Special Functions and their Application 8770226261, 9788770226264

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Special Functions and their Applications

RIVER PUBLISHERS SERIES IN MATHEMATICAL AND ENGINEERING SCIENCES Series Editors MANGEY RAM Graphic Era University, India TADASHI DOHI Hiroshima University, Japan ALIAKBAR MONTAZER HAGHIGHI Prairie View Texas A&M University, USA Mathematics is the basis of all disciplines in science and engineering. Especially applied mathematics has become complementary to every branch of engineering sciences. The purpose of this book series is to present novel results in emerging research topics on engineering sciences, as well as to summarize existing research. It engrosses mathematicians, statisticians, scientists and engineers in a comprehensive range of research fields with different objectives and skills, such as differential equations, finite element method, algorithms, discrete mathematics, numerical simulation, machine leaning, probability and statistics, fuzzy theory, etc. Books published in the series include professional research monographs, edited volumes, conference proceedings, handbooks and textbooks, which provide new insights for researchers, specialists in industry, and graduate students. Topics covered in the series include, but are not limited to: • • • • • • • • • • • • • • • • • • •

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The NEC and You Perfect Together: Special Functions and A Comprehensive Study ofApplications the their National Electrical Code

Bipin Singh Koranga Kirori Mal College, Delhi University, India

Gregory P. Bierals

Electrical Design Institute, USA

Sanjay Kumar Padaliya

S.G.R.R. (P.G.) College, India

Vivek Kumar Nautiyal Babasaheb Bhimrao Ambedkar University, India

River Publishers

Published 2021 by River Publishers

River Publishers Alsbjergvej 10, 9260 Gistrup, Denmark www.riverpublishers.com Distributed exclusively by Routledge

4 Park Square, Milton Park, Abingdon, Oxon OX14 4RN 605 Third Avenue, New York, NY 10017, USA

Special Functions and their Applications / by Bipin Singh Koranga, Sanjay Kumar Padaliya, Sanjay Kumar Padaliya. 2021 River Publishers. All rights reserved. No part of this publication may be reproduced, stored in a retrieval systems, or transmitted in any form or by any means, mechanical, photocopying, recording or otherwise, without prior written permission of the publishers. ©

Routledge is an imprint of the Taylor & Francis Group, an informa business

ISBN 978-87-7022-626-4 (print) While every effort is made to provide dependable information, the publisher, authors, and editors cannot be held responsible for any errors or omissions.

Contents

Preface

ix

List of Tables

xi

1

The Gamma Function

1

1.1

Definition of Gamma Function . . . . . . . . . . . . .

1

1.2

Gamma Function and Some Relations . . . . . . . . .

3

1.3

The Logarithmic Derivative of the Gamma Function . .

6

1.4

Asymptotic Representation of the Gamma Function

2

for Large |z| . . . . . . . . . . . . . . . . . . . . . . .

10

1.5

Definite Integrals Related to the Gamma Function . . .

11

1.6

Exercises . . . . . . . . . . . . . . . . . . . . . . . .

12

The Probability Integral and Related Functions

15

2.1

The Probability Integral and its Basic Properties . . . .

15

2.2

Asymptotic Representation of Probability Integral for Large |z| . . . . . . . . . . . . . . . . . . . . . . . . .

17

2.3

The Probability Integral of Imaginary Argument . . . .

18

2.4

The Probability Fresnel Integrals . . . . . . . . . . . .

20

v

vi Contents

3

4

5

2.5

Application to Probability Theory . . . . . . . . . . .

23

2.6

Application to the Theory of Heat Conduction . . . . .

24

2.7

Application to the Theory of Vibrations . . . . . . . .

26

2.8

Exercises . . . . . . . . . . . . . . . . . . . . . . . .

28

Spherical Harmonics Theory

31

3.1

Introduction . . . . . . . . . . . . . . . . . . . . . . .

31

3.2

The Hypergeometric Equation and its Series Solution .

32

3.3

Legendre Functions . . . . . . . . . . . . . . . . . . .

35

3.4

Integral Representations of the Legendre Functions . .

37

3.5

Some Relations Satisfied by the Legendre Functions . .

39

3.6

Workskian of Pairs of Solutions of Legendre’s Equation 40

3.7

Recurrence Relations for the Legendre Functions . . .

42

3.8

Associated Legendre Functions . . . . . . . . . . . . .

44

3.9

Exercises . . . . . . . . . . . . . . . . . . . . . . . .

46

Bessel Function

49

4.1

Bessel Functions . . . . . . . . . . . . . . . . . . . .

49

4.2

Generating Function . . . . . . . . . . . . . . . . . .

54

4.3

Recurrence Relations . . . . . . . . . . . . . . . . . .

57

4.4

Orthonormality . . . . . . . . . . . . . . . . . . . . .

59

4.5

Application to the Optical Fiber . . . . . . . . . . . .

60

4.6

Exercises . . . . . . . . . . . . . . . . . . . . . . . .

62

Hermite Polynomials

65

5.1

Hermite Functions . . . . . . . . . . . . . . . . . . .

65

5.2

Generating Function . . . . . . . . . . . . . . . . . .

69

6

Contents

vii

5.3

Recurrence Relations . . . . . . . . . . . . . . . . . .

70

5.4

Rodrigues Formula . . . . . . . . . . . . . . . . . . .

73

5.5

Orthogonality and Normalilty . . . . . . . . . . . . . .

74

5.6

Application to the Simple Harmonic Oscillator . . . .

76

5.7

Exercises . . . . . . . . . . . . . . . . . . . . . . . .

78

Laguerre Polynomials

81

6.1

Laguerre Functions . . . . . . . . . . . . . . . . . . .

81

6.2

Generating Function . . . . . . . . . . . . . . . . . .

85

6.3

Recurrence Relations . . . . . . . . . . . . . . . . . .

87

6.4

Rodrigues Formula . . . . . . . . . . . . . . . . . . .

91

6.5

Orthonormality . . . . . . . . . . . . . . . . . . . . .

92

6.6

Application to the Hydrogen Atom . . . . . . . . . . .

94

6.7

Associated Laguerre Polynomials . . . . . . . . . . .

98

6.7.1 6.8

Properties of Associated Laguerre Polynomials 102

Exercises . . . . . . . . . . . . . . . . . . . . . . . . 102

Bibliography

105

Index

107

About the Authors

109

Preface

We feel great pleasure in bringing out the first edition of the book “Special Functions and their Applications.” This book has been written especially in accordance with the latest and modified syllabus framed for UG and PG students. A reasonably wide coverage in sufficient depth has been attempted. The book contains sufficient number of problems. We hope that if a student goes through all these, he or she would appreciate and enjoy the subject. The work is dedicated to our past students whose inspiration motivated us to do this work without any stress and strain. We wish to express our indebtedness to numerous authors of those books that were consulates during the preparation of the matter. We feel great pleasure to express deepest sense of gratitude, respect, and honor to Prof. Uma Sankar, IIT Bombay and Prof. Mohan Narayan, Mumbai University for providing valuable guidance and for their patronly behavior. We would like to thank several of our colleagues at Kirori Mal College and SGRR(PG) College, Dehradun. We would like to thank Dr. Vinod Kumar, Assistant Prof. University of Lucknow, Dr. Imran Khan, Assistant Prof. Ramjas College and Dr. Shushil, Assistant Prof. Hindu College, for making useful suggestions during the preparation of this manuscript.

ix

x

Preface

Errors might have crept in here and there in spite of care to avoid them. We will be very grateful for bringing them to our notice. Suggestions or criticisms toward further improvement of the book shall be gratefully acknowledged. We shall appreciate receiving comments and suggestions, which can be sent to the following emails: [email protected] [email protected] [email protected] Thanks are due to Prof. Angelo Galanty, University of L’Aquila and Dr. Rakesh Joshi, Australia’s Global University for encouragement. We are deeply grateful to our family members who always have been a source of inspiration for us. Last, but not the least, we are thankful to the publisher of this book.

Dr. Bipin Singh Koranga Dr. Vivek Kumar Nautiyal Dr. Sanjay Kumar Padaliya

List of Tables

Table 5.1

Some Hermite Polynomials . . . . . . . . . . .

70

Table 6.1

Some Laguerre Polynomials . . . . . . . . . . .

88

xi

1 The Gamma Function

1.1 Definition of Gamma Function One of the simplest and most important special functions is the gamma function. Its properties are used for the study of many other special functions, the cylinder functions, and the hypergeometric function. The gamma function is usually studied in courses on complex variable theory and in advanced calculus. The gamma function is defined by the following: ˆ∞ e−t tz−1 dt,

Γ(z) =

Re z  0,

(1.1)

0

whenever the complex variable z has a positive real part Re z. We can write (1.1) as the sum of two integrals, ˆ1

ˆ∞ e−t tz−1 dt +

Γ(z) = 0

e−t tz−1 dt,

(1.2)

1

where it can easily be shown that the first integral defines a function P (z), which is analytic in the half plane Re z > 0, while the second integral defines an entire function. If follows that the function Γ(z) = P (z) + Q(z) is analytic in the half-plane Re z > 0. The values of Γ(z)

1

2

The Gamma Function

in the rest of the complex plane can be found by analytic continuation of the function defined by (1.1). First, we replace the exponential in the integral for P (z) by its power expansion and we integrate term by term, we obtaining ˆ1 P (z) =

=

t 0 ∞ X k=0

z−1

dt

∞ X (−1)k k=0

k|

k

t =

ˆ1 ∞ X (−1)k k=0

k|

tk+z−1 dt

0

k

(−1) 1 , k| z + k

(1.3)

where it is permissible to reverse the order of integration and summation since ˆ1 |t 0

ˆ ˆ ∞ k ∞ X X t (−1)k k x−1 t | = t dt = et tx−1 dt < ∞ |dt | k| k| k=0 k=0 1

z−1

1

0

0

The last integral converges for x = Re z > 0. The terms of the series (1.3) are analytic functions of z, if z 6= 0, −1, −2.... In the region |z + k| > δ > 0,

k = 0, 1, 2....,

(1.3) is expressed by the convergent series ∞ X 1 , k|δ k=0

and, hence, it is uniformly convergent in this region. We conclude that the sum of the series (1.3) is a meromorphic function with simple poles at the points z = 0, −1, −2.... For Re z > 0, this function concides with the integral P (z) and, hence, is the analytic continuation of P (z). The function Γ(z) differs from P (z) by the term Q(z), which

1.2 Gamma Function and Some Relations

3

is an entire function. Therefore, Γ(z) is a meromorphic function of the complex variable z, with simple poles at the points z = 0, −1, −2.... An analytic expression for Γ(z), suitable for defining Γ(z) in the whole complex plane, is given by Γ(z) =

∞ X (−1)k

k|

k=0

1 + z+k

ˆ∞ e−t tz−1 dt,

z 6= 0, −1, −2, .... (1.4)

1

It follows from (1.4) that Γ(z) that has the representation Γ(z) =

∞ X (−1)n k=0

n|

1 + Ω(z + n) z+n

(1.5)

in a neighborhood of the pool z = −n (n = 0, 1, 2...), with regular part Ω(z + n). 1.2 Gamma Function and Some Relations We now consider three basic relations satisfied by the gamma function Γ(z + 1) = zΓ(z),

(1.6)

π , (1.7) sinπz √ 1 22z−1 Γ(z)Γ(z + ) = πΓ(2z). (1.8) 2 These expressions play an important role in various transformations Γ(z)Γ(1 − z) =

and calculations involving Γ(z). To prove (1.1), we assume that Re > 0 and use the integral representation (1.1). An integration by parts gives ˆ∞

ˆ∞ e−t t2 dt = −e−t t2 |∞ 0 +z

Γ(z + 1) = 0

e−t tz−1 dt = zΓ(z). (1.9) 0

4

The Gamma Function

The validity of this result for arbitrary complex z 6= 0, −1, −2, ... is an immediate consequence of the principle of analytic continuation; both sides of the expression are analytic everywhere except at the points z = 0, −1, −2... To derive (1.7), we assume that 0 < Re z < 1, and, again, we use (1.6); we get ˆ∞ ˆ∞ e−(s+t) s−z tz−1 dsdt.

Γ(z)Γ(1 − z) = 0

(1.10)

0

We introduce the new variables t v= . s

u = s + t, We find that ˆ∞ ˆ∞ Γ(z)Γ(1 − z) = 0

dudv e−u v z−1 = 1+v

0

ˆ∞

π v z−1 dv = . 1+v sinπz

0

The above expression is valid in the complex plane except at the points z = 0, ±1, ±2, ... To prove (1.8), we assume that Re z > 0 and then use (1.6), and we get ˆ∞ ˆ∞ 2z−1

2

Γ(z)Γ(z + 1) = 0

√ e−(s+t) (2 st 2z−1 t−1/2 dsdt

0

ˆ∞ ˆ∞ 2 +β 2 )

e−(α

=4 0

(2αβ)2z−1 αdαdβ,

0

where we have introduced new variables α =

√ √ s, β = t. To this

expression, we add the similar expression by permuting α, β. This

1.3 The Logarithmic Derivative of the Gamma Function

5

gives the more symmetric expression ˆ∞ ˆ∞ 1 2 2 2z−1 2 Γ(z)Γ(z + ) = 2 e−(α +β ) (2αβ) 2z−1 (α + β)dαdβ 2 0

0

ˆ∞ ˆ∞ e−(α

=4 0

2 +β 2 )

(2αβ)2z−1 (α + β)dαdβ,

0

where the last integral is over the sector σ; 0 6 α < ∞, 0 6 β 6 α. Introducing new variable u = α2 + β 2 ,

v = 2αβ,

we find that 2z−1

2

1 Γ(z)Γ(z + ) = 2

ˆ∞

ˆ∞ v

2z−1

dv

0

ˆ∞ 0

e−u du u−v

0

ˆ∞ 2

e−v v 2z−1 dv

=2

√

e−w dw =

√

π Γ(2z).

0

This result can be extended to arbitrary complex values z 6= 0, − 21 , −1, − 23 , ... by using the principle of analytic continuation. We now use (1.6), and noting that Γ(z) = 1, we find by mathematical induction that Γ(n + 1) = n|,

n = 0, 1, 2...

(1.11)

Putting z = 21 , we obtain 1 Γ( ) = 2

ˆ∞

ˆ∞ −t −1/2

e t 0

2

e−u du =

dt = 2

√

π,

(1.12)

0

and then (1.6) implies 1 1.3.5........(2n − 1) √ Γ(n + ) = π, 2 2n

n = 1, 2, ...

(1.13)

6

The Gamma Function

1.3 The Logarithmic Derivative of the Gamma Function The theory of the gamma function is intimately related to the theory of another special function, the logarithmic derivative of Γ(z) 0

ψ(z) =

Γ (z) . Γ(z)

(1.14)

Since Γ(z) is a meromorphic function with no zeros, ψ(z) can have no singular points other than the poles z = −n (n = 0, 1, 2....) of Γ(z). It follows from (1.5) that ψ(z) has the representation ψ(z) = −

1 + Ω(z + n) z+n

(1.15)

in a neighbourhood of the point z = −n and ψ(z) is a meromorphic function with simple poles at the point z = 0, −1, −2... The function ψ(z) satisfied relations obtained from expression (1.6)–(1.8) by taking logarithmic derivatives. In this way, we find that ψ(z + 1) =

1 + ψ(z), z

(1.16)

ψ(1 − z) − ψ(z) = πcotπz, (1.17) 1 ψ(z) + ψ(z + ) + 2log2 = 2ψ(2z). (1.18) 2 These formulas can be used to calculate ψ(z) for special values of z. For example, writing 0

ψ(1) = Γ (1) = −Υ,

(1.19)

where Γ = 0.57721566... is Euler’s constant, and we obtain ψ(n + 1) = −γ +

n X 1 k=1

k

,

n = 1, 2........

(1.20)

1.3 The Logarithmic Derivative of the Gamma Function

Moreover, substituting z =

1 2

7

into (1.18), we find that

1 ψ( ) = −γ − 2log2, 2

(1.21)

which then gives n X 1 1 ψ(n + ) = −γ − 2log2 + 2 , 2 2k − 1 k=1

n = 1, 2, .......

(1.22)

The function ψ(z) has simple representations in the form of definite integrals involving the variable z as a parameter. To derive these representations, we first note that ˆ∞ 0 Γ (z) = e−t tz−1 logtdt,

Re z > 0.

(1.23)

0

If we replace the logarithm in the integrand by ˆ∞ −x e − e−xt dx, Re t > 0, logt = x

(1.24)

0

we find that ˆ∞

0

Γ (z) =

dx x

0

ˆ∞ =

ˆ∞ (e−x − e−xt )e−t tz−1 dt, 0

dx −x [e Γ(z) − x

ˆ∞ e−t(x+1) tz−1 dt]. 0

0

Introducing the new variable of integration u = t(x + 1), we find that the integral in brackets equals (x + 1)−z Γ(z). This leads to the following integral representation of ψ(z) : ˆ∞ ψ(z) = [e−x − 0

1 dx ] , 2 (x + 1) x

Re z > 0.

(1.25)

8

The Gamma Function

To obtain another integral representation of ψ(z), we write (1.25) in the form ˆ∞ e−x −

ψ(z) = limδ→0 [

1 dx ] (x + 1)2 x

0

ˆ∞ −x ˆ∞ e 1 = limδ→0 [ dx − ], x (x + 1)2 x 0

0

and change the variable of integration in the second integral by setting x + 1 = et . This gives ˆ∞ ψ(z) = limδ→0 [ 0

ˆ∞

ˆ∞

et dt − t

log(1+δ)

e−tz e−t − )dt − ( t 1 − e−t

limδ→0 [

e−tz dt] 1 − e−t

log(1+δ)

ˆ∞

e−tz dt], 1 − e−t

log(1+δ)

and therefore, since the second integral integral approaches zero as δ → 0,

ˆ∞ t e e−tz ψ(z) = ( − )dt, t 1 − e−t

Re z > 0.

(1.26)

0

Setting z = 1 and subtracting the result from (1.26), we find that ˆ∞ ψ(z) = −γ +

e−t − etz dt, 1 − e−t

Re z > 0,

(1.27)

1 − xz−1 dx, 1 − x−

Re z > 0,

(1.28)

0

or

ˆ1 ψ(z) = −γ + 0

1.4 Asymptotic Representation of the Gamma Function for Large |z|

9

where we have introduced the new variable of integration x = e−t . From expression (1.28), we can deduce an important representation of ψ(z) as an analytic expression valid for all z 6= 0, −1, −2, ....... To obtain this representation, we substitute the power series expansion (1 − x)−1 = 1 + x + x2 + x3 + ......xn ......,

0≤x 0,

n = 0, 1, 2........and |z| < a,

since 1 a+1 1 − |< n+1 n+z (n + 1)(n − a) for n > N > a, and the series |

∞ X n=N

a+1 (n + 1)(n − a)

converges. Therefore, since δ is arbitrarily small and a arbitrarily small and a arbitrarily large. The following infinite product representation of the gamma function: ∞ X 1 z γz e−z/n (1 + ). =e Γ(z + 1) n n=1

(1.30)

This formula can be made the starting point for the theory of the gamma function.

10 The Gamma Function 1.4 Asymptotic Representation of the Gamma Function for Large |z| To describe the behavior of a given function f (z) as |z| → ∞ within a sector α 6 arg z 6 β, it is, in many cases, sufficient to derive an expression of the form f (z) = φ(z)[1 + r(z)].

(1.31)

Here, φ(z) is a function of a simpler structure than f (z), and r(z) converges uniformly to zero as |z| → ∞ within the given sector. Formulas of this type are called asymptotic representations of f (z) for large |z|. It follows from (1.31) that the ratio f (z)φ(z) converges to unity as |z| → ∞; the two functions f (z) and φ(z) are asymptotically a fact we indicates by writing |z| → ∞,

f (z) ≈ φ(z),

α 6 arg z 6 β. (1.32)

An estimate of |r(z)| gives the size of the error committed when f (z) is replaced by φ(z) for large but finite |z|. We now look for a description of the behavior of the function f (z) as |z| → ∞, which is more exact than by (1.31). Suppose we derive the expression " N # X f (z) = φ(z) an z −n + rN (z) , ao = 1,

N = 1, 2, ......,

n=0

(1.33) where z N rN (z) converges uniformly to zero as |z| → ∞,

α 6

arg z 6 β. Then we write f (z) ≈

N X n=0

an z −n ,

|z| → ∞,

α 6 arg z 6 β,

(1.34)

1.5 Definite Integrals Related to the Gamma Function

11

and the right-hand side is called an asymptotic series or asymptotic expansion of f (z) for large |z|. It should be noted that this definition does not exist that the given series converges in the ordinary sense, and, on the contrary, the series will usually diverge. Asymptotic series are very useful, since, by taking a finite number of terms, we can obtain an arbitarily good approximation to the function f (z) for sufficiently large |z|. 1.5 Definite Integrals Related to the Gamma Function The class of integrals that can be expressed in terms of the gamma function is very large. Here, we consider few examples with the intent of deriving some formulas that will be needed later. Our first result is the expression ˆ ∞ Γ(z) e−pt tz−1 dt = z , p 0

Re p > 0,

Re z > 0, (1.35)

which is easily proved for positive real p by making the change of variables s = pt, and then using the integral ˆ∞ e−t tz−1 dt,

Γ(z) =

Re z  0.

0

The extension of (1.35) to arbitrary complex p with Re p > 0 is accomplished by using the principle of analytic continuation. Now consider the integral ˆ 1 B(x, y) = tx−1 (1 − t)y−1 dt, 0

Re x > 0,

Re y > 0, (1.36)

12 The Gamma Function known as the beta function. It is easy to see that (1.36) represents an analytic function in each of the complex variables x and y. If we introduce the new variable on integration u = t/(1 − t), then (1.36) becomes

ˆ

B(x, y) = 0

∞

ux−1 du, (1 + u)x+y

Re x > 0,

Re y > 0. (1.37)

Setting p = 1 + u, z = x + y in (1.35), we find that 1 1 = (1 + u)x+y Γ(x + y)

ˆ∞ e−(1+u)t−x+y−1 dt,

(1.38)

0

and substituting the result into (1.37), we obtain 1 B(x, y) = Γ(x + y)

ˆ∞

ˆ∞ e−t tx+y−1 dt

0

Γ(x) = Γ(x + y)

e−ut ux−1 du 0

ˆ∞ e−t ty−1 dt =

Γ(x)Γ(y) . Γ(x + y)

(1.39)

0

Thus, we have derived the formula B(x, y) =

Γ(x)Γ(y) , Γ(x + y)

(1.40)

relating the beta function to the gamma function, which can be used to derive all the properties of the beta function. 1.6 Exercises 1. Prove that ˆ π/2 ˆ v cos θdθ = 0

0

π/2

√

π Γ( v+1 ) 2 sin θdθ = , v 2Γ ( 2 + 1) v

Re v − 1,

1.6 Exercises ˆ

π/2

cosv θsinv θdθ =

v+1 1Γ( µ+1 2 )Γ( 2 )

2Γ( µ+v 2 + 1)

0

,

Re µ > −1,

13

Re v > −1.

2. Verify the formula Γ(3z) =

33z−1/2 1 2 Γ(z)Γ(z + )Γ(z + ) 2π 3 3

3. Prove that |Γ(iy)|2 =

π , ysinh πy

1 π |Γ( + iy)|2 = , 2 ycosh πy

for real y. 4. Derive the formula 2 1 3Ψ(3z) = Ψ(z) + Ψ(z + ) + Ψ(z + ) + 3log3. 3 3 5. Derive the asymptotic formula |Γ(x + iy)| =

√

2π e−1/2π|y| |y|x−1/2 [1 + r(x, y)],

where as |t| → ∞, r(x, y) → 0 uniformly in the strip |x| 6 α (α is a constant). 6. Derive the expressions γ(z + 1, α) = zγ(z, α) − e−α α2 Γ(z + 1, α) = zΓ(z, α) − e−α α2

2 The Probability Integral and Related Functions

2.1 The Probability Integral and its Basic Properties The probability integral is meant to be the function defined for any complex z by the integral 2 φ(z) = √ π

ˆz 2

e−t dt,

(2.1)

0

calculated along an arbitrary path joining the origin to the point t = z. The form of this path does not matter. The integrand is an entire function of the complex variable t. We can assume that the integration is along the line segment joining the point t = 0 and t = z. According to a familiar theorem of complex variable theory, φ(z) is an entire function and hence can be expanded in convergent power series for 2

any value of z. To find this expansion, we need to only replace e−t by its power series in (2.1) and then integrate the term obtaining 2 φ(z) = √ π

ˆz X ∞ 0

k=0

∞ (−1)k t2k 2 X (−1)k z 2k+1 dt = √ , k| π k=0 k|(2k + 1)

|z| < ∞. (2.2)

15

16 The Probability Integral and Related Functions It follows from (2.2) that φ(z) is an odd function of z. For real values of z, φ(z) is a real monotonically increasing function. At zero, we have φ(0) = 0, and as z increases, φ(z) rapidly approaches the limiting value φ(∞) = 1, since ˆ

∞

e

−t2

0

√ π . dt = 2

(2.3)

The difference between φ(z) and this limit can be written in the form 2 1 − φ(z) = √ π

ˆ

∞

e

−t2

0

2 dt − √ π

ˆ

z

−t2

e 0

2 dt = √ π

ˆ

∞

2

e−t dt. 0

(2.4)

The probability integral is encountered in many branches of applied mathematics, probability theory of errors, the theory of heat conduction, and various branches of mathematical physics. The two functions related to the probability integral, the error function √ ˆ z π −t2 φ(z), e dt = Erf z = 2 0

(2.5)

and its complement ˆ Erf c z =

∞

√ −t2

e 0

dt =

π [1 − φ(z)]. 2

(2.6)

Many more complicated integrals can be expressed in terms of the probability integral. For example, by differentiation of the parameter z, it can be shown that ˆ 2 √ 2 ∞ e−zt z dt = e [1 − φ( z)]. π 0 1 + t2

(2.7)

2.2 Asymptotic Representation of Probability Integral for Large |z|

17

2.2 Asymptotic Representation of Probability Integral for Large |z| To find an asymptotic representation of the function φ(z) for large |z|, We apply repeated integration by parts to the given integral ˆ ∞ ˆ ˆ 2 2 1 ∞ 1 −t2 1 ∞ e−t e−z −t2 e dt = − d(e ) = − dt 2 z t 2z 2 z t2 z ˆ 2 2 2 e−z 1.3 ∞ e−t e−z − 2 3+ 2 dt = 2z 2z z z t4  1 1 1.3 1.3.5 −z 2 =e − 2 3 + 3 5 − 4 7 + .......... 2z 2 z 2z 2z  n 1.3........(2n − 1) +(−1) 2n+1 z 2n+1 +

1.3........(2n − (−1)n+1 2n+1 z 2n+1

1)

ˆ

∞

z

2

e−t dt. t2n+1

It follows that " # 2 n X e−z 1.3.....(2k − 1) 1 − φ(z) = √ 1+ (−1)k + rn (z) , 2 )k (2z πz k=1

(2.8)

where

rn (z) =

1.3......(2n (−1)n+1 2n

+ 1)

ˆ ze

z2 z

∞

2

e−t dt. t2n+2

(2.9)

where δ is an arbitrarily small positive number, and choose the path of integration in (2.9) to be the infinite line segment beginning at the point t = z and parallel to the real axis. If z = x + iy = reiΦ , then

18 The Probability Integral and Related Functions this segment has the equation t = u + iy (x 6 u < ∞), and on the segment, we have |ez

2 −t2

| = ex

2 −u2

,

|t|−(2n+3) 6 |z|−(2n+3) ,

so that 1.3......(2n + 1) secφ |rn (z)| < 2n |z|2n+2

ˆ

∞

2

ex

−u2

udu =

z

|t| 6 usecφ. 1.3......(2n + 1) secφ, (2|z|2 )n+1

which implies |rn (z)| 6

1.3......(2n + 1) 1.3......(2n + 1) secφ 6 . 2 n+1 (2|z| ) (2|z|2 )n+1 sinδ

(2.10)

It follows from (2.10) that as |z| → ∞ the product z 2n rn (z) converges uniformly to zero in the indicated sector, # " 2 n X e−z 1.3.....(2n − 1) , 1 − φ(z) ≈ √ (−1)n 1+ (2z 2 )n πz k=1

(2.11)

π − δ. 2 Thus, the series on the right is the asymptotic series of the function |z| → ∞,

|arg z| 6

1−φ(z), and a bound on the error committed in approximating 1−φ(z) by the sum of a finite number of terms of the series is given by (2.10). For positive real z, this error does not exceed the first neglected term is absolute value. 2.3 The Probability Integral of Imaginary Argument In the applications, one often encounters the case where the argument of the probability integral is a complex. We now examine the particular simple case where z = ix is a purely imaginary. Choosing a segment

2.3 The Probability Integral of Imaginary Argument

19

of the imaginary axis as the path of integration, and making the substitution t = iu, we find from (2.11) that ˆ x Φ(ix) 2 2 =√ eu du. (2.12) i π 0 The integration in the right increases without limit as x → ∞, and, therefore, it is more convenient to consider the function ˆ z 2 −z 2 F (z) = e eu du,

(2.13)

0

which remains bounded for all real z. In the general case of complex z, F (z) is an entire function, and the choice of the path of integration in (2.13) is complete. To expand F (z) in power series, we note that F (z) satisfies the linear differential equation 0

F (z) + 2zF (z) = 1,

(2.14)

with initial condition F (0) = 0. Substituting the series F (z) =

∞ X

ak z k

k=0

into (2.14), and comparing coefficients if identical powers of z, we obtain recurrence relation a0 = 0,

a1 = 1,

(k + 1 = 1)ak+1 + 2ak−1 = 0.

After some simple calculations, this leads to the expansion ∞ X (−1)k z 2k+1 , F (z) = 1.3.....(2k + 1) k=0

|z| < ∞.

(2.15)

To study the behavior of F (z) as z → ∞ for real z, we apply L’Hospital’s rule twice to the ratio ´z 2 2z 0 eu du , ez 2

20 The Probability Integral and Related Functions and then use (2.14) to deduce that limz→∞ 2zF (z) = 1, and 1 , z → ∞. (2.16) 2z The maximum of the function occurs at z = 0.924... and equals F (z) ≈

Fmax = 0.541... The function F (z) comes up in the theory of propagation of electromagnetic waves along the earth’s surface, and in other problems of mathematical physics. 2.4 The Probability Fresnel Integrals Another interesting case from the stand point of the application occurs, when the argument of the probability integral is the complex number z=

√

x ix = √ (1 + i), 2

where x is real. In this case, we choose the path of the integration in (2.11) to be a segment of the bisector of the angle between the real and √ imaginary axes. Then, using the formula t = iu to introduce the new variable u, we find from (1.1) that √ ˆ x ˆ x ˆ x φ( ix) 2 2 2 −iu2 2 √ =√ e du = √ cosu du − i √ sinu2 du. π 0 π 0 π 0 ( i) (2.17) The integrals on the right can be expressed in terms of the functions ˆ x ˆ x πt2 πt2 C(z) = cos dt, S(z) = sin dt, (2.18) 2 2 0 0

2.4 The Probability Fresnel Integrals

21

where the integration is along any path joining the origin to the point t = z. The function C(z) and S(z) are known as the Fresnel integrals. Since the integrands in (2.28) are functions of the complex variable t, the choice of the path of integration does not matter, and both C(z) and S(z) are entire functions of z. For real z = x, the Fresnel integrals are real. Both C(x) and S(x) vanish for x = 0, and have an oscillatory character, as follows from the expressions πx2 πx2 0 , S (x) = sin , 2 2 √ which shows that C(x) has extrema at x = ± 2n + 1, which S(x) √ has extrema at x = ± 2n(n = 0, 1, 2...). The largest maxima are √ C(1) = 0.779893... and S( 2) = 0.713972.., respectively. As x → 0

C (x) = cos

∞, each of the functions approaches the limit 1 C(∞) = S(∞) = , 2 as implied by the familiar expression √ ˆ ∞ ˆ ∞ π 2 2 cost dt = sint dt = √ . (2.19) 2 2 0 0 Replacing the trigonometric functions in the integrands in (2.58) by their power series expansions, and integrating term by term, we obtain the following series expansions for the Fresnel integrals ˆ zX ∞ (−1)k  π 2k z 4k+1 , C(z) = 2 4k + 1 0 k=0 (2k)| ˆ

∞ zX

 2 2k+1 (−1)k πt S(z) = dt = C(z) 2 0 k=0 (2k + 1)| ∞ X (−1)k  π 2k+1 z 4k+3 = . (2k + 1)| 2 4k + 3 k=0

22 The Probability Integral and Related Functions The relation between the Fresnel integrals and the probability integral is given by the formula ˆ

z

ˆ √π/2ze±πi/4

r

2 ±πi/4 dt = e π 0 r  0 1 π ±πi/4 = √ e±πi/4 φ ze , 2 2 2 /2

e±πt

C(z) ± iS(z) =

2 du

e−u

(2.20)

which implies  r   r 1 π −πi/4 π πi/4 −πi/4 πi/4 ze ze −e Φ , S(z) = √ e Φ 2 2 2i 2 (2.21)  r  r  1 π π C(z) = √ eπi/4 Φ ze−πi/4 + e−πi/4 Φ zeπi/4 . 2 2 2 2 (2.22) Using (2.21) and (2.22), we can derive the properties of C(z) and S(z) from the corresponding properties of C(z) and S(z) of the probability integral. The following asymptotic representations of the Fresnel integrals, valid for large |z| in the sector |argz| 6   1 πz 2 1 πz 2 C(z) = − B(z)cos , − A(z)sin 2 πz 2 2   1 1 πz 2 πz 2 S(z) = − A(z)cos + B(z)sin , 2 πz 2 2 where A(z) =

k=n X (−1)k α2k k=0

B(z) =

k=n X k=0

(πz 2 )2k

+ O(|z|−4N −4 ),

(−1)k α2k+1 + O(|z|−4N −6 ), (πz 2 )2k+1

αk = 1.3......(2k − 1),

α0 = 1.

π 4

− δ, (2.23) (2.24)

2.5 Application to Probability Theory

23

The Fresnel integrals come up in various branches of physics and engineering, diffraction, and theory of vibrations. Many integrals of more complicated type can be expressed in terms of the functions C(z) and S(z). 2.5 Application to Probability Theory Normal or Gaussian random variable with mean m and standard deviation σ is meant to be a random variable ξ such that the probability of ξ lying in the interval [x, x + dx] is given by the expression √

1 2 2 e−(x−m) /2σ dx. 2πσ

(2.25)

The probability P {a 6 ξ − m 6 b}

(2.26)

that ξ − m lies in the interval [a, b] is just the integral ˆ a/√2σ 1 2 e−t dt e dx = √ √ π a/ 2σ a+m      1 b a = Φ √ −Φ √ , (2.27) 2 2σ 2σ where Φ(x) is the probability integral. Setting a = −δ, b = δ, we 1 √ 2πσ

ˆ

b+m

−(x−m)2 /2σ 2

obtain the probability that |ξ − m| does not exceeds δ : δ P {|ξ − m| 6 δ} = Φ( √ ). 2σ

(2.28)

The the probability that |ξ − m| exceed δ is just δ P {|ξ − m| > δ} = 1 − Φ( √ ). 2σ

(2.29)

24 The Probability Integral and Related Functions The value δ = δp for which (2.28) and (2.29) are equal is called the probable error, and it clearly satisfies the equation δp 1 Φ( √ ) = . 2 2σ Using a table of the function Φ(x) to solve the equation, we find that δp = 0.67449σ.

(2.30)

2.6 Application to the Theory of Heat Conduction Consider the following problem in the theory of heat conduction. An object occupying the half-surface x > 0 is initially heated to temperature To . It then cools off by radiating heat through its surface x = 0 into the surrounding medium which is at zero temperature. We want to find the temperature T (x, t) of the object as a function of position x and time t. Let the object have thermal conductivity k, heat capacity c, density ρ, and emissivity λ, and let τ = kt/cρ. Then our problem reduces to the solution of the equation of heat conduction a2 T aT = , aτ ax2

(2.31)

subject to the initial condition T |τ =0 = T0

(2.32)

and the boundary conditions (

aT − hT )|x=0 = 0, aτ

where h = λ/k>0.

T |x→∞ ,

(2.33)

2.6 Application to the Theory of Heat Conduction

25

To solve the problem, we introduce the Laplace transformation T¯ = T¯(x, p) of T = T (x, τ ), defined by the expression ˆ∞ T¯ =

e−pτ T dτ,

Re p > 0.

(2.34)

0

A system of equations determining T¯ can be obtained from (2.31) to (2.33) if we multiply the first and third equations by e−pτ and integrate from 0 to ∞, taking the second equation into account. The result is d2 T¯ = pT¯ − To , dx2 dT¯ − hT¯|x=0 , dx Equation (2.35) has the solution √ To h T¯ = (1 − √ e− px ), p h+ p

To T¯|x→∞ = . p

Re p > 0,

√ Re p > 0.

(2.35)

(2.36)

We can now solve for T by inverting (2.34). This can be done by Laplace transform or Fourier–Millin inversion theorem, which state ˆ a+i∞ 1 (2.37) epτ T¯dp, T = 2πi a−i∞ where a is constantly greater than the real part of all the singular points of T¯. that

The quantity of greatest interest is the surface temperature of the object. If we, put x = 0 in (2.36), we find that   To 1 1 h ¯ T |x=0 = √ √ = To − √ . p( p + h) p − h2 p − h2 p

(2.38)

The simplest way to solve (2.38) for the original function T |x=0 is to use the convolution theorem, which states that if f1 and f2 are

26 The Probability Integral and Related Functions the Laplace transforms of f1 and f2 , then f˙ = f˙1 f˙2 is the Laplace transform of the function

ˆ

τ

f1 (t)f2 (τ − t)dt.

f (τ ) =

(2.39)

0

Since it easily verified that h f˙1 = √ , p

f˙2 =

1 p − h2

are the Laplace transform of h f1 = √ , πτ

2

f2 = eh τ ,

(2.39) implies  ˆ ∞  dt h 2 h2 τ √ = To eh τ T |x=0 = To e − √ π 0 t 2τ

T |x=0 = To eh



2 1− √ π

√  1 − Φ(h τ ) ,

ˆ

√ h τ

e

! −s2

ds ,

0

(2.40)

where Φ(x) is the probability integral. It follows from the asymptotic formula (2.11) that for large τ, the surface temperature falls off like √ 1/ τ : To T |x=0 ≈ √ , τ → ∞. (2.41) h πτ The temperature inside the object (x 6= 0) can also be expressed in closed form in terms of the probability integral. 2.7 Application to the Theory of Vibrations Consider an infinite rod of linear density ρ and Young’s modulus E, lying along the positive x-axis. Let I be the moment of inertia of a cross section of the rod about a horizontal axis through the center of

2.7 Application to the Theory of Vibrations

mass of the section, and let τ =

27

p EI/ρt. Suppose the end x = 0

satisfies a sliding condition while the end x = ∞ is clamped, and suppose a constant force Q is suddenly applied at the end x = 0. The displacement u = u(x, t) at an arbitrary point x > 0 of the rod is described by the system of equation a2 u a4 u + = 0, aτ 2 ax2 au |τ =0 = 0, (2.42) u|τ =0 = aτ au Q au a3 u |x=0, |x=0 = , u|x→∞ = 0, |x→∞ = 0. 3 ax ax EI ax To solve this system, we use the Laplace transform, as in the preceding section. Writing

ˆ

∞

e−pτ udτ,

u¯ =

Re p > 0,

(2.43)

0

we obtain the following equations for u¯: a4 u ¯ 0, + p2 u = 4 ax Q a3 u¯ |x=0 = , 3 ax EIp a¯ u ¯ u| |x→∞ = 0. x=0 = 0, ax Simple calculation then shows that ! √ √ p e− pix e− pix Q √ , Re p > 0, Re ±pi > 0. u¯ = − √ 2 2EIp i −pi pi (2.44) a¯ u |x=0, ax

To find u, we again use the convolution theorem, √

Q f¯1 = , EIp2

1 f¯2 = 2i

e− √

−pix

√

e− pix − √ −pi pi

!

28 The Probability Integral and Related Functions are the Laplace transforms of Q f1 = τ, EI

1 f2 = 2i

  x2 x2 sin + cos , 4τ 4τ

(2.39) implies u=

Q √

ˆ

EI 2π

0

τ



x2 x2 sin + cos 4t 4t



τ −t Qxτ √ dt = f EI t



 x √ , 2 τ (2.45)

where

ˆ ∞ x2 1 − 1 y2 f (x) = √ (siny 2 + cosy 2 ) 2 dy. (2.46) y 2π 0 The function f (x) can be expressed in terms of the Fresnel integrals C(z) and S(z), introduced in Section 2.4. Integrating (2.45) by parts twice, we find that  " 2 2 1 f (x) = 1 + x −C 3 2

r

2 x π

!#  r !# " 2 2 1 2 − 1− x −S 3 2 π

  2 2 2 2 cosx 2 sinx + (1 − x ) + √ . (1 + x ) x x 3 2π

(2.47)

2.8 Exercises (1) Show that the function √ π z2 φ(z) = e Φ(z) 2 0

satisfies the differential equation φ − 2zφ = 1, and use this fact to derive the expansion ∞

2z −z2 X (2z 2 )k √ , Φ(z) = e 1.3......(2k + 1) π k=0

|z| < ∞.

2.8 Exercises

29

(2) Use integration by parts to show that ˆ 1 2 Φ(x)dx = xΦ(x) + √ e−x + C. π ¯ be the laplace transform of the probability integral, (3) Let Φ ˆ ∞ ¯ e−px Φ(x)dx. Φ(p) = 0

Prove that 1 2 h p i ¯ Φ(p) = ep /4 1 − Φ( ) . 4 2 (4) Derive the integral representations ˆ ˆ ∞ 2 ∞ −t2 sin2zt −t2 e dt. F (z) = e sin2ztdt, Φ(z) = π 0 t 0 (5) Derive the following integral representations for the square of the probability integral: ˆ 2 2 4 1 e−z (1+t ) Φ (z) = 1 − dt, π 0 1 + t2 ˆ 2 2 4 ∞ e−z (1+t ) π 2 [1 − Φ(z)] = dt, |argz| 6 . 2 π 1 1+t 4 (6) Derive the formulas ˆ 2 −z2 1 −t2 −2zt 1 − Φ(z) = √ e e dt, π 0 ˆ 4 −2z2 ∞ −t2 −2√2zt 2 [1 − Φ(z)] = √ e e Φ(t)dt. π 0 2

3 Spherical Harmonics Theory

3.1 Introduction By spherical harmonics, we mean solutions of the linear differential equation  µ2 (1 − z )u − 2zu + v(v + 1) − u = 0, 1 − z2 2

00

0



(3.1)

where z is complex variable, and µ, v are parameters that can take arbitrary real or complex values. Equation(3.1) is encountered in mathematical physics when a system of orthogonal curvilinear coordinates to solve the boundary value problems of potential theory for certain special kinds of domains like the sphere, spheroid, and the simplest of these domains is the sphere, which gives rise to the term “spherical harmonics.” In the spherical case, the variable z takes real values in the interval (−1, 1), and the parameters µ and v are nonnegative integers, but boundary value problems with more complicated geometries lead to the consideration of more general values of z, µ, and v. For most applications, it is sufficient to assume that z is either a real variable in the interval (−1, 1) or a complex variable in the plane cut along the segment [ − ∞, 1], while v is an arbitrary real or complex number and µ = m is a non-negative integer (m = 0, 1, 2, 3...). 31

32 Spherical Harmonics Theory 3.2 The Hypergeometric Equation and its Series Solution Before considering the theory of spherical harmonics, it is appropriate to consider the problem of solving the linear differential equation 00

0

z(1 − z)u + [γ − (α + β + 1)z]u − αβu = 0,

(3.2)

where z is complex variable, and α, β, and γ are parameters which can take various real or complex values. Equation (3.1) is called the hypergeometric equation, and contains as special cases many differential equations encountered in the applications. Reducing (3.1) 00

to a standard form by dividing it by the coefficient of u , we obtain an equation whose coefficients are analytic functions of z in the domain 0 < |z| < 1 and have the point z = 0 as a simple pole or a regular point, depending on the values of the parameters α, β, and γ. It follows from the general theory of linear differential equations that (3.1) has a particular solution of the form u=z

s

∞ X

ck z k ,

(3.3)

k=0

where co 6= 0, s is a suitable chosen number, and the power series converges for |z| < 1. Substituting (3.2) into (3.1), we find that ∞ X

ck z

s+k−1

(s+k)(s+k−1+γ)−

k=0

∞ X

ck z s+k (s+k+α)(s+k+β) = 0,

k=0

(3.4) which gives the following system of equations for determining the exponent s and the coefficients ck : c0 s(s − 1 + γ) = 0,

3.2 The Hypergeometric Equation and its Series Solution

33

ck (s + k)(s + k − 1 + γ) − ck−1 (s + k − 1 + α)(s + k − 1 + β) = 0, k = 1, 2, ... Solving the first equation, we obtain s = 0 or s = 1 − γ. Suppose γ 6= 0, −1, −2, ... and choose s = 0. Then the coefficients ck can be calculated from the recurrence relation ck =

(k − 1 + α)(k − 1 + β) ck−1 , k(k − 1 + γ)

k = 1, 2...,

if we set co = 1, this implies ck =

(α)k (β)k , k|(γ)k

k = 0, 1, 2...,

where we have introduced the abbreviation (λ)o = 1,

(λ)k = λ(λ + 1)..........(λ + k − 1),

k = 1, 2... (3.5)

Thus, if γ 6= 0, −1, −2, ..., a particular solution of (3.2) is u = u1 = F (α, β; γ; z) =

∞ X (α)k (β)k k=0

k|(γ)k

zk ,

|z| < 1,

(3.6)

where the series on the right is known as the hypergeometric series. The convergence of this series for |z| < 1 follows from the general theory of linear differential equations. However, by using the ratio test, it can easily be proved without recourse to this theory that the radius of convergence of the series (3.6) is unity, except when the parameter α, β equals zero or a negative integer, in which case it reduces to polynomial. Similarly, choosing s = 1 − γ and assuming that γ 6= 2, 3, 4, ..., we obtain ck =

(k − γ + α)(k − γ + β) ck−1 , k(k + 1 − γ)

k = 1, 2, ...,

(3.7)

34 Spherical Harmonics Theory or ck =

(1 − γ + α)k (1 − γ + β)k , k|(2 − γ)k

k = 0, 1, 2, ...,

(3.8)

If we set co = 1. Thus if γ 6= 2, 3, 4, ..., a particular solution of (3.2) is u = u2 = z

1−γ

∞ X (1 − γ + α)k (1 − γ + β)k k=0

k|(2 − γ)k

= z 1−γ F (1−γ+α, 1−γ+β; 2−γ; z),

zk

|z| < 1, |argz| < π. (3.9)

Therefore, if γ 6= 0, 1, 2, ......., the two solutions (3.8) and (3.9) exist simultaneously and are linearly independent. Then the general solution of (3.1) can be written in the following form: u = AF (α, β; γ; z) + Bz 1−γ F (1 − γ + α, 1 − γ + β; 2 − γ; z), (3.10) where |z| < 1, |argz| < π, and A, B are arbitrary constants. We can obtain a number of other differential equations whose solutions can be expressed in terms of hypergeometric series. Thus, for example, setting z = t2 , we arrive at the differential equation t(1 − t2 )

1 1 2 du d2 u + 2[y − − (α + β + )t ] − 4αβtu = 0, (3.11) dt2 2 2 dt

with particular solutions u = u1 = F (α, β; γ; t2 ),

γ 6= 0, −1, −2, ...,

(3.12)

u = u2 = t2−2γ F (1 − γ + α, 1 − γ + β; 2 − γ; t2 ) |t| < 1, |argt| < π, γ 6= 2, 3, 4, ...,

(3.13)

which for nonintegral γ constitute a pair of linearly independent solutions of (3.11) in the domain 0 < |t| < 1.

3.3 Legendre Functions

35

3.3 Legendre Functions The simplest class of spherical harmonics consists of the Legendre polynomials. The solution of (3.2) for µ = 0 and arbitrarily real or complex solutions of the equation 00

0

(1 − z 2 )u − 2zu + v(v + 1)u = 0,

(3.14)

known as Legendre’s equation. To determine these functions, we first note that (3.14) can be reduced to the hypergeometric equation by making suitable changes to variables. In particular, the substitution t = 21 (1 − z) converts (3.14) into the equation 00

0

t(1 − t)u − 2zu + v(v + 1)u = 0,

(3.15)

which is a special case of (3.2) corresponding to α = −v,

β = v + 1,

γ = 1,

(3.16)

while the substitution t = z −2 , u = z −v−1 v converts (3.14) into the equation d2 v t(1−t) 2 + dt



     v 1 3 5 dv  v v+ − v+ t v = 0, − +1 + 2 2 dt 2 2 2 (3.17)

which is a special case of (3.14) corresponding to α=

v + 1, 2

β=

v 1 + , 2 2

3 γ=v+ . 2

Therefore, it follows from the result of the section that two particular solutions of (3.14) are u = u1 = F (−v, v + 1; 1;

1−z ), 2

|z − 1| < 2,

(3.18)

36 Spherical Harmonics Theory   √ v v 1 3 1 πΓ(v + 1) u = u2 = F + 1, + ; v + ; 2 , 2 2 2 2 z Γ(v + 32 )(2z)v+1 |z| > 1,

|arg z| < π, v 6= −1, −2, .........,

(3.19)

where F (α, β; γ; z) is the hypergeometric series. These solutions are called the Legendre of degree v of the first and second kinds, denoted by Pv (z) and Qv (z), respectively. Thus, we have 1−z Pv (z) = F (−v, v + 1; 1; ), |z − 1| < 2, (3.20) 2   √ πΓ(v + 1) v 1 3 1 v + 1, + ; v + ; 2 , Qv (z) = F 2 2 2 2 z Γ(v + 32 )(2z)v+1 |z| > 1,

|arg z| < π, v 6= −1, −2, .........,

(3.21)

The functions Pv (z) and Qv (z) are defined in certain restricted regions of the complex z-plane. To make the analytic continuation of Pv (z), the legendre function of the first kind, we use the formula 2 π

ˆπ/2 ( 21 )k 2k sin φdφ = , k|

k = 0, 1, 2...

(3.22)

0

to write (3.20) as  k ∞ X (−v)k (v + 1)k 1 − z Pv (z) = (k|)2 2 k=0  k ˆπ/2 ∞ 2 X (−v)k (v + 1)k 1 − z = sin2k φdφ. π k=0 2 ( 12 )k k|

(3.23)

0

To make the analytic continuation of Qv (z), the Legendre function of the second kind, we start the formula

√ ˆ ∞ πΓ(v + 1) v2 + 12 k v2 + 1 k dt


, = 3√ Γ v + 32 v2 + 34 k v2 + 45 k t2k+v+ 2 t − 1 1

(3.24)

3.4 Integral Representations of the Legendre Functions

Re v > −1,

37

k = 0, 1, 2...,

and we have



√ ∞ πΓ(v + 1) X v2 + 1 k v2 + 21 k 1

Qv (z) = , z 2k Γ v + 32 (2z)v+1 k=0 v + 32 k k|

ˆ ∞ ∞ v X + 34 k v2 + 54 k 1 dt 1 2
, = 3√ 3 v+1 2k 2k+v+ (2z) z v + 2 k k| 2 t t−1 1 k=0  ˆ ∞  1 v 3 v 5 3 1 dt = F + , + ; v + ; 2 2 v+ 3 √ . v+1 (2z) 2 4 2 4 2 z t t 2 t−1 1 (3.25)

The general solution u of the differential equation (3.2) can be written as a linear combination of Legendre functions of the first and second kinds, u = AQv (z) + BPv (z),

(3.26)

where |arg(z-1)| 0) and introducing a new variable of integration in (3.29) by setting α θ sinh = sinh sinΦ, 2 2 we find that ˆ cosh(v + 21 )Θ 2 α √ Pv (coshα) = dΘ. (3.30) π 0 2coshα − 2coshΘ For any real or complex value of the degree v, We write (3.30) in the form 1 Pv (coshα) = π and then setting

ˆ

α

−α

1

e−(v+ 2 )Θ √ dΘ, 2cosα − 2coshΘ

eΘ = coshα + sinhαcosψ. The Legendre function of the first kind, ˆ 1 π dψ Pv (coshα) = , π 0 (coshα + sinhαcosψ)v+1 where v is arbitrary.

(3.31)

The integral representations of Qv (z), the Legendre function of the second kind. Assuming that z = coshα(α > 0) and introducing a new variable of integration by setting Θ α sinh = sinh coshψ, 2 2

3.5 Some Relations Satisfied by the Legendre Functions

we find that 1 Qv (coshα) = π

ˆ

∞

α

39

1

e−(v+ 2 )Θ √ dΘ, 2coshΘ − 2coshα

(3.32)

for Re v > −1. Then writing eΘ = coshα + sinhαcosφ, we reduce (3.32) to the form ˆ ∞ dφ Qv (coshα) = v+1 , α > 0, Re v > −1. α (2coshφ − sinhαcoshφ) (3.33) 3.5 Some Relations Satisfied by the Legendre Functions The differential equation (3.2) does not change if we replace v by −v− 1 or z by −z, and, hence, it has solutions P−v−1 (z), Q−v−1 (z), Pv (−z), and Qv (−z), as well as Pv (z) and Qv (z). Since every three solutions of a second-order linear differential equation are linearly dependent, there must be certain functional relation between the solutions. The simplest such relation is given by the formula P−v−1 (z) = Pv (z). To obtain a relation connecting Pv (z), Qv (z),

(3.34) and Q−v−1 (z), we

assume temporarily that z > 1 and −1 < Re v < 0. In this case, −1 < Re(−v − 1) < 0, we have Qv (coshα) − Q−v−1 (coshα) = πcotvπPv (coshα), or sinvπ [Qv (z) − Q−v−1 (z)] = πcotvπPv (z).

(3.35)

40 Spherical Harmonics Theory Formula (3.35) remains valid for all z in the plane cut along [-∞, 1]. Similarly, the given relation for Legendre functions Qn−1/2 (z) = Q−n−1/2 (z).

(3.36)

Qv (−z) = e±vπi Qv (z), v 6= −1, −2, ..., (3.37) 2sinvπ Qv (z) = Pv (z)e±vπi − Pv (−z), (3.38) π where v 6= −1, −2, ..., and the upper limit is chosen if Im z > 0 and the lower sign if im z < 0. The relations (3.35)–(3.38) play an important role in the theory of spherical harmonics. In particular, it follows from (3.38) that 2sinvπ Qv (x + i0) = Pv (x)e−vπi − Pv (−x), π 2sinvπ Qv (x − i0) = Pv (x)evπi − Pv (−x), π if -11. (2) Derive the following representation of the Legendre function Pv (z) in terms of hypergeometric series:   v+1 v 2 , − ; 1; 1 − z , |1−z 2 |, |arg(z+1)| < π, Pv (z) = F 2 2  v   z+1 z−1 Pv (z) = F −v, −v; 1; , Re z > . 2 z+1 (3) Prove that if v is not an integer, then the asymptotic behavior as z = −1 of the Legendre function of the first kind and its derivatives is described by the formulas Pv (z) ≈

z+1 sinvπ log , π 2

sinvπ 1 . π z+1 (4) Derive the integral representations ˆ ∞ 1 e−tcoshα Io (tsinhα)tv dt, Pv (coshα) = Γ(v + 1) 0 ˆ ∞ 1 Qv (coshα) = e−tcoshα Ko (tsinhα)tv dt, Γ(v + 1) 0 ˆ ∞ 1 Pv (cosΘ) = e−tcosΘ Jo (tsinΘ)tv dt, Γ(v + 1) 0 0

Pv (z) ≈

3.9 Exercises

where |Imα| 6

π , 2

Re v > −1,

0 6 Θ < π,

and Jo (x), Io (x), and Ko (x) are Bessels functions. (5) Prove the formulas ˆ

1

Plm (x)Pnm (x)dx = 0, −1

ˆ

1

[Pnm (x)]2 dx = −1

2 (n + m)| , 2n + 1 (n − m)|

m = 0, 1, 2, ....., l = m, m + 1, ....., n = m, +1, ...... (6) Prove that ˆ 1 x2m P2n (x)dx = 22n+1 −1

Γ(2m + 1)Γ(m + n + 1) . Γ(m − n + 1)Γ(2m + 2n + 2)

47

4 Bessel Function

4.1 Bessel Functions Bessel function is the solution of second-order differential equation that has the form x2

dy d2 y + x + (x2 − n2 )y = 0 2 dx dx

(4.1)

and equation has an essential singularity at x = ∞ and n is any integer. Since x = 0, is a regular singularity of the equation. We can find the complementary function of the above equation by using the most generalized power series or Frobenius series: y=

∞ X

ak xk+p .

(4.2)

k=0

Taking the first and second derivative of (4.2), we get ∞

dy X = ak (k + p)xk+p−1 dx k=0

(4.3)

∞

d2 y X = ak (k + p)(k + p − 1)xk+p−2 dx2 k=0

49

(4.4)

50 Bessel Function Putting (4.2), (4.3), and (4.4) into (4.1), we get ∞ X

ak (k + p)(k + p − 1)x

k+p

+

k=0

∞ X

ak (k + p)xk+p

k=0

+ (x2 − n2 )

∞ X

ak xk+p = 0

(4.5)

k=0

Multiplying the above by x−p , we get ∞ X

k

ak (k +p)(k +p−1)x +

∞ X

k

2

2

ak (k +p)x +(x −n )

∞ X k=0

k=0

k=0

ak x k = 0 (4.6)

∞ X

ak xk (k + p)(k + p − 1) + (k + p) + x2 − n

k=0 ∞ X


2

=0

(4.7)


ak xk (k + p)(k + p − 1) + (k + p) + x2 − n2 = 0

(4.8)

k=0 ∞ X

∞
X ak xk (k + p)(k + p) − n2 + ak xk+2 = 0

k=0

(4.9)

k=0

0

Taking x cofficient in (4.9) we obtain, p2 − n2 = 0

(4.10)

(4.10) is called Indicial equation and it has two repeated roots, i.e, p = ±n. And for taking x1 coefficient in eq.(4.9) we obtain,
(p + 1)2 − n2 a1 = 0

(4.11)

=⇒ a1 = 0

(4.12)

Now (4.9) gives ∞ X k=0



ak xk (k + p)(k + p) − n2 + ak xk+2 = 0

(4.13)

4.1 Bessel Functions

51

Making (4.13) in xk by changing k → k − 2 in second term, we get ∞ X



xk ak (k + p)(k + p) − n2 + ak−2 = 0

(4.14)

k=0 ∞ X



xk ak (k + p)2 − n2 + ak−2 = 0

(4.15)

k=0

Again, (4.15) demands that each cofficients of xk must vanishes. We get,
ak (k + p)2 − n2 + ak−2 = 0

(4.16)

Taking root of (4.10) i.e. p = ±n, we get our recurrence relation. ak = −

1 ak−2 k(k ± 2n)

k = 2, 3, 4, 5, ... Now, take the integer values of k and using (4.12), we find a2 = −

k=2 k = 3,

a3 =

1 a0 , 2(2 ± 2n)

1 a1 , 3(3 ± 2n)

k = 3,

k = 4,

a3 = 0 1 1 a4 = a0 , 4(4 ± 2n) 2(2 ± 2n)

k = 5, a5 = 0, 1 1 1 k = 6, a6 = − a0 , 6(6 ± 2n) 4(4 ± 2n) 2(2 ± 2n) and so on. Putting value of a0 , a1 , a2 , a3 , ... in (4.2), we get  1 1 1 y =a0 1 − x2±n + x4±n 2(2 ± 2n) 4.(4 ± 2n) 2.(2 ± 2n)

(4.17)

52 Bessel Function  1 1 1 6±n − x + ... 6(6 ± 2n) 4(4 ± 2n) 2(2 ± 2n) y = a0 x

±n

 1−

(4.18)

1 1 1 x2 + x4 2(2 ± 2n) 4.(4 ± 2n) 2.(2 ± 2n)

 1 1 1 6 − x + ... (4.19) 6(6 ± 2n) 4(4 ± 2n) 2(2 ± 2n) y = a0 x

±n

 1−

 x 2  x 4 1 1 1 + (1 ± n) 2 2!(2 ± n) (1 ± n) 2

  x 6 1 1 1 − + ... 3!(3 ± n) (2 ± n) (1 ± n) 2 Now, choosing the value of a0 = denoted by Jn (x) J±n (x) =

we get the Bessel function

 x 2  x 4 1 1 1 1 ±n x (1 − + − 2±n Γ(1 ± n) (1 ± n) 2 2!(2 ± n) (1 ± n) 2

− J±n (x) =

1 , 2±n Γ(1±n)

(4.20)

 x 6 1 1 1 + ...) 3!(3 ± n) (2 ± n) (1 ± n) 2

(4.21)

 x ±n  x 2  x 4 1 1 1 1 (1 − + − Γ(1 ± n) 2 (1 ± n) 2 2!(2 ± n) (1 ± n) 2

 x 6 1 1 1 + ...) − 3!(3 ± n) (2 ± n) (1 ± n) 2 ∞ X

 x ±n+2m (−1)m J±n (x) = m! Γ(1 ± n + m) 2 m=0

(4.22)

(4.23)

where Γn is a Gamma function and it is given by Γn = n(n − 1)!.

4.1 Bessel Functions

53

The general solution of eq.(4.1) has the form, y = AJn (x) + BJ−n (x)

(4.24)

where, Jn (x) and J−n (x) are called Bessel function of first kind and it is governed by (4.23). Since Bessel polynomials (J±n (x)) are solutions of (4.1). Hence, it satisfies the (4.1), and we get 00

0

x2 J±n (x) + xJ±n (x) + (x2 − n2 )J±n (x) = 0.

(4.25)

For examples, for n = 1/2, (4.1) becomes x2

d2 y dy 1 +x + (x2 − )y = 0 2 dx dx 4

(4.26)

The solution of (4.26) is found by putting the value of n = 1/2 in (4.23) and we get J±1/2 (x) =

∞ X

 x ±1/2+2m (−1)m m! Γ(±1/2 + m + 1) 2 m=0

(4.27)

(4.27) contains two series. The series form of Bessel function is found √ by (4.27) using the relation, i.e., Γ(m+1) = mΓm ; and Γ1/2 = π J1/2 (x) =

∞ X

 x 1/2+2m (−1)m m! Γ(1/2 + m + 1) 2 m=0

J1/2 (x) = J1/2 (x) = J1/2 (x) =

∞ X

 x 1/2+2m (−1)m m! Γ(m + 23 ) 2 m=0

x1/2 x5/2 x9/2 x13/2 − + − + ... 21/2 Γ 32 1!25/2 Γ 52 2!29/2 Γ 27 3!213/2 Γ 29

x1/2 21/2 12 Γ 21

−

x5/2 1!25/2 32 12 Γ 21

+

x9/2 2!29/2 52 23 12 Γ 21

−

x13/2 3!213/2 72 25 32 21 Γ 12

+ ...

54 Bessel Function J1/2 (x) =

x5/2 x9/2 x13/2 x1/2 − + − √ √ √ √ + ... 1 3 5 1 3 1 21/2 2 π 25/2 2 2 π 2.29/2 2 2 2 π 3.2.213/2 72 52 23 12 π

  x1/2 x2 x4 x6 J1/2 (x) = 1/2 1 √ 1 − 2 3 + − + ... 2 2 π 2 2 2.24 25 32 3.2.26 72 52 32   x1/2 x2 x4 x6 J1/2 (x) = 1/2 1 √ 1 − + − + ... 3.2 5.4.3.2 7.6.5.4.3.2. 2 2 π   x6 x1/2 x2 x4 + − + ... J1/2 (x) = 1/2 1 √ 1 − 3! 5! 7!. 2 2 π   x1/2 x7 x3 x5 J1/2 (x) = 1/2 1 √ + − + ... x− 3! 5! 7!. 2 2 πx sinx

1 √ x1/2 21/2 π

J1/2 (x) =

r J1/2 (x) =

2 sinx. πx

(4.28)

Similarly, for n = −1/2, r J−1/2 (x) =

2 cosx. πx

(4.29)

4.2 Generating Function The generating function of Bessel function Jn (x) is defined as g(x, t) = e

x (t− 1t ) 2

=

∞ X

Jn (x)tn

n=0

We know that, x

e =

∞ X xn n=0

n!

=1+

x x2 x3 + + + ... 1! 2! 3!

(4.30)

4.2 Generating Function

55

Now, the left-hand side of (4.30) has the form x t 2

e .e x

e 2 t = 1+

x − 2t

∞ X ( xt )n ( −x )n 2 = . 2t n! n! n=0

( x2 t) ( x2 t)2 ( x2 t)3 ( x t)n ( x2 t)n+1 ( x2 t)n+2 + + +...+ 2 + + +... 1! 2! 3! n! (n + 1)! (n + 2)! (4.31)

x 2 x ( x )3 ( x )n ( x )n+1 ( x )n+2 ( 2t ) ( 2t ) + − 2t +...+ 2t − 2t + 2t −... 1! 2! 3! n! (n + 1)! (n + 2)! (4.32) Multiplying (4.31) and (4.32), we get x

e− 2t = 1−

 =⇒

1+

 ( x2 t) ( x2 t)2 ( x t)3 ( x t)n ( x t)n+1 ( x t)n+2 + + 2 + ... + 2 + 2 + 2 + ... × 1! 2! 3! n! (n + 1)! (n + 2)!

  ( x )3 ( x )n ( x )n+1 ( x )n+2 ( x ) ( x )2 + 2t − ... 1 − 2t + 2t − 2t + ... + 2t − 2t 1! 2! 3! n! (n + 1)! (n + 2)! (4.33)

=⇒ ...

−

( x )n+1 tn−1 ( x )n+2 tn−2 ( x )n+1 tn+1 ( x2 )n tn − 2 + 2 − ... + 2 n! 1!n! 2!n! (n + 1)! ( x2 )n+2 tn ( x )n+3 tn−1 + 2 − ... 1!(n + 1)! 2!(n + 1)!

... +

( x2 )n+2 tn+2 ( x )n+1 tn+1 ( x )n+4 tn ( x )n+5 tn−1 − 2 + 2 − 2 + ... (4.34) (n + 2)! 1!(n + 2)! 2!(n + 2)! 3!(n + 2)!

Collecting the coefficient of tn from (4.34), we get =⇒

( x2 )n ( x )n+2 ( x )n+4 − 2 + 2 − ... n! 1!(n + 1)! 2!(n + 2)! =⇒

∞ X (−1)r ( x )n+2r 2

r=0

r!(n + r)!

56 Bessel Function using the fact m! = Γ(m + 1), =⇒

∞ X (−1)r ( x2 )n+2r r!Γ(n + r + 1) r=0

(4.35)

from (4.30) and (4.35), we can write ∞ X (−1)r ( x2 )n+2r Jn (x) = r!Γ(n + r + 1) r=0

(4.36)

Now change the sign of n by − n, ∞ X (−1)r ( x2 )−n+2r J−n (x) = r!Γ(−n + r + 1) r=0

and replace variable, i.e., r → n + p, J−n (x) =

∞ X (−1)n+p ( x )n+2p 2

(n + p)!Γ(p + 1) r=0

∞ X (−1)p ( x2 )n+2p J−n (x) = (−1) p!Γ(n + p + 1) p=0 n

J−n (x) = (−1)n Jn (x). Now, putting n = 0, 1, .... in (4.36), we get J0 (x) =

∞ X (−1)r ( x )2r 2

r=0

J0 (x) = 1−

r!Γ(r + 1)

( x2 )2 ( x2 )4 ( x2 )6 + − + ... 1! 2!2! 3!3!

and J1 (x) =

∞ X (−1)r ( x )1+2r 2

r=0

r!Γ(r + 2)

(4.37)

4.3 Recurrence Relations

J1 (x) =

x 2

57

( x2 )3 ( x2 )5 ( x2 )7 + − + ... 1! 2! 2!3! 3!4! −

4.3 Recurrence Relations We know that

∞ X (−1)r ( x2 )n+2r Jn (x) = r!Γ(n + r + 1) r=0

(4.38)

multiplying the above equation by xn , ∞ X (−1)r ( x2 )n+2r x Jn (x) = x r!Γ(n + r + 1) r=0 n

n

xn Jn (x) =

∞ X (−1)r ( 1 )n+2r x2n+2r 2

r=0

(4.39)

r!Γ(n + r + 1)

Differentiating (4.39) w.r.t. x, we get d d n (x Jn (x)) = dx dx

∞ X (−1)r ( 1 )n+2r x2n+2r

!

2

r=0

r!Γ(n + r + 1)

∞

X (2n + 2r)(−1)r ( 1 )n+2r x2n+2r−1 d n 2 (x Jn (x)) = dx r!Γ(n + r + 1) r=0 ∞ X (2n + 2r)(−1)r ( 12 )n+2r xn+2r−1 d n n (x Jn (x)) = x dx r!(n + r)! r=0 ∞ X (n + r)(−1)r ( 12 )n−1+2r xn−1+2r d n (x Jn (x)) = xn dx r!(n + r)(n + r − 1)! r=0 ∞ X (−1)r ( x2 )(n−1)+2r d n (x Jn (x)) = xn dx r!((n − 1) + r)! r=0

58 Bessel Function ∞ X (−1)r ( x2 )(n−1)+2r d n (x Jn (x)) = xn dx r!Γ ((n − 1) + r + 1) r=0

d dx

(xn Jn (x)) = xn Jn−1 (x).

Similarly, x−n Jn (x) =

(4.40)

∞ X (−1)r ( 1 )n+2r x2r 2

r!Γ(−n + r + 1) r=0


d d x−n Jn (x) = dx dx

∞ X (−1)r ( 1 )n+2r x2r

!

2

r=0

r!Γ(n + r + 1)

∞
X (2r)(−1)r ( 21 )n+2r x2r−1 d −n x Jn (x) = dx r(r − 1)!Γ(n + r + 1) r=0 ∞ X
(2)(−1)r ( 12 )n+2r x2r−1 d x−n Jn (x) = xn dx (r − 1)!Γ(n + r + 1)xn r=0 ∞ X
(−1)r ( 21 )n+2r−1 x2r+n−1 d −n −n x Jn (x) = x dx (r − 1)!Γ(n + r + 1) r=0

change r → p + 1 ∞ X
(−1)p+1 ( 12 )n+2p+1 x2p+n+1 d x−n Jn (x) = x−n dx (p)!Γ(n + p + 2) r=0 ∞ X (−1)p ( x2 )(n+1)+2p d n −n (x Jn (x)) = −x dx p!Γ((n + 1) + p + 1) r=0

d dx


x−n Jn (x) = −xn Jn+1 (x).

(4.41)

4.4 Orthonormality

59

Now, multiplying (4.40) by x−n , we get x−n

d n (x Jn (x)) = Jn−1 (x) dx

o n 0 x−n xn Jn (x) + Jn (x)nxn−1 = Jn−1 (x) 0

x−n xn Jn (x) + x−n Jn (x)nxn−1 = Jn−1 (x) 0

n

Jn (x) = Jn−1 (x). x Similarly, multiplying (4.41) by x−n , we get Jn (x) +

x−n

(4.42)

d −n (x Jn (x)) = −Jn+1 (x) dx

o n 0 x−n x−n Jn (x) + Jn (x)(−n)xn−1 = −Jn+1 (x) 0

x−n xn Jn (x) − x−n Jn (x)nxn−1 = −Jn+1 (x) 0

Jn (x) −

n x

Jn (x) = −Jn+1 (x).

(4.43)

4.4 Orthonormality The orthogonality of Bessel function is defined by a multiplicative weight factor u = x, i.e., ˆ 1 xJn (αx)Jn (βx)dx = 0 0

if α 6= β

(4.44)

60 Bessel Function and the normality is defined as, ˆ 1 1 2 xJn (αx)Jn (βx)dx = Jn+1 (α) 2 0

if α = β. (4.45)

4.5 Application to the Optical Fiber Bessel function is frequently comes into the solution of problem possesses cylindrical symmetry (r, φ, z) like fiber optical cable. Optical fiber is based on the principle of total internal reflections. An optical fiber has a core of refractive index n1 and cladding of refractive index n2 . Core refractive index is always greater than the cladding refractive index. This refractive index variation decides the type of optical fibers. Here, for simplicity, we are considering the step index optical fiber whose refractive indexes are constant, i.e.,   n1 0