) and Problem 3.13.65, Item (6.).] (b) If n ∈ N, prove lim

0 2, 4n 4n (n!)2 n=0 n=0 (See and compare with Application 7.) 3.13.67 Project on Derivatives of the Gamma function. Find and consult appropriate bibliography to study the derivative and/or the logarithmic derivative of the Gamma function. Fill in the missing details and answer the questions in the following. We have already proved that Γ0 (1) "= −γ [see Problem 3.5.18 and # n X 1 also property (Γ, 2)], where γ = lim − ln(n) ' 0.57721566 . . . n→∞ k k=1 is the Euler-Mascheroni constant.

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369

We let d {ln [Γ(p)]} := ψ(p) dp

⇐⇒

d [Γ(p)] = Γ(p) ψ(p). dp

This function is called polygamma function or the psi function of Gauß . We observe that ψ(1) = Γ0 (1) = −γ. Also, with the help of Γ(p+1) = p Γ(p) and Γ0 (p+1) = Γ(p)+p Γ0 (p), we find the recursive relation ψ(p + 1) =

1 + ψ(p). p

Using the above pieces of information and bibliography, we find that for any p > 0, we have  ∞  ∞ X X 1 p 1 1 1 − = ψ(p) = − − γ + =− −γ+ p n n + p p n(n + p) n=1 n=1  ∞  ∞ X X 1 1 p−1 −γ + − = −γ + . n + 1 n + p (n + 1)(n + p) n=0 n=0 Hence, we find: (A)

∞ ∞ X 1 1 d d2 1 X [ψ(p)] = 2 {ln [Γ(p)]} = 2 + = > 0, dp dp p n=1 (p + n)2 (p + n)2 n=0

and so ψ(p) is increasing and ln[Γ(p)] is strictly convex, a fact that is also true for all p ∈ R − {0, −1, −2, . . . }. [See also Problems 3.13.55, (e) and 3.13.65, Item (5).] (B) By Problem 3.5.32, we also have Z ψ(p) = −γ + 0

1

1 − xp−1 dx. 1−x

[To justify some points so far, we may need bibliography that exposes the asymptotic behavior of the function ψ(p), as p → ∞. We may also use Stirling’s Formula or property (B, 12). Or, we may use integral representations of ψ(p). The above integral form  of ψ(p) can be evalu1 ated explicitly for certain values of p. E.g., ψ = −γ − 2 ln(2).] 2

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Improper Riemann Integrals

Then, ∀ p > 0, the derivative of the Gamma function is " # ∞  X d 1 1 1 [ Γ(p)] = Γ(p) ψ(p) = Γ(p) − − γ + − = dp p n n+p n=1 " " # # ∞ ∞ X X 1 p−1 p Γ(p) − − γ + = Γ(p) −γ + , p n(n + p) (n + 1)(n + p) n=0 n=1 etc. [See also property (Γ, 2).] Next, from the first derivative formula and Problem 3.5.6, prove the formula of the second derivative of Gamma   d2 ∀ p > 0, [ Γ(p)] = Γ(p) ψ 2 (p) + ψ 0 (p) = dp 2  " #2 ∞  ∞   1 X X 1 1 1 1 − + 2+ = Γ(p) − −γ+  p n n+p p (n + p)2  n=1 n=1 "  #2 ∞ ∞   X X p−1 1 Γ(p) −γ + + > 0.  (n + 1)(n + p) (n + p)2  n=0 n=0 So, Γ(p) is strictly convex in (0, ∞). Now, investigate the convexity of the Gamma function in (−∞, 0) − {−1, −2, −3, . . . }. [See also property (Γ, 4).] d3 [ Γ(p)] , ∀ p > 0. Now, find dp 3 Also, answer the following items: (a) Using the above formulae check that Γ 0 (1) = −γ. (b) Check that Γ 00 (1) =

π2 + γ2. 6

(c) Use Γ(p + 1) = pΓ(p) and (a) to prove the following derivative recurrence relations Γ0 (p + 1) = Γ(p) + pΓ0 (p),

Γ00 (p + 1) = 2Γ0 (p) + pΓ00 (p)

and so on. Now find Γ0 (2) = 1 − γ, Γ00 (2), Γ0 (3), Γ00 (3), etc. (d) Since Γ0 (1) = −γ < 0

and

Γ0 (2) = 1 − γ > 0,

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371

there is a number 1 < r < 2 such that Γ0 (r) = 0. Since Γ 00 (p) > 0, for all p > 0, Γ(r) is a local minimum value of the Gamma function. Justify why this is also global minimum. [A numerical method to approximate r is the bisection method. For   3 > 0 and so 1 < r < 1.5. See example, we can compute and find Γ0 2 also property (Γ, 7).] (e) Use (a) to prove    1 lim n − Γ = γ. n→∞ n So,  lim

n→∞

and then

1 ·Γ n

  1 =1 n

     1 1 1+ ·Γ − n = 1 − γ. n→∞ n n lim

[The last limit follows also from Γ0 (2) = 1 − γ, in (c).] [See also Problem 3.13.3, (a) and compare.] (f) Prove   1 = Γ 2 " # ∞  X √ 1 1 − = π −2 − γ + 2 2n 2n + 1 n=1 √ − π · [γ + 2 ln(2)],     5 3 and Γ0 . and then find Γ0 2 2 0

(g) Use (f ) and u = x2 to prove   Z ∞ 2 1 1 ln(x) e−x dx = Γ 0 = 4 2 0 √ − π [γ + 2 ln(2)] . 4

372

Improper Riemann Integrals

3.13.68 Project on the Real Incomplete Gamma Functions. Find and consult appropriate bibliography to study the incomplete Gamma functions, defined as follows: (a) The Upper Incomplete Gamma Function Z ∞ For s > 0 and x > 0, Γ(s, x) = ts−1 e−t dt. x

(b) The Lower Incomplete Gamma Function Z x ts−1 e−t dt. For s > 0 and x > 0, γ(s, x) = 0

(I) Notice that for x > 0, the integral in (a) exists for all s ∈ R. Also, for s > 0 and x > 0, we have the following five immediate relations: Γ(s, 0) = Γ(s), γ(s, ∞) = Γ(s), γ(s, x) + Γ(s, x) = Γ(s). ∂Γ(s, x) = −xs−1 e−x , ∂x ∂γ(s, x) = xs−1 e−x . ∂x (II) By integration by parts we find the following two recurrence relations Γ(s, x) = (s − 1)Γ(s − 1, x) + xs−1 e−x , and γ(s, x) = (s − 1)γ(s − 1, x) − xs−1 e−x . (III) Use repeated integration by parts or mathematical induction to prove that for n ≥ k ≥ 0 integers and given x ≥ 0, we have that Fx (n) :=

n X xk e−x k=0

k!

=

Γ(n, x) . n!

[This relation is useful in computing the values of the cumulative probability distribution function Fx (n), n = 0, 1, 2, . . . , of a Poisson random variable whose Poisson probability distribution with parameter x ≥ 0 is given by P (k, x) =

xk e−x , k!

k = 0, 1, 2, . . . ,

in terms of the incomplete Gamma function.]

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373

Important Note In the theory and the problems of the sections covered thus far, we have seen that when computing improper integrals, we must justify: 1. Switching limit and integral. See Theorem 3.1.1 Part I and Section 3.3 Part I, etc. 2. Switching derivative and integral. See Theorem 3.1.1 Part II and Section 3.3 Part II, etc. 3. Switching the order of iterated integrals in a double integral. See Section 3.6, etc. 4. Changing coordinates in a double integral. See Problem 3.7.5, etc. We must keep these points in mind. Otherwise, errors may ensue.

3.14

Appendix

In this appendix we will prove the identity    2n X 2m 2n (−1)n−l = m+n−l l l=0

2m 2n m n
m+n n





=

(2m)! (2n)! , m! n! (m + n)!

(3.11)

where m ≥ n ≥ 1 integers, in two ways. Such identities are often used in computing integrals, especially involving powers of sine and cosine, and values of the Gamma and Beta functions. (For example, find and study II Section 1.8.)   k k! [For simplifying expressions, we use = for 0 ≤ r ≤ k r r!(k − r)!   k integers. For any other integer r, the combination number is 0. r Also, observe that this identity is symmetrical with respect to m and n.47 ] 47 These kinds of results have to do with the theory of Hypergeometric Forms and Kummer’s Summation Formulae. These types of sums are very important in mathematics and applications, and so they have been standardized and tabulated. Also, some computer programs, such as Mathematica, can evaluate them.

374

Improper Riemann Integrals

Way 1: To give an elementary proof, we let L :=

   2n X 2m 2n (−1)n−l m+n−l l l=0

"

2n X = (−1)n−l l=0

# (2m)!(2n)! , (m + n − l)!(m − n + l)!l!(2n − l)! 2m 2n m n
m+n n




and so we want to prove L = So,

m+n 2 n

2m 2n L m n




=




 (2m)! (2n)! = . m! n! (m + n)!

[(m + n)!]2 L= (2m)!(2n)!

2n X (−1)n−l

[(m + n)!]2 = l!(2n − l)!(m + n − l)!(m − n + l)!    2n X m+n m+n (−1)n−l . l 2n − l

l=0

(3.12)

l=0

By the identity (1 − t2 )m+n = (1 + t)m+n (1 − t)m+n and the Binomial Theorem, we find m+n X

   m+n X m + n m+n X m+n k 2k k1 k2 m + n (−1) t = t (−1) tk2 = k1 k k2 k1 =0 k=0 k2 =0   m+n X m+n X m + n m + n k1 +k2 (−1)k2 t = (let k1 + k2 = k) k1 k2 k1 =0 k2 =0 " k   # 2(m+n) X X m+n k−k1 m + n (−1) tk . k1 k − k1 k=0

k1 =0

By equating the coefficients of t2n , we obtain      2n X m+n m+n m+n (−1)n = (−1)2n−k1 . n k1 2n − k1 k1 =0

(3.13)

Real Analysis Techniques

375

So, by simplifying the (−1)n and replacing k1 with l in (3.13)
  m+n 2 m+n n and using (3.12) above, we find 2m
2n
L = . Therefore, n m n

2m 2n L=

m n
m+n n

, which finishes the proof of identity (3.11).

Way 2: For those who have sufficient knowledge on the hypergeometric functions and Kummer’s formulae, but without wanting to go into a substantial exposition of this big chapter of the mathematical field of Special Functions, we would like to present a proof of identity (3.11), by using just what we need from the hypergeometric functions and Kummer’s formulae. With m ≥ n ≥ 1 and l ≥ 0 integers, we let    2m 2n n−l cl = (−1) = m+n−l l (2n)! (2m)! · . (−1)n−l (m + n − l)!(m − n + l)! l!(2n − l)! We compute and simplify the ratio (l − m − n)(l − 2n) cl+1 = ... = − . cl (l + m − n + 1)(l + 1) That is, we have the recursive formula cl+1 = −

(l − m − n)(l − 2n) · cl (l + m − n + 1)(l + 1)

with starting term (for l = 0) c0 = (−1)n

(2n)! (2m)! (2m)! · = (−1)n . (m + n)!(m − n)! 0!(2n)! (m + n)!(m − n)!

Using this recursive formula, the expression of c0 , the rising or ascending factorial of a real number x of order an integer k [defined in property (B, 8) in Subsection I 2.6.2], and the hypergeometric function48 ∞ X [a]n [b]n z n F (a, b; c; z) = · , for |z| < 1, 2 1 [c]n n! n=0 48 The hypergeometric differential equation is the differential equation : z(1 − z)u00 (z) + [c − (a + b + 1)z]u0 (z) − ab u(z) = 0, where z is a complex variable, and a, b, c are parameters with values complex numbers, in general. The hypergeometric functions are the power series solutions of this differential equations about the regular singular points 0, 1, and ∞. The hypergeometric function 2 F1 (a, b; c; z), above, is the power series solution about z = 0 and converges for all z such that |z| < 1, and also for z = −1.

376

Improper Riemann Integrals

which also converges for z = −1 (e.g., by the alternating series test, see one footnote of Example 1.7.12 or a book of calculus or mathematical analysis), we get: ∞ X

cl = c0

l=0

(−1)n

∞ X [−m − n]l [−2n]l l=0

[m − n + 1]l

·

(−1)l = l!

(2m)! · 2 F1 (−m − n, −2n; m − n + 1; −1). (m + n)!(m − n)!

We see that a = −m−n, b = −2n a negative integer, c = m−n+1 ≥ 1 integer, a − b + c = −m − n + 2n + m − n + 1 = 1, and z = −1. Then, with these parameters, the following Kummer’s formula   Γ (|b|) Γ(b − a + 1) bπ  F (a, b; c; −1) = 2 cos ·    2 1 b |b| 2 Γ −a+1 Γ 2 2 applies. (Find and examine some related bibliography. E.g., Whittaker and Watson 1927–1996, XIV, Lebedev 1972, Chapter 9, etc.)  Chapter  2n Since for l > 2n, = 0, from this formula, we finally find l    X    2n ∞ X 2m 2n 2m 2n n−l n−l (−1) = (−1) = m+n−l l m+n−l l l=0

l=0

(2m)! (−1) · 2 F1 (−m − n, −2n; m − n + 1; −1) = (m + n)!(m − n)!   (2m)! −2nπ Γ(2n)Γ(m − n + 1) (−1)n · 2 cos · = (m + n)!(m − n)! 2 Γ(n)Γ(m + 1) (2m)! (2n − 1)!(m − n)! (−1)n · 2(−1)n · = (m + n)!(m − n)! (n − 1)!m! (2m)! 2n[(2n − 1)!] (2m)! (2n)! · = , (m + n)! n[(n − 1)!m!] m! n! (m + n)! n

which finishes the proof of identity (3.11).

3.15

Problems

3.15.1 Can you prove the identity (3.11) combinatorially?

Real Analysis Techniques

377

3.15.2 Use any convenient test for power series convergence to prove that the radius of convergence of 2 F1 (a, b; c; z), defined above, is R = 1. 3.15.3 Prove the equalities   X   √ m 1 1 π Γ(m + 1) (−1)l m
= B m + 1, = = · 2 2 2l + 1 l 2 Γ m + 32 l=0 22m (m!)2 22m = · (2m + 1)! 2m + 1

1


2m = m

22m (m!)2 · = 2m + 1 (2m)!

  2m+1 2m + 1 2 · 4 . . . (2m) (−1)m X (−1)k = 2m+1 . k 1 · 3 . . . (2m + 1) 2 2m + 1 − 2k k=0

Chapter 4 Laplace Transform

In this brief overview, we are not going to develop the theory and its applications of the improper integral that is called Laplace transform, fully. There are a few of books that develop this important subject of mathematics and application, fully or nearly fully. Here, we want to give the basic ingredients of this very important subject, so that with the definitions, theory, examples, applications and problems provided anybody interested could acquire a taste and motivation on this subject, and to enable him/her to study a specialized book on his/her own, if they want or need to do so. The theory and remarks that we have included provide many points that are neglected in most books, e.g., the uniqueness theorem for the Laplace transform, examples in which Laplace transform cannot give a complete answer, the powerful use of Abel’s test in the Laplace transform, and more. So, this chapter is an important supplement to the material provided in any other book on this subject. The Laplace transform is a very powerful tool for solving applied problems, initial value and boundary value-problems of ordinary and partial differential equations, etc. Therefore, plenty of good books on this subject and extensive tables containing hundreds of evaluated Laplace transforms of very important and frequently encountered functions have been published. Among all these functions, we see many special ones. These include the Bessel1 functions, Heaviside2 step functions, shift functions, Dirac3 impulse functions, various other special functions, convolutions of functions (a very interesting, elegant and important topic), etc. Also, the inverse Laplace transform for analytic functions is a nice topic of complex analysis, in which we see the inverse Mellin4 and Bromwich5 inversion Theorem. 1 Friedrich

Wilhelm Bessel, German astronomer and mathematician, 1784–1846. Heaviside, English engineer and mathematician, 1850–1925. 3 Paul Adrien Maurice Dirac, English physicist and mathematician, 1902–1984. 4 Robert Hjalmar Mellin, Finnish mathematician, 1854–1933. 5 Thomas John l’Anson Bromwich, English mathematician, 1875–1929. 2 Oliver

DOI: 10.1201/9781003433477-4

379

380

4.1

Improper Riemann Integrals

Laplace Transform, Definitions, Theory

Definition 4.1.1 Given a nice real function y = f (x) defined on [0, ∞), we define its Laplace transform to be the following improper integral with one parameter s: Z ∞ L{f (x)}(s) = e−sx f (x) dx. 0

If y = f (x) is a nice real function defined on A ⊆ [0, ∞), then to find its Laplace transform we use this definition again, after setting f (x) = 0 for all x ∈ [0, ∞) − A. For instance, if y = f (x) is defined on [a, ∞) or (a, ∞), with 0 < a < ∞, then to find its Laplace transform we set f (x) = 0 on [0, a) or [0, a], respectively. Let us call D the set of all s ∈ R for which this improper integral converges. If D 6= ∅, then this improper integral defines a real-valued function of s on the set D. The parameter s is the independent variable of this newly obtained function, which we write by L{f (x)}(s), and we call it the Laplace transform of f (x). If D = ∅, then the Laplace transform of f (x) does not exist. Since the Laplace transform of a function is an improper Riemann integral with one parameter, we can use and apply the theory and the results about improper integrals with parameters along with all the different related techniques when we study and compute Laplace transforms. Theorem 3.1.1 and the Theorems exposed in Section 3.3 can be used with regard to questions on the continuity and differentiability of the Laplace transform. In this brief exposition, we mostly deal with existence and computation questions. To this end, we continue with some definitions, results and problems. First, we observe that if A = [a, ∞) or A = (a, ∞), where a ≥ 0, and f (x) is defined on A and is absolutely integrable, i.e., Z |f (x)|dx < ∞, A

then the Laplace transform of f (x) exists (converges absolutely) ∀ s ≥ 0. In this case L{f (x)}(s) is differentiable. This follows immediately from the Absolute Convergence Test, 1.7.7, since ∀ s ≥ 0 and ∀ x ∈ A it holds: |e−sx f (x)| ≤ |f (x)| and Theorem 3.1.1, (II). (The Cauchy Test, 1.7.11, can also be used.) Concerning the existence of the Laplace transform for larger classes of functions along with some very important properties, see Problem 4.2.7 and provide its solution.

Laplace Transform

381

Next, we examine the set of functions of exponential order in [0, ∞) or (0, ∞) and conditions under which their Laplace transforms exist. They form a very large class of functions, sufficient for most needs of applications. Hence, we have: Definition 4.1.2 A real function y = f (x) on [0, ∞) or (0, ∞) is defined to be of exponential order u if there are constants u ≥ 0, M ≥ 0 and A ≥ 0, such that |f (x)| ≤ M eux , for all x ∈ [A, ∞). Example 4.1.1 All bounded functions on [0, ∞) are of exponential order. M is the bound of the |f (x)| on [0, ∞) (so, A = 0) and u = 0 (the smallest u). In particular all constant functions are of exponential order. If f , g : [0, ∞) −→ R are of exponential order, then so are f ± g and f · g. (Problem 4.2.3.)6 All powers xn with n ∈ N, and then by the previous claim all polynomials P (x) with x ∈ [0, ∞) are of exponential order. This is so because xn = en ln(x) ≤ e−n enx for all x ∈ [0, ∞) [since ln(x) ≤ x − 1 for all x > 0]. Therefore, M = e−n and u = n, etc. Also, all functions C ekx on [0, ∞) with C and k any real constants are of exponential order. Take M = |C| and u = |k|. The functions exp(xα ) for all real α > 1 and exp(ex ) are not of exponential order as it can easily be verified, etc. (See also Problem 4.2.1.) N Now consider a function f (x) of exponential order u in [0, ∞) with constants M ≥ 0 and A ≥ 0 as in the above definition. We have Z ∞ Z A Z ∞ e−sx f (x) dx = e−sx f (x) dx + e−sx f (x) dx. 0

0

A

Z

∞

But for all s > u, we get that the second integral verges absolutely because: Z ∞ Z −sx e |f (x)| dx ≤ A

e−sx f (x) dx con-

A ∞

e−sx M eux dx =

A

e(u−s)x M u−s 

∞ =M A

e(u−s)A < ∞. s−u

Hence, we conclude the following Result: 6 By

this property the set of functions of exponential order on [0, ∞) is an algebra.

382

Improper Riemann Integrals

“For a function f (x) of exponential order u in [0, ∞) with constants M ≥ 0 and A ≥ 0, as in Definition 4.1.2, Z ∞ ∃ µ > 0 such that L{f (x)}(s) = e−sx f (x) dx exists, ∀ s > µ, 0

iff ⇐⇒

Z ∃ v > 0 such that

A

e−sx f (x) dx

exists,

∀ s > v.

(4.1)

0

In such a case, we can choose a constant µ > max{u, v}.” (Compare this result with Example 4.1.2 below.) Since the interval [0, A] is closed and bounded, we observe that these integrals exist when, for instance, f (x) is continuous, or bounded, or absolutely integrable, etc., on [0, A]. (See also Problems 4.2.7 and 4.2.8 and solve them.) Remark: Similarly: If a real function y = f (x) is defined on [0, ∞) [or (0, ∞)] and its restriction on [a, ∞), with a > 0, is absolutely integrable, then for some c > 0, Z a L {f (x)}(s), exists ∀ s ≥ c, iff e−cx f (x) dx exists. 0

Example 4.1.2 The function 1 with x ∈ (0, ∞) x is of exponential order, since |f (x)| ≤ x for all x ∈ [1, ∞). (So, choose: M = 1, u = 1 and A = 1, for Definition 4.1.2.) But,   Z ∞ 1 1 (s) = e−sx dx does not exist, L x x 0 since, on account of the singularity at x = 0, we have Z 1 1 ∀ s ∈ R, e−sx dx = ∞. x 0 f (x) =

If we now choose a constant B > 0 (e.g., B = 1) and define   0, if 0 ≤ x < B,   g(x) =   1, if B ≤ x < ∞, x then, even though g(x) is not absolutely integrable in (0, ∞) (check it), its Laplace transform Z ∞ Z ∞ 1 L{g(x)}(s) = e−sx g(x) dx = e−sx dx x 0 B exists (in fact, converges absolutely) for all s > 0. (Check it!)

N

Laplace Transform

383

The Laplace transform, as an integral operator, is a linear operator in the set of functions for which it exists. That is: If y = f (x) and y = h(x) on [0, ∞) [or (0, ∞)] are any nice real functions whose Laplace transforms exist for all s > k, for some constant k ≥ 0, and if c ∈ R is a constant, then the Laplace transform of each of the following functions (f + h)(x) = f (x) + h(x) and (cf )(x) = cf (x) exists, for all s > k, and satisfies the two linearity properties: (a) L{(f + h)(x)}(s) = L{f (x) + h(x)}(s) = L{f (x)}(s) + L{h(x)}(s), (b) L{(cf )(x)}(s) = L{cf (x)}(s) = c L{f (x)}(s). (The proof is straightforward. See Problem 4.2.5.) The following theorem is of fundamental importance. Theorem 4.1.1 Consider a continuous real function y = f (x) defined on [0, ∞) or (0, Z ∞), and suppose that there is a constant c ≥ 0 such that ∞

f (x)e−cx dx exists. Then we have:

L{f (x)}(c) = 0

(a) For all s ≥ c, the Laplace transform Z ∞ L{f (x)}(s) = f (x)e−sx dx

exists.

0

Z

∞

f (x)e−sx dx = 0.

(b) lim L{f (x)}(s) = lim s→∞

s→∞

0 ∞

Z

f (x)e−sx dx is

(c) On the infinite interval [c, ∞), L{f (x)}(s) = 0

continuous and on account of (b), it is bounded, uniformly continuous, and the convergence of the improper integral is uniform. Z (d) On the infinite interval (c, ∞), L{f (x)}(s) = continuously differentiable.

∞

f (x)e−sx dx is

0

Proof (a) We apply Abel’s Test for convergence of improper integrals, The−cx −rx orem 1.7.12, to f (x)e with x ∈ [0, ∞), or x ∈ (0, ∞), Z , e x

f (t)e−ct dt. All the conditions of the test

and r ≥ 0 and h(x) = 0

are satisfied. h(x) is differentiable and there is M ≥ 0 such that

384

Improper Riemann Integrals |h(x)| ≤ M . As in the proof os Abel’s Test, we apply integration by parts and we find that, Z ∞ Z ∞   −(c+r)x ∀ r > 0, f (x)e f (x)e−cx e−rx dx = dx = 0 0 Z ∞ Z ∞   ∞ −rx −rx h(x)e−rx dx = +r e d[h(x)] = h(x)e 0 0 0 Z ∞ Z ∞ 0−0+r h(x)e−rx dx = r h(x)e−rx dx. 0

0

But for any r > 0, the last integral converges absolutely since Z ∞ Z ∞ h(x)e−rx dx = r |h(x)|re−rx dx ≤ 0 0 Z ∞ M re−rx dx = M · 1 = M, 0

and so it exists. Therefore, the first integral also exists (regardless if it converges absolutely or conditionally). That is with s = c+r > c and also with s = c (by hypothesis), we find that Z ∞ ∀ s ≥ c, L{f (x)}(s) = f (x)e−sx dx exists. 0

(b) For any x ≥ 0, we have that max re−rx = e−x and r∈[0,∞) ∞

Z 0

|h(x)|re−rx dx ≤

0

∞

Z

∞

Z

h(x)re−rx dx =

−x

|h(x)|e

∞

Z

e−x dx = M · 1 = M.

dx ≤ M ·

0

0

Next, we have h(0) = 0 and ∀ x ∈ (0, ∞), lim h(x)re−rx ≤ M lim re−rx = M · 0 = 0. r→∞

r→∞

Therefore, by Theorem 3.3.11, we find Z ∞ Z ∞ −(c+r)x ≤ f (x)e dx |h(x)re−rx |dx −→ 0, 0

as r −→ ∞.

0

Hence, with s = c + r, Z lim L{f (x)}(s) = lim

s→∞

s→∞

0

∞

f (x)e−sx dx = 0.

Laplace Transform

385

(c) Let F (s) = L{f (x)}(s), with s ≥ c. For any s1 , s2 ≥ c, we have Z ∞   f (x) e−s1 x − e−s2 x dx = |F (s1 ) − F (s2 )| = 0 Z ∞ Z ∞ −s x −s x 1 2 ≤ |h(x)|[s1 e−s1 x − s2 e−s2 x ]dx ≤ h(x)[s e − s e ]dx 1 2 0 0 Z ∞ −s x s1 e 1 − s2 e−s2 x dx. M 0

Also, for any s1 ≥ c and s2 ≥ c, we have Z ∞ Z ∞ −s x  −s1 x  s1 e 1 − s2 e−s2 x dx ≤ s1 e + s2 e−s2 x dx < 0 0 Z ∞  −x  e + e−x dx = 2, 0

and lim |s1 e−s1 x − s2 e−s2 x | = 0. Then by Theorems 3.1.1 or s1 →s2

3.3.11, we find Z

∞

lim

s1 →s2 ∞

Z

0

−s x s1 e 1 − s2 e−s2 x dx =

0

lim s1 e−s1 x − s2 e−s2 x dx = 0.

s1 →s2

Finally, lim |F (s1 ) − F (s2 )| = 0

or

s1 →s2

lim F (s1 ) = F (s2 ) and

s1 →s2

so F (s) is continuous on [c, ∞). This result along with (b) imply that F (s) = L{f (x)}(s) is bounded and uniformly continuous and the convergence of the improper integral is uniform (by the results of the basic mathematical analysis). (d) For any u > c and s > c, by the mean value theorem we have e−ux − e−sx = (u − s)xe−vx with v between u and s. Then,

Z

∞

F (u) − F (s) = Z ∞
−ux −sx f (x) e −e dx = f (x)(u − s)(−x)e−vx dx.

0

0

So, F (u) − F (s) = u−s

Z

∞

f (x)(−x)e−vx dx.

0

Applying again Abel’s Test with the same h(x) and xe−vx , we

386

Improper Riemann Integrals prove that this integral exists and is continuous in v. Then, taking the limits as u −→ s, we get v −→ s and we can switch limit and integral to find that Z ∞ d F (s) = F 0 (s) = − f (x)xe−sx dx, ds 0 exists and it is continuous in (c, ∞). 

Remarks: (a) In Problems 4.2.7 and 4.2.8, we prove similar results in an easier way due to the extra conditions posed. (b) The Theorem is also true if the function y = f (x) is piecewise continuous with finitely or countably many discontinuities. [See Theorem 1.7.12, Remark (d).] It is rather intuitively obvious that: if y = f (x) is a real nice function defined on [0, ∞) [or (0, ∞)] and its Laplace transform exists, then its Laplace transform is unique. That is, the Laplace transform is a one-to-one (injective) linear operator on the set of continuous functions for which it exists. This is proven in the following theorem of Lerch.7 The theorem simply requires the Laplace transform to exist regardless of what the class of the underlying function is. (Exponential order, L1 , Lp , p > 1, etc.) Theorem 4.1.2 (Lerch) Let f : [0, ∞) −→ R be a continuous function, such that for some constant c ≥ 0, the Laplace transform of f is zero on [c, ∞), i.e., L{f (x)}(s) = 0, ∀ s ≥ c. Then, f ≡ 0 on [0, ∞). Equivalently, by the linearity of the Laplace transform, we have: If L{f1 (x)} = L{f2 (x)}, then f1 = f2 . Or, if f1 6= f2 , then L{f1 (x)} 6= L{f2 (x)}. Proof By hypothesis, for s = c + 1(≥ max{1, c}) we have Z ∞ L{f (x)}(c + 1) = e−(c+1)x f (x) dx = 0. 0

Then for any s > c + 1 ≥ 1, we let s0 = s − (c + 1) > 0 and we have that, ∀ s0 > 0, Z ∞ Z ∞ h i 0 −sx L{f (x)}(s) = e f (x) dx = e−s x e−(c+1)x f (x) dx = 0. 0 7 Maty´ aˇs

0

Lerch, Czech mathematician, 1860–1922.

Laplace Transform

387

We will prove that e−(c+1)x f (x) = 0 on [0, ∞) and so f (x) = 0 on [0, ∞). For convenience, we let g(x) := e−(c+1)x f (x) and we will prove that g ≡ 0 on [0, ∞). Z ∞

g(x) dx ∈ R exists (here it will be

Since g is continuous and 0

proven to be zero), the function Z

x

h(x) :=

g(t) dt 0

is differentiable and bounded. So, h(0) = 0 and there is a constant M ≥ 0, such that, Z x ∀ x ≥ 0, |h(x)| = g(t) dt ≤ M. 0

By the Result that follows Example 4.1.2 (and Abel’s Test, Theorem 1.7.12), we have that the L{g(x)}(s0 ) exists ∀ s0 ≥ 0 (and it is continuous), and by hypothesis, it is zero. That is, Z ∞ 0 0 0 ∀ s ≥ 0, L{g(x)}(s ) := e−s x g(x) dx = 0. 0

Then, for any 0 < θ < 1 constant, we make the change of variables e−x ⇐⇒ x = − ln(θu) and we get u= θ Z θ1 Z θ1 0 0 0 0 θs us −1 g[− ln(θu)] du = θs us −1 g[− ln(θu)] du = 0 0

0

Z and so

1 θ

0

us −1 g[− ln(θu)] du =

0 1

Z

0

us −1 g[− ln(θu)] du +

0

Z

1 θ

0

us −1 g[− ln(θu)] du = 0.

1

1 < ∞, 0 < x = − ln(θu) < ∞ and θ Z 1 Z θ1 0 0 us −1 g[− ln(θu)] du = − us −1 g[− ln(θu)] du.

Thus, we have 1
0, Z 1 Z 1 v=ln(u) s0 −1 s0 −1 u g[− ln(θ) − ln(u)] du = u g[− ln(θu)] du = 0 Z ∞ 0Z ∞ 0 −s0 v −s v e d{h[− ln(θ) + v]} = e g[− ln(θ) + v] dv = 0 0h iv=∞ Z ∞ −s0 v −s0 v e h[− ln(θ) + v] − h[− ln(θ) + v] d(e ) = v=0 0 Z ∞ −s0 v 0 − h[− ln(θ)] − h[− ln(θ) + v] d(e ) ≤ 0 Z ∞ 0 |h[− ln(θ)]| + |h[− ln(θ) + v]| d(e−s v ) ≤ 0 Z ∞ −s0 v M +M d(e ) = M + M · [0 − (−1)] = 2M. 0

Hence 0

∀ s > 0,

Z 1 θ 0 s −1 u g[− ln(θu)] du ≤ 2M. 1

That is, the absolute value of this integral is uniformly bounded by the constant 2M ≥ 0, ∀ s0 > 0. Therefore, by the Theorem of Phragm´ en, Problem 3.5.24, (c), we get   1 . g[− ln(θu)] = 0 for u ∈ 1, θ But 0 ≤ x = − ln(θu) < ∞ and

0 0 constant and s > 0 is the variable.

N

Example 4.1.4 Assume that F (s) := L{f (x)}(s)

and G(s) := L{g(x)}(s)

are the Laplace transforms of two real functions f (x) and g(x) on [0, ∞), for s > (≥)a ≥ 0, where a is constant.

Laplace Transform

391

Then, we multiply to get   Z ∞ Z ∞ −sv −su e g(v) dv = e f (u) du · F (s) · G(s) = 0 Z ∞ Z ∞0 e−s(u+v) f (u) du. g(v) dv 0

0

Now we let u + v = t to get Z ∞ Z F (s) · G(s) = g(v) dv 0

∞

e−st f (t − v) dt.

v

We switch the order of integration (use Figure 4.1 with analogous labels) and obtain Z t  Z ∞ −st F (s) · G(s) = e f (t − v)g(v) dv dt. 0

0

We denote the particular integral that has appeared here, by Z t (f ∗ g)(t) := f (t − v)g(v) dv, ∀ t ≥ 0. 0

This integral is a new function on [0, ∞) and is called convolution of the functions f and g in the context of the Laplace transform, that is, functions defined on [0, ∞). [See also and compare with the definition of convolution in the context of the Fourier8 transform, defined in Subsection II 1.7.6, property (7).] Hence, we have proved the convolution rule for the Laplace transform L{(f ∗ g)(x)}(s) = L{f (x)}(s) · L{g(x)}(s). That is, the Laplace transform of the convolution of two functions (as defined above) is the product of their Laplace transforms. This rule has many applications. For example, an application of this rule is the following: If we let f ≡ 1, then we obtain Z x  Z x  ∀ x ≥ 0, L g(t) dt (s) = L 1 · g(t) dt (s) = 0

0

L {(1 ∗ g)(x)} (s) = L{1}(s) · L{g(x)}(s) =

1 · L{g(x)}(s). s

This is Rule (6.) of the table in Problem 4.2.21. 8 Jean

Baptiste Joseph Fourier, French mathematician, 1768-1830.

N

392

Improper Riemann Integrals

Example 4.1.5 Some properties of the convolution are the following: (a) It is commutative, i.e., f ∗ g = g ∗ f . This is immediately obtained by the change of variables t − v = u, through which we get Z t Z t f (t − v)g(v) dv = f (u)g(t − u) du. 0

0

(b) It is straightforward that a(f ∗ g) = (af ) ∗ g = f ∗ (ag), ∀ a ∈ R. In particular, 0 ∗ g = 0 = f ∗ 0. (c) The convolution is linear with respect to each function position. I.e., for any real constants a and b, we have: (af1 + bf2 ) ∗ g = a(f1 ∗ g) + b(f2 ∗ g), and f ∗ (ag1 + bg2 ) = a(f ∗ g1 ) + b(f ∗ g2 ). The proof is immediate from the definition. (d) The convolution is associative, i.e., (f ∗ g) ∗ h = f ∗ (g ∗ h). The proof goes as follows: Z t Z [(f ∗ g) ∗ h](t) = 0

t−r

 f (t − r − s)g(s)ds h(r) dr.

0

We let s + r = w, and we have  Z t Z t [(f ∗ g) ∗ h](t) = f (t − w)g(w − r)dw h(r) dr. 0

r

We switch the order of integration, and we obtain associativity Z w  Z t [(f ∗g)∗h](t) = f (t−w) g(w − r)h(r) dr dw = [f ∗(g∗h)](t). 0

0

(e) Notice that in general x

Z (f ∗ 1)(x) =

f (x − w) dw 6= f (x). 0

Z For example,

(sin ∗1)(x) =

x

sin(x − w) · 1 dw = [cos(x − w)]x0 =

0

cos(0) − cos(x) = 1 − cos(x) 6= sin(x).

Laplace Transform

393

r

r=w

t

Triangle of double integration O t

w

FIGURE 4.2: For switching integration in Example 4.1.5, (d)

Also, (f ∗ f )(x) may not be positive or non-negative. For example: with f (x) = sin(x), we find (f ∗ f )(x) = (sin ∗ sin)(x) =

sin(x) − x cos(x) . 2

(f ) If f (x) and g(x) are absolutely integrable and one of them continuous, then (f ∗ g)(x) is continuous. The proof follows from the definition of continuity and Theorem 3.1.1 Part (I) or Theorem 3.3.11 and its remark. For the derivative of the convolution, see Problem 4.2.31. (g) Suppose f (x) and g(x) are continuous real functions on [0, ∞). Then (f ∗ g)(x) = 0 for all x ∈ [0, ∞) if and only if f (x) = 0 or g(x) = 0 for all x ∈ [0, ∞). I.e., If f and g are real continuous functions on [0, ∞), then: f ∗ g ≡ 0 ⇐⇒ f ≡ 0 or g ≡ 0.

394

Improper Riemann Integrals This result is a famous Theorem proved by E. C. Titchmarsh9 in 1926. The proof on the set of real continuous (or continuous almost everywhere) functions on [0, ∞) is lengthy and technical. [See also Problem 4.5.20 and compare with Problem 4.5.21. Additionally, see Problem 4.2.3 and Subsection II 1.7.6, (7), (h).] N

Example 4.1.6 We consider the set of functions C = {f : [0, ∞) −→ R continuous}. In the previous Example we have seen that the operation convolution is a closed operation and behaves like a nice algebraic multiplication in this set. The only property left is the unit or neutral element for this operation. As we have seen above this is not the constant function f (x) ≡ 1. Here we explain that the unit or neutral element of convolution is an object that we call the Dirac delta function. Let us see how we arrive at its definition in the context of the Laplace transform and the convolution we study here. For every  > 0, we consider the step function  1   if 0 ≤ x ≤ , , D (x) =    0, if  < x, and any function f : [0, ∞) −→ R which is continuous in [0, α), for some α > 0. Then we have: (a) pw

lim D (x) =

→0

  ∞,

if x = 0,

  0,

if x 6= 0.

9 Edward Charles “Ted” Titchmarsh, English mathematician, 1899–1963. In fact, Titchmarsh proved the following more general result: If f (x) and g(x) are real integrable functions of the real variable x ∈ R, such that Z x (f ∗g)(x) = f (v)g(x−v) dv = 0, almost everywhere in the interval [0, κ], (κ > 0), 0

then there exist λ ≥ 0 and µ ≥ 0 satisfying λ + µ ≥ κ, such that f (x) = 0, almost everywhere in [0, λ], and g(x) = 0, almost everywhere in [0, µ]. For κ = ∞, we achieve the above-stated result on [0, ∞).

Laplace Transform (b) For every  > 0 Z ∞ D (x) dx = 1

395

Z and so

0

→0

∞

D (x) dx = 1.

lim

0

(c) For 0 <  < α, we use the Mean Value Theorem for integrals to get Z  Z ∞ 1 1 f (x) dx =  f (x ) = f (x ) , D (x)f (x) dx =  0  0 for some 0 ≤ x ≤ . Then, by the continuity of f (x) in [0, α), we obtain   Z ∞ lim+ D (x)f (x) dx = f lim+ x = f (0). →0

→0

0

So, (a), (b) and (c) suggest to define the symbol δ(x), ∀ x ∈ R, by  6= 0, (= ∞?), ?, if x = 0,  δ(x) =   0, if x 6= 0, and by stipulating that it satisfies the following two properties: R∞ (1) 0 δ(x) dx = 1. (2) For every f : [0, ∞) −→ R continuous in [0, α), for some α > 0, Z ∞

f (x)δ(x) dx = f (0). 0

We use this symbol by means of these two properties! We call δ(x) Dirac delta function or unit impulse function, even though, as we saw above, it is not a function in the classical meaning of the word but rather an operator derived from function processes.10 Now, for any a > 0, we consider the shift δ(x − a) of δ(x). Since δ(x) = 0 on [−a, 0), we obtain Z ∞ Z ∞ ∀ a > 0, f (x)δ(x − a) dx = f (u + a)δ(u) du = 0 −a Z ∞ f (u + a)δ(u) du = f (0 + a) = f (a). 0 10 The Dirac delta function is not a function in the classical sense. Here it is used as an operator on functions via integrals. Advanced mathematical theories show that the Dirac delta function can be viewed as a point-measure or as a generalized function or distribution. Then compositions of the delta function with other functions and its derivatives are considered in the sense that the theories develop.

396

Improper Riemann Integrals

Also, for all x ∈ [0, ∞) and any f (x) ∈ C, since δ(x) = 0 on (x, ∞), we obtain Z x Z ∞ (f ∗δ)(x) = f (x−u)δ(u) du = f (x−u)δ(u) du = f (x−0) = f (x). 0

0

Therefore, f ∗ δ = f = δ ∗ f , that is, the unit element for the commutative operation of convolution in the set C is the Dirac delta function. From these properties, we can easily find the Laplace transforms of δ(x) and its shifts δ(x − a) for a > 0. Namely: Z ∞ L{δ(x)}(s) = e−sx δ(x) dx = e−0s = 1 0

and

Z L{δ(x − a)}(s) =

∞

e−sx δ(x − a) dx = e−as .

0

N Example 4.1.7 Prove that √ L{sin( x)}(s) =

√

−1

πe 4s 3

2s 2

.

We use the power series of sin(x) to obtain √ 1 ∞ ∞ X X √ (−1)n ( x)2n+1 (−1)n xn+ 2 sin( x) = = . (2n + 1)! (2n + 1)! n=0 n=0 We use Rule (6.) of the table in Problem 4.2.14 and the result of Problem 3.13.14, (a) to find
∞ X √ (−1)n Γ n + 1 + 12 L{sin( x)}(s) = = 1 (2n + 1)! sn+1+ 2 n=0  n √ √ X ∞ ∞ π 1 X (−1)n [2(n + 1)]! π (−1)n 2(n + 1) 1 = = 3 3 4s s 2 n=0 (2n + 1)! 4n+1 sn (n + 1)! s 2 n=0 n!4(n + 1)  n √ X √ −1 ∞ (−1)n 1 π π e 4s = . 3 3 n! 4s 2s 2 n=0 2s 2 √ We can do analogous work with cos( x). We use the power series of cos(x) to obtain √ ∞ ∞ X X √ (−1)n ( x)2n (−1)n xn cos( x) = = . (2n)! (2n)! n=0 n=0

Laplace Transform

397

Then by Rule (5.) of the table in Problem 4.2.14, we obtain √ L{cos( x)}(s) = " # ∞ ∞ X X 1 (−1)n n! 1 (−1)n = 1+ . (2n)! sn+1 s (n + 1)(n + 2) . . . (2n) sn n=0 n=1 N Example 4.1.8 Let a ≥ 0 be a constant and f (x) a real function defined on [0, ∞), or (0, ∞), whose Laplace transform exists. Then Z ∞ Z ∞ u=x+a e−s(u−a) f (u) du = e−sx f (x + a) dx = L{f (x + a)}(s) = a 0 Z ∞  Z ∞ Z a as −su as −su e e f (u) du = e e f (u) du − e−su f (u) du = a 0 0 Z a as as −su e L{f (u)}(s) − e e f (u) du. 0

So, a ≥ 0, we have the Rule ∀ a ≥ 0,

L{f (x + a)}(s) = eas L{f (u)}(s) − eas

Z

a

e−su f (u) du.

0

(See also Problems 4.2.22, 4.2.23 and II 1.7.162, solve them and compare them with this rule.) N Application: We can use known Laplace transforms to compute improper integrals efficiently. We analyze four examples below. (a) We consider sinh(x) = Z

ex − e−x and any k > 0 constant. Then 2

∞

e−kx sinh(x) sin(x) dx = 0 Z Z 1 ∞ −kx x 1 ∞ −kx −x e e sin(x) dx − e e sin(x) dx = 2 0 2 0 1 1 L{ex sin(x)}(k) − L{e−x sin(x)}(k) = 2 2 [use (9.) of the table in Problem 4.2.14]   1 1 1 2k . − = 4 2 2 2 2 2 (k − 1) + 1 (k + 1) + 1 k +4 Z ∞ 6 For example, e−3x sinh(x) sin(x) dx = . 85 0

398

Improper Riemann Integrals

(b) Working similarly we obtain Z ∞ sinh(x) sin(x) e−kx dx = x 0 Z Z 1 ∞ −kx e−x sin(x) 1 ∞ −kx ex sin(x) e e dx − dx = 2 0 x 2 0 x [use (9.) of the table in Problem 4.2.21] Z ∞ Z 1 ∞ 1 x L{e sin(x)}(u) du − L{e−x sin(x)}(u) du = 2 k 2 k Z Z 1 ∞ 1 1 ∞ 1 du − du = 2 k (u − 1)2 + 12 2 k (u + 1)2 + 12 1 1 [ arctan(v) ]k+1 k−1 = [arctan(k + 1) − arctan(k − 1)] 2 2   1 2 = arctan . 2 k2 Z ∞ √ 1 π sinh(x) sin(x) For example, e− 2x dx = arctan(1) = . x 2 8 0 (c) By Problem 1.6.15 or rule (4.) of the table in Problem 4.2.14, we have that Z ∞ s ∀ β ∈ R, L[cos(βx)](s) = e−sx cos(βx) dx = 2 , ∀s > 0. s + β2 0 s |s=0 = 0, the integral We must remark that even though 2 s + β2 Z ∞ Z ∞ e−0x cos(βx) dx = cos(βx) dx does not exist. 0

0

This phenomenon occurs in many Laplace transforms. Now by rule (8.) of the table in Problem 4.2.21, we get that, Z ∞ ∀s > 0, L[x cos(βx)](s) = e−sx x cos(βx) dx =  0  d d s s2 − β 2 = − L[cos(βx)](s) = − 2. 2 2 ds ds s + β (s2 + β 2 ) Z ∞ s2 − β 2 1 Again | = − , but x cos(βx) dx does not exist. 2 s=0 β2 (s2 + β 2 ) 0 Now we find that Z ∀ α > 0, and ∀ β ∈ R, 0

∞

e−αx x cos(βx) dx =

α2 − β 2 . (α2 + β 2 )2

Laplace Transform Z ∞ Similarly: ∀ α > 0, and ∀ β ∈ R, e−αx x sin(βx) dx = 0

399 2αβ . (α2 + β 2 )2

(See also Problem 4.2.35.) (d) For a > 0 and b > 0, by rule (2.) of the table in Problem 4.2.14, we have   L e−bx − e−ax (s) =

1 1 − , s+b s+a

∀ s > max{−a, −b}.

Then, by rule (9.) of the table in Problem 4.2.21, we obtain   −bx e − e−ax (s) = ∀ s > max{−a, −b}, L x Z ∞ Z ∞   e−bx − e−ax dx = e−sx L e−bx − e−ax (u) du = x s    Z0 ∞  1 1 s+a − du = (take limits) = ln . u+b u+a s+b s Plugging s = 0 both sides and so we get the Frullani integral Z ∞exists a e−bx − e−ax dx = ln . of Example 3.8.1 x b 0 Note: To find the Laplace transform of a given function, we use: 1. The definition, for primitive cases. 2. The linearity Properties (a) and (b) above. 3. Already known Laplace transforms of other functions. 4. The rules in the table of Problem 4.2.14. 5. The rules in the table of Problem 4.2.21. 6. Other rules not listed in the two tables above 7. Limit processes with known results, especially when parameters are involved. 8. Double integration and switching order. 9. Power series.

400

Improper Riemann Integrals

10. Tables of Laplace transforms with or without adjustments, if we can trust them, of course. (Sometimes, there are human errors and/or typos in tables, and so we may need to check the readily available answers, especially when we use them for crucial applications.) 11. Special computer packages, if we can trust them, of course. (Keep in mind that human errors are always possible, even with computers and/or computer packages. Therefore, we must check the answers provided carefully.)

4.2

Problems

4.2.1 Provide all the missing details in Example 4.1.1. 4.2.2 Prove that f (x) = ln(x), where x > 0 is of exponential order. 4.2.3 If f , g : [0, ∞) −→ R are functions of exponential order, then so are f ± g, f · g and f ∗ g. Hence, the set of functions of exponential order is an algebra under + and · and an integral domain under + and ∗. 4.2.4 2

(a) Show that f (x) = ex on [0, ∞), is not of exponential order and does not have Laplace transform. n

(b) Show that ∀ n ∈ N, the function e−x is of exponential order and its Laplace transform exists on [0, ∞). 4.2.5 Provided that all integrals involved exist, prove the two linearity properties (a) and (b) of the Laplace transform. 4.2.6 (a) Construct an example of a continuous function y = f (x) defined on [0, ∞), such that it is absolutely integrable, i.e., Z ∞

|f (x)|dx < ∞, but not of exponential order. 0

(b) Give an example of a continuous function y = f (x) defined on Z ∞ [0, ∞) of exponential order and such that |f (x)|dx = ∞. 0

Laplace Transform

401

(c) Give an example of a continuous function y = f (x) defined on Z ∞ |f (x)|dx < ∞. [0, ∞) of exponential order and such that 0

(d) Give an example of a continuous function y = f (x) defined on [0, ∞) which is neither of exponential order inte Z x nor absolutely f (t)dt ≤ M for all grable and there is M > 0, such that 0 x ∈ [0, ∞). 4.2.7 Consider y = f (x) “nice” function defined on [a, ∞), where a ≥ 0. {We assume that f (x) = 0 on the interval [0, a) whenever we write the integral of the Laplace transform of such an f .} Prove: (a) If kf k1 < ∞, the Laplace transform of f (x) converges absolutely (and therefore it exists) ∀ s ≥ 0. (b) If kf kp < ∞, with 1 < p, the Laplace transform of f (x) converges absolutely (and therefore it exists) ∀ s > 0 (s = 0 may not be included in this case). (c) In both cases (a) and (b) prove that lim L{f (x)}(s) = 0, and the s→∞ convergence is uniform. That is, ∀  > 0, ∃ K > 0 : { ∀ s ∈ R, [ s > K =⇒ |L{f (x)}(s)| < ] .} Hence, if under the posed conditions L{f (x)}(s) = c constant, then c = 0 and so f ≡ 0, a.e. (d) Prove the results claimed in (a), (b) and (c) for s ≥ u, for some u ≥ 0, if we respectively replace kf k1 < ∞ and kf kp < ∞ (1 < p) with the conditions Z ∞ Z ∞ −ux e |f (x)|dx < ∞, and e−ux |f (x)|p dx < ∞. a

a

[Hint: You may use the Lebesgue Dominated Convergence Theorem, 3.3.11, H¨ older’s inequality, project Problem 3.13.65 and the Cauchy criterion for convergence as it was done in the Cauchy Test, Theorem 1.7.11.]

402

Improper Riemann Integrals

4.2.8 (a) If f (x) is a “nice” function of exponential order in [0, ∞) that satisfies Condition 4.1. (See also Example 4.1.2.) Prove that L{f (x)}(s) exists, in fact converges absolutely, on the interval (q, ∞), where q is the infimum of all possible u’s> 0 that may be used in Definition 4.1.2 and so q ≥ 0. (For some functions the Laplace transform exists even for q = 0. Give an example for either case.) (b) Then also prove: lim L{f (x)}(s) = 0 and the convergence is unis→∞ form. That is, ∀  > 0, ∃ K > 0 : {∀ s ∈ R, [ s > K =⇒ |L{f (x)}(s)| < ] .} Hence, if under the posed conditions L{f (x)}(s) = c constant, then c = 0 and so f ≡ 0 a.e. 4.2.9 Look at the definition of the Gamma function Γ(p). Notice that it may be considered as the value of the Laplace transform of a certain function for a special choice of s. Identify this function and the choice of s. 4.2.10 Give Laplace transform interpretations to the following five improper integrals with parameters (found in Problems 1.6.13, 1.6.15, 1.6.17, 1.6.18 and 3.7.11). That is, find the functions and their corresponding Laplace transforms hidden in these integrals. Justify your answers. (a) If α > 0 and β ∈ R real constants, then Z ∞ e−αx sin(βx) dx = 0

(b) If α > 0 and β ∈ R real constants, then Z ∞ e−αx cos(βx) dx = 0

α2

β . + β2

α . α2 + β 2

(c) If α > 0 and β ∈ R real constants, then   Z ∞ β −αx sin (βx) e dx = arctan . x α 0 (d) If α > 0 and −α < β < α constants, then Z ∞ β e−αx sinh(βx) dx = 2 . α − β2 0

Laplace Transform

403

(e) If α > 0 and −α < β < α constants, then Z ∞ α e−αx cosh(βx) dx = 2 . α − β2 0 4.2.11 The following results can be used to find the Laplace transforms of some functions. In each one, find the function and its corresponding Laplace transform. Explain your answers. (a) Z

∞

I(β) =

e

−αx2

0

1 cos(βx) dx = 2

r

π −β2 e 4α α

for any −∞ < β < ∞ and any α > 0 (found in Example 3.1.14). (b) Z 0

∞

2

e−αx dx =

1 2

r

π , α

where α > 0 (found in Problem 2.3.11). [Hint: Let x2 = t and then observe and interpret what you get.] 4.2.12 The three results listed below, found in Problem 3.9.23, can be used to find the Laplace transforms of some functions. In each one, find the function and its corresponding Laplace transform. Explain your answers. For β ≥ 0 and a, b real constants, we have found:  2  Z ∞ 1 β + b2 −βx cos(ax) − cos(bx) (a) e dx = ln . x 2 β 2 + a2 0     Z ∞ a sin(ax) + sin(bx) b dx = arctan (b) e−βx + arctan . x β β 0     Z ∞ a b sin(ax) − sin(bx) dx = arctan − arctan . (c) e−βx x β β 0 4.2.13 Prove

√ s n o π s2 −x2 · e 4 · erfc . L e (s) = 2 2 (See also Problem 4.2.22.)

4.2.14 Verify the results of the following table:

404

Improper Riemann Integrals

Function h(x), x ∈ [0, ∞)

Laplace transform L {h(x)} (s)

1.

a (=constant)

a , s

2.

eax

1 , s−a

3.

sin(bx)

b , s2 + b2

s>0

4.

cos(bx)

s , + b2

s>0

5.

6.

xn ,

n = 0, 1, 2, 3, ....

xp ,

p > −1

s2 n! sn+1

,

s>0

s > 0,

Γ(p + 1) , sp+1

s>a

[ 0! = Γ(1) = 1 ]

s > 0,

[ p ! = Γ(p + 1) ]

s2

b , − b2

s > |b|

s2

s , − b2

s > |b|

7.

sinh(bx)

8.

cosh(bx)

9.

eax sin(bx)

b , (s − a)2 + b2

s>a

10.

eax cos(bx)

s−a , (s − a)2 + b2

s>a

11.

xn eax ,

n = 0, 1, 2, 3, ....

n! , (s − a)n+1

s>a

Laplace Transform

405

4.2.15 Use the table in Problem 4.2.14 to compute the Laplace transforms of the following functions: √ √ −1 −2 (a) (b) 5x + (10x) 3 . + e−5x + 3 sin(2x) − 7 cos( 5x). 3 2 (c) 2x3 − 5x 3 + e3x cos(5x) − e−2x sin(12x) + 5. (d)

(e) x3 e−3x − 5e2x sin(−5x) − 6.

5 cosh(−3x) + 7 sinh(2x).

4.2.16 Use the table in Problem 4.2.14 and prove that ∀ a, b ∈ R, (a) L {sin(ax + b)} (s) =

a · cos(b) + s · sin(b) . s2 + a2

(b) L {cos(ax + b)} (s) =

s · cos(b) − a · sin(b) . s2 + a2

4.2.17 Modify the results of Problem 3.2.33 (and you can also use some rules of Table in Problem 4.2.21 to prove that for any a > 0 constant, we have:   hπ i o 1 1n (a) L cos(as) − Si(as) − sin(as) Ci(as) . (s) = x2 + a2 a 2   hπ i x (b) L (s) = sin(as) − Si(as) + cos(as) Ci(as). x2 + a2 2 s hπ  s i  s   s o 1n cos − Si − sin Ci . (c) L {arctan(ax)}(s) = s a 2 a a a 
(d) L ln x2 + a2 (s) = hπ i o 2 ln(a) 2 n + sin(as) − Si(as) + cos(as) Ci(as) . s s 2 4.2.18 (a) If y = f (x) defined on [0, ∞) Z p is periodic with period p and for some u ≥ 0 the integral e−sx f (x) dx exists, ∀ s ≥ u, then 0

prove that, ∀ s ≥ u, L{f (x)}(s) =

1 1 − e−sp

Z 0

p

e−sx f (x) dx =

esp esp − 1

Z

p

e−sx f (x) dx.

0

(b) Use this rule to find the Laplace transforms of sin(ax) and cos(ax), where a > 0 constant [as they appear in the table of Problem, 2π 4.2.14, Rules (3.) and (4.).] (Their period is p = .) Then find a 2 2 the Laplace transforms of sin (ax) and cos (ax). (Their period is π p = .) a

406

Improper Riemann Integrals

(c) Explain why this rule does not work with tan(ax) and cot(ax), where a > 0 constant? 4.2.19 (a) For any b ∈ R constant, prove directly that the Heaviside unit step function defined by   0, if x < b, Hb (x) := H(x − b) :=   1, if x ≥ b, [so, H0 (x) := H(x − 0) := H(x)] has Laplace transform for all s ≥ 0,  −bs e   if b ≥ 0,  Z ∞  s , −sx L{Hb (x)}(s) := e Hb (x) dx =  0   1, if b ≤ 0. s (b) For b ≥ 0, this rule also follows by Rule (1.) of the table in Problem 4.2.14 and Rule (5.) of the table in Problem 4.2.21. Check this! (c) Find the Laplace transforms of 3H(x − 5) and − 7H(x + 5).   if 0 ≤ a < (≤) x < (≤) b, 1, If p(x) = (d)   0, otherwise, prove that

L{p(x)}(s) =

−e−bs + e−as e−as − e−bs = . s s

(e) Find the Laplace transforms of the following two functions √  if 2.45 < x ≤ 12.34,  5.78, q(x) =   0, otherwise. √ 3  if 0 ≤ x < 123.45,  5.78, r(x) =   0, otherwise. (f) Prove that for any x ∈ R, its integer part or floor function satisfies [[x]] =

∞ X n=1

Hn (x) =

∞ X n=1

H(x − n).

Laplace Transform (g) Generalize (f ) to   X ∞ ∞ X x = Hpn (x) = H(x − pn), p n=1 n=1

407

∀ x ∈ R and p 6= 0.

(h) Prove that for n ∈ N the convolutions of H(x) := H0 (x) with itself n times are tn−1 tn−1 = . H n (t) := (H ∗ H ∗ . . . ∗ H)(t) = {z } | (n − 1)! Γ(n) n times (i) Use (h) to justify: If m, n ∈ N, H m (t) ∗ H n (t) = H m+n (t) and tm−1 ∗ tn−1 = B(m, n) · tm+n−1 . Now derive a general formula for the convolution of two polynomials. (See also Problems 3.13.35, 3.13.41, 4.2.26 and 4.2.30.) 4.2.20 Z ∞ (a) Use the known result Γ 0 (1) = e−x ln(x) dx = −γ, where 0 " n # X1 γ = lim − ln(n) ' 0.57721566... > 0 is the Eulern→∞ k k=1 Mascheroni constant (see Problem 2.3.32 and its footnote and Problem 3.5.18), to prove that for anyZs > 0 the Laplace ∞

transform of ln(x) is given by

e−sx ln(x) dx =

L{ln(x)}(s) = 0

−[γ + ln(s)] . s ∞

π2 + γ 2 , prove that for 6 0 any s > 0 the Laplace transform of ln2 (x) is given by Z ∞  2 [γ + ln(s)]2 π2 + . L ln (x) (s) = e−sx ln2 (x) dx = 6s s 0 Z ∞ (c) Given that Γ 0 (2) = e−x x ln(x) dx = 1 − γ, prove that for any

(b) Given that Γ 00 (1) =

Z

e−x ln2 (x) dx =

0

s > 0 the Laplace transform of x ln(x) is given by Z ∞ 1 − γ − ln(s) L {x ln(x)} (s) = e−sx x ln(x) dx = . s2 0

408

Improper Riemann Integrals


(d) Find the Laplace transforms of: f (x) = 6 ln 10 x2 ,

g(x) = 6 ln2 10 x2 , h(x) = 6x ln 10 x2 and p(x) = x ln2 (x). 4.2.21 Verify the following general properties-rules of the Laplace transform in the following table:

Function h(x) x ∈ [0, ∞) or x ∈ (0, ∞)

Laplace transform L {h(x)} (s), for s > k ≥ 0, k constant

1.

f 0 (x)

sL{f (x)}(s) − f (0+ )

2.

f 00 (x)

s2 L{f (x)}(s) − sf (0+ ) − f 0 (0+ )

3.

f (n) (x) n−1

s 4.

eax f (x) with a constant

5.

Hb (x) · f (x − b) shift of f (x) by constant b ≥ 0 [Hb (x) = Heaviside function]

sn L{f (x)}(s)− f (0+ ) − ... − f (n−1) (0+ )

L{f (x)}(s − a) for s > k + a (shift by a in the ) Laplace transform e−bs L{f (x)}(s) for s > k

x

Z 6.

1 L{f (x)}(s) s

f (t) dt 0

Z

∞

7.

f (t) dt x

8.

xn f (x),

n = 0, 1, 2, 3, ... f (x) x

9. 10.

1 s

∞

Z

 f (t) dt − L{f (x)}(s)

0

(−1)n

dn L{f (x)}(s) dsn

∞

Z

L{f (x)}(u) du s

f (ax) a > 0 constant

11. (−x)n f (x),

n = 0, 1, 2, 3, ...

s 1 L{f (x)} a a dn L{f (x)}(s) dsn

Laplace Transform

409

(Notice that the results of some previous and/or following problems and of some examples provide important rules that can be added to the two tables in Problems 4.2.14 and 4.2.21. Check these problems and examples and attach their additional rules to the tables, thus creating a more complete collection of Laplace transforms rules.) 4.2.22 Combine the Rules (5.) and (10.) of the table in Problem 4.2.21 to prove that for a > 0 and b ≥ 0 constants, we have the rule L{H b (x) · f (ax − b)}(s) = a

s 1 −bs . e a L{f (x)} a a

Compare this rule with the rule of Example 4.1.8. n o 2 Now use Problem 4.2.13 to find L H b (x) · e−(ax−b) (s). a

4.2.23 Let a > 0 and b ∈ R and suppose that f (x) is defined on all respected intervals in the formula below and the involved integrals exist.    s  Z 0 −su 1 −bs a a L{f (x)} f (u) du . Prove: L{f (ax − b)}(s) = e + e a a −b Compare this rule with the rule of Example 4.1.8. Find conditions and derive an analogous formula, if a < 0 and b ∈ R. 4.2.24 (a) In Rule (9.) of the table in Problem 4.2.21, justify why we f (x) need f (0) = 0 for the Laplace transform of to exist. x (b) Prove Rule (9.) of the table in Problem 4.2.21 by letting f (x) g(x) = ⇐⇒ xg(x) = f (x) and use Rule (8.) and Probx lems 4.2.7 and 4.2.8. (We need functions such that their Laplace transform at ∞ is 0.) (c) Use Rule (9.) of the table in Problem 4.2.21 to prove that the sin (βx) is Laplace transform of f (x) = x     π s β L{f (x)}(s) = − arctan = arctan , 2 β s as it was already referred in Problem 4.2.10.

410

Improper Riemann Integrals Z ∞ Z ∞ f (x) (d) Prove that in general, dx = L{f (x)}(s) ds. x 0 0 Z ∞ sin(x) π Use this rule to show dx = (the result of Example x 2 0 3.1.8). Z

x

δ(t − a) dt = Ha (x) for any a ∈ R. [So, in a sense

4.2.25 Prove that −∞

the derivative of Ha (x) is the Dirac delta function δ(x − a).] 4.2.26 (a) For a continuous function f : R −→ R, use the equality {H ∗ . . . ∗ [H ∗ (H ∗ f )]} = (H ∗ H ∗ . . . ∗ H) ∗ f, with n Heaviside functions H := H0 in each side, to prove the following Cauchy formula that changes the stated n-tuple integral based at 0 to a convolution integral: ∀ n ∈ N and x ∈ R, we have: Z un−2 Z un−1   Z x Z u1 ... f (un ) dun dun−1 . . . du2 du1 = 0 0 0 0 Z x n−1 1 (x − τ ) f (τ ) dτ = g ∗ f, where g(x) = xn−1 . (n − 1)! (n − 1)! 0 [See also Problem 4.2.19 (h) and (i). We can also derive this formula in other ways! Can you find one? The formula is also valid if all lower limits 0 are replaced by a ∈ R.] (b) Use the convolution rule to prove that the Laplace transform of L{f (x)}(s) . this n-tuple integral is sn 2

4.2.27 For any c > 0 constant, let fc (x) = e−cx and gc (x) = e−cx , with x ∈ [0, ∞). Prove: e−bx − e−ax = (fb ∗ fa )(x) −→ 0, as x −→ ∞. (a) (fa ∗ fb )(x) = a−b Z √bx a+b −abx2 2 1 a+b e (b) (ga ∗ gb )(x) = √ e−w dw = (gb ∗ ga )(x) −→ 0. x−→∞ −ax a+b √ a+b (See also Problem II 1.7.111 and compare.) 4.2.28 Use the two tables in Problems 4.2.14 and 4.2.21 to compute the Laplace transforms of the following ten functions for which we

Laplace Transform

411

assume that x > 0 or x ≥ 0. If some function does not have Laplace transform, then explain why. (a) x2 cos(5x), e−3x sin(−5x) , Z x x sin(t) (e) dt, t Z0 ∞ cos(t) (g) dt, t x (i) − 7e3x H(x + 5), (c)

(−x)3 sinh(2x),

(b)

sin(x) , x cos(x) , x

(d) (f)

3e−2x H(x − 5),

(h)

(j) H(x + 5) cos(x + 10).

4.2.29 (a) Take a > 0 constant and define the functions f (x) and g(x)

f (x) =

  0, if 0 ≤ x < a,      cos(x) , if x ≥ a, x

g(x) =

 0,    Z   

∞

x

if 0 ≤ x < a, cos(t) dt, if x ≥ a. t

Find the Laplace transforms of f (x) and g(x). (If you cannot find them in closed form, at least justify why they exist, then leave them in integral form and/or look at an advanced Laplace transform table.) (b) Prove that      Z ∞ 1 − cos(x) t 1 1 1 L (s) = − dt = ln 1 + . x t 1 + t2 2 s2 s (Careful: We cannot use the additive property of Laplace transform and of the integral here! Why?) 4.2.30 Let f (x) = xp and g(x) = xq , where p > −1 and q > −1 are real numbers. Prove that (f ∗ g)(x) = xp+q+1 B(p + 1, q + 1) =

Γ(p + 1)Γ(q + 1) xp+q+1 . Γ(p + q + 2)

[For p and q integers, we write this result with the corresponding three factorials. See also Problem 3.13.36 (a).]

412

Improper Riemann Integrals

4.2.31 If f and g are real functions on [0, ∞), and f or g is differentiable, and assume that the convolutions involved exist, prove that d (f ∗ g)(x) = f (0) · g(x) + (f 0 ∗ g)(x) = f (x) · g(0) + (f ∗ g 0 )(x). dx We can use this rule and the fact that (f ∗g)(0) = 0, to compute (f ∗g)(x). So, for example, find (f ∗ g)(x), if f (x) = ex and g(x) = x2 . 4.2.32 (a) Prove that: if f : [0, ∞) −→ R continuous function and L {f (x)} (s) = c constant, then c = 0 and f ≡ 0. (b) If c ∈ R constant, prove that: L {φ} (s) = c ⇐⇒ φ = c · δ, where δ is the Dirac Delta “function”. 4.2.33 Let a > 0. Define

f (x) =

 1   x,

if x ≥ a,

   0,

if 0 ≤ x < a.

1. Prove that L {f (x)} (s) exists for all s > 0, but it cannot be found in closed form from its definition. 2. Find the derivative

d L {f (x)} (s). ds

3. Use what you have found in the previous item to prove, ∀ s > 0, L {f (x)} (s) = − ln(s) −

∞ X (−1)n an sn + c, where c constant. n! n n=1

4. Prove that an expression of the constant c is Z ∞ −x ∞ X e (−1)n an c= dx + . x n · n! a n=1 4.2.34 (Extension of the previous Problem.) Let a > 0 and any k = 2, 3, 4, . . . . Define  1   if x ≥ a,  xk , f (x) =    0, if 0 ≤ x < a.

Laplace Transform

413

1. Prove that L {f (x)} (s) exists for all s ≥ 0. 2. Find the k th derivative of L {f (x)} (s). 3. For s > 0, find an expression of L {f (x)} (s) depending on k constants. 4. Explain how you can determine the k constants. 4.2.35 For n ∈ N0 , b ≥ 0 and s > 0, justify why   Z ∞ dn b (a) xn sin(bx) e−sx dx = (−1)n n . ds s2 + b2 0   Z ∞ n s n −sx n d (b) x cos(bx) e dx = (−1) . dsn s2 + b2 0 At s = 0 the integrals have discontinuity. [See also Application, (c).] 4.2.36 For given real constants a and b, prove that    ax Z ∞ s − b e − ebx 1 1 . (s) = − dt = ln L x t−a t−b s − a s 4.2.37 For given real constants a > 0 and b ≥ 0, find the simplest expression of L {ln(ax + b)} (s). [See Example 4.1.8, Problem 4.2.20, (a) and Rule 10 of Table in Problem 4.2.21.] 4.2.38 Project: Search the bibliography on Laplace transform and study the Laplace transforms of the: (a) Bessel functions. (b) Legendre polynomials. (c) Shifted functions. (d) Dirac impulse functions. (e) Convolutions of functions.

414

Improper Riemann Integrals

4.3

Inverse Laplace Transform

As we have already seen, Lerch’s Theorem, 4.1.2, on continuous functions the Laplace transform is a one-to-one operator. That is, the Laplace transforms of two different continuous functions are different. The same is essentially true on nice discontinuous functions in which we may allow two functions to be different at “a few” exceptional points (points of discontinuity). So, we can give the following definition: Definition 4.3.1 Given f (x) with Laplace transform g(s) = L{f (x)}(s), we call f (x) the inverse Laplace transform of g(s), and we write L−1 {g(s)}(x) = f (x). Since L−1 is the inverse of the linear operator L, it is a linear operator itself. (See Problem 4.5.1.) Generically speaking, if we know the Laplace transform of a function, then by applying to it the inverse Laplace transform we recover the function. For this purpose, many extensive tables and computer libraries have been created, so that from the Laplace transform we can find the function readily. In theory and application, we essentially have and use the following scheme: L

L−1

f (x) −→ L{f (x)}(s) = g(s) −→ L−1 {g(s)}(x) = f (x).

Examples Example 4.3.1 By Rule (2.) of the table in Problem 4.2.14, we have  1 . L e−5x (s) = s+5 Therefore,   1 −1 f (x) = L (x) = e−5x . s+5 Or, according to the above scheme,  L e−5x −→ L e−5x (s) =

1 L−1 −→ L−1 s+5



1 s+5



(x) = e−5x . N

3 Example 4.3.2 Let L{f (x)}(s) = 2 . s +4

Laplace Transform

415

Then by adjusting Rule (3.) of the table in Problem 4.2.14, we find that 3 f (x) = sin(2x). 2 So, in this example the above scheme is   3 3 3 L L−1 sin(2x) −→ L sin(2x) (s) = 2 −→ 2 2 s +4

−1

L



3 s2 + 4

 (x) =

3 sin(2x). 2 N

Example 4.3.3 Using the partial fraction decomposition, we find 3 s 4s2 + 12 = + 2 . 2 s(s + 4) s s +4 Then by using linearity and Rules (1.) and (4.) of the table in Problem 4.2.14, we find that the given expression is the Laplace transform of the function f (x) = 3 + cos(2x). That is,  2  4s + 12 −1 (x) = 3 + cos(2x). L s(s2 + 4) Now, by Rule (4.) of the table in Problem 4.2.21, we find the Laplace transform of the function g(x) = e−5x f (x) = e−5x [3 + cos(2x)].  L e−5x [3 + cos(2x)] (s) =

Therefore, L

−1



4(s + 5)2 + 12 . (s + 5)[(s + 5)2 + 4]

4(s + 5)2 + 12 (s + 5)[(s + 5)2 + 4]



(x) = e−5x [3 + cos(2x)]. N

Example 4.3.4 Prove n √ o −3 −a2 a L−1 e−a s (x) = √ x 2 e 4x , 2 π where a > 0 constant. (See also Problem II 1.7.159.) If L{f (x)}(s) = g(s), then [by Rule (1.) of the table in Problem 4.2.21], we have L{f 0 (x)}(s) = sg(s) − f (0).

416

Improper Riemann Integrals

So, L−1 {sg(s)}(x) = f 0 (x).

if f (0) = 0, Then, with f (x) := erfc

Z √a a√  2 x 2 2 x =1− √ e−u du, 2 π 0

we have f (0) = 1 − 1 = 0 and, by Problem II 1.7.161 that involves complex analysis, we get √ s

e−a g(s) := L{f (x)}(s) = s

.

Finally, n √ o L−1 {s · g(s)}(x) = L−1 e−a s = f 0 (x) =

d dx

2 1− √ π

Z

!

a √ 2 x

e 0

−u2

du

−3 −a2 a = √ x 2 e 4x . 2 π

N

4.4

Applications

Application 1: Laplace Transform and ODE’s. Here we remark that the Laplace transform may be used to give complete solutions of initial value-problems with ordinary differential equations. E.g., see the last seven of the problems that follow. However, there are cases in which it does not find the complete solution or even the solution at all. We study these situations in the following two examples. (1) The power series J0 (x) =

∞ X (−1)n  x 2n (n!)2 2 n=0

converges absolutely for all real x’s (prove this!) and defines the so-called Bessel function of the first kind of order zero.

Laplace Transform

417

The function J0 (x) was originally derived as a power series solution of the Bessel’s differential equation of order 0 11 xy 00 + y 0 + xy = 0, which is a homogenous equation with x = 0 a regular singular point. Bessel derived this differential equation while studying problems of planetary motion. Ever since, the Bessel differential equations of all orders appear in mathematics, application and engineering, and their solutions are very rich of properties and are so important that they have been tabulated. We find the Laplace transform of J0 (x) by integrating term by term and summing up ) (∞ X (−1)n  x 2n (s) = L{J0 (x)}(s) = L (n!)2 2 n=0 ∞ X (−1)n 1  2n L x (s) = (n!)2 22n n=0

(use the table in Problem 4.2.14) =  ∞  ∞ X 1 X − 21 1 (−1)n 1 (2n)! = = = (n!)2 22n s2n+1 s n=0 n s2n n=0 (use the convergent binomial series for s > 1, Problem 3.13.59)  − 21  2  21 1 1 s 1 1 , ∀ s ≥ 0. 1+ 2 =√ = s s s 1 + s2 1 + s2 Then, by the uniqueness of the Laplace transform (or its real analyticity in this particular example), we have ∀ s ≥ 0,

1 L{J0 (x)}(s) = √ , 1 + s2

thus, L−1

 √

1 1 + s2

 (x) = J0 (x).

  d2 u 1 du ν2 + + 1 − 2 = 0. 2 dz z dz z It has a bounded solution which is called the Bessel function of the first kind of order ν and an unbounded solution, an expression of which is called the Bessel function of the second kind of order ν. So, with ν = 0, we obtain the equation we examine above. 11 Bessel’s

differential equation of order ν ≥ 0 is

418

Improper Riemann Integrals

We could also take the Laplace transform of the equation itself and using the appropriate rules in the table in Problem 4.2.21, we find L {xy 00 + y 0 + xy} (s) = 0, or

h

i 0 0 − (L {y 00 }) + L{y 0 } − (L{y}) (s) = 0.

For convenience, we put Y (s) = L{y}(s), use the table in Problem 4.2.21 and we find  0 − s2 Y (s) − sy(0) − y 0 (0) + sY (s) − y(0) − Y 0 (s) = 0, or

(s2 + 1)Y 0 (s) + sY (s) = 0.

The last equation is a homogeneous separable ordinary differential equation of first order for Y (s), which can be solved directly by separating the variables and integrating both sides, etc., to find c Y (s) = √ , 1 + s2

where c is an arbitrary constant.

Therefore, a solution y(x) of the ODE xy 00 + y 0 + xy = 0 is   c (x) = cJ0 (x), y(x) = L−1 {Y (s)} (x) = L−1 √ 1 + s2 which we can directly verify that it is a solution, by plugging it into the given differential equation and differentiating the power series term by term. So, with c = 1, both methods gave the same answer 1 L{J0 (x)}(s) = √ , 1 + s2

for all

s≥0

and one correct solution, namely J0 (x), to this Bessel differential equation. Since the above Bessel equation is a linear homogeneous differential equation of second order, it must have another solution linearly independent of J0 (x), which was not found by the Laplace transform method. This happened because, for the other solution, the Laplace transform of at least one of xy 00 or y 0 or xy does not exist. In fact, if we find the other solution by means of power series, we will see that it is unbounded and the Laplace transform of at least one of these expressions does not exist. In a course of differential equations, we learn other methods for finding the evasive second solution in which we use the solution found here. So, the Laplace transform method may not always give the complete solution to a problem with a differential equation but a partial one, which in turn helps in finding the complete solution.

Laplace Transform

419

We have just seen that the method of the Laplace transform did not find both solutions of an ordinary differential equation of second order. Whenever this happens, we may not be able to solve a particular initial value-problem by using the Laplace transform method. (2) The method may exhibit even more serious difficulty. Sometimes it is not able to find any solution at all. For instance, the initial valueproblem  0 t2  y (t) − y(t) = (2t − 1)e , for t > 0,   y(0) = 2, 2

has unique (complete) solution y(t) = et + et (found easily, as we can easily check it). However, the method of the Laplace transform cannot find this solution because Z ∞ Z ∞ n o t2 −st t2 L (2t − 1)e (s) = e (2t − 1)e dt = (2t − 1)et(t−s) dt 0

0

does not exist. (This integral does not converge for any s ≥ 0.) [Note: This solution, however, can be found by the Mikusinski ˙ 12 general operator theory (1950–1951). In this theory, the results in Examples 4.1.5 and 4.1.6 play a fundamental role, especially Titchmarsh Theorem. This separate chapter of mathematics lies outside the scope of this book.] Application 2: Laplace Transform and PDE’s. We consider the following initial boundary value-problem with a partial differential equation:  ∂ 2 u(x, t) ∂u(x, t)   =k , k > 0 constant, for x > 0 and t > 0,   ∂t ∂x2         u(x, 0) = 0, for x ≥ 0,     u(0, t) = a constant, for t > 0,         u(∞, t) = lim u(x, t) = 0, for t > 0. x→∞

This is a mathematical model of the problem of heat convection along 12 Jan

Mikusinski, ˙ Polish mathematician, 1913–1987.

420

Improper Riemann Integrals

an insulated and uniform rod of infinite length placed on the positive x-axis and with initial point at the origin. u(x, t) is the temperature at position x at time t. k is a positive constant that depends on the uniform material of the rod which is a good heat conductor. (In reality, we have a rod considered very long.) At the beginning, when t = 0, the temperature at every point of the rod is zero. Afterwards, a source emits heat at the beginning of the rod in such a way that: (1) The temperature at the beginning of the rod (x = 0) stays constant equal a for every t > 0. (2) The temperature at the endpoint of the rod (x = ∞) is always the initial temperature, which is zero. We are going to solve this problem by means of the Laplace transform. We let Z ∞ U (x, s) = L {u(x, t)} (s) = e−st u(x, t) dt 0

be the Laplace transform of u(x, t) with respect to t. Then, by the first initial boundary condition, Rule (1.) in the table in Problem 4.2.21 and differentiating under the integral sign with respect to x twice, from the partial differential equation we find sU (x, s) = k

∂ 2 U (x, s) . ∂x2

As we see, this has general solution √s √s U (x, s) = c1 (s) ex k + c2 (s) e−x k . The last two conditions change to U (0, s) =

a , s

and

U (∞, t) = lim U (x, s) = 0. x→∞

a From these, we conclude that we must have c1 (s) = 0 and c2 (s) = . s Therefore, the Laplace transform of the solution of this problem is √s a U (x, s) = e−x k . s Then, by Problem II 1.7.161, we obtain      x x −1 √ √ u(x, t) = L {U (x, s)} (t) = a · erfc = a 1 − erf , 2 kt 2 kt which is easily verified to be a solution satisfying the partial differential equation and the imposed conditions.

Laplace Transform

421

We remark that the function u(x, t)+v(x, t) with u(x, t) the solucx −x2 tion just found and v(x, t) = √ e 4kt , where c 6= 0 constant, is another t3 solution of this problem, as we can directly check. This solution was not found by the Laplace transform method. We also notice that it is not a bounded solution. E.g., if t = x2 → 0, then v(x, t) → ∞. If we require that the solutions of the problem are bounded, a thing natural to assume, then we obtain the uniqueness of the solution u(x, t) found above. Application 3: Laplace Transform and Pulses. We want to solve the initial value-problem  2 d y(t) d y(t)   +4 + 6 = C · δ(t − t0 ),  2  dt dt      y(0) = 0,          d y(0) = 0. dt Problems like this arise in studying electrical circuits or mechanical oscillators. At an “instant” t0 ≥ 0 a voltage or force pulse of size C, constant, is applied to the circuit or the oscillator, and we would like to find the electric current or the size of the oscillation, respectively, at time t ≥ t0 . We use the respected rules of the Laplace transform from the table in Problem 4.2.21 and the Laplace transform of the Dirac delta function from Example 4.1.6. If the Laplace transform of the solution of the problem is Y (s), then the differential equation together with the initial conditions give Y (s) =

Ce−t0 s C e−t0 s √ . = s2 + 4s + 6 (s + 2)2 + ( 2)2

By the inverse process and using the table in Problem 4.2.14, we first find ( ) √ √  C 2 C −1 √ · √ L (t) = √ · e−2t sin 2t . 2 [s − (−2)]2 + ( 2 )2 2 Then, by Rule (4.) in the table in Problem 4.2.21, the solution of the problem is h√ i C y(t) = L−1 {Y (s)}(t) = Ht0 (t) √ · e−2(t−t0 ) sin 2 (t − t0 ) , 2

422

Improper Riemann Integrals

or more explicitly  0,    y(t) =  √  C   √ · e−2(t−t0 ) sin 2 (t − t0 ) , 2

if t < t0 , if t ≥ t0 ,

a graph of which, with t0 = 3 and C = 5, is seen in Figure 4.3. y

1.0

0.8

0.6

0.4

0.2

O

2

4

6

8

x

FIGURE 4.3: Pulse in application 3, with t0 = 3 and C = 5

Note: When we solve differential equations via the Laplace transform, at times the Laplace transform does not alleviate the situation. For example, the homogeneous linear differential equation of order two d2 y(x) + x2 y(x) = 0 dx2

changes to

d2 Y (s) + s2 Y (x) = as + b, ds2

where a and b are constants and Y (s) := L[y(x)](s). (Check this.) The new differential equation is as hard as the original one (if not harder, since it is nonhomogeneous). So, we need a different method to solve such a differential equation.

Laplace Transform

423

Application 4: The computation of the convolution of two functions defined on [0, ∞) is not always easy. Using the Laplace and inverse Laplace transform we can compute the convolution, as we illustrate in the following easy example. Let f (x) = sin(x) and g(x) = x, with x ∈ [0, ∞). We will compute their convolution in two ways. (1) Straightforward way: Z x [sin(x − t)] · t dt = (f ∗ g)(x) = 0 Z x [sin(x) cos(t) − cos(x) sin(t)] · t dt = 0 Z x Z x sin(x) t cos(t) dt − cos(x) t sin(t) dt = 0 0 Z x Z x sin(x) t d sin(t) + cos(x) t d cos(t). 0

0

The last two integrals are found by integration by parts, and after computing them and simplifying we find (f ∗ g)(x) = sin(x) ∗ x = x ∗ sin(x) = x − sin(x). (We can also use the u-substitution u = x − t in the first integral, etc.) (2) Way using the Laplace and inverse Laplace transform: By the Convolution Rule (Example 4.1.4) and the rules in the table of Problem 4.2.14, we have L{(f ∗ g)(x)}(s) = L{f (x)}(s) · L{g(x)}(s) = L{sin(x)}(s) · L{x}(s) = 1 1 1 1 1 · 2 = 2 2 = 2− 2 . 2 s +1 s s (s + 1) s s +1 Now, by the inverse Laplace transform, we find (f ∗ g)(x) = sin(x) ∗ x = x ∗ sin(x) =     1 1 −1 L−1 (x) − L (x) = x − sin(x). s2 s2 + 1 Remark: We can also use Problem 4.2.31. We conclude this section on the inverse Laplace transform with the following note: To find the inverse Laplace transform of a given Laplace transform, we use: 1. The definition (especially for primitive cases).

424

Improper Riemann Integrals

2. For direct computations of the inverse Laplace transforms (see Subsection II 1.7.10 and its problems). 3. The linearity properties. 4. Already known inverse Laplace transforms. 5. The inverse rules in the table in Problem 4.2.14. 6. The inverse rules in the table in Problem 4.2.21 7. Other rules not listed in the two tables above. 8. Limit processes with known results, especially when parameters are involved. 9. Tables of Laplace and inverse Laplace transforms with or without adjustments, if we can trust them, of course. (Sometimes there are human errors and/or typos in tables, and so you may need to check the readily available answers.) 10. Special computer packages, if we can trust them, of course. (Keep in mind that human errors are always possible with computers and computer packages.)

4.5

Problems

4.5.1 Prove that the Inverse Laplace transform is a linear operator defined on the range of the Laplace transform. Do this by proving the following general result, encountered in Linear Algebra and elsewhere: Let V and W be two vector spaces over a field F and T : V −→ W a one-to-one and onto linear operator. Then prove that the set-theoretic inverse T −1 : W −→ V is also a one-to-one and onto linear operator. 4.5.2 Compute the inverse Laplace transforms of the following four functions and write the corresponding four schemes as in Examples 4.3.2 and 4.3.3: (a) (c)

−2 2 2s + 2 − 2 , s 3s + 5 s + 8 3 10 + 6, 2s − 5 s

(b) (d)

4 3 2s + 2 , − 2 5 (s + 2) s −6 s −9 2s s−5 − . s2 + 5s − 2 s2 − 3s + 9

Laplace Transform

425

4.5.3 Use partial fraction decomposition to find the inverse Laplace transforms of (a)

8s2 − 4s + 12 , s(s2 + 4)

and

(b)

(s2

s2 + 2s + 3 . + 4s + 5)(s2 + 2s + 10)

4.5.4 Find the inverse Laplace transforms of (a)

(s − 5)e−2s , s2 − 5s + 6

and

(b)

2e−3s − 5e−4s . s

4.5.5 (a) Prove: (1) L−1



ln(s) s

 (x) = −[ln(x) + γ].

 π2 ln2 (s) . (x) = [ln(x) + γ]2 − s 6   ln(s) −1 (3) L (x) = x[1 − γ − ln(x)]. s2  2  ln (s) −1 (x). (b) Find L s2 (2) L−1



(c) Explain what happens with L−1 { ln(s)}(x). [Hint: Use Problem 4.2.20.] 4.5.6 Collect all the Laplace transforms computed in the previous section and the problems and rewrite them as a table of inverse Laplace transforms. 4.5.7 Let p (s) and q(s) be two polynomials and the degree of q(s) is n ≥ 1. Suppose also that q(s) has n distinct one another (real, complex) roots r1 , r2 , r3 , . . . , rn . (a) Prove that the partial faction decomposition n

X Ai p (s) A1 A2 A3 An = + + + ... + = q(s) s − r1 s − r2 s − r3 s − rn s − ri i=1 has coefficients Ai =

p (ri ) , q 0 (ri )

i = 1, 2, 3, . . . , n.

426

Improper Riemann Integrals

(b) Prove that L−1



p (s) q(s)

 (x) = n

p (r1 ) r1 x p (r2 ) r2 x p (rn ) rn x X p (ri ) ri x e + 0 e + ... + 0 e e . = 0 q (r1 ) q (r2 ) q (rn ) q 0 (ri ) i=1 4.5.8 Use Application 1 and Laplace transform to show that J0 (x) =

1 π

Z

π

cos[x cos(θ)]dθ = 0

2 π

Z

π 2

cos[x cos(θ)]dθ. 0

4.5.9 The modified Bessel function of order zero is I0 (x) = 1 +

x4 x2 + . . . = J0 (ix), 2·2 2·4·2·4

where i =

Prove that its Laplace transform is 1 L {I0 (x)} (s) = √ . 2 s −1 (See Application 1.) 4.5.10 Verify that the initial value-problem  0  y (x) + p y(x) = r(x), p is a constant,   y(0) = y0 , has solution y(x) = L−1

(y0 is constant), 

L{r(x)}(s) + y0 s+p

 (x).

4.5.11 Find the solution of the initial value-problem  0 x  y (x) − 5 y(x) = e cos(x),   y(0) = −2.

√

−1.

Laplace Transform

427

4.5.12 Verify that the initial value-problem  00 0   y (x) + p y (x) + q y(x) = r(x), p, q are constants,     y(0) = y0 , (y0 is constant),       y 0 (0) = y 0 , (y 0 is constant), 0

has solution y(x) = L

0

−1



L{r(x)}(s) + (s + p)y0 + y00 s2 + ps + q

4.5.13 Find the solution of the initial value-problem   y 00 (x) − 4 y 0 (x) + 3 y(x) = 2 sin(x) + 5 cos(2x),        y(0) = 1,        y 0 (0) = 1 . 2 4.5.14 Find the solution of the initial value-problem   y 00 (x) + 2 y 0 (x) + y(x) = 4e−x ,       y(0) = 2,       y 0 (0) = −1. 4.5.15 Find the solution of the initial value-problem    0 ≤ x < π,  3,    00  y (x) + 4y(x) =        0, π ≤ x < ∞,     y(0) = 2,         0 y (0) = −1.

 (x).

428

Improper Riemann Integrals

4.5.16 (a) Find the general formula for the solution of the following third order initial value-problem:  000 y (x) + p y 00 (x) + q y 0 (x) + c y(x) = r(x), p, q, c are constants,           (y0 is constant), y(0) = y0 ,    y 0 (0) = y00 ,         00 y (0) = y000 ,

(y00 is constant), (y000 is constant).

(b) Next, solve the two initial value-problems  000 00 0 x  y (x) − 2y (x) + 2y + 3y(x) = 4e ,   y(0) = y 0 (0) = y 00 (0) = 1, and

 000 0 4x  y (x) + y (x) = e ,   y(0) = y 0 (0) = y 00 (0) = 0.

4.5.17 Use the convolution rule to prove that the solution of the initial value problem y 00 (x) + y(x) = f (x) and y(0) = y 0 (0) = 0, Z x y(x) = (f ∗ sin)(x) = f (u) sin(x − u)du.

is

0

4.5.18 Use the convolution rule to prove that the solution of the integral equation Z x y(x) = x + (y ∗ sin)(x) = x + y(u) sin(x − u)du, 0

is y(x) = x +

x3 , 6

and so (y ∗ sin)(x) =

x3 . 6

Laplace Transform

429

4.5.19 Use the Laplace transform and the Inverse Laplace transform to find two different or equal continuous real functions f and g on [0, ∞), such that f ∗ g = 1 (or = c 6= 0 constant). 4.5.20 Suppose that f ∗ g ≡ 0, where f and g are continuous real functions defined on [0, ∞). Find what could be wrong with the following short “proof” of Titchmarsh Theorem of convolution (as convolution is defined in Example 4.1.4): “By hypothesis we get L{f (x)}(s) · L{g(x)}(s) = L{(f ∗ g)(x)}(s) = L{0}(s) ≡ 0

and so

L{f (x)}(s) ≡ 0 or L{g(x)}(s) ≡ 0. Therefore, f ≡ 0 or g ≡ 0.” However, explain why this proof is valid if the Laplace transforms of f and g exist and are analytic, that is, they are power series locally at every point at which they exist.13 4.5.21 Project: Put together and prove the results in Examples 4.1.5 and 4.1.6, especially Titchmarsh Theorem of convolution, to demonstrate that the set of the continuous functions C = {f : [0, ∞) −→ R continuous} equipped with the usual operation of addition (+) and the operation of convolution (∗) is an algebraic integral domain with unit element the Dirac delta function and so it can be extended to an algebraic field. This is a field of operators . Then also prove that f ∗f =f

⇐⇒

f = 0,

or f = δ

the Dirac delta function.

(Compare with Problem II 1.7.126. Find bibliography, e.g., Erd´elyi 1962, etc.)

13 For example, L{f (x)}(s) is analytic in the region in which the Laplace transform is absolutely convergent. This is a consequence of Tonelli-Fubini’s Theorem (see Section 3.6, Conditions II, III and IV), together with Morera’s* Theorem, in complex analysis. (See Theorem II 1.5.5.) *Giacinto Morera, Italian mathematician and engineer, 1856–1909.

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432

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[13] Brown, J. W. and R. V. Churchill. 2008.Complex Variables and Applications, eighth edition. McGraw-Hill. [14] Budak, B. M. and S. V. Fomin. 1973. Multiple Integrals, Field Theory and Series, translated from Russian by V. M. Volosov, D. Sc. Moscow: Mir Publishers. [15] Cartan, H. 1973. Elementary Theory of Analytic Functions of One or Several Complex Variables, second edition. Paris: Hermann and Addison Wesley. [16] Churchill, R. V. 1972. Operational Mathematics, third edition. New York: McGraw Hill. [17] Copson, E. T. 1948. An Introduction to the Theory of Functions of a Complex Variable. Oxford University Press. [18] Danese, A. E. 1965. Advanced Calculus, An Introduction to Applied Mathematics (two volumes). Boston: Allyn and Bacon. [19] De Barra, G. 1974. Introduction to Measure Theory. Van Nostrand Reinhold. [20] De Silva, N. 2010. A Concise, Elementary Proof of Arzel` a’s Bounded Convergence Theorem. Am. Math. Mon. 117(10): 918–920. [21] Demidovich, B. editor. 1973. Problems in Mathematical Analysis, translated from Russian by G. Yankovsky. Moscow: Mir Publishers. [22] Dickson, L. E. 1949. New First Course in the Theory of Equations. John Wiley. [23] Dodge, C. W. 1972. Euclidean Geometry and Transformations. Dover. [24] Erd´elyi, A. 1962. Operational Calculus and Generalized Functions. Holt, Rinehart and Winston. [25] Franklin, Ph. 1946. A Treatise on Advanced Calculus. John Wiley & Sons. [26] Fisher, S. D. 1999. Complex Variables, second edition. Dover. [27] Furdui, O. 2013. Limits, Series, and Fractional Part Integrals, Problems in Mathematical Analysis, Problem Books in Mathematics. Springer. [28] Furdui, O. and A. Sˆınt˘ am˘ arian, 2021. Sharpening Mathematical Analysis Skills, Problem Books in Mathematics. Springer.

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[29] Gelbaum, R. B. 1992. Problems in Real and Complex Analysis, Problem Books in Mathematics. Springer. [30] Gradshteyn, I. S. and I. M. Ryzhik. 2007. Table of Integrals, Series and Products, seventh edition, translated from Russian by Scripta Technica, edited by Alan Jeffrey and Daniel Zwillinger. Academic Press-Elsevier. [31] Gray, A. 1998. Modern Differential Geometry of Curves and Surfaces with Mathematica, second edition. CRC Press. [32] Hardy, G., J. E. Littlewood & G. P´ olya. 1988. Inequalities, second edition. Cambridge Mathematical Library. [33] Helms, L. L. 1975. Introduction to Potential Theory, Pure and Applied Mathematics Volume XXII. Krieger. [34] Hildebrand, F. B. 1976. Advanced Calculus for Applications. Englewood Cliffs: Prentice Hall. [35] Hitt, R. and I. M. Roussos. 1991. Computer Graphics of Helicoidal Surfaces with Constant Mean Curvature. An. Acad. Bras. Ci. 63 (3): 211–228. [36] Knopp, K. 1990. Theory and Application of Infinite Series, translated and revised from German by R. C. H. Young. New York: Dover. [37] Kumchev A. V. 2013. On the Convergence of Some Alternating Series. Ramanujan J. 30: 101–116. [38] Lang, S. 1983. Undergraduate Analysis. Springer-Verlag. [39] Lebedev, N. N. 1972. Special Functions & Their Applications, translated from Russian and edited by R. A. Silverman. New York: Dover. [40] Marichev, O. I. 1982. Handbook of Integral Transforms of Higher Transcendental Functions, Theory and Algorithmic Tables, translated from Russian by L. W. Longdon. Shrivenham: Ellis Horwood. [41] Markushevich, A. I. 1977. Theory of Functions of a Complex Variable, revised English edition, translated from Russian and edited by R. A. Silverman (three volumes or one volume). Chichester: Ellis Horwood ltd. [42] Marsden, J. E. and M. J. Hoffman. 1987. Basic Complex Analysis, second edition. Freeman.

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[43] Marsden, J. E. and M. J. Hoffman. 1993. Elementary Classical Analysis, second edition. Freeman. [44] Mead, D. G. 1961. Integration. Amer. Math. Monthly 68: 152–154. [45] Miller, I. and M. Miller. 1999–2014. John E. Freund’s Mathematical Statistics, eight edition. Prentice Hall. [46] Miller, K. S. 1960. Advanced Complex Calculus. New York: Harper. [47] Nahin, J. P. 2015. Inside Interesting Integrals. Undergraduate Lecture Notes in Physics. New York: Springer. [48] Needham, T. 1997. Visual Complex Analysis. Oxford: Clarendon Press. [49] Nikolsky, S. M. 1977. A Course of Mathematical Analysis (two volumes), translated from Russian by V. M. Volosov, D. Sc. Moscow: Mir Publishers. [50] Ostrowski, A. M. 1949. On Some Generalizations of the CauchyFrullani Integral. Proc. N. A. Sc. 35: 612–616. [51] Pedoe, D. 1988. Geometry, A Comprehensive Course. Dover. [52] Peirce, B. O. and R. M. Foster. 1956. A Short Table of Integrals, fourth edition. Ginn. [53] Pons, M. A. 2014. Real Analysis for the Undergraduate With an Invitation to Functional Analysis. Springer. [54] Prudnikov, A. P., Yu. A. Brychkov and O. I. Marichev, 1986. Integrals and Series: Volume 1: Elementary Functions; Volume 2: Special Functions, translated from Russian by N. M. Queen. Gordon and Breach Science. [55] Richards, I. J., H. K. Youn 1990. Theory of Distributions, A Nontechnical Introduction. Cambridge University Press. [56] Ritt, J. F. 1948. Integration in Finite Terms, Liouville Theory of Elementary Methods. Columbia University Press. [57] Rudin, W. 1976. Principles of Mathematical Analysis, third edition. McGraw-Hill. [58] Rudin, W. 1987. Real and Complex Analysis, third edition. McGraw-Hill.

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Index

Abel N. H., text and footnote, 67 Abel’s Lemma, 67 Abel’s Test for convergence of improper integrals, 81, 383 Absolute Convergence Test, 61, 380 Absolute Limit Comparison Test, example, 55 Absolute p-Test, 71 Absolute Ratio Test, 53 Absolute Ratio Test for convergence of improper integrals, 72 Absolute Ratio Test for uniform convergence of series of functions, 178 Absolute Root Test, 52 Absolute Root Test, example, 60 absolute value in integral form, example, 136 absolutely, converges , 61 absolutely, diverges, 61 alternating series Test, text and footnote, 67 arc-length in polar coordinates, 19 Archimedes of Syracuse, footnote, 9 area of (n − 1)-dimensional sphere, 320 area of (n − 1)-dimensional unit sphere, problem, 349 ascending factorial, 301, 375 average value of a continuous random variable, 103

Beppo-Levi Theorem, 195 Beppo-Levi Theorem, example, 196 Beppo-Levi, text and footnote, 195 Berboulli’s integral, problem, 338 Bernoulli J., 240 Bernoulli J., footnote, 10 Bernoulli J., problem and footnote, 338 Bessel F. W., text and footnote, 379 Bessel function of order zero, modified, problem, 426 Bessel function of the first kind of order ν, footnote, 417 Bessel function of the first kind of order zero, 416 Bessel function of the second kind of order ν, footnote, 417 Bessel functions, 379 Bessel functions, problem, 413 Bessel’s differential equation of order 0, 417 Bessel’s differential equation of order ν, footnote, 417 Beta and Gamma functions, 296, 299 Beta and Gamma functions, problem, 341 Beta function, 296 Beta function, basic properties, 296 Beta function, recursive formulae, 297 Beta function, relation to Gamma function, 298

437

438 Beta-Gamma function, Euler’s formula, 299 binomial series, real, 355 Bromwich T., text and footnote, 379 Brouncker series, 67 Brouncker W., text and footnote, 67 Bunyakovsky V. Y., text and footnote, 359 Bunyakovsky-H¨ older’s inequality, 359 Cantor G. F. L. Ph., footnote, 192 Cantor or Kantor set, 192 Cauchy, 52, 74, 76, 359 Cauchy A. L., text and footnote, 32 Cauchy Condensation Theorem, example, 60 Cauchy Criterion for convergence in the Real Line, general, text and footnote, 76 Cauchy criterion for convergence, 74 Cauchy formula for repeated integration, problem, 410 Cauchy Positive Series Condensation Theorem, 52 Cauchy principal value, 32 Cauchy sequence, footnote, 76 Cauchy Test for uniform convergence of improper integrals, 180 Cauchy Test for uniform convergence of sequences or series of functions, 177 Cauchy Test, example, 78–80 Cauchy’s Root Test, 52 Cauchy-Schwarz inequality, 359 characteristic function of the irrationals, 27 characteristic function of the rationals, 27

Index Chebyshev P. L., text and footnote, 305 Chebyshev’s conditions, 305 closed ball, n-dimensional, 320 clothoid, 240 commuting limits and integrals, 168 commuting limits and derivatives, 200 Comparison Test for convergence of non-negative series, Note, 50 Comparison Test with Non-negative Functions, for checking convergence, 49 complementary error function, Craig’s formula, problem, 116 complementary error function, erfc, definition, 102 complementary error function, Laplace transform, example, 390 complementary error function, problem, 115 complete elliptic integral of the first kind, Legendre form, application, 318 Condensation Theorem for Convergence of Non-negative Series, Cauchy, 52 conditionally, converges, 61 conditions of Chebychev, 305 conditions of Tschebyscheff, 305 cone, n-dimensional right circular, problem, 351 conjugate exponents, 359 continuous fractions, reference to, 92 converges absolutely, 61 converges conditionally, 61 convolution of functions, context of the Laplace transform, 391

Index convolution of two polynomials, problem, 407 convolution rule, Laplace transform, example, 391 convolutions of functions, 379 convolutions of functions, problem, 413 Cornu, M. A., text and footnote, 240 Craig J. W., problem, footnote, 116 Craig’s formula, complementary error function, problem, 116 Criteria of Existence, 47

439

of convolution, example, 396 Dirac P. A. M., text and footnote, 379 Dirichet sine integral, problem, 240 Dirichlet function, 27 Dirichlet function, example, 192 Dirichlet gamma integrals, 328 Dirichlet gamma integrals, problem, 351, 352 Dirichlet J. P. G. L., text and footnote, 27 Dirichlet sine integral, 130, 131, 135 Dirichlet Test for uniform converD’ Alembert J. Le R., text and gence of improper intefootnote, 53 grals, 182 D’ Alembert’s Ratio Test, 53 diverges absolutely, 61 Darboux J. G., footnote, 9 dominating function, 189 Darboux Sums, footnote, 9 Double Integral Technique, 224 density of the rth order statistics, double Riemann sums, 227 problem, 340 duplication formula of the Gamma descending factorial, 301 function, 307 Dilogarithm function, 237 duplication formula of the Gamma Dini U., problem and footnote, 165 function, problem, 339, Dini U., text and footnote, 332 343, 345 Dini’s Theorem for uniform convergence of sequences of e, e, footnote, 10 ellipsoid, n-dimensional, 323 functions, 175 Dini-Kummer criterion for conver- error function, an integral representation of, example, 139 gence of positive series, text and footnote, 332, error function, erf, definition, 102 error function, Laplace transform, 333 example, 389 Dini-Poisson integral, problem, 165 Dini-Poisson integral, problem, hint, error function, two integral representations of, example, 167 140 Dirac delta function, example, 394 Dirac delta function, Laplace trans- Euler constant, 284 Euler L., 240 form, example, 396 Dirac delta function, project prob- Euler L., footnote, 10 Euler L., text and footnote, 99 lem, 429 Dirac delta function, unit element Euler reflection formula of the Gamma function, 299

440

Index

Euler’s constant, definition, footnote, 117 Euler’s formula, Gamma-Beta function, 299 Euler’s integral, 124 Euler’s integral of the first kind, 296 Euler’s integral of the second kind, 280 Euler’s integral, problem, 145 Euler’s sum, example, 233, 236, 237 Euler-Mascheroni constant, 284 Euler-Mascheroni constant, definition, footnote, 117 Euler-Mascheroni constant, footnote, 347 Euler-Mascheroni constant, integral representation, footnote, 214 Euler-Mascheroni constant, integral representation, problem, 212–214 Euler-Mascheroni constant, integral representations, problem, 212 Euler-Mascheroni constant, problem, 219, 368, 407 Euler-Poisson-Gauß Integral, 99 even function, 35 even function about c, 36 even part of a function, 36 expected value of a continuous random variable, 103

Fisher distribution, problem and footnote, 355 Fisher R. A. Sir, footnote, 355 Fourier J. B. J., text and footnote, 391 Fresnel A. J., text and footnote, 79 Fresnel cosine integral, 233 Fresnel integral, 79, 81, 232 Fresnel integral, problem, 83 Fresnel integrals, example, 229 Fresnel integrals, problem, 242, 338 Fresnel sine integral, 233 Frullani G., text and footnote, 248 Frullani integral, example, 142 Frullani integral, problem, 158 Frullani Integrals, 248 Fubini Condition IV, double integrals, 227 Fubini Condition, double integrals, 227 Fubini G., text and footnote, 227 functions of exponential order, 381 functions of exponential order, problem, 402 Fundamental Theorem of Integral Calculus, 1

gamma distribution, 324 Gamma function, 279 Gamma function as infinite product, project problem, 362 Gamma function integral converges uniformly, problem, 333 Gamma function, a characterizaF-distribution, problem and foottion of, 362 note, 355 Gamma function, basic properties factorials and Gamma function, and values, 282 288 Gamma function, derivatives of, falling factorial, 301 368 Fatou P., text and footnote., 193 Gamma function, duplication forFatou Theorem, 193 mula, Legendre, 307 Fatou Theorem, example, 194 Gamma function, duplication for-

Index mula, problem, 339, 343, 345 Gamma function, Gauß formula, 361 Gamma function, general Euler’s formula, 352 Gamma function, Preliminary Estimates, 280 Gamma function, product representation of the reciprocal of, 347 Gamma function, reflection formula of, 299 Gamma function, relation to Beta function, 298 Gamma function, series expansion, problem, 356 Gamma function, triplication formula, problem, 347 Gamma-Beta function, Euler’s formula, 299 Gauß K. F. J., text and footnote, 99 Gauß K. F. J., footnote, 347 Gauß multiplication formula of the Gamma function, footnote, 347 Gauß multiplication formula of the Gamma function, proof, 363 Gauß formula for the Gamma function, 361 general Euler’s formula, Gamma function, 352 generalized or improper Riemann integral, 8 Gosset W. S., footnote, 353

441 H¨ older’s inequality, 359 H¨ older’s inequality, general, problem, 366 Heaviside O., text and footnote, 379 Heaviside unit step function, problem, 406 Hessenberg form of square matrices, footnote, 322 Hessenberg K. A., text and footnote, 322 hypergeometric differential equation, footnote, 375 Hypergeometric forms, footnote, 373 hypergeometric function, problem and footnote, 344 hypergeometric functions, text and footnote, 375

improper or generalized Riemann integral, 8 impulse functions, 379 impulse functions, problem, 413 Incomplete Gamma Function, project problem, 372 integer part or floor function, problem, 406 integral cosine, problem, 160 integral of Laplace, 137, 211 integral sine, problem, 160 Integral Test, example, 60 Integral Test, for checking convergence, 52 Integrals Dependent on Parameters, 119 inverse Laplace transform, 414 inverse Laplace transform, probH¨older continuity of some order, lem, 424 problem and footnote, 93 H¨older continuous, problem and Jacobi C. G. J., text and footnote, 322 footnote, 93 H¨older O. L., text and footnote, Jones W., footnote, 9 359

442 Kantor G. F. L. Ph., footnote, 192 Kantor or Cantor set, 192 Kronecker L., text and footnote, 30 Kummer E. E., footnote, 373 Kummer E. E., text and footnote, 332 Kummer’s formula, 376 Kummer’s formulae, 375 Kummer’s Summation Formulae, footnote, 373 Kummer-Dini criterion for convergence of positive series, text and footnote, 332 L’ Hˆopital G. F. A. M. de, text and footnote, 16 L’ Hˆopital’s rule, 16, 18, 132 L’ Hˆopital’s rule, problem, 109 Landen J., footnote, 238 Landen’s formula, example, 238 Laplace P. S., text and footnote, 137 Laplace transform, 379, 380 Lebesgue Criterion for Riemann Integrability, 192 Lebesgue Dominated Convergence Theorem, 189 Lebesgue Dominated Convergence Theorem, example, 190, 191 Lebesgue Dominated Convergence Theorem, problem, hint, 218 Lebesgue H. L., text and footnote, 122 Lebesgue Monotone Convergence Theorem, 188 Lebesgue Monotone Convergence Theorem, example, 190, 191 Lebesgue Monotone Convergence Theorem, problem, hint, 219

Index Lebesgue theory of integration, 122 Legendre A. - M., application, 318 Legendre A. - M., text and footnote, 307 Legendre polynomials, problem, 413 Leibniz G. W., von, text and footnote, 67 Leibniz rule, 122 Leibniz Rule for differentiation of an improper integral, 183 Leibniz rule, differentiation of integrals , 122 Leibniz rule, general, differentiation of integrals, 122 Leibniz’s alternating series Test, text and footnote, 67 Leibniz’s Theorem for alternating series, footnote, 92 Lerch M., text and footnote, 386 Lerch’s Theorem, 386 Limit Comparison Test, 53 Limit Comparison Test, example, 58 linear operator, 383 linear operator, problem, 424 Liouville J., text and footnote, 51 Liouville’s theory for antiderivatives in terms of elementary functions, 51 Lipschitz constant, 92 Lipschitz continuity, 92 Lipschitz continuity, problem and footnote, 93 Lipschitz continuous, 92 Lipschitz continuous, problem and footnote, 93 Lipschitz R., problem and footnote, 93 Lipschitz R., text and footnote, 92 logarithmic spirals, 18 Lower Incomplete Gamma Function, project problem, 372

Index Mascheroni L., footnote, 117 Mathematica, computer program, footnote, 373 maximum of two functions, text and footnote, 95 mean of a continuous random variable, 103 mean value of a continuous random variable, 103 measurable functions, 48 Mellin R. H., text, 379 Mikusinski ˙ J., Note and footnote, 419 minimum of two functions, text and footnote, 95 Minkowski H., text and footnote, 365 Minkowski’s inequality, 365 modified Bessel function of order zero, problem, 426 modified Dirichlet or Riemann function, 30 moment-generating function of a continuous random variable, 104 moment-generating function of a continuous random variable, example, 141, 199 moments about the origin of a continuous random variable, 103 Morera G., footnote, 429 multiplication formula of the Gamma function, Gauß footnote, 347 Napier J., footnote, 10 negative part of a real function, 226 negative part of a real function, problem, 149 Newton I. Sir, text and footnote, 20 Newton’s binomial series, real, 355

443 Newton’s sum for π, 129 nice function, definition for this text, 48 nice functions, domain of definition of, 49 non-degeneracy of norm function, 365 norm function, 365 norm, Ls -norm of function, 365 normal distribution, 104 normal distribution, generating function, example, 141, 199 Number theory, reference to, 92 odd function, 35 odd function about c, 36 odd part of a function, 36 order statistics, problem, 340 p -Test, example, 56 Pascal B., problem and footnote, 155 pendulum, mathematical, 317 pendulum, simple, 317 period of a real function, 28 periodic function, 26 Phragm´en L. E., problem and footnote, 215 Phragm´en, Theorem, 388 pi, π, footnote, 9 ´ a Theorem of, footnote, Picard E., 92 ´ footnote, 92 Picard E., Pochhammer L. A., text and footnote, 295 Pochhammer symbol, 301 point-wise convergence of sequence of functions, 171 point-wise limit function, 171 point-wise limit of improper integral, 176 point-wise limit of series of functions, 176

444

Index

Poisson cumulative probability distribution function, project problem, 372 Poisson probability distribution, project problem, 372 Poisson S. D., text and footnote, 99 Poisson-Dini integral, problem, 165 Poisson-Dini integral, problem, hint, 167 polygamma function, 369 Polylogarithmic function, 238 positive part of a real function, 226 positive part of a real function, problem, 149 Positivity Condition I, double integrals, 226 positivity of norm function, 365 principal value, 32 probability density function, 103 product representation of the reciprocal of the Gamma function, 347 proper Riemann integral, 2 psi function of Gauß, 369 Raabe L. J., footnote, 352 Raabe’s integral, 352 Ramanujan S., problem and footnote, 279 Ramanujan’s integral, 279 real continuous random variable, 103 reflection formula of the Gamma function, 299 regular singular point, example, 417 Riemann Dirichlet function, example, 193 Riemann G. F. B., text and footnote, 2 Riemann or modified Dirichlet function, 30 Riemann sums, 2

Riemann Zeta function, complex, Note, 216 Riemann Zeta function, real, integral representations, problem, 337 Riemann Zeta function, real, problem, 216 Riemann-Darboux Sums, footnote, 9 rising factorial, 301, 375 Root Test for convergence of improper integrals, 73 Root Test, example, 60 rule of shift of a real function, 36 rule of translate of a real function, 36 scaled complementary error function, 102 Schwarz H. A., text and footnote, 359 semi-linearity of norm function, 365 set of Lebesgue measure zero, 191 shift functions, 379 shift functions, problem, 413 shift of a real function, 36 Snedecor distribution, problem and footnote, 355 Snedecor G. W., footnote, 355 sphere, (n − 1)-dimensional, 320 spiral of Cornu, 240 standard deviation (from the mean)of a continuous random variable, 104 standard normal distribution, 104 standard normal probability density, problem, 354 step functions, 379 Stirling J., text and footnote, 289 Stirling’s formula, 289 Stirling’s formula, general, problem, 367

Index

445

Student distribution, problem and Tschebyscheff’s conditions, 305 footnote, 353 uniform convergence of sequence of t-distribution, problem and footfunctions, 172 note, 353 uniform limit function, 172 Taylor B., text and footnote, 284 uniform limit of improper integral, Taylor Power Series Theorem, 284 176 Taylor remainder, 284 uniform limit of series of functions, Technique combining Continuity 176 and Differentiability, ex- uniformly continuous function, ample, 130 definition, 93 Technique of Continuity, example, uniformly continuous functions, 124 facts, 94 Technique of Differentiability, ex- unit impulse function, example, ample, 137 395 Technique of Switching Order of Upper Incomplete Gamma FuncIntegration in Double Intion, project problem, 372 tegrals, 224 Techniques of Continuity, 120, 183 variance of a continuous random variable, 104 Techniques of Continuity and Difvolume of n-dimensional right cirferentiability, 119 cular cone, problem, 351 Techniques of Differentiability, 121, volume of n-dimensional ball, 320 183 volume of n-dimensional ellipsoid, Theorem of Lerch, 386 323 Thomae C. J., footnote, 30 Titchmarsh E. C., text and footWeierstraß M(x)-Test for uniform note, 394 convergence of improper Titchmarsh Theorem of convoluintegrals, 181 tion, 394 Weierstraß M-Test for uniform Tonelli Condition II, double inteconvergence of series of grals, 227 functions, 178 Tonelli Condition III, double inteWeierstraß M-Test, example, 179, grals, 227 196 Tonelli Conditions, double inteWeierstraß K. T. W., text and footgrals, 227 note, 122 Tonelli conditions, example, 232 Wolstenholme J., problem and Tonelli L., text and footnote, 227 footnote, 272 translate of a real function, 36 triangle inequality of norm funcZeta function, complex, Note, 216 tion, 365 Zeta function, real, integral repretriplication formula of the Gamma sentations, problem, 337 function, problem, 347 Zeta function, real, problem, 216 Tschebyscheff P. L., text and footnote, 305