The Riemann hypothesis and the distribution of prime numbers 9781536194821, 1536194824

Citation preview

MATHEMATICS RESEARCH DEVELOPMENTS

THE RIEMANN HYPOTHESIS AND THE DISTRIBUTION OF PRIME NUMBERS

No part of this digital document may be reproduced, stored in a retrieval system or transmitted in any form or by any means. The publisher has taken reasonable care in the preparation of this digital document, but makes no expressed or implied warranty of any kind and assumes no responsibility for any errors or omissions. No liability is assumed for incidental or consequential damages in connection with or arising out of information contained herein. This digital document is sold with the clear understanding that the publisher is not engaged in rendering legal, medical or any other professional services.

MATHEMATICS RESEARCH DEVELOPMENTS Additional books and e-books in this series can be found on Nova’s website under the Series tab.

MATHEMATICS RESEARCH DEVELOPMENTS

THE RIEMANN HYPOTHESIS AND THE DISTRIBUTION OF PRIME NUMBERS

NAJI ARWASHAN, PHD, PE

Copyright © 2021 by Nova Science Publishers, Inc. All rights reserved. No part of this book may be reproduced, stored in a retrieval system or transmitted in any form or by any means: electronic, electrostatic, magnetic, tape, mechanical photocopying, recording or otherwise without the written permission of the Publisher. We have partnered with Copyright Clearance Center to make it easy for you to obtain permissions to reuse content from this publication. Simply navigate to this publication’s page on Nova’s website and locate the “Get Permission” button below the title description. This button is linked directly to the title’s permission page on copyright.com. Alternatively, you can visit copyright.com and search by title, ISBN, or ISSN. For further questions about using the service on copyright.com, please contact: Copyright Clearance Center Phone: +1-(978) 750-8400 Fax: +1-(978) 750-4470 E-mail: [email protected].

NOTICE TO THE READER The Publisher has taken reasonable care in the preparation of this book, but makes no expressed or implied warranty of any kind and assumes no responsibility for any errors or omissions. No liability is assumed for incidental or consequential damages in connection with or arising out of information contained in this book. The Publisher shall not be liable for any special, consequential, or exemplary damages resulting, in whole or in part, from the readers’ use of, or reliance upon, this material. Any parts of this book based on government reports are so indicated and copyright is claimed for those parts to the extent applicable to compilations of such works. Independent verification should be sought for any data, advice or recommendations contained in this book. In addition, no responsibility is assumed by the Publisher for any injury and/or damage to persons or property arising from any methods, products, instructions, ideas or otherwise contained in this publication. This publication is designed to provide accurate and authoritative information with regard to the subject matter covered herein. It is sold with the clear understanding that the Publisher is not engaged in rendering legal or any other professional services. If legal or any other expert assistance is required, the services of a competent person should be sought. FROM A DECLARATION OF PARTICIPANTS JOINTLY ADOPTED BY A COMMITTEE OF THE AMERICAN BAR ASSOCIATION AND A COMMITTEE OF PUBLISHERS. Additional color graphics may be available in the e-book version of this book.

Library of Congress Cataloging-in-Publication Data ISBN: ϵϳϴͲϭͲϱϯϲϭϵͲϰϴϮͲϭ;ĞŽŽŬͿ

Published by Nova Science Publishers, Inc. † New York

A significant progress in writing this book was made during the national lockdown due to the coronavirus. Therefore, I dedicate this book to all the people who were affected by the virus and those who continue to risk their lives.

Naji Arwashan [email protected] Troy, Michigan January, 2021

CONTENTS Preface

ix

Chapter 1

The Zeta Function in the World of Real Numbers

1

Chapter 2

A Guided Tour in the World of Complex Numbers

17

Chapter 3

The Eta Series and the First Extension of Zeta

57

Chapter 4

The Functional Equation and the Second Extension of Zeta

73

The Roots of the Zeta Function and the Riemann Hypothesis

91

Chapter 5 Chapter 6

The Zeta Function and Counting the Prime Numbers

109

Chapter 7

Von Mangoldt Formula to the Rescue

145

Appendix A

Fourier Transform

169

Appendix B

The Functional Equation of the Jacobi Theta Function

179

The Functional Equations of the Gamma Function (Legendre’s Duplication and Euler’s Reflection)

189

Appendix C

viii Appendix D

Contents Calculation of the Integral 𝒂+𝒊∞ 𝒚𝒔 𝟏 ∫ 𝟐𝝅𝒊 𝒂−𝒊∞ 𝒔

𝒅𝒔 (𝒚 > 𝟎, 𝒂 > 𝟎)

203

References

213

Index

215

PREFACE The writing of this book was inspired from my personal desire and quest to understand the Riemann Hypothesis, and to understand all the mathematical derivations, and proofs that surround and lead to its formulation. Aficionados of math have appreciation for understanding the proof of any mathematical statement; they are not satisfied by the end formulation of the problem, but instead see beauty in its derivation. In this book I put together, in accessible style and format, all the pieces of math necessary for the derivation and understanding of one of the most important and famous unsolved questions in math today. Bernard Riemann formulated his hypothesis in 1859 in his seminal paper “On the Number of Prime Numbers less than a Given Quantity.” His paper established a bridge between two separate branches of mathematics: complex analysis and number theory, and that bridge eventually led to great advancements in the latter field including the proof of the Prime Number Theorem. At the beginning of the twentieth century the celebrated German mathematician David Hilbert listed the Riemann Hypothesis on his list of 23 most important unsolved problems in mathematics. One hundred years later, following the same tradition, the private Clay Institute listed

x

Naji Arwashan

the Riemann Hypothesis as one of its seven millennial problems and allocated a million-dollar prize for the proof of any of them. The Riemann Hypothesis is considered by many accounts the single most important and difficult question in math today. A huge amount of numerical data supports the hypothesis, but still a proof has eluded the greatest minds in mathematics. If the Riemann Hypothesis is proved true, then a better understanding will be gained on the distribution of prime numbers (an abstract branch of mathematics that has fascinated mathematicians for centuries and received unexpected attention in the last fifty years due to its relation to encryption). In addition, numerous theorems and PhD theses start with the assumption of the Riemann Hypothesis so its proof will lead immediately to the proof of a significant amount of mathematical work at once. Besides, the monetary prize assigned by the Clay Institute to its proof adds fame and hype in the media about this hypothesis. With all the interest in the Riemann Hypothesis, one would expect several books, online lectures and courses written about it. Unfortunately, the reality is that most of the writing and presentations on it are very simply addressed to the general public without going deeper into the mathematical derivations, and don’t cover all the mathematical details behind it. The other category of writing on the hypothesis is mostly geared towards math professionals and researchers and might be difficult to read or follow. If you are an undergraduate student in math, science or engineering; or you are an engineer or a scientist with a penchant for math, and are someone who appreciates a complete and clear derivation of every step along the way, and have a desire to understand one of the most important open questions in math today, then this book is for you. Finally, special thanks to Nikhil Shankar for his review of the transcript of this book, and for his valuable comments and feedback.

Chapter 1

THE ZETA FUNCTION IN THE WORLD OF REAL NUMBERS 1.1. INTRODUCTION In his 1859 seminal paper, “On the Number of Prime Numbers less than a Given Quantity”, Bernard Riemann conjectured his famous hypothesis that all nontrivial zeros (roots) of the zeta function are all on the critical line. Well, what does that mean? First let’s look at the zeta function: ∞

𝜁(𝑠) = ∑ 𝑛=1

1 1 1 1 1 = + + + … 𝑛𝑠 1𝑠 2𝑠 3𝑠 4𝑠

It is practically the infinite sum of the reciprocals of all natural numbers raised to the power 𝑠. The very first observation is that all terms of the series defining the zeta function seem positive, so the immediate question arises: how can such a function ever be zero?! The answer to this is that the power of the reciprocals, 𝑠, is not a real number but a complex number. This answer unfortunately might turn

2

Naji Arwashan

off some readers and discourage their interest in understanding this hypothesis! But no worries! That is what this book is written for. Before we introduce and start talking about complex numbers, let’s examine in this chapter the zeta function in the world of real numbers since this function was well known and received a good share of attention from mathematicians before Bernard Riemann extended its use to complex numbers. But before we do that, we need to introduce couple of tools that will be used several times in this book.

1.2. CONVERGENCE TESTS OF A SERIES There are many tests that can be used to check the convergence of an infinite series ∑∞ 𝑛=1 𝑎𝑛 here we will list only the three tests that will be used in this book. But before we do that, we need to define the absolute convergence of a series. The series ∑∞ 𝑛=1 𝑎𝑛 is said to be ∞ absolutely convergent if the series ∑𝑛=1|𝑎𝑛 | is convergent (more about the absolute convergence of series in Chapter 2).

Ratio Test We define the ratio 𝑟 as 𝑟 = lim | 𝑛→∞

𝑎𝑛+1 𝑎𝑛

|

If 𝑟 < 1, then the series is absolutely convergent. If 𝑟 > 1, then the series diverges, and if 𝑟 = 1, then the test is inconclusive.

Direct Comparison Test If the series ∑∞ 𝑛=1 𝑏𝑛 is absolutely convergent and |𝑎𝑛 | ≤ |𝑏𝑛 | then ∞ the series ∑𝑛=1 𝑎𝑛 converges absolutely.

The Zeta Function in the World of Real Numbers

3

Alternate Series Test If the following three conditions are satisfied 1- 𝑎𝑛 is a sequence of positive real numbers 2- lim 𝑎𝑛 = 0 𝑛→∞

3- 𝑎𝑛+1 < 𝑎𝑛 for every n 𝑛 Then ∑∞ 𝑛=𝑘(−1) 𝑎𝑛 is convergent.

1.3. THE GEOMETRIC SERIES The geometric series is a very common and important series in calculus; it is defined as 𝑀(𝑟) = 1 + 𝑟 + 𝑟 2 + 𝑟 3 + 𝑟 4 + ⋯ Where every term in the series is obtained by multiplying the previous term by 𝑟 (called the ratio of the geometric series). If we truncate the series to the first n terms, we get 𝑀𝑛 (𝑟) = 1 + 𝑟 + 𝑟 2 + ⋯ + 𝑟 𝑛−1 Multiplying both sides by the ratio, 𝑟, gives 𝑟𝑀𝑛 (𝑟) = 𝑟 + 𝑟 2 + 𝑟 3 … + 𝑟 𝑛 Subtracting the two equations yields: (1 − 𝑟)𝑀𝑛 (𝑟) = 1 − 𝑟 𝑛 Provided 𝑟 ≠ 1, we can divide both sides by (1 − 𝑟) and obtain

4

Naji Arwashan 𝑀𝑛 (𝑟) =

1 − 𝑟𝑛 1−𝑟

Splitting the right-hand sides into two terms gives 𝑀𝑛 (𝑟) =

1 𝑟𝑛 − 1−𝑟 1−𝑟

When −1 < 𝑟 < 1, 𝑟 𝑛 goes to zero when 𝑛 goes to infinity, and the 1

geometric series converges to 1−𝑟. We should mention that the derivation and result obtained here puts no condition on 𝑟 being positive. In other words, if 𝑟 is negative, then the terms of the series will be alternating between positive and negative but the same conclusion about the convergence of the series will be valid.

1.4. THE ZETA SERIES Keeping in mind that in this chapter we are limiting the argument of zeta to be a real number only, we start first by checking the convergence of the zeta series when the power, 𝑠, is equal to 1. In this case the function is the sum of the reciprocals of the natural numbers, which is called the harmonic series. ∞

𝜁(1) = ∑ 𝑛=1

1 1 1 1 1 1 1 1 1 = + + + + + + + … 1 𝑛 1 2 3 4 5 6 7 8

Unfortunately, in the case of the harmonic series, the ratio test is inconclusive because as shown below, the limit of the ratio 𝑟 is 1

The Zeta Function in the World of Real Numbers

5

1

𝑎𝑛+1 𝑛 1 𝑟 = lim | = lim = lim =1 | = lim 𝑛+1 1 𝑛→∞ 𝑎𝑛 𝑛→∞ 𝑛→∞ 𝑛 + 1 𝑛→∞ 1 + 1 𝑛

𝑛

Therefore, we need to rely on a different technique that goes back to the 14th century French philosopher and bishop of Lisieux, Nicole Oresme. If we group the terms of the sum as follows: ∞

𝜁(1) = ∑ 𝑛=1

1 1 1 1 1 1 1 1 1 = + + + + + + + ]+⋯ [ ] [ 𝑛1 1 2 3 4 5 6 7 8 1

1

1

1

1

1

1

1

1

1

We can see that [3 + 4] is greater than [4 + 4] and [5 + 6 + 7 + 8] is 1

1

greater than [8 + 8 + 8 + 8] etc… therefore ∞

𝜁(1) = ∑ 𝑛=1

1 1 1 1 1 1 1 1 1 > + + [ + ] + [ + + + ]… 1 𝑛 1 2 4 4 8 8 8 8 ∞

𝜁(1) = ∑ 𝑛=1

1 1 1 1 1 > + + + … 1 𝑛 1 2 2 2

Since we are adding ½ infinitely many times, we can conclude that 𝜁(1) diverges. Additionally, whenever 𝑠 < 1, we can easily see that 1 𝑛𝑠

1

> 𝑛 so all terms of the series will be greater than their corresponding

terms from the case of 𝑠 = 1, and therefore we can conclude that zeta diverges when 𝑠 ≤ 1. When 𝑠 > 1 the series converges, and the proof relies on a trick similar to the one used to prove the divergence of the harmonic series. It goes as follows: ∞

𝜁(𝑠) = ∑ 𝑛=1

1 1 1 1 1 1 1 1 = 𝑠 + [ 𝑠 + 𝑠] + [ 𝑠 + 𝑠 + 𝑠 + 𝑠] + ⋯ 𝑠 𝑛 1 2 3 4 5 6 7

6

Naji Arwashan 1

1

1

1

1

1

2

3

2

2

4

5𝑠

Note that[ 𝑠 + 𝑠 ] is less than [ 𝑠 + 𝑠 ] and [ 𝑠 + 1

1

1

+

1 6𝑠

1

+ 𝑠 ] is less 7

1

than [4𝑠 + 4𝑠 + 4𝑠 + 4𝑠 ] etc… ∞

𝜁(𝑠) = ∑ 𝑛=1

1 1 1 1 1 1 1 < 1 + + + + + + [ ] [ ]+⋯ 𝑛𝑠 2𝑠 2𝑠 4𝑠 4𝑠 4𝑠 4𝑠 ∞

𝜁(𝑠) = ∑ 𝑛=1 ∞

1 2 4 8 < 1 + + + [ ] [ ] [ ]… 𝑛𝑠 2𝑠 4𝑠 8𝑠

1 2 22 23 𝜁(𝑠) = ∑ 𝑠 < 1 + [ 𝑠 ] + [ 2𝑠 ] + [ 3𝑠 ] … 𝑛 2 2 2 𝑛=1 ∞

𝜁(𝑠) = ∑ 𝑛=1 ∞

𝜁(𝑠) = ∑ 𝑛=1

1 < 1 + 21−𝑠 + 22−2𝑠 + 23−3𝑠 … 𝑛𝑠

1 < 1 + 21−𝑠 + (21−𝑠 )2 + (21−𝑠 )3 … 𝑛𝑠

Figure 1-1. The value of 𝜻(𝒔) when 𝒔 is a real number > 1.

The Zeta Function in the World of Real Numbers

7

Note that every term of the last series can be obtained by multiplying the previous term by 21−𝑠 . This is a geometric series with a ratio of 21−𝑠 and when 𝑠 > 1 the ratio is less than one and consequently, the zeta series converges, and its values are shown in Figure 1-1.

1.5. THE EULER PRODUCT FORMULA The zeta function received the attention of Leonhard Euler (17071783), one of the most brilliant and prolific mathematicians of all time. Euler marvelously discovered that the zeta series can be written as an infinite product as follows: ∞

1 𝑠 1 𝜁(𝑠) = ∑ ( ) = ∏ ( 1) 𝑛 1− 𝑠 𝑛=1 𝑝 𝑝𝑖

Among natural numbers there is a special set called prime numbers, 𝑝, which are the numbers that can be divided by themselves only (excluding the number 1) like 2, 3, 5, 7, … A more detailed discussion of prime numbers will be given in Chapter 6 but for now, we accept that primes are the building blocks for all natural numbers and any natural number can be written as a unique product of primes raised to different powers (this is the fundamental theorem of arithmetic). For example, 4 = 22 , 6 = 2 × 3, 8 = 23 , 9 = 33 , 10 = 2 × 5 etc. …. This extremely important theorem allows every term of the zeta series to be replaced by its prime composition as follows: 1 𝑠

1 𝑠

1 𝑠

1 𝑠

1 𝑠

1 𝑠

1 𝑠 2

1 𝑠 5

1 𝑠 1 𝑠 2 3

1 𝑠

1 𝑠

1 + (2) + (3) + (4) + (5) + (6) + (7) + (8) + (9) + ⋯ = 1 𝑠 2

1 𝑠 3

1 𝑠 7

1 𝑠 2

1+( ) + ( ) + ( 2 ) + ( ) + ( ) ( ) + ( ) + ( 3 ) + ⋯

8

Naji Arwashan

A very smart move now is to write the right-hand side as the product of infinite series as follows: 1 𝑠 1 𝑠 1 𝑠 1 𝑠 1 𝑠 1+( ) +( ) +( ) +( ) +( ) …= 2 3 4 5 6 1 𝑠 1 𝑠 1 𝑠 1 𝑠 [1 + ( ) + ( 2 ) + ( 3 ) + ( 4 ) + ⋯ ] 2 2 2 2 𝑠 𝑠 𝑠 1 1 1 1 𝑠 [1 + ( ) + ( 2 ) + ( 3 ) + ( 4 ) + ⋯ ] 3 3 3 3 𝑠 𝑠 𝑠 1 1 1 1 𝑠 [1 + ( ) + ( 2 ) + ( 3 ) + ( 4 ) + ⋯ ] … 5 5 5 5 1

𝑠

Every series on the right hand side is of the form [1 + ( ) + 𝑝 𝑖

𝑠

1

1

𝑠

1

𝑠

(𝑝 2 ) + (𝑝 3 ) + (𝑝 4 ) + ⋯ ] where 𝑝𝑖 is a prime number and the 𝑖

𝑖

𝑖

product of the series is over all the primes. Therefore, every term on the left-hand side is equal to the product of infinite terms on the right-hand 1 𝑠

side. For instance, the term (6) on the left-hand side is equal to 1 𝑠

1 𝑠

(2) (3) which is the product on the right hand-side of the second term in the series of the prime number 2, the second term from the series of the prime number 3, and the first term (which is 1) from the series of all 1 𝑠

other prime numbers. Take another example: (8) , which is 1

𝑠

(23) , which is the product of the fourth term of the 2 series and the first term from all other series. The logic in the last step is simple but leads to a very interesting and powerful equality. If we examine carefully we can see that each of the series on the right hand side is a geometric 1

𝑠

series with the ratio equal to ( ) , and because the ratio is between -1 𝑝 𝑖

and 1, the series are convergent and can be replaced by their values as follows:

The Zeta Function in the World of Real Numbers

9

1 𝑠 1 𝑠 1 𝑠 1 𝑠 1 𝑠 1+( ) +( ) +( ) +( ) +( ) … 2 3 4 5 6 =[

1 1−

1] 2𝑠

1 1 1 [ 1] [ 1][ 1]… 1− 𝑠 1− 𝑠 1− 𝑠 3

5

7

And hence we have Euler product ∞

1 1 =∏ ( 1) 𝑠 𝑛 1 − 𝑛=1 𝑝 𝑝𝑠

𝜁(𝑠) = ∑

𝑖

The above discussion is more of an explanation of the Euler product formula; a formal proof goes as follows: 𝜁(𝑠) = 1 +

1 1 1 1 + 𝑠+ 𝑠+ 𝑠… 𝑠 2 3 4 5

Let us multiply both sides by

1 2𝑠

1 1 1 1 1 𝜁(𝑠) = 𝑠 + 𝑠 + 𝑠 + 𝑠 … 𝑠 2 2 4 6 8 Subtracting the last two equations and isolating 𝜁(𝑠) gives: 𝜁(𝑠) (1 −

1 1 1 1 1 )= 1+ 𝑠+ 𝑠+ 𝑠+ 𝑠… 𝑠 2 3 5 7 9

As you can see, the multiplier of 2 is removed from all the bases of the denominators of the right-hand side. Now let us multiply the last 1

equation by 3𝑠 1 1 1 1 1 1 𝜁(𝑠) (1 − 𝑠 ) = 𝑠 + 𝑠 + 𝑠 + 𝑠 … 𝑠 3 2 3 9 15 21

10

Naji Arwashan 1

Subtracting the last two equations and isolating 𝜁(𝑠) (1 − 𝑠 ) gives 2 𝜁(𝑠) (1 −

1 1 1 1 ) (1 − 𝑠 ) = 1 + 𝑠 + 𝑠 + ⋯ 𝑠 2 3 5 7

Note that all the multipliers of 2 and 3 are removed from all the bases of the denominators of the right-hand side. Repeat the same process for all the remaining prime numbers 5, 7, 11… and we can end up with: 𝜁(𝑠) (1 −

1 1 1 1 ) (1 − 𝑠 ) (1 − 𝑠 ) (1 − 𝑠 ) … = 1 𝑠 2 3 5 7

And now we can write: 𝜁(𝑠) =

1 1

1

1

1

(1 − 2𝑠 ) (1 − 3𝑠 ) (1 − 5𝑠 ) (1 − 7𝑠 ) … 1 𝜁(𝑠) = ∏ ( 1) 1− 𝑠 𝑝 𝑝𝑖

And if we expand both sides, we get 1 𝑠 1 𝑠 1 𝑠 1 𝑠 1+( ) +( ) +( ) +( ) +⋯ 2 3 4 5 =[

1

1]

1 − 2𝑠

1 1 1 [ 1] [ 1][ 1]… 1 − 3𝑠 1 − 5𝑠 1 − 7𝑠

The beauty of Euler product formula is that it relates all natural numbers (on the left) to all prime numbers on the right. It is important to mention that it is an equality between an infinite sum (on the left) and

The Zeta Function in the World of Real Numbers

11

infinite product (on the right). The formula is considered a cornerstone in number theory and was the starting point for Riemann in his seminal paper mentioned in the introduction.

1.6. IMPORTANT FORMULAE DERIVED FROM EULER PRODUCT Two important formulae can be derived from Euler product: the reciprocal of zeta and the derivative of zeta

The Reciprocal of Zeta

𝟏 𝜻(𝒔)

Start with the Euler product 1 𝜁(𝑠) = ∏ ( 1) 1 − 𝑝𝑠 𝑝 𝑖

Let us take the reciprocal of both sides 1 1 = ∏ (1 − 𝑠 ) 𝜁(𝑠) 𝑝𝑖 𝑝

Let us expand the right-hand side 1 1 1 1 1 = (1 − 𝑠 ) (1 − 𝑠 ) (1 − 𝑠 ) (1 − 𝑠 ) … 𝜁(𝑠) 2 3 5 7 And if we multiply the factors of the right-hand side, we obtain

12

Naji Arwashan 1 1 1 1 1 1 1 =1− 𝑠− 𝑠− 𝑠+ 𝑠− 𝑠+ 𝑠… 𝜁(𝑠) 2 3 5 6 7 10

Note that the numbers that appear in the denominators of the righthand side are only the numbers that have prime factors with powers equal to 1, like 6 = 2 × 3 and 10 = 2 × 5 but numbers like 8 = 23 and 9 = 32 cannot appear. The sign next to them is positive if their number of prime factors is even and negative if it is odd. For instance, 1 6𝑠

1

1

will result from the product of (− 2𝑠 ) from the first term, and (− 3𝑠 )

from the second term, and (1) from all the other terms. The last equality can be written formally as: ∞

1 𝜇(𝑛) =∑ 𝑠 𝜁(𝑠) 𝑛 𝑛=1

The function 𝜇(𝑛) is called the Mobius function and is often seen in papers and textbooks related to number theory. It will be addressed and studied again in detail in Chapter 6 of this book. By definition, the Mobius function takes the following values: 𝜇(1) = 1 when n = 1 𝜇(𝑛) = (−1)𝑘 if n is the product of k different primes 𝜇(𝑛) = 0 if any of n factors has a power higher than one.

The Derivative of Zeta Start again with Euler product 1 𝜁(𝑠) = ∏ ( 1) 1 − 𝑝𝑠 𝑝 𝑖

The Zeta Function in the World of Real Numbers

13

But this time let us take the natural logarithm of both sides: 1 1 log 𝜁(𝑠) = log ∏ ( 1 ) = ∑ log ( 1) 1− 𝑠 1− 𝑠 𝑝𝑖

𝑝

𝑝

= ∑ [log(1) − log (1 − 𝑝

𝑝𝑖

1 )] 𝑝𝑖𝑠

Obviously, log(1) = 0, so that gives us

log 𝜁(𝑠) = − ∑ log (1 − 𝑝

1 ) 𝑝𝑖𝑠

Let us take the derivative of both sides with respect to 𝑠 [remember if 𝑦 = log(𝑔(𝑥)), then using the chain rule of differentiation we get 𝑑𝑦

𝑑𝑦 𝑑𝑔

1

1

𝑦 ′ = 𝑑𝑥 = 𝑑𝑔 𝑑𝑥 = 𝑔(𝑥) 𝑔′(𝑥)], and writing (1 − 𝑝𝑠 ) as (1 − 𝑝𝑖−𝑠 ) gives 𝑖

(1 − 𝑝𝑖−𝑠 )′ 𝜁′(𝑠) = −∑ 1 𝜁(𝑠) (1 − ) 𝑝

𝑝𝑖𝑠

𝜁′(𝑠) 0 − (−1)log (𝑝𝑖 )𝑝𝑖−𝑠 = −∑ 1 𝜁(𝑠) (1 − ) 𝑝

𝑝𝑖𝑠

Simplifying gives us 𝜁′(𝑠) 1 1 = − ∑ log 𝑝𝑖 𝑠 𝜁(𝑠) 𝑝𝑖 (1 − 1 ) 𝑝 𝑝𝑠 𝑖

And

1 1 (1− 𝑠 ) 𝑝𝑖

1

is the sum of the geometric series with the ratio 𝑝𝑠 𝑖

14

Naji Arwashan 𝜁′(𝑠) 1 1 1 1 1 = − ∑ log 𝑝𝑖 𝑠 (1 + 𝑠 + 2𝑠 + 3𝑠 + 4𝑠 + ⋯ ) 𝜁(𝑠) 𝑝𝑖 𝑝𝑖 𝑝𝑖 𝑝𝑖 𝑝𝑖 𝑝

Multiplying the terms of the geometric series by the factor

1 𝑝𝑖𝑠

gives

𝜁′(𝑠) 1 1 1 1 1 = − ∑ log 𝑝𝑖 ( 𝑠 + 2𝑠 + 3𝑠 + 4𝑠 + 5𝑠 … ) 𝜁(𝑠) 𝑝𝑖 𝑝𝑖 𝑝𝑖 𝑝𝑖 𝑝𝑖 𝑝

And now we have in parenthesis a new series ∞

𝜁′(𝑠) 1 = − ∑ log 𝑝𝑖 ∑ 𝑚𝑠 𝜁(𝑠) 𝑝𝑖 𝑝

𝑚=1

Expanding the terms of the right-hand side gives 𝜁′(𝑠) log 2 log 2 log 2 log 2 =− − 2𝑠 − 3𝑠 − 4𝑠 − ⋯ 𝜁(𝑠) 2𝑠 2 2 2 log 3 log 3 log 3 log 3 − 𝑠 − 2𝑠 − 3𝑠 − 4𝑠 − ⋯ 3 3 3 3 log 5 log 5 log 5 log 5 − 𝑠 − 2𝑠 − 3𝑠 − 4𝑠 − ⋯ 5 5 5 5 ⋮ Note that the denominator of every term on the right-hand side is a distinct natural number raised to the power 𝑠, so the right-hand side can be written as follows ∞

𝜁′(𝑠) 𝛬(𝑛) = −∑ 𝑠 𝜁(𝑠) 𝑛 𝑛=2

The Zeta Function in the World of Real Numbers

15

Where 𝛬(𝑛) = log 𝑝 if 𝑛 is a prime number 𝑝, or a power of 𝑝, otherwise 𝛬(𝑛) = 0

1.7. SPECIAL VALUES OF ZETA The Euler product formula was a major breakthrough, but it doesn’t provide the value of the zeta series. The calculation of the zeta series when 𝑠 is greater than 1 is normally done numerically, except for certain values where an explicit formula is derived, like in the case of 𝜁(2) ∞

1 2 1 2 1 2 1 2 1 2 1 2 𝜁(2) = ∑ ( ) = 1 + ( ) + ( ) + ( ) + ( ) + ( ) … 𝑛 2 3 4 5 6 𝑛=1

𝜁(2) is the infinite sum of the square of the reciprocals of natural numbers. The calculation of the sum of this series was known as the Basel problem and was solved by Euler himself. He proved that the value of 𝜁(2) is

𝜋2 6

(one might wonder how 𝜋 found its way to the sum

of such series! Hopefully the following chapters will shed some light on this observation), and this proof brought him his early fame. We should also mention that

𝜋2 6

is the reciprocal of the probability of two natural

numbers to be coprime (having no common divisor).

Chapter 2

A GUIDED TOUR IN THE WORLD OF COMPLEX NUMBERS 2.1. INTRODUCTION A complex number is a mathematical entity defined as 𝑧 = 𝑥 + 𝑖𝑦 where 𝑥 and 𝑦 are real numbers and 𝑖 is a special imaginary constant with a value equal to √−1. We call 𝑥 and 𝑦 real and imaginary parts of 𝑧. The conjugate of a given complex number 𝑧 = 𝑥 + 𝑖𝑦 is defined to be 𝑧 = 𝑥 − 𝑖𝑦 Therefore, if we add a complex number to its conjugate the outcome is real and equal to 2𝑥. Furthermore, if a complex number is multiplied by its conjugate the outcome is real number because 𝑧𝑧 = (𝑥 + 𝑖𝑦)(𝑥 − 𝑖𝑦) = 𝑥 2 − (𝑖𝑦)2 = 𝑥 2 − 𝑖 2 𝑦 2 = 𝑥 2 + 𝑦 2

18

Naji Arwashan

2.2. A WILD IDEA AT THE ORIGIN OF COMPLEX NUMBERS The story behind the origin of complex numbers [4] goes as follows. It is well known, and easy to prove, that the quadratic equation 𝑥 2 = 𝑚𝑥 + 𝑏 has the following solutions −𝑏 ± √𝑏 2 − 4𝑎𝑐 𝑥= 2𝑎 And if the discriminant (𝑏 2 − 4𝑎𝑐) is positive the equation has two solutions and if zero it has one solution and if negative the equation has no solution. This can be explained geometrically by looking at the solution of the quadratic equation as a solution to the two equations 𝑦 = 𝑥 2 and 𝑦 = 𝑚𝑥 + 𝑏 which is the intersections of the parabola and the straight line representing the two equations (as shown below); and if the line doesn’t intersect the parabola then there is no solution.

Figure 2-1. Graphic representation of the solution of a quadratic equation.

A Guided Tour in the World of Complex Numbers

19

If we apply the same logic to the cubic equation 𝑥 3 = 𝑝𝑥 + 𝑞 then there should always be a solution, because 𝑥 3 goes from −∞ to + ∞ (when 𝑥 goes from −∞ to +∞) and therefore an intersection with the line 𝑦 = 𝑝𝑥 + 𝑞 is guaranteed.

Figure 2-2. Graphic representation of the solution of a cubic equation.

In early sixteen century an analytic solution to the above cubic equation was found as follows:

3

3 𝑞2 𝑝3 𝑞 𝑞 2 𝑝3 𝑞 √ 𝑥= √ − + − √√ − − 4 27 2 4 27 2

𝑞2

However, when (

4

−

𝑝3 27

) is negative, we face a contradiction since

the above formula yields no solution! For instance, if 𝑥 3 = 15𝑥 + 4, 3

3

the solution is 𝑥 = √2 + √−121 + √2 − √−121 and obviously there is no value for the square root of -121. The Italian mathematician

20

Naji Arwashan

Rafael Bombelli (1526-1572) had a wild idea; he discovered that if he 3

treats √−1 as an imaginary constant then he can expand (2 + √−1) as follows: 3

2

3

(2 + √−1) = 23 + 3 × 22 × (√−1) + 3 × 2 × (√−1) + (√−1) 3

2

3

(2 + √−1) = 8 + 12(√−1) + 6(√−1) + (√−1) 3

2

2

Writing (√−1) = (√−1) (√−1) , and replacing (√−1) = −1 gives 3

(2 + √−1) = 8 + 12(√−1) + 6(−1) + (−1)(√−1) Simplifying 3

(2 + √−1) = 2 + 11(√−1) Writing 11 𝑎𝑠 √121, we get 3

(2 + √−1) = 2 + √−121 3

Replacing 2 + √−121 by (2 + √−1) in 𝑥 gives: 3

3

3

3

𝑥 = √(2 + √−1) + √(2 − √−1) 𝑥 = 2 + √−1 + 2 − √−1 = 4

and hence the analytic formula yields the real solution of 𝑥 = 4. Bombelli pioneering and creative work entitled him to be considered the inventor of complex numbers.

A Guided Tour in the World of Complex Numbers

21

The idea outlined here of using complex numbers for the convenience and flexibility they provide, then later in the process eliminating the imaginary constant in order to obtain results with real and physical meaning is quite encountered in physics and mechanics, and in particular it will be seen in chapters 6 and 7 of this book.

2.3. GEOMETRIC REPRESENTATION Complex numbers have geometric representation to them. The same way a real number can be represented as a point on a simple axis, a complex number can be represented by the point (𝑥, 𝑦) in the cartesian plane.

Figure 2-3. Graphic represenation of a complex number and its conjugate.

The distance between the origin and the complex number is equal to |𝑧| = √𝑥 2 + 𝑦 2

22

Naji Arwashan

And it can be referred to by different name: absolute value, magnitude or modulus. And actually, this definition of the absolute value of a complex number coincides with the definition of the absolute value of a real number because when the imaginary part 𝑦 is zero then |𝑧| = |𝑥| = √𝑥 2 The argument of a complex number is the angle 𝜑 (as shown below). As we know from geometry, we can add 2𝜋𝑘 to an angle, and the angle will still point to the same direction, therefore a complex number has infinite number of arguments, and is important to define its principal argument as −𝜋 < 𝜑 ≤ 𝜋

Figure 2-4. Graphic represenation of the absolute value and the argument of a complex number.

The sum of two complex numbers is similar to the sum of two vectors as shown in the Figure 2-5, and from that graph we can easily see that if 𝑧 = 𝑧1 + 𝑧2 then |𝑧| ≤ |𝑧1 | + |𝑧2 |

(1)

A Guided Tour in the World of Complex Numbers

23

Figure 2-5. The sum of two complex numbers.

Another important property if 𝑧 = 𝑧1 𝑧2 then |𝑧| = |𝑧1 ||𝑧2 |

(2)

To prove the last property, we write 𝑧1 = 𝑥1 + 𝑦1 and 𝑧2 = 𝑥2 + 𝑦2 we can easily find 𝑧 = 𝑧1 𝑧2 = (𝑥1 + 𝑖𝑦1 )(𝑥2 + 𝑖𝑦2 ) = 𝑥1 𝑥2 − 𝑦1 𝑦2 + 𝑖(𝑥1 𝑦2 + 𝑥2 𝑦1 ) The square of the absolute value of 𝑧 can be calculated as follows: |𝑧|2 = (𝑥1 𝑥2 − 𝑦1 𝑦2 )2 + (𝑥1 𝑦2 + 𝑥2 𝑦1 )2 = 𝑥12 𝑥22 + 𝑦12 𝑦22 + 𝑥12 𝑦22 + 𝑥22 𝑦12 And also, we can calculate |𝑧1 |2 |𝑧2 |2 = (𝑥12 + 𝑦12 )(𝑥22 + 𝑦22 ) = 𝑥12 𝑥22 + 𝑦12 𝑦22 + 𝑥12 𝑦22 + 𝑥22 𝑦12 And hence, |𝑧|2 = |𝑧1 |2 |𝑧2 |2

24

Naji Arwashan

2.4. COMPLEX MAPPING A real single variable function 𝑦 = 𝑓(𝑥) is nothing else than a mapping of a point from the 𝑥 axis to the 𝑦 axis, and the mapping can be visualized easily as in the example of the function 𝑦 = 𝑥 2 shown below.

Figure 2-6. The mapping of 𝒙 by the real function of 𝒚 = 𝒙𝟐 .

In the case of a function with complex variable, the function becomes the mapping of a point from the 𝑥 − 𝑦 to another 𝑢 − 𝑣 plane, and the full mapping becomes difficult to visualize. However, to help visualize the mapping, we can select certain lines on the 𝑥 − 𝑦 plane and see how they map in the 𝑢 − 𝑣 plane. We illustrate this by the following example: consider the complex function 𝑤 = 𝑓(𝑧) = 𝑧 2 where 𝑧 = 𝑥 + 𝑖𝑦 𝑤 = 𝑓(𝑧) = 𝑧 2 = (𝑥 + 𝑖𝑦)2 = 𝑥 2 + (𝑖𝑦)2 + 2𝑥𝑖𝑦 = (𝑥 2 − 𝑦 2 ) + 𝑖2𝑥𝑦

A Guided Tour in the World of Complex Numbers

25

We call 𝑢, and 𝑣 the real and imaginary part of 𝑤 𝑢 = 𝑥 2 − 𝑦 2 , 𝑣 = 2𝑥𝑦 So a point 𝑧 = 𝑥 + 𝑖𝑦 with the magnitude 𝑟 = √𝑥 2 + 𝑦 2 and argument 𝜑 (where sin 𝜑 =

𝑦 √𝑥 2 +𝑦 2

and cos 𝜑 =

𝑦 √𝑥 2 +𝑦 2

) will be

mapped to a new point 𝑤 with the magnitude 𝑅 and argument ∅ as shown below

Figure 2-7. The mapping of a circle by the complex function of 𝒇(𝒛) = 𝒛𝟐 .

𝑅 = √(𝑥 2 − 𝑦 2 )2 + (2𝑥𝑦)2 = √𝑥 4 + 𝑦 4 − 2𝑥 2 𝑦 2 + 4𝑥 2 𝑦 2 𝑅 = √𝑥 4 + 𝑦 4 + 2𝑥 2 𝑦 2 = √(𝑥 2 + 𝑦 2 )2 𝑅 = 𝑥2 + 𝑦2 = 𝑟2 2𝑥𝑦 𝑥 𝑦 sin ∅ = 2 =2 = 2(cos𝜑)(sin𝜑) 2 𝑥 +𝑦 √𝑥 2 + 𝑦 2 √𝑥 2 + 𝑦 2 sin ∅ = sin2𝜑 ∅ = 2𝜑 Therefore a circle with radius 𝑟 will be mapped to a new circle with radius 𝑟 2 , and the meaning of ∅ = 2𝜑 is that when 𝑧 rotates once around a circle with radius 𝑟, 𝑤 will rotate twice around a circle with

26

Naji Arwashan

radius 𝑟 2 . This result can be derived in much simpler way by using the polar coordinates of a complex numbers, which we will introduce later in this chapter.

2.5. EULER FORMULA When one studies complex numbers, he runs quickly into one of the most important formulas in math and is called Euler Formula because it was discovered by no one else than the same great Leonard Euler we encountered in first chapter. 𝑒 𝑖𝑥 = cos 𝑥 + 𝑖 sin 𝑥 When 𝑥 is replaced by 𝜋 then the formula becomes: 𝑒 𝑖𝜋 + 1 = 0 And this is considered the most beautiful equation in math because it relates together the four most common and important constants in math:1, 𝜋, 𝑒, and 𝑖. Since Euler formula is such an important formula and it is a key for the math involving the Riemann Hypothesis, we believe one would appreciate to see its proof. There are several proves for it, the most common one is the one given by Euler himself. It is a proof that relies mainly on Taylor expansion (a technique widely used in calculus). So, let us start by introducing and proving that expansion.

Taylor Expansion Taylor expansion or series is a very powerful and often used tool in calculus. It is named after Brook Taylor who introduced them in 1715.

A Guided Tour in the World of Complex Numbers

27

It is an infinite sum of terms that are expressed in terms of the function’s derivatives at a single point. For most common functions, the function and the sum of its Taylor series are equal near this point 𝑓(𝑥) = 𝑓(𝑎) + 𝑓 ′ (𝑎)(𝑥 − 𝑎) + ……+

𝑓(𝑛) (𝑎) 𝑛!

𝑓′′ (𝑎) 2!

(𝑥 − 𝑎)2 +

𝑓(3) (𝑎) 3!

(𝑥 − 𝑎)3 +

(𝑥 − 𝑎)𝑛 +……

To prove the above equality, we start by applying the fundamental theorem in calculus (a proof of that theorem will be presented later in this chapter) to the function 𝑓(𝑥) 𝑥

𝑓(𝑥) = 𝑓(𝑎) + ∫ 𝑓 ′ (𝑡1 )𝑑𝑡1 𝑎

Let us apply the fundamental theorem again to 𝑓 ′ (𝑡1 ) 𝑡1

𝑓 ′ (𝑡1 ) = 𝑓 ′ (𝑎) + ∫ 𝑓 ′ ′(𝑡2 )𝑑𝑡2 𝑎

Replacing 𝑓 ′ (𝑡1 ) in the above equation of 𝑓(𝑥) gives 𝑥

𝑓(𝑥) = 𝑓(𝑎) + ∫ [𝑓

′ (𝑎)

𝑡1

+ ∫ 𝑓 ′ ′(𝑡2 )𝑑𝑡2 ] 𝑑𝑡1

𝑎

𝑎 𝑥

𝑥

𝑡1

𝑓(𝑥) = 𝑓(𝑎) + ∫ 𝑓 ′ (𝑎) 𝑑𝑡1 + ∫ ∫ 𝑓 ′ ′(𝑡2 )𝑑𝑡2 𝑑𝑡1 𝑎

𝑎

𝑎

Applying the fundamental theorem to 𝑓 ′′ (𝑡2 ) 𝑡2

𝑓 ′′ (𝑡2 ) = 𝑓 ′′ (𝑎) + ∫ 𝑓 ′ ′′(𝑡3 )𝑑𝑡3 𝑎

28

Naji Arwashan And replacing 𝑓 ′′ (𝑡2 ) in the last equation of 𝑓(𝑥) gives 𝑥

𝑓(𝑥) = 𝑓(𝑎) + ∫ 𝑓 ′ (𝑎) 𝑑𝑡1 𝑎 𝑥

𝑡1

𝑡2

+ ∫ ∫ [𝑓 ′′ (𝑎) + ∫ 𝑓 ′ ′′(𝑡3 )𝑑𝑡3 ] 𝑑𝑡2 𝑑𝑡1 𝑎

𝑎

𝑎 𝑥

𝑥

𝑡1

𝑓(𝑥) = 𝑓(𝑎) + ∫ 𝑓 ′ (𝑎) 𝑑𝑡1 + ∫ ∫ 𝑓 ′′ (𝑎)𝑑𝑡2 𝑑𝑡1 𝑎 𝑡1

𝑥

𝑎

𝑎

𝑡2

+ ∫ ∫ ∫ 𝑓 ′ ′′(𝑡3 )𝑑𝑡3 𝑑𝑡2 𝑑𝑡1 𝑎

𝑎

𝑎

And one can continue like this for all the derivatives of the function 𝑥

𝑓(𝑥). But now let us evaluate ∫𝑎 𝑓 ′ (𝑎) 𝑑𝑡1 [the second term in the expansion of 𝑓(𝑥) ] 𝑥

𝑥

∫ 𝑓 ′ (𝑎) 𝑑𝑡1 = 𝑓 ′ (𝑎) ∫ 𝑑𝑡1 = 𝑓 ′ (𝑎)(𝑥 − 𝑎) 𝑎

𝑎 𝑥

𝑡

And the same for the third term ∫𝑎 ∫𝑎 1 𝑓 ′′ (𝑎)𝑑𝑡2 𝑑𝑡1 𝑥

𝑡1

∫ ∫ 𝑎

𝑥

𝑓 ′′ (𝑎)𝑑𝑡2 𝑑𝑡1

=𝑓

𝑎

′′ (𝑎)

∫ 𝑑𝑡1 ∫ 𝑑𝑡2 = 𝑓 𝑎 2

= 𝑓 ′′ (𝑎)

𝑡1

𝑥 ′′ (𝑎)

𝑎

∫ (𝑡1 − 𝑎) 𝑑𝑡1 𝑎

(𝑥 − 𝑎) 2

And in general: 𝑥

𝑡1

𝑡𝑛−1

∫ ∫ …∫ 𝑎

𝑎

𝑓 (𝑛) 𝑑𝑡𝑛 … 𝑑𝑡2 𝑑𝑡1 = 𝑓 (𝑛) (𝑎)

𝑎

And hence the proof for Taylor expansion.

(𝑥 − 𝑎)𝑛 𝑛!

A Guided Tour in the World of Complex Numbers

29

If a real function is infinitely differentiable (smooth is another term used to describe this property) and can be expanded in Taylor series, then it is called analytic function. We should stress here that not every smooth function is analytic. A classic example is the following function 1

𝑓(𝑥) = 𝑒 −𝑥 ∶ 𝑥 > 0 & 𝑓(𝑥) = 0 ∶ 𝑥 ≤ 0 The function is infinitely differentiable (smooth), but when 𝑥 = 0 the function and all its derivatives are equal to 0, and therefore the Taylor series of 𝑓(𝑥) at the origin converges to zero and is not equal to 𝑓(𝑥) for 𝑥 > 0 and consequently 𝑓(𝑥) is not analytic at the origin. In the special case when 𝑎 = 0 Taylor expansion is called the Maclaurin series (after Colin Maclaurin):

𝑓(𝑥) = 𝑓(0) + 𝑓

′ (0)𝑥

𝑓 ′′ (0) 2 𝑓 (3) (0) 3 𝑓 (𝑛) (0) 𝑛 + 𝑥 + 𝑥 +⋯+ 𝑥 2! 3! 𝑛!

And now back to Euler Formula, the Maclaurin series for 𝑒 , sin(𝑥) and cos(𝑥)are 𝑥

𝑥2 𝑥3 𝑥4 𝑥5 + + + … 2! 3! 4! 5! 𝑥3 𝑥5 𝑥7 𝑥9 sin(𝑥) = 𝑥 − + − + … 3! 5! 7! 9! 𝑥2 𝑥4 𝑥6 𝑥8 cos(𝑥) = 1 − + − + … 2! 4! 6! 8! 𝑒𝑥 = 1 + 𝑥 +

And for log(1 + 𝑥)

log(1 + 𝑥) = 𝑥 −

𝑥2 𝑥3 𝑥4 𝑥5 + − + −⋯ 2 3 4 5

30

Naji Arwashan Replacing 𝑥 by 𝑖𝑥 in the expansion of 𝑒 𝑥 yields: 𝑒 𝑖𝑥 = 1 + 𝑖𝑥 +

(𝑖𝑥)2 (𝑖𝑥)3 (𝑖𝑥)4 (𝑖𝑥)5 (𝑖𝑥)6 (𝑖𝑥)7 + + + + + … 2! 3! 4! 5! 6! 7!

Taking into account that: 𝑖 0 = 1, 𝑖1 = 𝑖, 𝑖 2 = −1, 𝑖 3 = −𝑖 𝑖 4 = 1, 𝑖 5 = 𝑖, 𝑖 6 = −1, 𝑖 7 = −𝑖 ⋮ ⋮ ⋮ ⋮ And replacing in 𝑒 𝑖𝑥 gives 𝑒 𝑖𝑥 = 1 + 𝑖𝑥 −

𝑥 2 𝑖𝑥 3 𝑥 4 𝑖𝑥 5 𝑥 6 𝑖𝑥 7 𝑥 8 − + + − − + … 2! 3! 4! 5! 6! 7! 8!

Rearranging the terms of the series allows to write: 𝑒 𝑖𝑥 = (1 −

𝑥2 𝑥4 𝑥6 𝑥8 𝑥3 𝑥5 𝑥7 + − + − ⋯ ) + 𝑖 (𝑥 − + − + ⋯ ) 2! 4! 6! 8! 3! 5! 7! 𝑒 𝑖𝑥 = cos 𝑥 + 𝑖 sin 𝑥

However, the rearrangement of the terms of Maclaurin series of 𝑒 𝑖𝑥 requires the series to be absolutely convergent. A topic that we will address in the next paragraph.

2.6. ABSOLUTE CONVERGENCE OF A SERIES A series (complex or real) is said to be absolutely convergent if the series of the absolute value of its terms is convergent. If a series is convergent but its absolute value series is not, then the convergence of

A Guided Tour in the World of Complex Numbers

31

the series is called conditional. The classic example of conditional convergence is the following series: 1 1 1 1 1− + − + −⋯ 2 3 4 5 This series meets the definition of the alternating series because all its terms are positive and every other term has a negative sign, therefore the convergence test of alternating series can be applied here. The test says that if the summand goes to zero in infinity the alternating series 1

converges, and indeed the summand here is 𝑛 goes to zero when 𝑛 gores to infinity, and the series converges, and its sum is equal to log 2. This sum is very important, and we will run into it again later in the book, so let us show how it is calculated: We found in the previous section that Maclaurin series for log(1 + 𝑥) is

log(1 + 𝑥) = 𝑥 −

𝑥2 𝑥3 𝑥4 𝑥5 𝑥𝑛 + − + −⋯ 2 3 4 5 n

Replacing 𝑥 by 1 gives: log(2) = 1 −

12 13 14 15 + − + −⋯ 2 3 4 5

Now let us examine the series of the absolute values of the series we are studying, it can be written as: 1 1 1 1 1+ + + + … 2 3 4 5

32

Naji Arwashan

But this is the harmonic series that we found earlier in Section 1.4 to be divergent, and therefore the convergence of our series is conditional. The strange thing about conditionally convergent series is that their sum varies if we change the order of the series. To illustrate that let us consider again the alternating harmonic series 1 1 1 1 1 1 1 1 1 1 1 1− + − + − + − + − + − … = log 2 2 3 4 5 6 7 8 9 10 11 12 Multiplying both sides by 2 gives: 2 1 2 1 2 1 2 1 2 1 2−1+ − + − + − + − + − + ⋯ = 2 log 2 3 2 5 3 7 4 9 5 11 6 If we add the terms with the same denominator, like adding −1

2

1

3

; adding

and

−1 3

; and adding

2 5

and

−1 5

2 1

and

etc.. then rearranging the

terms, we get: 1 1 1 1 1 1 − + − + − = 2 log 2 2 3 4 5 6 But what we have on the left is nothing else than the alternating harmonic series that we started with, and its sum now is equal to 2 log 2. This puzzling result was discovered by Peter Lejeune-Dirichlet in 1827 and explained later by Bernard Riemann who proved that we can make a conditionally convergent series converge to any sum we want by rearranging the order of its terms! Not only that, but even with a certain rearrangement we can make a conditionally convergent series diverge. On the other hand, the good news is that an absolutely convergent series has only one sum no matter how we rearrange its terms. And that is the case of the Maclaurin series of 𝑒 𝑖𝑥 we ran into earlier and was:

A Guided Tour in the World of Complex Numbers 𝑒 𝑖𝑥 = 1 + 𝑖𝑥 −

33

𝑥 2 𝑖𝑥 3 𝑥 4 𝑖𝑥 5 𝑥 6 𝑖𝑥 7 𝑥 8 − + + − − + … 2! 3! 4! 5! 6! 7! 8!

The terms of this complex series are either on the real axis or the imaginary axis, and we defined earlier in Section 2.3 the absolute value of a complex number to be the distance from zero, therefore the series of the absolute value of the previous series is ∞

∑ 𝑛=0

|𝑥 𝑛 | 𝑛!

And using the ratio test for convergence introduced in Section 1.2 |𝑥 𝑛+1 |

𝑟=

(𝑛+1)! |𝑥 𝑛 |

=

|𝑥| 𝑛+1

𝑛!

Clearly the ratio goes to zero when n goes to infinity, and therefore the series is absolutely convergent.

2.7. POLAR COORDINATES OF COMPLEX NUMBERS We defined earlier the modulus of a complex number 𝑧 = 𝑥 + 𝑖𝑦 to be: 𝑟 = |𝑧| = √𝑥 2 + 𝑦 2 And we defined its argument to be the angle 𝜑 (as shown in Figure 2-4).

34

Naji Arwashan We can write z as: 𝑥 𝑦 𝑧 = 𝑥 + 𝑖𝑦 = √𝑥 2 + 𝑦 2 ( +𝑖 ) √𝑥 2 + 𝑦 2 √𝑥 2 + 𝑦 2 𝑧 = 𝑥 + 𝑖𝑦 = 𝑟(cos𝜑 + 𝑖 sin𝜑) 𝑧 = 𝑥 + 𝑖𝑦 = 𝑟𝑒 𝑖𝜑

And therefore 𝑧 can be expressed in terms of 𝑟 and 𝜑 instead of 𝑥 and 𝑦. 𝑟 and 𝜑 are called the polar coordinates of z, and in certain situations it is more convenient to use polar coordinate, like for instance if we wanted to calculate 𝑤 as the square of 𝑧; 𝑤 = 𝑧 2 = (𝑟 𝑒 𝑖𝜑 )2 = 𝑟 2 𝑒 𝑖2𝜑 = 𝑅𝑒 𝑖 ∅ Where the modulus and argument of 𝑤 are 𝑅 = 𝑟 2 and ∅ = 2𝜑 (that is what we found earlier in Section 2.4 when we addressed the mapping of complex functions). Also, we should mention that as shown in Section 2.3 a given complex number has infinite arguments 𝜑 + 2𝜋𝑘 (𝜑 is the principal one) therefore in terms of polar coordinates a complex number can be written as: 𝑧 = 𝑟𝑒 𝑖𝜑 = 𝑟𝑒 𝑖(𝜑+2𝜋𝑘)

2.8. COMPLEX LOGARITHM The natural next step after we discussed the complex exponential is to address the complex logarithm. If a complex number is represented by its polar coordinates, 𝑧 = 𝑟𝑒 𝑖(𝜑+2𝜋𝑘)

A Guided Tour in the World of Complex Numbers

35

Where 𝜑 is the principal argument −𝜋 < 𝜑 ≤ 𝜋, then we can take the natural logarithm (base e) log 𝑧 = log(𝑟𝑒 𝑖(𝜑+2𝜋𝑘) ) = log 𝑟 + log 𝑒 𝑖(𝜑+2𝜋𝑘) = log 𝑟 + 𝑖(𝜑 + 2𝜋𝑘) So, for every value of k we obtain a different value for the logarithm, or in other words because a given complex number has infinite arguments then it has infinite logarithms, and consequently the complex logarithm is a multi-valued function. This is usually a big source of ambiguity when it comes to dealing with complex logarithm. To make the complex logarithm a single-value function that we can deal with like other functions, we define branches of logarithm where every value of k defines a different branch. When 𝑘 = 0 (principal argument) we obtain the principal branch of the complex logarithm.

2.9. COMPLEX DIFFERENTIATION AND ANALYTICITY We found in Section 2.5 when we talked about Taylor expansion that a real function is said to be analytic if it is infinitely differentiable and can be expanded by Taylor series. What about complex functions? Well, a complex function is called analytic if it is infinitely differentiable. But the question now is how do we differentiate a complex function? Actually, it is done in a similar fashion to the differentiation of a real function. If 𝑓(𝑧) is a complex function with complex argument 𝑧 = 𝑥 + 𝑖𝑦. Normally it can be written as: 𝑓(𝑧) = 𝑢(𝑥, 𝑦) + 𝑖𝑣(𝑥, 𝑦)

36

Naji Arwashan By definition the derivative of 𝑓(𝑧) at 𝑧0 is 𝑓(𝑧) − 𝑓(𝑧0 ) 𝑧→𝑧0 𝑧 − 𝑧0

𝑓 ′ (𝑧0 ) = lim

But it is important now to notice that 𝑧 belong to the complex plane and therefore it can approach 𝑧0 in infinite ways. And for the derivative to exist all the ways should lead to the same limit. The same definition can also be written as: 𝑓(𝑧0 + ℎ) − 𝑓(𝑧0 ) ℎ→0 ℎ

𝑓 ′ (𝑧0 ) = lim

And again, ℎ can approach zero following any direction. When ℎ goes to zero along the 𝑥 axis, then ℎ = ∆𝑥, and the limit is: 𝑓(𝑧0 + ∆𝑥) − 𝑓(𝑧0 ) ∆𝑥→0 ∆𝑥

𝑓 ′ (𝑧0 ) = lim We can write

𝑓(𝑧0 ) = 𝑢(𝑥0 , 𝑦0 ) + 𝑖𝑣(𝑥0 , 𝑦0 ) 𝑓(𝑧0 + ∆𝑥) = 𝑢(𝑥0 + ∆𝑥, 𝑦0 ) + 𝑖𝑣(𝑥0 + ∆𝑥, 𝑦0 ) Replacing gives us 𝑓 ′ (𝑧0 ) = [𝑢(𝑥0 + ∆𝑥, 𝑦0 ) + 𝑖𝑣(𝑥0 + ∆𝑥, 𝑦0 )] − [𝑢(𝑥0 , 𝑦0 ) + 𝑖𝑣(𝑥0 , 𝑦0 )] lim ∆𝑥→0 ∆𝑥 𝑢(𝑥0 + ∆𝑥, 𝑦0 ) − 𝑢(𝑥0 , 𝑦0 ) ∆𝑥→0 ∆𝑥 𝑣(𝑥0 + ∆𝑥, 𝑦0 ) − 𝑣(𝑥0 , 𝑦0 ) + 𝑖 lim ∆𝑥→0 ∆𝑥

𝑓 ′ (𝑧0 ) = lim

A Guided Tour in the World of Complex Numbers

37

𝜕𝑢 𝜕𝑣 +𝑖 | 𝜕𝑥 𝜕𝑥 𝑥=𝑥0 ,𝑦=𝑦0

𝑓 ′ (𝑧0 ) =

And when ℎ goes to zero along the imaginary axis, then ℎ = 𝑖∆𝑦, and the limit becomes 𝑓(𝑧0 + 𝑖∆𝑦) − 𝑓(𝑧0 ) ∆𝑥→0 𝑖∆𝑦

𝑓 ′ (𝑧0 ) = lim And in this case

𝑓(𝑧0 + 𝑖∆𝑦) = 𝑢(𝑥0 , 𝑦0 + ∆𝑦) + 𝑖𝑣(𝑥0 , 𝑦0 + ∆𝑦) Replacing gives 𝑓 ′ (𝑧0 ) = lim

[𝑢(𝑥0 ,𝑦0 +∆𝑦)+𝑖𝑣(𝑥0 ,𝑦0 +∆𝑦)]−[𝑢(𝑥0 ,𝑦0 )−𝑖𝑣(𝑥0 ,𝑦0 )] 𝑖∆𝑦

∆𝑦→0

𝑓 ′ (𝑧0 ) =

1 𝑢(𝑥0 , 𝑦0 + ∆𝑦) − 𝑢(𝑥0 , 𝑦0 ) lim 𝑖 ∆𝑦→0 ∆𝑦 1 𝑣(𝑥0 , 𝑦0 + ∆𝑦) − 𝑣(𝑥0 , 𝑦0 ) + 𝑖 lim 𝑖 ∆𝑥→0 ∆𝑦 1 𝜕𝑢 𝜕𝑣 𝑓 ′ (𝑧0 ) = + | 𝑖 𝜕𝑦 𝜕𝑦 𝑥=𝑥 ,𝑦=𝑦 0

0

𝑖

Multiplying the first term by 𝑖 gives 𝑓 ′ (𝑧0 ) =

𝑖 1 𝜕𝑢 𝜕𝑣 + | 𝑖 𝑖 𝜕𝑦 𝜕𝑦 𝑥=𝑥

0

𝑓 ′ (𝑧0 ) =

𝜕𝑣 𝜕𝑢 −𝑖 | 𝜕𝑦 𝜕𝑦 𝑥=𝑥

0

,𝑦=𝑦0

,𝑦=𝑦0

38

Naji Arwashan And for the derivative to exist at 𝑧0 the two limits should be equal 𝑓 ′ (𝑧0 ) =

𝜕𝑢 𝜕𝑣 𝜕𝑣 𝜕𝑢 +𝑖 = −𝑖 | 𝜕𝑥 𝜕𝑥 𝜕𝑦 𝜕𝑦 𝑥=𝑥

0

,𝑦=𝑦0

which leads to 𝜕𝑢 𝜕𝑣 𝜕𝑣 𝜕𝑢 = & =− ∶ 𝑥 = 𝑥0 , 𝑦 = 𝑦0 𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦 The last two equations are called Cauchy-Riemann equations. Normally the partial derivatives are designated as follows: 𝑢𝑥 =

𝜕𝑢 𝜕𝑢 𝜕𝑣 𝜕𝑣 , 𝑢𝑦 = , 𝑣𝑥 = , 𝑣𝑦 = 𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦

And with this we conclude that if a complex function has a derivative at 𝑧0 then the following Cauchy-Riemann equations should be satisfied at 𝑧0 𝑢𝑥 = 𝑣𝑦

&

𝑣𝑥 = −𝑢𝑦

The question now, is whether the converse holds true? Or in other words, if the two equations are satisfied would that be a sufficient condition for the complex function to be differentiable. Well the answer is the two equations should be satisfied and the four partial derivatives (𝑢𝑥 , 𝑢𝑦 , 𝑣𝑥 , 𝑣𝑦 ) should be continuous for the complex function to be differentiable. Furthermore, it can be proved that if the above two conditions are satisfied then the complex function is infinitely differentiable and analytic.

A Guided Tour in the World of Complex Numbers

39

2.10. COMPLEX INTEGRATION We covered so far all aspects of complex numbers and complex analysis that we need for this book and the only subject left is how to integrate in the complex plane. But first let us start with the real numbers.

The Fundamental Theorem of Calculus (Real Numbers) When a function is real, let us call it, 𝑓(𝑡), and is continuous on the interval from 𝑎 to 𝑏, then if we divide the interval from 𝑎 to 𝑏 into (𝑛 − 1) segments, where 𝑎 = 𝑡0 < 𝑡1 < 𝑡2 1

The final question for this chapter is zeta function analytic on the domain of 𝜎 > 1? Or in other words is

𝜕𝑢 𝜕𝜎

=

𝜕𝑣

,

𝜕𝑡

𝜕𝑣 𝜕𝜎

=−

𝜕𝑢 𝜕𝑡

and the four

partial derivatives continuous when 𝜎 > 1? Well, let us see. From equation (7) we found the summand of zeta to be

1 𝑛𝑠

1

therefore, zeta can be written as ∞

∞

∞

𝑛=1

𝑛=1

𝑛=1

1 1 1 𝑒 𝑖(−𝑡 log 𝑛) 𝜁(𝑠) = ∑ 𝑠 = ∑ 𝜎 𝑖(𝑡 log 𝑛) = ∑ 𝑛 𝑛 𝑒 𝑛𝜎

Figure 2-16. The domain of convergence of 𝜁(𝑠) (𝝈 > 𝟏).

1

= 𝑛𝜎 𝑒 𝑖(𝑡 log 𝑛) and

A Guided Tour in the World of Complex Numbers

55

Applying Euler formula, we obtain ∞

𝜁(𝑠) = ∑ [ 𝑛=1

cos(−𝑡 log 𝑛) sin(−𝑡 log 𝑛) +𝑖 ] 𝜎 𝑛 𝑛𝜎

And the formulation of the zeta function when the exponent 𝑠 is a complex number becomes as follows 𝜁(𝑠) = ∑∞ 𝑛=1

cos(𝑡 log 𝑛) 𝑛𝜎

− 𝑖 ∑∞ 𝑛=1

sin(𝑡 log 𝑛) 𝑛𝜎

(8)

Let us calculate the partial derivatives of the real and imaginary parts of zeta as follows: ∞

𝜕𝑢 cos(𝑡 log 𝑛) = 𝑢𝜎 = − ∑ log 𝑛 𝜕𝜎 𝑛𝜎 𝑛=1 ∞

𝜕𝑢 sin(𝑡 log 𝑛) = 𝑢𝑡 = − ∑ log 𝑛 𝜕𝑡 𝑛𝜎 𝑛=1 ∞

𝜕𝑣 sin(𝑡 log 𝑛) = 𝑣𝜎 = ∑ log 𝑛 𝜕𝜎 𝑛𝜎 𝑛=1 ∞

𝜕𝑣 cos(𝑡 log 𝑛) = 𝑣𝑡 = − ∑ log 𝑛 𝜕𝑡 𝑛𝜎 𝑛=1

We can see that 𝑢𝜎 = 𝑣𝑡 , 𝑢𝑡 = −𝑣𝜎 and the four partial derivatives are continuous functions and therefore Cauchy’s conditions (from Section 2.9) are satisfied, and consequently zeta is analytic when 𝜎 > 1.

Chapter 3

THE ETA SERIES AND THE FIRST EXTENSION OF ZETA 3.1. THE ETA SERIES 𝜼(𝒔) We found in last chapter that the zeta series converges whenever 𝜎 > 1, in this chapter will extend this convergence to include the region of 0 < 𝜎 ≤ 1. How is this possible? Well, let us start by defining the eta function 1 1 1 1 1 + − + − … 2𝑠 3𝑠 4𝑠 5𝑠 6𝑠 ∞ 1 𝜂(𝑠) = ∑(−1)𝑛−1 𝑠 𝑛

𝜂(𝑠) = 1 −

𝑛=1

It is the same zeta function but with every other term negated, so the eta series is often called the alternating zeta series, and denoted by 𝜁𝑎 (𝑠), or 𝜁 ∗ (𝑠).

58

Naji Arwashan

3.2. CONVERGENCE OF THE ETA SERIES One nice thing about the eta series is that, contrary to the zeta series (which converges only when 𝜎 > 1), eta converges for all 𝜎 > 0. To prove this fact, first consider the case when 𝑠 = 𝜎 is real: ∞

𝜂(𝜎) = ∑(−1)𝑛−1 𝑛=1

1 𝑛𝜎

When 𝜎 > 0, this series meets the three conditions of alternating series test (Chapter 1, Section 1.2), and therefore converges. What if 𝑠 = 𝜎 + 𝑖𝑡 is complex? Well the standard proof of that convergence relies on the theory of Dirichlet series. A Dirichlet series is one of the form ∞

∑ 𝑛=1

𝑎𝑛 𝑛𝑠

where 𝑠 and 𝑎𝑛 are complex number. The series was introduced by the German Mathematician Peter Gustav Lejeune Dirichlet and named after him. There is an important theorem [2] which states that if Dirichlet series is convergent at a specific point 𝑠0 = 𝜎0 + 𝑖𝑡0 , then it is convergent for all 𝑠 with 𝜎 > 𝜎0 . In our case eta series is a Dirichlet series with 𝑎𝑛 = (−1)𝑛−1, and we just showed that it is convergent when 𝑡 = 0 and 𝜎 > 0 therefore it is convergent for all 𝑠 with 𝜎 > 0 However, the proof of the theorem mentioned above regarding the convergence of Dirichlet series is not simple, and instead we will show here a different proof for the convergence of eta given by J. Sondow [11]. We start by defining a new series, 𝜂∗ (𝑠), by grouping the terms of eta as follows:

The Eta Series and the First Extension of Zeta 𝜂 ∗ (𝑠) = (1 −

59

1 1 1 1 1 ) + ( 𝑠 − 𝑠) + ⋯ + ( 𝑠 − )+⋯ 𝑠 (𝑛 + 1)𝑠 2 3 4 𝑛

Which can be written as 𝜂∗ (𝑠) = ∑ ( 𝑛 𝑜𝑑𝑑

1 1 − ) 𝑠 (𝑛 + 1)𝑠 𝑛

One can verify that 𝑛+1 1 1 𝑑𝑥 − = 𝑠 ∫ 𝑠 𝑠 (𝑛 + 1) 𝑛 𝑥 𝑠+1 𝑛

therefore 𝑛+1 1 1 𝑑𝑥 |𝑠| = | 𝑠− | |∫ | 𝑠 (𝑛 + 1) 𝑛 𝑥 𝑠+1 𝑛

An integral is really an infinite sum, and as we have seen in Chapter 2 (Section 2.3) if 𝑠 = 𝑠1 + 𝑠2 then |𝑠| ≤ |𝑠1 | + |𝑠2 | 𝑛+1

|∫ 𝑛

𝑛+1 𝑛+1 𝑑𝑥 𝑑𝑥 𝑑𝑥 ≤ ∫ = ∫ | | | |𝑥 𝑠+1 | 𝑥 𝑠+1 𝑥 𝑠+1 𝑛 𝑛

𝑠 is a complex number and can be written as 𝑠 = 𝜎 + 𝑖𝑡 and 𝑥 𝑠+1 becomes 𝑥 𝑠+1 = 𝑥 (𝜎+1)+𝑖𝑡 = 𝑥 𝜎+1 𝑥 𝑖𝑡 𝑥 can be written as 𝑥 = 𝑒 log 𝑥 𝑖𝑡

𝑥 𝑠+1 = 𝑥 𝜎+1 (𝑒 log 𝑥 ) = 𝑥 𝜎+1 𝑒 i(𝑡 log 𝑥)

60

Naji Arwashan And since |𝑒 i(𝑡 log 𝑥) | = |cos(𝑡 log 𝑥) + 𝑖 sin(𝑡 log 𝑥)| = 1 then |𝑥 𝑠+1 | = 𝑥 𝜎+1 Therefore, 𝑛+1

∫ 𝑛

We know that

1

𝑛+1 𝑑𝑥 𝑑𝑥 =∫ 𝑠+1 |𝑥 | 𝑥 𝜎+1 𝑛

1

𝑥 𝜎+1

≤ 𝑛𝜎+1 when 𝑛 ≤ 𝑥 ≤ 𝑛 + 1 therefore from the

ML inequality (Chapter2, Section 2.11) we have 𝑛+1

∫ 𝑛

𝑑𝑥 1 (𝑛 + 1 − 𝑛) < 𝑥 𝜎+1 𝑛𝜎+1

thus, we can conclude that the absolute value of the summand of the series 𝜂∗ (𝑠) is 1 1 1 | 𝑠− | < |𝑠| 𝜎+1 𝑠 (𝑛 + 1) 𝑛 𝑛 1

Note that the series ∑∞ 𝑛=1 𝑛𝜎+1 is nothing else than zeta series, and therefore it converges absolutely when (𝜎 + 1) > 1 or when 𝜎 > 0. Applying the comparison test (Chapter 1, Section 1.2), we can conclude from the inequality above that 𝜂∗ (𝑠) converges absolutely when 𝜎 > 0, and thus the original eta series converges when 𝜎 > 0. We should mention that eta’s convergence is absolute only when 𝜎 > 1 and conditional when 0 < 𝜎 ≤ 1 , because the absolute value of eta is zeta which diverges when 𝜎 ≤ 1 . The convergence of eta when

The Eta Series and the First Extension of Zeta

61

0 < 𝜎 ≤ 1 is a big advantage to the zeta series, and the benefit becomes evident in the next section.

3.3. THE RELATION BETWEEN THE ETA AND THE ZETA FUNCTIONS Let us write zeta and eta again: 1 1 1 1 1 + 𝑠+ 𝑠+ 𝑠+ 𝑠… 𝑠 2 3 4 5 6 1 1 1 1 1 𝜂(𝑠) = 1 − 𝑠 + 𝑠 − 𝑠 + 𝑠 − 𝑠 … 2 3 4 5 6 𝜁(𝑠) = 1 +

Subtraction eta from zeta yields: 𝜁(𝑠) − 𝜂(𝑠) =

2 2 2 2 + + + … 2𝑠 4𝑠 6𝑠 8𝑠

2

Factoring out 2𝑠 = 21−𝑠 we have 𝜁(𝑠) − 𝜂(𝑠) = 21−𝑠 (1 +

1 1 1 1 + 𝑠 + 𝑠 + 𝑠 …) 𝑠 2 3 4 5

But what we got between parenthesis is nothing else than zeta, and therefore 𝜁(𝑠) − 𝜂(𝑠) = 21−𝑠 𝜁(𝑠) And now we can easily extract zeta and obtain 𝜁(𝑠) =

1 𝜂(𝑠) 1 − 21−𝑠

62

Naji Arwashan

This is now a new definition for zeta that has values in the domain 0 < 𝜎 ≤ 1, and yields values in the domain 𝜎 > 1 identical to the original definition. This kind of extension is called in mathematics analytic continuation. More formally, analytic continuation can be defined as: given functions 𝑓1 , analytic on domain 𝐷1 , and 𝑓2 , analytic on domain 𝐷2 , such that 𝐷1 ∩ 𝐷2 is not empty and 𝑓1 = 𝑓2 on 𝐷1 ∩ 𝐷2 , then 𝑓2 is called analytic continuation of 𝑓1 to 𝐷2 . Moreover, if it exists, the analytic continuation of 𝑓1 to 𝐷2 is unique. The relationship between zeta and eta is easier to illustrate and understand when 𝑠 is a real number as shown in Figure 3-1.

Figure 3-1. The 𝜁(𝑠) and 𝜂(𝑠) when 𝑠 is a real number > 0.

Let us go back to the world of complex numbers and let us examine the points in the complex plane where the denominator in the equation relating zeta to eta is zero. 1 − 21−𝑠 = 0

The Eta Series and the First Extension of Zeta

63

Replacing 𝑠 by 𝑠 = 𝜎 + 𝑖𝑡 21−𝜎−𝑖𝑡 = 1 21−𝜎 2−𝑖𝑡 = 1 Expanding 2−𝑖𝑡 gives 21−𝜎 (cos(𝑡 log 2) − 𝑖 sin(𝑡 log 2)) = 1 The real and imaginary parts should be equal on both side of the equality giving 21−𝜎 cos(𝑡 log 2) = 1 and 21−𝜎 sin(𝑡 log 2) = 0 And the solution to these two equations is: 𝜎 = 1, and 𝑡 = 𝑛

2𝜋 log 2

(𝑛 = 0, 1, 2, 3, … ) or it can be written in a 2𝜋

format of a complex number 𝑠𝑛 = 1 + 𝑖 𝑛 log 2 So how can we define zeta at these points? For the first point (𝑛 = 0), 𝑡 is equal to zero and eta at that point (1 + 𝑖0) is equal to 𝜂(1) = 1 −

1 1 1 1 + − + … 2 3 4 5

We found in Chapter 2 (Section 2.6) that the above series is equal to log 2, and consequently zeta at that point is simply not defined because we are dividing log 2 by a zero. How about the remaining points (𝑛 = 1,2,3, … )? This question was raised by E. Landau in 1909 and was answered 40 years later by D.V. Widder who proved that eta is equal to zero at these points. A second proof was published by J. Sondow in 2003 [10], and it goes as follows:

64

Naji Arwashan

3.4. SONDOW PROOF OF 𝜼(𝒔) = 0 AT 𝒔𝒏 = 𝟏 + 𝒊 𝒏

𝟐𝝅 𝐥𝐨𝐠 𝟐

(N = 1, 2, 3, …) We call 𝜁2𝑁 and 𝜂2𝑁 the sum of the first 2𝑁 terms of zeta and eta series as follows 1 1 1 1 1 1 + 𝑠+ 𝑠+⋯ 𝑠+ + … 𝑠 𝑠 (𝑁 + 1) (𝑁 + 2)𝑠 2 3 4 𝑁 1 + (2𝑁)𝑠 1 1 1 1 1 1 𝜂2𝑁 (𝑠) = 1 − 𝑠 + 𝑠 − 𝑠 + ⋯ 𝑠 − + … 𝑠 (𝑁 + 1) (𝑁 + 2)𝑠 2 3 4 𝑁 1 − (2𝑁)𝑠 𝜁2𝑁 (𝑠) = 1 +

And if we calculate 𝜂2𝑁 (𝑠) − 𝜁2𝑁 (𝑠) we find it equal to 𝜂2𝑁 (𝑠) − 𝜁2𝑁 (𝑠) = −2

1 1 1 − 2 − ⋯ 2 (2𝑁)𝑠 2𝑠 4𝑠

1

And factoring out (−2 2𝑠 ) yields 1

1

1

1

𝜂2𝑁 (𝑠) − 𝜁2𝑁 (𝑠) = −21−𝑠 (1 + 2𝑠 + 3𝑠 + 4𝑠 + ⋯ 𝑁𝑠)

(1)

But what we have in parenthesis is the sum of the first N terms of zeta which can be written as 2𝑁

𝜁𝑁 (𝑠) = 𝜁2𝑁 (𝑠) − ∑ 𝑛−𝑠 𝑛=𝑁+1

The Eta Series and the First Extension of Zeta

65

Replacing 𝜁𝑁 (𝑠) in (1) 2𝑁 1−𝑠

𝜂2𝑁 (𝑠) − 𝜁2𝑁 (𝑠) = −2

[𝜁2𝑁 (𝑠) − ∑ 𝑛−𝑠 ] 𝑛=𝑁+1

Rearranging gives −𝑠 𝜂2𝑁 (𝑠) = (1 − 21−𝑠 )𝜁2𝑁 (𝑠) + 21−𝑠 [∑2𝑁 𝑛=𝑁+1 𝑛 ]

(2)

And the trick now is to write the term in brackets of (2) as 2𝑁

𝑁

∑ 𝑛

−𝑠

𝑁

= ∑(𝑁 + 𝑘)

𝑛=𝑁+1

−𝑠

𝑘=1

=𝑁

−𝑠

𝑘 −𝑠 = ∑ [𝑁 (1 + )] 𝑁 𝑘=1

𝑁

𝑁

𝑘=1

𝑘=1

𝑘 −𝑠 1 𝑘 −𝑠 ∑ (1 + ) = 𝑁1−𝑠 ∑ (1 + ) 𝑁 𝑁 𝑁

And replacing in (2) gives 1

𝑘 −𝑠

𝜂2𝑁 (𝑠) = (1 − 21−𝑠 )𝜁2𝑁 (𝑠) + 21−𝑠 𝑁1−𝑠 [𝑁 ∑𝑁 𝑘=1 (1 + 𝑁 )

]

(3) 1

At this point we hold on equation (3) and study the integral ∫0 (1 + 𝑥)−𝑠 𝑑𝑥 . We start by changing the variable to 𝑦 = 1 + 𝑥, and the integral becomes 1

2

∫ (1 + 𝑥)−𝑠 𝑑𝑥 = ∫ (𝑦)−𝑠 𝑑𝑦 0

1

When 𝑠 = 1 the integral value is 2

∫ 1

1 𝑑𝑦 = [log 𝑦]12 = log(2) − log(1) = log 2 𝑦

66

Naji Arwashan And when s≠ 1 the integral becomes: 𝑦 −𝑠+1

2

2

2−𝑠+1

1−𝑠+1

∫1 (𝑦)−𝑠 𝑑𝑦 = [ −𝑠+1 ] = [ −𝑠+1 − −𝑠+1] = 1

1−21−𝑠 𝑠−1

(4)

That same integral can be calculated as follows: As shown in Figure 3-2, we divide the interval from 0 to 1 into 𝑁 segments, the width of 1

𝑘

each segment is 𝑁, and the value of 𝑥 at the end of each segment is 𝑁 where 𝑘 = 1, 2, 3, … 𝑁, and the value of the function (1 + 𝑥)−𝑠 at the 𝑘 −𝑠

end of each segment is (1 + 𝑁) .

1

𝑘 −𝑠

𝑁

𝑁

Figure 3-2. The function (𝟏 + 𝒙)−𝒔 and the sum ∑𝑁 𝑘=1 (1 + )

The area 𝐴(𝑁) of the rectangular segments is equal to 𝑁

𝐴(𝑁) = ∑ 𝑘=1

1 𝑘 −𝑠 (1 + ) 𝑁 𝑁

The Eta Series and the First Extension of Zeta

67

Which, if we pay attention, is equal to the term in brackets of equation (3). The above sum is called Riemann sum of the integral 1

∫0 (1 + 𝑥)−𝑠 𝑑𝑥 (yes, it is the same Bernard Riemann). The difference, 1

𝑑𝑁 (𝑠), between the integral ∫0 (1 + 𝑥)−𝑠 𝑑𝑥 and its Riemann sum goes to 0 when 𝑁 goes to infinity, and is equal to 1

1

𝑘 −𝑠

𝑑𝑁 (𝑠) = ∫0 (1 + 𝑥)−𝑠 𝑑𝑥 − ∑𝑁 (1 + 𝑁) 𝑁 𝑘=1

(5)

Now we can go back to equation (3) and replace the term in the brackets by its value from (4) 1

𝜂2𝑁 (𝑠) = (1 − 21−𝑠 )𝜁2𝑁 (𝑠) + 21−𝑠 𝑁1−𝑠 [∫ (1 + 𝑥)−𝑠 𝑑𝑥 − 𝑑𝑁 (𝑠)] 0

Our goal is to calculate eta when 𝑠 = 𝑠𝑛 = 1 + 𝑖 𝑛

2𝜋 ln 2

(𝑛 =

1,2,3, … ), so let us replace 𝑠 by 𝑠𝑛 keeping in mind that 𝑠𝑛 satisfies the equation 1 − 21−𝑠𝑛 = 0, and replacing the integral by its value from (4) since 𝑠𝑛 ≠ 0 we obtain 2𝜋 0 𝜂2𝑁 (𝑠𝑛 ) = (0)𝜁2𝑁 (𝑠𝑛 ) + (1)𝑁 − 𝑖 𝑛ln 2 [ − 𝑑𝑁 (𝑠𝑛 )] 𝑠𝑛 − 1 𝑑𝑁 (𝑠𝑛 ) 𝜂2𝑁 (𝑠𝑛 ) = − 2𝜋 𝑁 𝑖 𝑛ln 2

When 𝑁 goes to infinity 𝜂2𝑁 (𝑠𝑛 ) becomes 𝜂(𝑠𝑛 ); 𝑑𝑁 (𝑠) goes to 0, 2𝜋

and 𝑁 𝑖 𝑛ln 2 is bounded by 1, and hence we have eta equal to 0 when 2𝜋

𝑠 = 𝑠𝑛 = 1 + 𝑖 𝑛 log 2 (𝑛 = 1,2,3, … )

68

Naji Arwashan

3.5. ZETA AT 𝒔𝒏 = 𝟏 + 𝒊 𝒏

𝟐𝝅 𝐥𝐨𝐠 𝟐

(𝒏 = 𝟏, 𝟐, 𝟑, … )

Now that we established that eta is zero at 𝑠𝑛 , with the exception of 𝑠 = 1 or (or 𝑛 = 0), then we can write 𝜂(𝑠) 𝑠→𝑠𝑛 1 − 21−𝑠

𝜁(𝑠𝑛 ) = lim

Both numerator and denominator go to zero when 𝑠 goes to 𝑠𝑛 , and to remove this kind of indeterminacy we resort to Hospital Rule. It is a rule that will be used several times in this book, and therefore we decided to show its proof in the following:

Hospital Rule If we have two functions 𝑓(𝑥) and 𝑔(𝑥), and both are zero at 𝑥0 , 𝑓(𝑥0 ) = 𝑓0 = 0, and 𝑔(𝑥0 ) = 𝑔0 = 0. Hospital Rule, named after the 17th century French mathematician Guillaume de l’Hôpitale, states that 𝑓(𝑥) 𝑓 ′ (𝑥) = lim ′ 𝑥→𝑥0 𝑔(𝑥) 𝑥→𝑥0 𝑔 (𝑥) lim

And the same result also holds if both 𝑓(𝑥) and 𝑔(𝑥) go to infinity when 𝑥 goes to 𝑥0 To prove this important rule, we start by the first case when 𝑓(𝑥0 ) = 𝑓0 = 0, and 𝑔(𝑥0 ) = 𝑔0 = 0. In this case since 𝑓(𝑥0 ) = 𝑔(𝑥0 ) = 0, we can write 𝑓(𝑥) 𝑓(𝑥) − 𝑓(𝑥0 ) 𝑓(𝑥) − 𝑓(𝑥0 ) 𝑥 − 𝑥0 = lim = lim 𝑥→𝑥0 𝑔(𝑥) 𝑥→𝑥0 𝑔(𝑥) − 𝑔(𝑥0 ) 𝑥→𝑥0 𝑥 − 𝑥0 𝑔(𝑥) − 𝑔(𝑥0 ) lim

The Eta Series and the First Extension of Zeta

69

𝑓(𝑥) 𝑓(𝑥) − 𝑓(𝑥0 ) 𝑥 − 𝑥0 = lim lim 𝑥→𝑥0 𝑔(𝑥) 𝑥→𝑥0 𝑥→𝑥0 𝑔(𝑥) − 𝑔(𝑥0 ) 𝑥 − 𝑥0 lim

𝑓(𝑥)−𝑓(𝑥 )

lim 𝑥−𝑥 0 𝑓(𝑥) 𝑥→𝑥 𝑓 ′ (𝑥) 0 0 lim = = lim 𝑔(𝑥)−𝑔(𝑥0 ) 𝑥→𝑥0 𝑔(𝑥) 𝑥→𝑥0 𝑔 ′ (𝑥) lim 𝑥→𝑥0

𝑥−𝑥0

Now let us consider the second case where 𝑓(𝑥) and 𝑔(𝑥) go to infinity when 𝑥 goes to 𝑥0 . In this case we can write: 1

𝑓(𝑥) 𝑔(𝑥) lim = lim 1 𝑥→𝑥0 𝑔(𝑥) 𝑥→𝑥0 𝑓(𝑥)

Both

1 𝑔(𝑥)

and

1 𝑓(𝑥)

go to zero when 𝑥 goes to 𝑥0 and therefore we can

apply the first case of Hospital rule 1

[

′

]

0−𝑔′ (𝑥)

] 𝑓(𝑥) 𝑔′ (𝑥)𝑓(𝑥)2 𝑔(𝑥) 𝑔(𝑥)2 lim = lim = lim 0−𝑓′ (𝑥) = lim ′ 1 ′ 𝑥→𝑥0 𝑔(𝑥) 𝑥→𝑥0 𝑥→𝑥0 𝑥→𝑥0 𝑓 (𝑥)𝑔(𝑥)2 [ ] [ ] 𝑓(𝑥)

[

𝑓(𝑥)2

𝑓(𝑥) 𝑔′ (𝑥) 𝑓(𝑥) 𝑓(𝑥) = lim ′ lim lim 𝑥→𝑥0 𝑔(𝑥) 𝑥→𝑥0 𝑓 (𝑥) 𝑥→𝑥0 𝑔(𝑥) 𝑥→𝑥0 𝑔(𝑥) lim

Eliminating lim

𝑓(𝑥)

𝑥→𝑥0 𝑔(𝑥)

from both sides leaves us with

𝑓(𝑥) 𝑓 ′ (𝑥) = lim ′ 𝑥→𝑥0 𝑔(𝑥) 𝑥→𝑥0 𝑔 (𝑥) lim

The same Hospital rule applies also when 𝑥 goes to ∞ instead of 𝑥0 . 1

To prove that we change the variable to 𝑦 = 𝑥 , and when x goes to infinity 𝑦 goes to 0. Therefore, we can apply what we have already proved and write:

70

Naji Arwashan 𝑑𝑓

𝑓(𝑥) 𝑓(𝑦) 𝑑𝑦 lim = lim = lim 𝑑𝑔 𝑥→∞ 𝑔(𝑥) 𝑦→0 𝑔(𝑦) 𝑦→0 𝑑𝑦

Applying the chain rule of differentiation, we can write: 𝜕𝑓 𝜕𝑥

𝑓(𝑥) 𝜕𝑥 𝜕𝑦 lim = lim 𝜕𝑔 𝜕𝑥 𝑥→∞ 𝑔(𝑥) 𝑦→0 𝜕𝑥 𝜕𝑦 𝜕𝑥

And eliminating 𝜕𝑦 from numerator and denominator gives 𝑓(𝑥) 𝑓 ′ (𝑥) lim = lim ′ 𝑥→∞ 𝑔(𝑥) 𝑥→∞ 𝑔 (𝑥) Going back to zeta and applying Hospital rule, we take the derivative of the numerator and denominator with respect to s. 𝜂(𝑠) 𝜂′(𝑠) 𝜂′(𝑠) = lim = lim 𝑠→𝑠𝑛 1 − 21−𝑠 𝑠→𝑠𝑛 (1 − 21−𝑠 )′ 𝑠→𝑠𝑛 0 − (−1)21−𝑠 log 2

𝜁(𝑠𝑛 ) = lim

And we know that 21−𝑠𝑛 = 1, therefore 𝜁(𝑠𝑛 ) =

𝜂′(𝑠𝑛 ) log 2

And with that, now zeta is defined everywhere 𝜎 > 0 except for (1 + 𝑖0).

The Eta Series and the First Extension of Zeta

71

3.6. RESIDUE OF ZETA AT 1 The point (1 + 𝑖0) where zeta is not defined is called pole (Chapter 2, Section 2.12), and its residue is calculated as follows 𝑅𝑒𝑠(𝜁, 1) = lim(𝑠 − 1)𝜁(𝑠) 𝑠→1

Replacing zeta by its value as function of eta gives (𝑠 − 1) (𝑠 − 1) 𝜂(𝑠) = [lim ] [lim 𝜂(𝑠)] 1−𝑠 𝑠→1 1 − 2 𝑠→1 1 − 21−𝑠 𝑠→1

𝑅𝑒𝑠(𝜁, 1) = lim

(𝑠−1)

For the first limit lim 1−21−𝑠 both numerator and denominator go to 𝑠→1

0 when 𝑠 goes to 1, and therefore we can apply Hospital rule as follows: (𝑠 − 1)′ 𝑠−1 1−0 1 = lim = lim = 1−𝑠 1−𝑠 ′ 1−𝑠 𝑠→1 1 − 2 𝑠→1 (1 − 2 𝑠→1 0 − (−2 ) log 2) log 2

lim

And for the second limit lim 𝜂(𝑠) = 𝜂(1), and we found earlier in 𝑠→1

Section 3.4 that 𝜂(1) = log 2 therefore 𝑅𝑒𝑠(𝜁, 1) = 1

Chapter 4

THE FUNCTIONAL EQUATION AND THE SECOND EXTENSION OF ZETA Bernard Riemann discovered a functional equation which expands the definition of the zeta function to the whole plane of complex numbers. But before we rederive that equation, we must introduce a new and very important function called the gamma function

4.1. THE GAMMA FUNCTION By definition the gamma function is ∞

Γ (𝑠) = ∫ 𝑥 𝑠−1 𝑒 −𝑥 𝑑𝑥 0

Where 𝑠 is a complex number and equal to 𝑠 = 𝜎 + 𝑖𝑡 and 𝑥 is a real number. Let us evaluate Γ (𝑠 + 1)

74

Naji Arwashan ∞

Γ (𝑠 + 1) = ∫ 𝑥 𝑠 𝑒 −𝑥 𝑑𝑥 0

Applying integration by part (∫ 𝑢𝑑𝑣 = 𝑢𝑣 − ∫ 𝑣𝑑𝑢) , calling 𝑣 = −𝑒 ( 𝑑𝑣 = 𝑒 −𝑥 𝑑𝑥 ) and 𝑢 = 𝑥 𝑠 (𝑑𝑢 = 𝑠𝑥 𝑠−1 ) we get −𝑥

∞

Γ (𝑠 + 1) =

[−𝑥 𝑠 𝑒 −𝑥 ]∞ 0

− ∫ −𝑠𝑥 𝑠−1 𝑒 −𝑥 𝑑𝑥 0 ∞

Γ (𝑠 + 1) = lim (−𝑥 𝑠 𝑒 −𝑥 ) − (−0𝑠 𝑒 −0 ) + 𝑠 ∫ 𝑥 𝑠−1 𝑒 −𝑥 𝑑𝑥 𝑥→∞

0

We can show that lim (−𝑥 𝑠 𝑒 −𝑥 ) = 0 by repeated application of 𝑥→∞

Hospital’s rule, and the integral on the right-hand side is nothing else than Γ (𝑠) so we have the recursion Γ (𝑠 + 1) = 𝑠Γ (𝑠)

(1)

This is the functional equation of the gamma function. Let’s calculate Γ (1) ∞

−∞ Γ (1) = ∫0 𝑡 1−1 𝑒 −𝑡 𝑑𝑡 = [−𝑒 −𝑡 ]∞ − (−𝑒 −0 ) = 1 0 = −𝑒

And using the functional equation of gamma we just found we can write Γ (2) = 1 × Γ (1) = 1 Γ (3) = 2 × Γ (2) = 2 × 1 Γ (4) = 3 × Γ (3) = 3 × 2 × 1 ⋮

The Functional Equation and the Second Extension of Zeta

75

Γ (𝑛) = (𝑛 − 1) … .3 × 2 × 1 = (n − 1)! And therefore, when 𝑠 is equal to a positive integer 𝑛, gamma of 𝑛 is equal to the factorial of (𝑛 − 1) Γ (𝑛) = (𝑛 − 1)!

(2)

So really the gamma function can be seen as an extension of the factorial function to the rest of the complex plane. Let us now check if there are any values where the gamma function is not defined. We know that n! =

(𝑛+1)! 𝑛+1 1!

by extension 0! =

1

and therefore 3! =

4! 4

, 2! =

3! 3

2!

, 1! = 2 , and finally

= 1 and if we wanted to extend it one more time

we get division by zero, and consequentially no factorial for the negative integers. Therefore, from equation 2, we can conclude that Γ (𝑠) is not defined when (𝑠 − 1) is a negative integer, or in other words when 𝑠 is equal to 0, −1, −2, −3, −4, …

4.2. THE RELATION BETWEEN THE GAMMA AND THE ZETA FUNCTIONS We defined the gamma function as follows: ∞

Γ (𝑠) = ∫ 𝑥 𝑠−1 𝑒 −𝑥 𝑑𝑥 0

Let us replace 𝑥 = 𝑛𝑢 where 𝑛 is a constant natural number and 𝑢 is a real variable, and therefore, 𝑑𝑥 = 𝑛𝑑𝑢

76

Naji Arwashan ∞

Γ (𝑠) = ∫ (𝑛𝑢)𝑠−1 𝑒 −𝑛𝑢 (𝑛𝑑𝑢) 0

Simplifying gives 1 𝑛𝑠

∞

Γ (𝑠) = ∫0 𝑢 𝑠−1 𝑒 −𝑛𝑢 𝑑𝑢

(3)

Now summing over the natural numbers ∞

∞

∞

1 ∑ 𝑠 Γ (𝑠) = ∑ ∫ 𝑢 𝑠−1 𝑒 −𝑛𝑢 𝑑𝑢 𝑛

𝑛=1

𝑛=1 0 ∞

∞

∞

1 Γ (𝑠) ∑ 𝑠 = ∫ 𝑢 𝑠−1 (∑ 𝑒 −𝑛𝑢 ) 𝑑𝑢 𝑛 𝑛=1

𝑛=1

0

1

When 𝜎 > 1, ∑∞ 𝑛=1 𝑛𝑠 is nothing else than our zeta function, and for −𝑛𝑢 the sum ∑∞ we know that the following geometric series 𝑛=1 𝑒

1 + 𝑒 −𝑢 + 𝑒 −2𝑢 + 𝑒 −3𝑢 + ⋯ converges

1

to

−𝑛𝑢 ∑∞ = 𝑛=1 𝑒

1−𝑒 −𝑢 1

1−𝑒 −𝑢

(from

−1=

𝑒 −𝑢 1−𝑒 −𝑢

Chapter1, =

1 𝑒 𝑢 −1

Section

1.3),

and after substituting

∞

1 𝑢 𝑠−1 ζ(𝑠) = ∫ 𝑑𝑢 Γ (𝑠) 𝑒 𝑢 − 1

therefore

( 𝜎 > 1)

0

This last formula is often used as a new definition of zeta.

The Functional Equation and the Second Extension of Zeta

77

4.3. THE RELATION BETWEEN THE GAMMA AND THE ETA FUNCTIONS Before we derive the functional equation of zeta, it is important to show the relationship between gamma and eta functions. We follow similar steps to the previous section and write again equation (3): ∞

1 Γ (𝑠) = ∫ 𝑢 𝑠−1 𝑒 −𝑛𝑢 𝑑𝑢 𝑛𝑠 0

But before we proceed to take the sum over all 𝑛, we will multiply both sides by (−1)𝑛−1 ∞

1 (−1)𝑛−1 𝑠 Γ (𝑠) = (−1)𝑛−1 ∫ 𝑢 𝑠−1 𝑒 −𝑛𝑢 𝑑𝑢 𝑛 0

And now we sum over all the natural numbers on both sides ∞

∞

∞

(−1)𝑛−1 Γ (𝑠) = ∑(−1)𝑛−1 ∫ 𝑢 𝑠−1 𝑒 −𝑛𝑢 𝑑𝑢 ∑ 𝑛𝑠

𝑛=1

∞

Γ (𝑠) ∑ 𝑛=1

𝑛=1 ∞

0

∞

(−1)𝑛−1 = ∫ 𝑢 𝑠−1 (∑(−1)𝑛−1 𝑒 −𝑛𝑢 ) 𝑑𝑢 𝑛𝑠

When 𝜎 > 0, ∑∞ 𝑛=1

0

(−1)𝑛−1 𝑛𝑠

𝑛=1

is eta function, and let us expand the sum

𝑛−1 −𝑛𝑢 ∑∞ 𝑒 𝑛=1(−1) ∞

∑(−1)𝑛−1 𝑒 −𝑛𝑢 = 𝑒 −𝑢 − 𝑒 −2𝑢 + 𝑒 −3𝑢 − ⋯ 𝑛=1

78

Naji Arwashan

we know that the following geometric series 1 − 𝑒 −𝑢 + 𝑒 −2𝑢 − 𝑒 −3𝑢 + ⋯ converges to

1 1+𝑒 −𝑢

(from Chapter1, Section 1.3) and therefore 𝑒 −𝑢

1

1

𝑛−1 −𝑛𝑢 ∑∞ 𝑒 = − (1+𝑒 −𝑢) + 1 = 1+𝑒 −𝑢 = 𝑒 𝑢+1 𝑛=1(−1)

And replacing yields: ∞

1 𝑢 𝑠−1 η(𝑠) = ∫ 𝑑𝑢 Γ (𝑠) 𝑒 𝑢 + 1

(𝜎 > 0 )

0

4.4. THE FUNCTIONAL EQUATION OF THE ZETA FUNCTION The functional equation of zeta function can be derived in several methods [12]. Here we will explain the original derivation done by Riemann himself. The method starts by establishing a second relationship between gamma and zeta, and for that purpose we will evaluate gamma at 𝑠/2 ∞

𝑠 𝑠 Γ ( ) = ∫ 𝑥 2−1 𝑒 −𝑥 𝑑𝑥 2

0

Replace 𝑥 by 𝜋𝑛2 𝑢 where 𝑛 is a constant natural number and 𝑑𝑥 = 𝜋𝑛2 𝑑𝑢 ∞

𝑠 𝑠 2 Γ ( ) = ∫(𝜋𝑛2 𝑢)(2−1) 𝑒 −𝜋𝑛 𝑢 𝜋𝑛2 𝑑𝑢 2

0

The Functional Equation and the Second Extension of Zeta

79

Simplifying gives ∞

𝑠 𝑠 𝑠 2 Γ ( ) = ∫ 𝜋 2 𝑛 𝑠 𝑢(2−1) 𝑒 −𝜋𝑛 𝑢 𝑑𝑢 2

0

𝑠

The term (𝜋 2 𝑛 𝑠 ) is constant and can be taken out of the integral, and then dividing both sides by it gives ∞

𝑠 1 −𝑠 𝑠 2 𝜋 2 Γ ( ) = ∫ 𝑢2−1 𝑒 −𝜋𝑛 𝑢 𝑑𝑢 𝑠 𝑛 2

0

Let us take the sum over all natural numbers ∞

∞

∞

𝑠 𝑠 1 𝑠 2 ∑ 𝑠 𝜋 −2 Γ ( ) = ∑ ∫ 𝑢2−1 𝑒 −𝜋𝑛 𝑢 𝑑𝑢 𝑛 2

𝑛=1

𝑛=1 0 ∞

∞

∞

𝑠 𝑠 1 2 𝜋 Γ ( ) ∑ 𝑠 = ∫(𝑢)2−1 ∑ 𝑒 −𝜋𝑛 𝑢 𝑑𝑢 2 𝑛

−

𝑠 2

𝑛=1

0

𝑛=1

1

We recognize ζ(𝑠) = ∑∞ 𝑛=1 𝑛𝑠 (when 𝜎 > 1) and we define the 2

−𝜋𝑛 𝑢 function 𝜓(𝑢) = ∑∞ ) and the last equation becomes 𝑛=1(𝑒 𝑠

𝑠

∞

𝑠

𝜋 −2 Γ (2) ζ(𝑠) = ∫0 𝑢 2−1 𝜓(𝑢)𝑑𝑢

(4)

Let us put (4) on hold for a moment. And let us consider the sum −𝜋𝑛2 𝑢 ∑∞ −∞ 𝑒

which can be be divided into 3 terms as follows:

80

Naji Arwashan 𝑛=−1

∞

∑𝑒

−𝜋𝑛2 𝑢

= ∑ 𝑒

−∞

+ [𝑒

−𝜋𝑛2 𝑢

]𝑛=0 + ∑ 𝑒 −𝜋𝑛

−∞

𝑒 −𝜋𝑛

But ∑∞ 𝑛=1 𝑒 we get

∞ −𝜋𝑛2 𝑢

−𝜋𝑛2 𝑢

2𝑢

= ∑−∞ 𝑛=−1 𝑒

2𝑢

𝑛=1

= 𝑒 −𝜋(−𝑛)

−𝜋𝑛2 𝑢

2𝑢

and obviously 𝑒

∞

and

therefore,

−𝜋02 𝑢

= 1, then replacing

∞

∑𝑒

−𝜋𝑛2 𝑢

= 2 ∑ 𝑒 −𝜋𝑛

−∞

2𝑢

+1

𝑛=1 2

−𝜋𝑛 𝑢 If we call 𝜙(𝑢) = ∑∞ ) and remember we called earlier −∞(𝑒 −𝜋𝑛 𝜓(𝑢) = ∑∞ 𝑛=1 𝑒

2𝑢

then we have

𝜙(𝑢) = 2 𝜓(𝑢) + 1

(5)

𝜙(𝑢) has a functional equation relating it to 𝜙(1/𝑢) as follows: 𝜙(𝑢) =

1 √𝑢

𝜙(1/𝑢)

(6)

The above functional equation of 𝜙(𝑢) was known at the time of Riemann, its proof is quite interesting and involving, and it will be shown in Appendix B of this book. But for now, let us just use it and continue with the derivation of the functional equation of zeta. Replacing the value of 𝜙(𝑢) from (5) into (6) gives: 2 𝜓(𝑢) + 1 =

1 √𝑢

[2 𝜓(1/𝑢) + 1]

Rearranging and simplifying gives: 𝜓(𝑢) =

1 √𝑢

1

𝜓 (𝑢) + 2

1 √𝑢

1

−2

(7)

The Functional Equation and the Second Extension of Zeta

81

Let us go back to where we left equation (4) and write it again ∞

𝑠 𝑠 𝜋 Γ ( ) ζ(𝑠) = ∫ 𝑢2−1 𝜓(𝑢)𝑑𝑢 2

−

𝑠 2

0

Let us split the integral from 0 to ∞ into two integrals from 0 to 1 and from 1 to ∞. 1

∞

𝑠 𝑠 𝑠 𝜋 Γ ( ) ζ(𝑠) = ∫(𝑢)2−1 𝜓(𝑢)𝑑𝑢 + ∫ 𝑢2−1 𝜓(𝑢)𝑑𝑢 2

−

𝑠 2

0

1

Replacing 𝜓(𝑢) in the first integral by its value from equation (7) gives 𝑠 𝑠 𝜋 −2 Γ ( ) ζ(𝑠) = 2

1

∫(𝑢)

𝑠 −1 2

0

∞

1

𝑠 1 1 1 − ] 𝑑𝑢 + ∫ 𝑢 2−1 𝜓(𝑢)𝑑𝑢 [ 𝜓( )+ 𝑢 2√𝑢 2 √𝑢

1

Expanding the first integral into 3 integrals gives 𝑠 𝑠 𝜋 −2 Γ ( ) ζ(𝑠) = 2

1

∫𝑢 0

𝑠 −1 2

1

1

1

𝑠 1 1 1 𝑠 𝜓 ( ) 𝑑𝑢 + ∫ 𝑢2−1 𝑑𝑢 − ∫ 𝑢2−1 𝑑𝑢 𝑢 2 2√𝑢 √𝑢

0

∞ 𝑠

+ ∫ 𝑢 2−1 𝜓(𝑢)𝑑𝑢 1

0

82

Naji Arwashan Replacing √𝑢 by u1/2 and simplifying 𝑠 𝑠 𝜋 −2 Γ ( ) ζ(𝑠) = 2

1

1

∫𝑢

𝑠 3 − 2 2

0

1

∞

𝑠 1 1 𝑠 3 1 𝑠 𝜓 ( ) 𝑑𝑢 + ∫ 𝑢2−2 𝑑𝑢 − ∫ 𝑢2−1 𝑑𝑢 + ∫ 𝑢2−1 𝜓(𝑢)𝑑𝑢 𝑢 2 2

0

0

1

The second and third integrals can be easily calculated 𝑠 𝑠 𝜋 −2 Γ ( ) ζ(𝑠) = 2

1

1

∫𝑢

𝑠 3 − 2 2

0

1 1 1 𝜓 ( ) 𝑑𝑢 + [ 𝑠 1 𝑢 𝑢 (2 − 2) 2

𝑠 1 − 2 2

1

∞

𝑠 11 𝑠 ] − [ 𝑠 𝑢 2 ] + ∫ 𝑢 2−1 𝜓(𝑢)𝑑𝑢 2

0

2

0

1

Simplifying gives 𝑠 𝑠 𝜋 −2 Γ ( ) ζ(𝑠) = 2

1

∞

∫𝑢 0

𝑠 3 − 2 2

𝑠 1 1 1 𝜓 ( ) 𝑑𝑢 + − + ∫ 𝑢 2−1 𝜓(𝑢)𝑑𝑢 𝑢 𝑠−1 𝑠

1

Adding the second and third terms on the right-hand side together gives 𝑠 𝑠 𝜋 −2 Γ ( ) ζ(𝑠) = 2

1

1

𝑠 3

1

∞

𝑠

+ ∫0 𝑢2−2 𝜓 ( ) 𝑑𝑢 + ∫1 𝑢 2−1 𝜓(𝑢)𝑑𝑢 𝑠(𝑠−1) 𝑢

(8)

Let us hold on equation (8) and examine the integral 1

𝑠 3 − 2

∫0 (𝑢)2

1

1

1

𝜓 (𝑢) 𝑑𝑢. Let us call 𝑣 = 𝑢 then 𝑢 = 𝑣 and 𝑑𝑢 = −𝑣 −2 𝑑𝑣.

The Functional Equation and the Second Extension of Zeta

83

And when 𝑢 goes to 0, 𝑣 goes to ∞ and when u goes to 1, 𝑣 goes to 1 as well, therefore the integral becomes equal to 1

∫𝑢

1

𝑠 3 − 2 2

0

𝑠 3 1 𝜓 ( ) 𝑑𝑢 = ∫(𝑣 −1 )2−2 𝜓(𝑣)(−𝑣 −2 𝑑𝑣) 𝑢

∞

Simplifying the right-hand side gives 1

∞

∫𝑢

𝑠 3 − 2 2

0

𝑠 1 1 𝜓 ( ) 𝑑𝑢 = ∫ 𝑣 −2−2 𝜓(𝑣)𝑑𝑣 𝑢

1

In calculus the value of a definite integral is independent of the 𝑏

𝑏

variable used, for instance we can say that ∫𝑎 𝑥 2 𝑑𝑥 = ∫𝑎 𝑦 2 𝑑𝑦 This is kind of obvious if 𝑥 and 𝑦 are unrelated variables, but actually the 1

equality still hold even if 𝑥 and 𝑦 are related like 𝑦 =𝑥 2 or 𝑦 = 𝑥 or any other relation. Applying that idea to the right-hand side of the last equation allows us to replace 𝑣 by 𝑢, and get 1

∫𝑢 0

∞

𝑠 3 − 2 2

𝑠 1 1 𝜓 ( ) 𝑑𝑢 = ∫ 𝑢−2−2 𝜓(𝑢)𝑑𝑢 𝑢

1

1

𝑠 3

1

Finally replacing the new value for∫0 𝑢2−2 𝜓 ( ) 𝑑𝑢 in equation (8) 𝑢 gives:

∞

𝑠 𝑠 𝜋 −2 Γ ( ) ζ(𝑠) = 2

∞

𝑠 1 𝑠 1 + ∫ 𝑢 −2−2 𝜓(𝑢)𝑑𝑢 + ∫ 𝑢 2−1 𝜓(𝑢)𝑑𝑢 𝑠(𝑠 − 1)

1

1

84

Naji Arwashan

And combining the second and the third terms on the right-hand side gives: ∞

𝑠 1 𝑠 𝑠 1 𝜋 Γ ( ) ζ(𝑠) = + ∫ (𝑢−2−2 + 𝑢2−1 ) 𝜓(𝑢)𝑑𝑢 2 𝑠(𝑠 − 1)

−

𝑠 2

1

One interesting thing about the right-hand side is that it stays the same if we replace 𝑠 by (1 − 𝑠) so the left-hand side shouldn’t change either and consequently we have 𝑠

𝑠

𝜋 −2 Γ (2) ζ(𝑠) = 𝜋 −

(1−𝑠) 2

Γ(

1−𝑠 2

) ζ(1 − 𝑠)

(9)

And the last equation is the functional equation of zeta as derived by Riemann. The important thing about the functional equation of zeta is that it extends the definition of zeta to the remaining region of the complex plane ( 𝜎 ≤ 0 ). To explain that, let’s consider a point 𝑠0 = 𝜎0 + 𝑖𝑡0 , where 𝜎0 ≤ 0, 𝑠0 is on the negative side of 𝜎, and (1 − 𝑠0 ) is simply equal to [(1 − 𝜎0 ) − 𝑖𝑡0 ] and since Re(1 − 𝑠0 ) is positive then (1 − 𝑠0 ) will be on the positive side of the complex plane and hence a value for 𝜁(1 − 𝑠0 ) can be calculated. Plugging this value in the functional equation allows us to calculate a value for 𝜁(𝑠0 ). As an example, let us calculate the value of zeta when 𝑠 = −1 −1

𝜋− 2 Γ (

(1−(−1)) −1 1 − (−1) ) ζ(−1) = 𝜋 − 2 Γ ( ) ζ(1 − (−1)) 2 2 𝜋 −1 Γ(1)𝜁(2) 𝜁(−1) = 1 1 𝜋 2 Γ (− ) 2

Γ(1) = 0! = 1, from the functional equation of gamma (1) we can 1

1

see that Γ (− ) is equal to -2 Γ ( ) and from the Euler reflection 2 2

The Functional Equation and the Second Extension of Zeta

85

formula (to be presented later in Section 4.8 and proved in Appendix C) 1

1

we can easily find that Γ (2) = √𝜋 and therefore Γ (− 2) is equal to −2√𝜋 ; and 𝜁(2) as stated in Chapter 1 (Section1.7) is equal to 𝜋 2 /6 and now replacing every term by its numerical value we get

𝜁(−1) =

𝜋 −1 (1)𝜋 2 /6 1 2

𝜋 (−2√𝜋)

=−

1 12

4.5. A MISLEADING STRETCH OF IMAGINATION The original definition of the zeta function is ∞

𝜁(𝑠) = ∑ 𝑛=1

1 1 1 1 1 = 𝑠+ 𝑠+ 𝑠+ 𝑠… 𝑠 𝑛 1 2 3 4

And we know that the zeta series in the original definition converges only when 𝑅𝑒(𝑠) > 1. And if we attempt to calculate zeta from the original definition when 𝑠 is equal to -1 we obtain: ∞

𝜁(−1) = ∑ 𝑛=1

1 1 1 1 1 = −1 + −1 + −1 + −1 … −1 𝑛 1 2 3 4

𝜁(−1) = 1 + 2 + 3 + 4 + ⋯ And if someone equates the value of zeta at -1 from the original definition and from the functional equation, he ends up having: −

1 = 1+2+3+4+⋯ 12

86

Naji Arwashan

This amusing and misleading result can be found quite often on the internet and can be the title or subject of many youtube videos. It was found also in the notes of the Indian mathematician Ramanujan even though at the time of his writing he didn’t know about zeta function or its continuation!

4.6. THE CALCULATION OF 𝜻(𝟎) What about the value of zeta when 𝑠 is equal to zero? To calculate 𝜁(0) from the functional equation we need to know 𝜁(1 − 0), but 1 is a pole and zeta is not defined at that point! To resolve this problem, we will use the result we found in Chapter 3 (Section 3.7) that zeta has a residue at 1 equal to lim(𝑠 − 1)𝜁(𝑠) = 1 𝑠→1

Let us multiply both sides of the functional equation by (1 − 𝑠) 𝜋𝑠 (1 − 𝑠)𝜁(𝑠) = 2𝑠 𝜋 𝑠−1 sin ( ) [(1 − 𝑠)Γ(1 − 𝑠)]𝜁(1 − 𝑠) 2 And from the functional equation of the gamma function (1) we know that Γ(2 − 𝑠) = [(1 − 𝑠)Γ(1 − 𝑠)] Replacing that in the last equation and taking the limit of both sides when 𝑠 goes to 1 𝜋𝑠 lim(1 − 𝑠)𝜁(𝑠) = lim [2𝑠 𝜋 𝑠−1 sin ( ) Γ(2 − 𝑠)𝜁(1 − 𝑠)] 𝑠→1 𝑠→1 2 From the residue of zeta at 1 we know that lim(1 − 𝑠)𝜁(𝑠) = − [lim(𝑠 − 1)𝜁(𝑠)] = −1 , therefore 𝑠→1 𝑠→1 𝜋 −1 = 21 𝜋 0 sin ( ) Γ(1)𝜁(0) 2

The Functional Equation and the Second Extension of Zeta

87

And we know that Γ(1) = 0! = 1 , and now replacing every term by its numerical value gives us 𝜁(0) = −

1 2

And with that final touch the zeta function is defined everywhere in the plane of complex numbers C except for one pole at 1.

4.7. EQUIVALENT FORM OF THE FUNCTIONAL EQUATION As mentioned earlier, equation (9) is the functional equation of zeta as derived by Riemann himself. However, a more popular form of the equation is 𝜋𝑠

𝜁(𝑠) = 2𝑠 𝜋 𝑠−1 sin ( 2 ) Γ(1 − 𝑠)𝜁(1 − 𝑠) )

(10)

The two equations are equivalent, but the equivalence is not obvious, and to show it we need to count on two additional functional formulae of the gamma function. The first is Euler Reflection formula: 𝜋

Γ(𝑠)Γ(1 − 𝑠) = sin (𝜋𝑠)

(11)

And the second one is Legendre Duplication formula: 1

Γ (𝑠 + ) = 21−2𝑠 √𝜋 2

Γ(2𝑠) Γ(𝑠)

(12)

The proof of the above two formulae will be presented in Appendix C, and here will simply make use of them. Start by the equation (9)

88

Naji Arwashan 𝑠 (1−𝑠) 𝑠 1−𝑠 𝜋 −2 Γ ( ) ζ(𝑠) = 𝜋 − 2 Γ ( ) ζ(1 − 𝑠) 2 2

Isolating ζ(𝑠) on the left hand-side gives

ζ(𝑠) =

1 𝜋− 𝑠

(1−𝑠) 2

Γ( ) 𝜋 2

−

𝑠 2

1−𝑠 Γ( ) ζ(1 − 𝑠) 2 𝑠

Evaluate Euler reflection formula (11) with 2 gives 𝑠 𝑠 𝜋 Γ ( ) Γ (1 − ) = 𝑠 2 2 sin (𝜋 ) 2

𝑠

Isolating Γ (2) gives 𝑠 1 𝜋 Γ( ) = 𝑠 𝑠 2 Γ (1 − ) sin (𝜋 ) 2

2

𝑠

Replacing Γ (2) in the last expression for ζ(𝑠) gives 𝜋−

1

ζ(𝑠) =

1

𝜋 𝑠 2

𝑠

Γ(1− ) sin (𝜋2)

(1−𝑠) 2

𝜋

−

𝑠 2

1−𝑠 Γ( ) ζ(1 − 𝑠) 2

Simplifying yields 𝑠 1 − 𝑠 1 𝑠−1 𝑠 ζ(𝑠) = Γ (1 − ) Γ ( ) 𝜋 sin (𝜋 ) ζ(1 − 𝑠) 2 2 √𝜋 2

The Functional Equation and the Second Extension of Zeta

89

Now let us call the duplication formula (12), and evaluate it with 𝑠

1

variable equal to (− 2 + 2) 𝑠

1

Γ (2 (− + )) 𝑠 1 𝑠 1 1 2 2 Γ (− + + ) = 21−2(−2+2) √𝜋 𝑠 1 2 2 2 Γ (− + ) 2

2

And simplifying we obtain 𝑠 1 𝑠 1 Γ (1 − ) Γ ( − ) = 2𝑠 Γ(1 − 𝑠) 2 2 2 √𝜋 And finally replacing in the last expression of ζ(𝑠) gives 𝜋𝑠 𝜁(𝑠) = 2𝑠 𝜋 𝑠−1 sin ( ) Γ(1 − 𝑠)𝜁(1 − 𝑠) 2

4.8. THE DEFINITION OF THE FUNCTION 𝝃(𝒔) After Riemann derived the functional equation 𝑠 (1−𝑠) 𝑠 1−𝑠 𝜋 −2 Γ ( ) ζ(𝑠) = 𝜋 − 2 Γ ( ) ζ(1 − 𝑠) 2 2

He took the left-hand side and multiplied it by

𝑠(𝑠−1) 2

and defined a

new function 𝜉(𝑠) as: 𝜉(𝑠) =

𝑠(𝑠−1) 2

𝑠

𝑠

𝜋 −2 Γ (2) ζ(𝑠)

(13)

90

Naji Arwashan Note that

𝑠(𝑠−1) 2

doesn’t change its value if we replace 𝑠 by (1 − 𝑠), 𝑠

𝑠

and we know from (9) that 𝜋 −2 Γ ( ) ζ(𝑠) doesn’t change when we 2 replace 𝑠 by (1 − 𝑠), therefore we can conclude that 𝜉(𝑠) = 𝜉(1 − 𝑠) The last equation is a functional equation of 𝜉(𝑠) and sometimes is also considered as the functional equation of zeta. This new functional equation means that the new function 𝜉(𝑠) is symmetrical around the axis 𝜎 = 1⁄2. This can be easily seen if we consider two points 𝑠1 and 𝑠2 symmetrical around 𝜎 = 1⁄2 so if we write 𝑠1 as 𝑠1 = (1⁄2 + 𝑎) + 𝑖𝑡 then 𝑠2 is equal to 𝑠2 = (1⁄2 − 𝑎) + 𝑖𝑡 and if we calculate 1-𝑠1 we find it equal to 𝑠2 as shown in the line below 1-𝑠1 = 1 − (1⁄2 + 𝑎) + 𝑖𝑡 = (1⁄2 − 𝑎) + 𝑖𝑡 = 𝑠2 With this symmetry in mind it is enough to find the values of 𝜉(𝑠) when ≥ 1/2. For that domain the value of 𝜉(𝑠) can be found from the definition of 𝜉(𝑠) in equation (13). All the terms of (13) are defined when 𝜎 ≥ 1/2 except for zeta when 𝜎 = 1. But fortunately, the factor (𝑠 − 1) eliminate that pole since lim(𝑠 − 1)𝜁(𝑠) = 1. So when 𝑠 = 1, 𝑠→1

1

−

1 2

1

1

𝜉(1) = 2 𝜋 Γ (2), Γ (2) can be found from the Euler reflection formula (used in Section 4.8 and is proved in Appendix C) to be equal to √𝜋, 1

and therefore 𝜉(1) = 2. And with that now 𝜉(𝑠) becomes defined everywhere in the C-plane. The reason for introducing the new function 𝜉(𝑠) will become evident in Chapter 6 (Section 6.4).

Chapter 5

THE ROOTS OF THE ZETA FUNCTION AND THE RIEMANN HYPOTHESIS The zeta function is now defined everywhere on the complex plane except at 1. We are ready to start searching for its zeros. In this search, it will be convenient to divide the complex plane into three regions. 1. 𝜎 > 1 2. 𝜎 < 0 3. 0 ≤ 𝜎 ≤ 1

5.1. THE FIRST REGION (𝝈 > 𝟏) Remember Euler product formula from Chapter 1: ∞

1 1 =∏ ( 1) 𝑠 𝑛 1 − 𝑝𝑠 𝑛=1 𝑝

𝜁(𝑠) = ∑

𝑖

92

Naji Arwashan

This formula was derived in Chapter 1 when 𝑠 was still a real number with a value greater than 1, but actually the same derivation is valid for complex 𝑠 as long as the zeta series converges (𝜎 > 1). From that formula we can see that when 𝜎 > 1, zeta is an infinite product in which no factor equal to 0. Furthermore, as we will show below, for large 𝑝𝑖 the factors approach 1. Let us examine

1 𝑝𝑖𝑠

where s= 𝜎 + 𝑖𝑡 −𝑖𝑡

1 1 𝑝𝑖−𝑖𝑡 (𝑒 log 𝑝𝑖 ) 𝑒 −𝑖𝑡 log 𝑝𝑖 = = = = 𝑝𝑖𝑠 𝑝𝑖𝜎+𝑖𝑡 𝑝𝑖𝜎 𝑝𝑖𝜎 𝑝𝑖𝜎 1 cos(𝑡 log 𝑝𝑖 ) sin(𝑡 log 𝑝𝑖 ) = −𝑖 𝑝𝑖𝑠 𝑝𝑖𝜎 𝑝𝑖𝜎

The sine and cosine are bounded between -1 and 1, and therefore when 𝑝𝑖 goes to infinity

1 𝑝𝑖𝑠

goes to zero and consequently (

to 1. Therefore, the product ∏𝑝 (

1 1

1− 𝑠 𝑝

) goes

𝑖

1 1

1− 𝑠 𝑝

) will not get small or close to

𝑖

zero, and consequently no zeros for zeta in that first region when 𝜎 > 1

5.2. THE SECOND REGION (𝝈 < 𝟎) AND THE TRIVIAL ZEROS OF ZETA The value of zeta in this region is determined by the functional equation of zeta, equation (10) in Chapter 4. 𝜋𝑠 𝜁(𝑠) = 2𝑠 𝜋 𝑠−1 sin ( ) Γ (1 − 𝑠)𝜁(1 − 𝑠) 2

The Roots of the Zeta Function …

93

From the above equation zeta can be zero only if one of the factors on the right-hand side is zero. The factor 𝜁(1 − 𝑠) cannot be zero because when 𝜎 < 0, the argument (1 − 𝑠) has the real value of (1 − 𝜎), and therefore (1 − 𝑠) is to the right of 1, and we just found in the previous section that zeta cannot be zero in that region. We know also that the factors Γ (1 − 𝑠), 2𝑠 and 𝜋 𝑠−1 cannot be zero. Then the only 𝜋𝑠

factor left is sin ( ) which is equal to zero when 𝑠 is a negative even 2 integers. Therefore 𝜁(2𝑘) = 0 when 𝑘 = −1, −2, −3, … These are called the trivial zeros of zeta (trivial because they are easy to find). 𝜋𝑠

One might think that sin ( 2 ) is also equal to zero when 𝑠 is positive even integer, couldn’t that lead to other zeros? But remember the functional equation, because of the gamma function, doesn’t hold for 𝑠 = 1, 2, 3, 4,…

5.3. THE THIRD REGION (𝟎 ≤ 𝝈 ≤ 𝟏) AND THE NON-TRIVIAL ZEROS OF ZETA The third and last region to examine is when 0 ≤ 𝜎 ≤ 1 and is called the critical strip and is shown in Figure 5-1. Riemann himself calculated the first four zeros (with no computer and no calculator!) in that region and found them all with real part 𝜎 equal to 1⁄2. In modern times, billions of other zeros (first ten trillions according to the website of Clay institute) have been calculated and they also all have real part 1⁄2 and hence Riemann Hypothesis predicts that all non-trivial zeros of zeta are on the critical line of 𝝈 = 1⁄2. In Chapter 3 (Section 3.3) we found that the value of zeta when 𝜎 > 1 is equal to 1 𝜁(𝑠) = ( ) 𝜂(𝑠) 1 − 21−𝑠

94

Naji Arwashan

Figure 5-1. The critical strip and critical line.

Where eta is defined in Chapter 3 (Section 3.1) as ∞

𝜂(𝑠) = ∑(−1)

𝑛−1

𝑛=1

1 𝑠 1 1 1 1 1 ( ) =1− 𝑠+ 𝑠− 𝑠+ 𝑠− 𝑠… 𝑛 2 3 4 5 6

And we found ∞

∞

𝜂(𝑠) = ∑(−1) 𝑛=1

𝑛−1

cos(𝑡 log 𝑛) sin(𝑡 log 𝑛) − 𝑖 ∑(−1)𝑛−1 𝜎 𝑛 𝑛𝜎 𝑛=1

1

Let us rewrite the term (1−21−𝑠 ) 1 2𝑠 2𝜎+𝑖𝑡 = = 1 − 21−𝑠 2𝑠 − 2 2𝜎+𝑖𝑡 − 2

The Roots of the Zeta Function …

95

Separating the real and imaginary parts in the numerator and denominator we get 2𝜎 cos(𝑡 log 2) + 𝑖 2𝜎 sin(𝑡 log 2) (2𝜎 cos(𝑡 log 2) − 2) +𝑖 2𝜎 sin(𝑡 log 2) Multiplying the numerator and denominator by the conjugate of the denominator gives [2𝜎 cos(𝑡 log 2) + 𝑖 2𝜎 sin(𝑡 log 2)][(2𝜎 cos(𝑡 log 2) − 2) −𝑖 2𝜎 sin(𝑡 log 2)] [(2𝜎 cos(𝑡 log 2) − 2) +𝑖 2𝜎 sin(𝑡 log 2)][(2𝜎 cos(𝑡 log 2) − 2) −𝑖 2𝜎 sin(𝑡 log 2)]

Expanding and simplifying the denominator, we get [2𝜎 cos(𝑡 log 2) + 𝑖 2𝜎 sin(𝑡 log 2)][(2𝜎 cos(𝑡 log 2) − 2) −𝑖 2𝜎 sin(𝑡 log 2)] (2𝜎 cos(𝑡 log 2) − 2)2 + (2𝜎 sin(𝑡 log 2))2

Finally separating the real and imaginary parts we get

1 1−21−𝑠

equal

to 22𝜎 − 2𝜎+1 cos(𝑡 log 2) 2𝜎+1 sin(𝑡 log 2) − 𝑖 4 − 4 × 2𝜎 4cos(𝑡 log 2) + 22𝜎 4 − 4 cos(𝑡 log 2) + 22𝜎 1

And now going back to zeta: 𝜁(𝑠) = (1−21−𝑠 ) 𝜂(𝑠) and substituting 1

for (1−21−𝑠 ) and 𝜂(𝑠) we get

𝜁 (𝑠 ) = [

22𝜎 −2𝜎+1 cos(𝑡 log 2) 4−4×2𝜎 cos(𝑡 log 2)+22𝜎

∞

[∑(−1) 𝑛=1

−𝑖

2𝜎+1 sin(𝑡 log 2) 4−4 ×2𝜎 cos(𝑡 log 2)+22𝜎

∞

𝑛−1

cos(𝑡 log 𝑛) sin(𝑡 log 𝑛) − 𝑖 ∑(−1)𝑛−1 ] 𝜎 𝑛 𝑛𝜎 𝑛=1

]

96

Naji Arwashan

Multiplying the two factors together we find the real part of zeta to be equal to 𝑅𝑒 (𝜁) = ∞

22𝜎 − 2𝜎+1 cos(𝑡 log 2) cos(𝑡 log 𝑛) [ ] [∑(−1)𝑛−1 ] 𝜎 2𝜎 4 − 4 × 2 cos(𝑡 log 2) + 2 𝑛𝜎 𝑛=1 ∞

𝜎+1

−[

2 sin(𝑡 log 2) sin(𝑡 log 𝑛) ] [∑(−1)𝑛−1 ] 𝜎 2𝜎 4 − 4 × 2 cos(𝑡 log 2) + 2 𝑛𝜎 𝑛=1

And the imaginary part 𝐼𝑚 (𝜁) = ∞

− 2𝜎+1 sin(𝑡 log 2) cos(𝑡 log 𝑛) [ ] [∑(−1)𝑛−1 ] 𝜎 2𝜎 4 − 4 × 2 cos(𝑡 log 2) + 2 𝑛𝜎 2𝜎

−[

𝑛=1 ∞

𝜎+1

2 −2 cos(𝑡 log 2) sin(𝑡 log 𝑛) ] [∑(−1)𝑛−1 ] 𝜎 2𝜎 4 − 4 × 2 cos(𝑡 log 2) + 2 𝑛𝜎 𝑛=1

And for zeta to be zero both real and imaginary parts need to be null 𝑅𝑒 (𝜁) = 0

and

𝐼𝑚 (𝜁) = 0

hence Riemann Hypothesis boils down to the solution of the above two equations with two unknowns: 𝜎 and 𝑡. Obviously the two equations are not simple ones, but still the concept is simple, and the hypothesis is now within the grasp of all; it is two equations with two unknowns and the solutions are always with one of the unknowns 𝜎 equal to 1⁄2. Figure 5-2 shows the first five intersections (zeros) of the above two equations.

The Roots of the Zeta Function …

97

Figure 5-2. The first five non-trivial zeros of 𝜁(𝑠) [first five intersections of 𝑅𝑒 (𝜁) = 0 & 𝐼𝑚 (𝜁) = 0].

However, instead of following the previous route we can see from the relation between zeta and eta

𝜁(𝑠) =

𝜂(𝑠)

(1)

1−21−𝑠

that the zeros of eta in the region 𝜎 > 0 include all the zeros of zeta in that same reion, and therefore we can replace the search for zeta’s zeros in the critical strip by searching for eta’s zeros. We know eta is equal to ∞

𝜂(𝑠) = ∑(−1) 𝑛=1

∞

𝑛−1

cos(𝑡 log 𝑛) sin(𝑡 log 𝑛) − 𝑖 ∑(−1)𝑛−1 𝜎 𝑛 𝑛𝜎 𝑛=1

98

Naji Arwashan Eta is zero when both real and imaginative parts are zeros: ∞

𝑅𝑒(𝜂) = ∑(−1)𝑛−1 𝑛=1

cos(𝑡 log 𝑛) =0 𝑛𝜎

∞

& 𝐼𝑚(𝜂) = − ∑(−1)𝑛−1 𝑛=1

sin(𝑡 log 𝑛) =0 𝑛𝜎

Figure 5-3. The coincidence of the first zero of 𝜻(𝒔) and 𝜼(𝒔) on the line 𝜎 =.

We found in Chapter 3 (Section 3.4) that eta has a set of zeros when 2𝜋

1 − 21−𝑠 = 0 and are equal to 𝑠𝑛 = 1 + 𝑖 𝑛 log 2 (all on the line of 𝜎 = 1). But these zeros are not of interest to us because the denominator of zeta (equation 1) at these locations is equal to zero and the limit of zeta at these locations as we found in Chapter3, Section 3.5 is equal to 𝜁(𝑠𝑛 ) =

𝜂′(𝑠𝑛 ) log 2

(except when n=0). However, eta has another set of zeros

The Roots of the Zeta Function …

99

that are common with zeta. In Figure 5-3 we show how the first zero of eta and zeta coincide at the point (0.5 + 𝑖14.135). And now Riemann Hypothesis, the most important unsolved problem in math, can be formulated as ∞

∞

cos(𝑡 log 𝑛) sin(𝑡 log 𝑛) = 0 & ∑(−1)𝑛 = 0 (0 < 𝜎 < 1) ∑(−1) 𝜎 𝑛 𝑛𝜎 𝑛

𝑛=1

𝑛=1

1

is possible only when 𝜎 = 2 Note that the power of (-1) became 𝑛 instead of (𝑛 − 1) because the real part is equal to zero and therefore we can multiply both sides by (-1); and for the imaginary part equation, a negative sign already existed before the sum. Also note that we justify the elimination of 𝜎 = 1 from the domain of 𝜎 because as we will see later in the book in Chapter 7 (Section 7-5) it was proved that zeta function has no zero on the line 𝜎 = 1. With regard to the line of 𝜎 = 0, we are certain of the absence of any zeros there because as we will see later in this chapter (Section 5.6) from the symmetry of the zeros if there was a zero on 𝜎 = 0 it will necessitate a zero on 𝜎 = 1 which we know it doesn’t exist. We should also mention that the above framed version of the Riemann Hypothesis appeared first in Havil’s book [7] but unfortunately, he omitted to specify the domain of applicability of 𝜎 [11]. Finally, one can certainly wish if all important math or physics problems could be formulated in a such simple and elegant format!

5.4. A CLOSER LOOK AT THE COMMON ZEROS OF THE ETA AND THE ZETA FUNCTIONS Let us focus on the real part of eta (the observation in this section will also apply to the imaginary part). Examining the sum

100 𝑛−1 ∑∞ 𝑛=1(−1)

Naji Arwashan cos(𝑡 log 𝑛) 𝑛𝜎

, we choose to define a related function 𝐴(𝑢) as

follows: 𝐴(𝑢) =

1 cos (𝑡 log 𝑢) 𝑢𝜎

The function 𝐴(𝑢) is a cosine function and therefore is a wave function, and the period of that function 𝑇 can be found as follows cos(𝑡 log (𝑢 + 𝑇)) = cos(𝑡 log u) 𝑡 log(𝑢 + 𝑇) = 𝑡 log u + 2𝜋 𝑢+𝑇 2𝜋 log ( )= 𝑢 𝑡 2𝜋

𝑇 = (𝑒 𝑡 − 1) 𝑢

Figure 5-4. The graph of 𝐴(𝑢) = 𝑛−1 cos(14.1347 log 𝑛) ∑∞ . 𝑛=1(−1) 𝑛0.5

1 𝑢0.5

cos(14.1347 log 𝑢) and the sum

The period is not constant but instead increases with 𝑢. We know that the first zero of zeta occurs at 0.5 + 𝑖14.1347, so let us plot 𝐴(𝑢)

The Roots of the Zeta Function …

101

when 𝜎 = 0.5 and 𝑡 = 14.1347 and on the same graph will plot the 𝑛−1 sum ∑∞ 𝑛=1(−1)

cos(14.1347 log 𝑛) 𝑛0.5

We see that the sum follows the same wave as the function 𝐴(𝑢), and to see better we zoom in the segment between 100 and 300 as shown in Figure 5-5.

Figure 5-5. A closer look at 𝐴(𝑢) = 𝑛−1 cos(14.1347 log 𝑛) ∑∞ 𝑛=1(−1) 𝑛0.5

1 𝑢0.5

cos(14.1347 log 𝑢) and the sum

.

Figure 5-6. A closer look at the start of 𝐴(𝑢) = 𝒏−𝟏 𝐜𝐨𝐬(𝟏𝟒.𝟏𝟑𝟒𝟕 𝐥𝐨𝐠 𝒏) ∑∞ . 𝒏=𝟏(−𝟏) 𝒏𝟎.𝟓

1 𝑢0.5

cos(14.1347 log 𝑢) and the sum

102

Naji Arwashan

Let us examine what is happening when 𝑢 is small and close to 1. In that region the period 𝑇 starts small (less than 1) and grows gradually. But the step of the sum is equal to 1 (𝑛 increases by 1), therefore, and as we can see in figure 5-6, it takes few terms for the sum before it starts following the waves of 𝐴(𝑢). The last point becomes even more obvious when we consider different roots of eta (or zeta) for which t is larger. For these roots, 𝐴(𝑢) starts with very small period and it needs larger 𝑢 for its period to grow and to become large enough for the sum of the real part of eta to follow. To demonstrate this, we repeat the same plots for the second and third zero of eta, and we can see in Figures 5-7 and 5-8 how it takes more terms for the sum before it starts following the waves of 𝐴(𝑢).

Figure 5-7. A closer look at the start of 𝐴(𝑢) = 𝑛−1 cos(21.022 log 𝑛) ∑∞ . 𝑛=1(−1) 𝑛0.5

1 𝑢0.5

cos(21.022 log 𝑢) and the sum

The Roots of the Zeta Function …

Figure 5-8. A closer look at the start of 𝐴(𝑢) = 𝑛−1 cos(25.0109 log 𝑛) ∑∞ . 𝑛=1(−1) 𝑛0.5

Figure 5-9. The graph of 𝐴(𝑢) = 𝑛−1 cos(14.1347 log 𝑛) ∑∞ . 𝑛=1(−1) 𝑛0.6

1 𝑢0.6

1 𝑢0.5

103

cos(25.0109 log𝑢) and the sum

cos(14.1347 log 𝑢) and the sum

104

Naji Arwashan

Finally, it is worth mentioning that the previous observations are not limited to the roots of eta but are true to any value of eta. For instance, if we calculate eta at the point (0.6 + 𝑖14.1347) [note that the imaginary part is equal to the first zero, but the real part is equal to 0.6 instead of 0.5] we find it to be equal to 0.1734-i0.01088. 1

And if we plot 𝐴(𝑢) = 𝑢0.6 cos (14.1347 log𝑢) and the sum 𝑛−1 ∑∞ 𝑛=1(−1)

cos(14.1347 log 𝑛) 𝑛0.6

(Figure 5-9) we find that the sum follows

the waves of 𝐴(𝑢), in similar way to what we saw in the graph for the first zero, but it is shifted now vertically by 0.1734.

5.5. GRAPHS OF THE CRITICAL LINE

Figure 5-10. The zeta mapping of the critical line.

Recall our discussion in Chapter 2 (Section 2.4), we treated complex functions as mappings from one plane to another, and to visualize the mappings we pick certain lines in the domain and see their image in the co-domain. Here we are going to numerically evaluate the map of the critical line for both the zeta and eta functions and show it in Figures 5-10 and 5-11. Note that we are mapping only the segment (0,

The Roots of the Zeta Function …

105

35) from the critical line, in that segment there are 5 zeros, the map of each of them will be a passing of the mapped line through the zero. It is worthwhile mentioning that the picture of the zeta mapping of the critical line is frequently seen in books and papers dealing with the Riemann Hypothesis.

Figure 5-11. The eta mapping of the critical line.

Figure 5-12. The real and imaginary parts of the points of the critical line.

106

Naji Arwashan

Figure 5-13. The magnitude of the points of the critical line.

Another popular graph related to the critical line is the plot of the 1

real part, imaginary part and the magnitude of 𝜁 (2 + 𝑖𝑡) versus 𝑡. In other words, zeta value is calculated for the points of the critical line; and their real part, imaginary part and magnitude are plotted against the critical line as shown in Figures (5-12) and (5-13).

5.6. SYMMETRY OF THE ZEROS If (𝜎 ∗ + 𝑖𝑡 ∗ ) is a non-trivial zero of zeta then its conjugate (𝜎 ∗ − 𝑖𝑡 ∗ ) is also a zero because ∞

∞

𝑛=1

𝑛=1

cos(−𝑡 ∗ log 𝑛) cos(𝑡 ∗ log 𝑛) 𝑛−1 𝑛−1 = ∑(−1) =0 ∑(−1) ∗ ∗ 𝑛𝜎 𝑛𝜎

and ∞

∞

𝑛=1

𝑛=1

sin(−𝑡 ∗ log 𝑛) sin(𝑡 ∗ log 𝑛) 𝑛−1 𝑛−1 = − [∑(−1) ∑(−1) ]=0 ∗ ∗ 𝑛𝜎 𝑛𝜎

The Roots of the Zeta Function …

107

and from the functional equation 𝜋𝑠 𝜁(𝑠) = 2𝑠 𝜋 𝑠−1 sin ( ) Γ (1 − 𝑠)𝜁(1 − 𝑠) 2 we know that if (𝜎 ∗ + 𝑖𝑡 ∗ ) is a zero for zeta then [1 − (𝜎 ∗ + 𝑖𝑡 ∗ )] = [(1 − 𝜎 ∗ ) − 𝑖𝑡 ∗ ] is also a zero. We conclude that if (𝜎 ∗ + 𝑖𝑡 ∗ ) is a zero of zeta then (𝜎 ∗ − 𝑖𝑡 ∗ ), (1 − 𝜎 ∗ + 𝑖𝑡 ∗ ) and (1 − 𝜎 ∗ − 𝑖𝑡 ∗ ) are also all zeros as shown below.

Figure 5-14. Symmetry of the nontrivial zeros of zeta.

5.7. THE ROOTS OF THE FUNCTION 𝝃(𝒔) Recall at the end of Chapter 4 (Section 4.8) we said that Riemann defined the function 𝜉(𝑠) as 𝜉(𝑠) =

𝑠(𝑠−1) 2

𝑠

𝑠

𝜋 −2 Γ ( ) ζ(𝑠) 2

(2)

108

Naji Arwashan And we found its functional equation to be 𝜉(𝑠) = 𝜉(1 − 𝑠)

(3)

And we said that the functional equation means the symmetry of 𝜉 around the critical axis of 𝜎 = 1/2 . And now keeping that symmetry in mind it is enough to search for the zeros of 𝜉 in the region of 𝜎 ≥ 1/2. In that region the value of 𝜉 is determined from its definition in equation (2) which indicates that 𝜉 can be zero only when zeta is zero 1

(remember we found 𝜉(1) = 2 because zeta has a pole at 1) and therefore the non-trivial zeros of zeta (normally designated as 𝜌) are the zeros of 𝜉. The relevance of the zeros of 𝜉 becomes evident in the next chapter.

Chapter 6

THE ZETA FUNCTION AND COUNTING THE PRIME NUMBERS The explanation of the Riemann Hypothesis can very well be stopped at the end of the previous chapter. However, that would rob the reader of the key motivation behind the interest in the zeros of the zeta function and their connection to number theory.

6.1. PRIME FASCINATION AND THE PRIME NUMBER THEOREM (PNT) Since early time mathematicians have been fascinated by prime numbers. These are natural numbers that can be divided only by themselves and by 1 like 2, 3, 5, 7, 11, 13, … Primes are the building blocks of any number since any natural number can be written as a unique product of primes, this is called the fundamental theorem of arithmetic, or the unique prime-factorization theorem. For instance, 36 = 23 × 32 and 50 = 2 × 52 . Note the number 1 is not prime, even though it can be divided only by itself! But including 1 in the primes

110

Naji Arwashan

will destroy the uniqueness characteristic of writing any natural number as a product of primes. For instance, if 1 is included in the primes, then 36 no longer can be written in only one way as 23 × 32 but also as 1 × 23 × 32 . Also there is a practical reason for excluding 1 from the primes; when 1 is included in the primes, many properties are true for the primes except for 1, and instead of always excluding the 1 when talking about these properties it was more convenient to exclude the 1 from the definition of primes. Two questions come to mind: How many primes are there and how are they distributed among natural numbers? The first question was answered by the ancient Greek mathematician Euclid (300 BC) who proved that the number of primes is infinite. A slightly modified version of his proof goes as follows: Suppose that the number of primes is finite and all the primes are: 𝑝1 , 𝑝2 , … 𝑝𝑖 , … 𝑝𝑛 then any natural number should be divisible by at least one of them. But the number 𝑁 = 𝑝1 𝑝2 … 𝑝𝑖 … 𝑝𝑛 + 1 (called Euclid number) is not, because when it is divided by any of them, say 𝑝𝑖 , it will have the remainder therefore the number of primes is infinite.

Figure 6-1. The staircase shape of the 𝝅(𝒙) function.

1 𝑝𝑖

and

The Zeta Function and Counting the Prime Numbers

111

The second question about the distribution of primes among natural numbers is a great mathematical mystery. In his inaugural lecture at Bonn University in 1975 [15], the mathematician Don Zagier famously said “They [Prime numbers] grow like weeds among the natural numbers, seeming to obey no other law than that of chance, and nobody can predict where the next one will sprout. The second fact is even more astonishing, for it states just the opposite: that the prime numbers exhibit stunning regularity, that there are laws governing their behavior, and that they obey these laws with almost military precision.” To understand Zagier’s observation, we should take a look at the prime counting function 𝜋(𝑥) which is defined to output the number of primes less than or equal to 𝑥. For instance, 𝜋(100) = 25 because there are 25 primes less or equal to 100. The prime counting function 𝜋(𝑥) has a staircase-like shape, because when we move along the 𝑥 axis each time we hit a new prime 𝜋(𝑥) jumps by one, as seen in Figure 6-1.

Figure 6-2. The 𝝅(𝒙) function looked at from a distance.

The randomness that Zagier talks about is in the distance between the rises of the staircase or in other words the gap between two primes which resists to follow any identifiable pattern and seems to defy any logic. Sometimes the primes appear close to each other like (11, 13),

112

Naji Arwashan

(41, 43), (101, 103) which are called twin primes, and other times we can go for a while before we hit a new prime. However, to appreciate the second part of Zagier statement concerning the laws governing the primes, we need to zoom out and look at 𝜋(𝑥) from a distance, then we can see the staircase becomes remarkably smooth (as shown in Figure 6-2) and seems to be following a well-defined curve. But what is that curve? That is a question that boggled mathematicians for many years. The first attempt to approximate 𝜋(𝑥) came from Carl Fredric Gauss, often titled the prince of mathematics for his ingenuity and great contribution to mathematics. In late eighteen century, and at the early 𝑥

age of 15 or 16, Gauss conjured that log(𝑥) is a good approximation for 𝜋(𝑥).

Figure 6-3. The

𝒙 𝐥𝐨𝐠(𝒙)

approximation of the 𝝅(𝒙) function.

For us, an approximation will be considered good if the relative error goes to zero when 𝑥 goes to infinity 𝑥

(log 𝑥) − 𝜋(𝑥) lim [ ]=0 𝑥 𝑥→∞ (log 𝑥) This is equivalent to

The Zeta Function and Counting the Prime Numbers

lim [1 −

𝑥→∞

113

𝜋(𝑥)

]=0 𝑥 (log 𝑥)

or 𝜋(𝑥) lim [ 𝑥 ] = 1 (log 𝑥)

𝑥→∞

Figure 6-4. The

𝒙 𝐥𝐨𝐠(𝒙)

and Li(x) approximation of 𝝅(𝒙).

Which means the ratio of the two functions goes to one in infinity. When two functions have this relationship, they are called asymptotically equivalent. Gauss’s conjecture about the asymptotic equivalence between 𝜋(𝑥) and

𝑥 log 𝑥

became known the Prime Number

Theorem (PNT) and was one of the most important and famous math problems in the entire 19th century. In 1838 Peter Gustav Lejeune 𝑥 𝑑𝑡

Dirichlet found that the offset logarithmic integral ∫2

log 𝑡

designated by

𝐿𝑖(𝑥) is a better approximation for 𝜋(𝑥) (see Figure 6-4). There are two justifications for his guess; the first one is if we accepted that the

114

Naji Arwashan 𝑥

number of prime under 𝑥 is log 𝑥 then the density of primes at 𝑥 is 1 log 𝑥

𝑥 log 𝑥

𝑥

=

and therefore to calculate the number of primes we take the integral

of the density. The other explanation is that

𝑥 log 𝑥

seems to be growing

too slow to catch up with 𝜋(𝑥), and it is easy to verify that the 𝑥

1

1

derivative of log 𝑥 is [log 𝑥 − (log 𝑥)2] . So if we enhance the derivative by removing the negative part from it, that would give the new curve a 𝑥 𝑑𝑡

better chance to grow faster and hence we obtain ∫0

log 𝑡

which is called

the logarithmic integral 𝑙𝑖(𝑥) (note the lower case l, to distinguish it 𝑥 𝑑𝑡

from the offset integral [𝐿𝑖(𝑥) = ∫2

log 𝑡

])

Let us examine the relationship between evaluating the limit lim

𝑥 log 𝑥

𝑥→∞ 𝐿𝑖(𝑥)

𝑥 log 𝑥

and 𝐿𝑖(𝑥) by

. For that we will use Hospital rule (from

Chapter3, Section 3.5) 𝑥

lim

𝑥→∞

log 𝑥

𝐿𝑖(𝑥)

[ = lim

′

𝑥

log 𝑥

]

𝑥→∞ [𝐿𝑖(𝑥)]′ 1

𝑥 ′ (1) log 𝑥 − 𝑥 𝑥 1 1 = − [ ] = 2 (log 𝑥) log 𝑥 log 𝑥 (log 𝑥)2 𝑥

′

𝑑𝑡 ] 2 log 𝑡 𝑑𝑡 If we call 𝐹(𝑡) = ∫ then log 𝑡 𝑥 𝑑𝑡 𝐿𝑖(𝑥) = ∫ = 𝐹(𝑥) − 𝐹(2) 2 log 𝑡 1 [𝐿𝑖(𝑥)]′ = [𝐹(𝑥)]′ − 0 = log 𝑥 [𝐿𝑖(𝑥)]′ = [∫

The Zeta Function and Counting the Prime Numbers

115

And therefore: 𝑥

lim

𝑥→∞

log 𝑥

𝐿𝑖(𝑥)

1

= lim

Which means

𝑥→∞

log 𝑥

1

− (log 1 log 𝑥

𝑥 log 𝑥

𝑥)2

= lim [1 − 𝑥→∞

1 ]=1 log 𝑥

and 𝐿𝑖(𝑥) are asymptotically equivalent, and

consequently if the prime number theorem is true then 𝜋(𝑥) and 𝐿𝑖(𝑥) are asymptotically equivalent too. Therefore, a different version of the prime number theorem emerged conjecturing the asymptotic equivalence between 𝐿𝑖(𝑥) and 𝜋(𝑥).

6.2. RIEMANN’S PAPER AND THE INTRODUCTION OF THE 𝑱(𝒙) FUNCTION Riemann presented his seminal paper in 1859, It was his first and only paper in number theory and considered a revolutionary paper because it established the connection between two separate branches in math; number theory and distribution of primes on one side and calculus and complex analysis on the other side. Riemann brilliantly, and intuitively showed in his paper an explicit formula describing the distribution of prime numbers and relating that distribution to the zeros of the zeta function. But first Riemann needed to make an adjustment to the definition of the prime counting function 𝜋(𝑥) in order to make it meet the requirements of the mathematical definition of function. The adjustment consists of making the value of 𝜋(𝑥) jumps by a half when 𝑥 is equal to a prime and then adds another half when 𝑥 passes the prime. So 𝜋(𝑥) is equal to zero until 𝑥 reaches 2 then 𝜋(2) is 1/2, and jumps to 1 after 2

116

Naji Arwashan 1

until it reaches 3 then 𝜋(3) is 1 2 and after 3 becomes 2 until it reaches the next prime, etc.… as shown in Figure 6-5

Figure 6-5. The mathematical definition of the 𝝅(𝒙) function.

Figure 6-6. The mathematical definition of the 𝑱(𝒙) function.

The Zeta Function and Counting the Prime Numbers

117

Establishing a direct relation between 𝜋(𝑥) and zeta was not possible, and Riemann needed to go through an alternative prime counting function, he called it 𝑓, but later that name changed by mathematicians into Π, because 𝑓 is normally reserved to designate a generic function. The mathematician H.M. Edwards proposed in his well-regarded book “Riemann’s Zeta Function” [6] to call the new counting function 𝐽(𝑥), and John Derbyshire followed that notation in his popular book “Prime obsession” [5]; and other authors did as well, and that is why we decided in this book to follow their suit and call it 𝐽(𝑥). The new function, 𝐽(𝑥), counts the primes and their powers in such a way that a prime counts as 1 and its nth power counts as 1⁄𝑛. So the function 𝐽(𝑥)starts by zero until 𝑥 hits the first prime 2 then its value jumps by 1 and stays 1 until 𝑥 his 3 then its values jumps by another one and becomes 2, and when it hits 4 =22 its value jumps by 1

1

½, and stays 2 2 until it hits 5 where it becomes 3 2, and when it hits 7 it 1

jumps to 4 2 and when it hits 8=23 it jumps by 1/3 and when it hits 9=32 it adds a 1/2, etc.… And the same adjustment we used for 𝜋(𝑥) we use also for 𝐽(𝑥), so the value of 𝐽(𝑥) at a prime or a prime power jumps by a half of the value of the step at that point. Therefore, the value of 1

1

𝐽(𝑥)at 2 is ½ and at 3 is 1 2 and at 4 is 2 4 as shown in Figure 6-6. Now we need to find a relationship between 𝜋(𝑥) and 𝐽(𝑥). To do that let us consider the fact that the number of primes under √𝑥 is equal to the number of prime squares under 𝑥, and number of primes under 3

√𝑥 is equal to the number of prime cube under 𝑥, etc... To illustrate this let us consider 𝑥 equal to 100, under √100=10 there are 4 primes (2, 3, 5, 7) and that is equal to the number of square primes under 100 which 3

are (4, 9, 25, 49); and under √100=4.6 there are 2 primes (2 and 3), and that is equal to the number of cube prime under 100 which are (8, 27). With this in mind we can write that:

118

Naji Arwashan 1 1 1 1 1 1 𝐽(𝑥) = 𝜋(𝑥) + 𝜋 (𝑥 2 ) + 𝜋 (𝑥 3 ) + 𝜋 (𝑥 4 ) + ⋯ 2 3 4

1

1

𝐽(𝑥) = ∑∞ 𝑚=1 𝑚 𝜋 (𝑥 𝑚 )

(1)

𝜋(𝑥) can be recovered from the last function by using Mobius Inversion formula which will be the subject of our next section.

6.3. THE MOBIUS FUNCTION AND INVERSION FORMULA The Mobius function 𝜇(𝑛) is a function quite often used in number theory. It was introduced in 1832 by the German mathematician August Ferdinand Mobius and was named after him. It takes only three values {−1,0,1} depending on the factorization of 𝑛 into prime factors as follows • • •

𝜇(𝑛) = 0 if n has a squared prime factor 𝜇(𝑛) = −1 if n has odd number of prime factors with no squared factor 𝜇(𝑛) = 1 if n has even number of prime factors with no squared factor

The above can be summarized as 𝜇(𝑛) is equal to 1 if 𝑛 = 1, to (−1)𝑘 if 𝑛 is the product of k distinct prime, and equal to zero otherwise. For instance when n is equal to 2 then 𝑘 = 1 and 𝜇(2) = (−1)1 = −1, and when 𝑛 = 2 × 2 we have 𝜇(4) =0 (because 4 is 22), and when 𝑛 = 6 = 2 × 3, 𝑘 is 2 and 𝜇(6) = (−1)2 = 1.

The Zeta Function and Counting the Prime Numbers

119

Main Property of the Mobius Function Before we address Mobius transform and inverse transform there are couple of formulae that we need to prove. The first one is a key property of Mobius function and it states that: 1, 𝑖𝑓 𝑢 = 1 ∑𝑑|𝑢 𝜇(𝑑) = { 0, 𝑖𝑓 𝑢 > 1

(2)

Where the summation is over all the divisors of 𝑛 including 1 and 𝑛. To explain this formula let us use an example like 𝑢 = 12, in this case the divisors of 12 are: 1, 2, 3, 4, 6, 12 ∑ 𝜇(𝑑) = 𝜇(1) + 𝜇(2) + 𝜇(3) + 𝜇(4) + 𝜇(6) + 𝜇(12) 𝑑|12

∑ 𝜇(𝑑) = 1 + (−1)1 + (−1)1 + (0) + (−1)2 + (0) = 0 𝑑|12

To prove the formula let us suppose 𝑢 has 𝑠 distinct prime factors and 𝑢 can be written as: 𝑠 𝛼

𝑢 = ∏ 𝑝𝑖 𝑖 𝑖=1

If 𝑑 is a divisor of 𝑢, then 𝜇(𝑑) is equal to •

𝜇(𝑑) = (−1)𝑘 if 𝑑 is the product of 𝑘 different primes out of the 𝑠 possible ones. The number of divisors with this characteristic is (𝑘𝑠 )

•

and for all other divisors 𝜇(𝑑) = 0 except for 𝜇(1) = 1

120

Naji Arwashan Therefore, we can write 𝑠

𝑠 ∑ 𝜇(𝑑) = ∑(−1)𝑘 ( ) 𝑘 𝑑|𝑢

𝑘=0

Note that when 𝑘 = 0 that covers the case of 𝑑 = 1 where 𝜇(1) = 1 The last formula can be expanded as: ∑ 𝜇(𝑑) = (−1)0 𝑑|𝑢

𝑠! 𝑠! 𝑠! + (−1)1 + (−1)2 0! (𝑠 − 0)! 1! (𝑠 − 1)! 2! (𝑠 − 2)!

𝑠! +⋯ 3! (𝑠 − 3)! 𝑠! 𝑠! 𝑠! + (+1) + (−1) ∑ 𝜇(𝑑) = 1 + (−1) 1! (𝑠 − 1)! 2! (𝑠 − 2)! 3! (𝑠 − 3)! + (−1)3

𝑑|𝑢

+⋯ Let us hold on the last equation for a moment and consider the expansion of (𝑥 + 𝑦)𝑠 𝑠! 𝑠! 𝑥 𝑠−1 𝑦 + 𝑥 𝑠−2 𝑦 2 1! (𝑠 − 1)! 2! (𝑠 − 2)! 𝑠! + 𝑥 𝑠−3 𝑦 3 + ⋯ 3! (𝑠 − 3)!

(𝑥 + 𝑦)𝑠 = 𝑥 𝑠 +

And replacing 𝑥 by 1 and 𝑦 by -1 we get: 𝑠! 𝑠! 1𝑠−1 (−1) + 1𝑠−2 (−1)2 1! (𝑠 − 1)! 2! (𝑠 − 2)! 𝑠! + 1𝑠−3 (−1)3 + ⋯ 3! (𝑠 − 3)!

(1 − 1)𝑠 = 1𝑠 +

The Zeta Function and Counting the Prime Numbers 0=1+

121

𝑠! 𝑠! 𝑠! (−1) + (+1) + (−1) + ⋯ 1! (𝑠 − 1)! 2! (𝑠 − 2)! 3! (𝑠 − 3)!

We can see that the right-hand side is equal to the expansion of ∑𝑑|𝑛 𝜇(𝑑) and therefore ∑𝑑|𝑢 𝜇(𝑑) = 0 when

𝑢>1

And when 𝑢 = 1 the only divisor of 𝑢 is 1 and we know by definition 𝜇(1) = 1 and therefore we can write ∑𝑑|𝑢 𝜇(𝑑) = 1 when

𝑢=1

The second formula we need to introduce is more of a point that we need to clarify. First let us consider the double summation ∞

∞

∑ ∑ 𝑛𝑚 𝑛=1 𝑚=1

Every natural number is going to appear in the above sum, but the question is how many times? Let us check for instance number 6 it is going to appear as (1 × 6), (2 × 3), (3 × 2), (6 × 1) so 6 will appear 4 times and that is equal to the number of all possible divisors of 6. In general, any number will appear number of times equal to the number of its divisors including 1 and the number itself. And therefore, the sum above can be written as ∞

∞

∞

∑ ∑ 𝑛𝑚 = ∑ ∑ 𝑢 𝑛=1 𝑚=1

𝑢=1 𝑑|𝑢

122

Naji Arwashan And in general, we can write: ∞ ∞ ∑∞ 𝑛=1 ∑𝑚=1 𝑓(𝑛𝑚) = ∑𝑢=1 ∑𝑑|𝑢 𝑓(𝑢)

(3)

The Mobius Inversion Formula Now with all this preparation, we can go back to the 𝐽(𝑥) function where we found it in the previous section (equation 1) equal to ∞

𝐽(𝑥) = ∑ 𝑚=1

1 1 𝜋 (𝑥 𝑚 ) 𝑚

As we stated at the end of last section the function 𝜋(𝑥) can be recovered from the above sum by the use of Mobius inversion formula shown below ∞

𝜋(𝑥) = ∑ 𝑛=1

1 𝜇(𝑛) 𝐽 (𝑥 𝑛 ) 𝑛

To prove the above inversion, let us start from the right-hand side and replace 𝐽(𝑥) by its value from equation (1) ∞

∞

𝑛=1

𝑛=1 ∞

∞

1 1 𝜇(𝑛) 𝜇(𝑛) 1 𝐽 (𝑥 𝑛 ) = ∑ ∑ ∑ 𝜋 (𝑥 𝑛𝑚 ) 𝑛 𝑛 𝑚

𝑚=1

∞

=∑∑ 𝑛=1 𝑚=1

1 𝜇(𝑛) 𝜋 (𝑥 𝑛𝑚 ) 𝑛𝑚

The Zeta Function and Counting the Prime Numbers

123

Now using the idea in equation (3), and keeping in mind that for a given 𝑢 = 𝑛𝑚 all possible values for 𝑛 are practically the divisors of 𝑢 and therefore we can write the right-hand side of last equation as ∞

∞

1 1 𝜇(𝑛) 𝜇(𝑑) 𝐽 (𝑥 𝑛 ) = ∑ ∑ 𝜋 (𝑥 𝑢 ) ∑ 𝑛 𝑢

𝑛=1

𝑢=1 𝑑|𝑢

∞

∞

𝑛=1

𝑢=1

1

𝜋 (𝑥 𝑢 ) 1 𝜇(𝑛) 𝐽 (𝑥 𝑛 ) = ∑ ∑ ∑ 𝜇(𝑑) 𝑛 𝑢 𝑑|𝑢

And according to the main property of the Mobius function (equation 2), the second sum on the right-hand side is equal to 0 for all 𝑢 except when 𝑢 is equal to 1 (in this case the sum is equal to 1), then replacing we get 1

∞

𝜋 (𝑥 𝑢 ) 𝜇(𝑛) 𝐽 (𝑥 ) = [ ] ∑ 𝑛 𝑢 1 𝑛

𝑛=1

𝑢=1

And when 𝑢 = 1, the term in brackets becomes 𝜋(𝑥) and hence we have ∞

∑ 𝑛=1

1 𝜇(𝑛) 𝐽 (𝑥 𝑛 ) = 𝜋(𝑥) 𝑛

Finally, let us mention that it is common to see the above equation written with the sum over 𝑛 from 1 to ∞. But in reality, no need for 𝑛 to log𝑥

go beyond the integer of [log2] because the function 𝐽(𝑥) is zero when 1

1

its argument is less than 2 therefore in 𝐽 (𝑥 𝑛 ) we need to keep 𝑥 𝑛 ≥ 2,

124

Naji Arwashan log 𝑥

and consequently 𝑛 ≤ log 2 which practically means 𝑛 goes from 1 to log 𝑥

the integer of [log 2 ]

The Relation between Mobius and Zeta Functions We should not leave Mobius function before we show its connection to the zeta function as illustrated in the following equation: ∞

𝜁(𝑠) ∑ 𝑚=1

𝜇(𝑚) =1 𝑚𝑠

Note that this equation was proved in Chapter 1 (Section 1.6) using Euler product but here will show it again using the equation (2). Let us start by the left-hand side and replace zeta by ∑∞ 𝑛=1 ∞

∞

∞

𝑚=1

𝑛=1

𝑚=1

1 𝑛𝑠

∞

∞

𝜇(𝑚) 1 𝜇(𝑚) 𝜇(𝑚) 𝜁(𝑠) ∑ = = ∑ ∑ ∑ ∑ 𝑚𝑠 𝑛𝑠 𝑚𝑠 𝑛𝑚𝑠 𝑛=1 𝑚=1

Using the idea in equation (3) and keeping in mind that 𝑚 is a divisor of 𝑢 = 𝑛𝑚 then we can write ∞

∞

∞

𝜇(𝑚) 𝜇(𝑑) 1 𝜁(𝑠) ∑ = ∑ ∑ 𝑠 = ∑ 𝑠 ∑ 𝜇(𝑑) 𝑠 𝑚 𝑢 𝑢 𝑚=1

𝑢=1 𝑑|𝑢

𝑢=1

𝑑|𝑢

And from (2) we found that ∑𝑑|𝑢 𝜇(𝑑) equal zero for all 𝑢 except when 𝑢 = 1 (in this case the sum is equal to 1), then we have: ∞

𝜁(𝑠) ∑ 𝑚=1

𝜇(𝑚) 1 = ( 𝑠 ) (1) = 1 𝑠 𝑚 1

The Zeta Function and Counting the Prime Numbers

125

6.4. RIEMANN EXPLICIT FORMULA FOR 𝑱(𝒙) Now we can turn our attention to the derivation of Riemann’s explicit formula of 𝐽(𝑥) as outlined in H.M. Edwards book [6]. For this derivation we will invoke Euler product formula we introduced and proved in Chapter 1 (Section 1.5) ∞

1 1 =∏ ( 1) 𝑠 𝑛 1− 𝑠 𝑛=1 𝑝

(𝜎 > 1 ∶ 𝑠 = 𝜎 + 𝑖𝑡)

𝜁(𝑠) = ∑

𝑝

John Derbyshire called this product formula in his popular and highly cited book “Prime Obsession” [5] the golden key that Riemann turned to open the door between the prime numbers and the complex analysis, and ultimately to find an explicit formula for the distribution of primes. Let us take the natural logarithm of both sides of Euler product formula. log 𝜁(𝑠) = ∑ − log (1 − 𝑝

1 ) 𝑝𝑠

(𝜎 > 1 )

Recall Maclaurin series for log (1 + 𝑥) from Chapter 2 (Section 2.5) where we found that log (1 + 𝑥) = 𝑥 −

𝑥2 𝑥3 𝑥4 𝑥5 + − + − 2 3 4 5

1

In our case 𝑥 = − 𝑝𝑠 and therefore ∞

1 1 1 1 1 1 log (1 − 𝑠 ) = − 𝑠 − 2𝑠 − 3𝑠 − 4𝑠 … . = − ∑ 𝑛𝑠 𝑝 𝑝𝑖 2𝑝𝑖 𝑛𝑝𝑖 4𝑝𝑖 3𝑝𝑖 𝑛=1

126

Naji Arwashan Replacing in log 𝜁(𝑠) gives ∞

log 𝜁(𝑠) = ∑ (∑ 𝑝

𝑛=1

1 ) 𝑛𝑝𝑛𝑠

(𝜎 > 1 )

1

Since all the terms of ∑∞ 𝑛=1 𝑛𝑝𝑛𝑠 are positive then the absolute series 𝑖

is equal to the series itself and therefore the series is absolutely convergent (check Section 2.6 from Chapter 2, for absolute convergence of a series) which is necessary to justify the next step ∞

log 𝜁(𝑠) = ∑ ∑ ( 𝑝 𝑛=1

1 1 ) 𝑝𝑛𝑠 𝑛

(𝜎 > 1 )

Which can be written as ∞

log 𝜁(𝑠) = ∑ ∑ (𝑝𝑛 )−𝑠 𝑝 𝑛=1

1 𝑛

(𝜎 > 1 )

Let us expand the above double sum: 1 1 1 1 (21 )−𝑠 + (22 )−𝑠 + (23 )−𝑠 + (24 )−𝑠 + ⋯ 1 2 3 4 1 1 1 1 (31 )−𝑠 + (32 )−𝑠 + (33 )−𝑠 + (34 )−𝑠 + ⋯ 1 2 3 4 1 1 1 1 (51 )−𝑠 + (52 )−𝑠 + (53 )−𝑠 + (54 )−𝑠 + ⋯ 1 2 3 4 1 1 1 1 (71 )−𝑠 + (72 )−𝑠 + (73 )−𝑠 + (74 )−𝑠 + ⋯ 1 2 3 4 ⋮

The Zeta Function and Counting the Prime Numbers

127

The expansion helps us realize that the double sum can be written as a single sum over all possible values of 𝑝𝑖𝑛 as follows 1

𝑛 −𝑠 log 𝜁(𝑠) = ∑∞ 𝑝𝑛 (𝑝 ) 𝑛

(𝜎 > 1 )

(4)

The next step is truly a crucial one, it allows us to convert the sum into integral and to bridge from the world of number theory into the world of complex analysis.

Riemann-Stieltjes Sum Let us start by defining the function 𝑓(𝑥) over an interval [𝑎, 𝑏], and let us partition that interval into 𝑛 number of subintervals by selecting 𝑥𝑖 in such way that 𝑎 = 𝑥0 < 𝑥1 < 𝑥2 < 𝑥3 … . < 𝑥𝑛 = 𝑏, and in every subinterval 𝑖 [𝑥𝑖 , 𝑥𝑖+1 ] we select a point 𝑡𝑖 as shown bellow

Figure 6-7. The partition of the interval [a, b].

128

Naji Arwashan

and now we can define Riemann sum as follows [among Riemann’s many other contributions to math is his formulation of the calculation of integral]: 𝑛−1

∑ 𝑓(𝑡𝑖 )(𝑥𝑖+1 − 𝑥𝑖 ) 𝑖=0

And when the norm of the partition ‖∆𝑥‖ (the largest length of the partition) goes to zero, the sum becomes the integral between 𝑎 and 𝑏 𝑛−1

𝑏

lim ∑ 𝑓(𝑡𝑖 ) (𝑥𝑖+1 − 𝑥𝑖 ) = ∫ 𝑓(𝑥)𝑑𝑥

‖∆𝑥‖→0

𝑎

𝑖−0

Stieltjes took the previous analysis a step further by introducing on the interval [𝑎, 𝑏] a second function 𝑔(𝑥), and now Riemann-Stieltjes sum becomes 𝑛−1

∑ 𝑓(𝑡𝑖 )[𝑔(𝑥𝑖+1 ) − 𝑔(𝑥𝑖 )] 𝑖=0

And when the norm ‖∆𝑔‖ goes to zero, 𝑏

lim ∑𝑛−1 𝑖=0 𝑓(𝑡𝑖 )[𝑔(𝑥𝑖+1 ) − 𝑔(𝑥𝑖 )] = ∫𝑎 𝑓(𝑥)𝑑𝑔(𝑥)

‖∆𝑔‖→0

(5)

Let us replace in the last equation the function 𝑔(𝑥) by 𝐽(𝑥), replace the function 𝑓(𝑥) by 𝑥 −𝑠 , and extend the interval [𝑎, 𝑏] to [0, ∞]

The Zeta Function and Counting the Prime Numbers 𝑛−1

129

∞

lim ∑ 𝑡𝑖−𝑠 [𝐽(𝑥𝑖+1 ) − 𝐽(𝑥𝑖 )] = ∫ 𝑥 −𝑠 𝑑𝐽(𝑥)

‖∆𝐽‖→0

0

𝑖=0

But we know that the function 𝐽(𝑥) is a step function and that for any small interval [𝐽(𝑥𝑖+1 ) − 𝐽(𝑥𝑖 )] is equal to zero except for the intervals where a jump takes place. And for those intervals [𝐽(𝑥𝑖+1 ) − 𝐽(𝑥𝑖 )] = 1/𝑛, and 𝑡𝑖 = 𝑝𝑖𝑛 therefore, replacing we obtain ∞

∑(𝑝 𝑝𝑛

𝑛 )−𝑠

∞ 1 = ∫ 𝑥 −𝑠 𝑑𝐽(𝑥) 𝑛 0

(𝜎 > 1 )

And comparing with equation (4) we obtain ∞

log 𝜁(𝑠) = ∫0 𝑥 −𝑠 𝑑𝐽

(𝜎 > 1 )

(6)

This is a significant result because now we left the world of natural numbers and we are in the world of integrals and calculus. One dazzling fact to be mentioned is that Stieltjes’s work came after Riemann publication of his paper! Our goal is to extract the function 𝐽(𝑥) from the last equation. We start by integrating the right-hand side by parts (∫ 𝑢 𝑑𝑣 = 𝑢𝑣 − ∫ 𝑣 𝑑𝑢 where 𝑢 = 𝑥 −𝑠 and 𝑑𝑣 = 𝑑𝐽) ∞ −𝑠−1 log 𝜁(𝑠) = [𝑥 −𝑠 𝐽(𝑥)]∞ 𝐽(𝑥) 𝑑𝑥 0 − ∫ −𝑠𝑥

(𝜎 > 1 )

0

From the definition of 𝐽(𝑥), 𝐽(𝑥) = 0 when 𝑥 < 2 and therefore 𝐽(𝑥) 𝑥𝑠

= 0 when 𝑥 → 0 One might argue that the denominator also goes to

0 when 𝑥 goes to 0, and therefore we have an indeterminacy! Well notice that as we said 𝐽(𝑥) and all its derivatives are zero when 𝑥 < 2,

130

Naji Arwashan

so a repeat of Hospital rule will bring the limit to zero. And when 𝑥 → ∞, 𝑥 −𝑠 𝐽(𝑥) → 0 because 𝐽(𝑥) 𝐽(𝑥) 𝐽(𝑥) = 𝜎+𝑖𝑡 = 𝜎 𝑠 𝑥 𝑥 𝑥 (cos𝑡 + 𝑖sin𝑡) 𝐽(𝑥) jumps only at 𝑝𝑛 and its jump is always equal to or less than 1, and therefore it will always be less than 𝑥, and since 𝜎 > 1 then

𝐽(𝑥) 𝑥𝜎

→

0 when 𝑥 → ∞ And as a result, the last equation can be written as: log 𝜁(𝑠) 𝑠

∞

= ∫0 𝐽(𝑥)𝑥 −𝑠−1 𝑑𝑥

(𝜎 > 1 )

(7)

At this point we will hold on equation (7) and open two brackets to introduce Mellin transform:

The Mellin Transform Mellin transform (named after the Finish Mathematician Hjalmar Mellin) of a real function𝑓(𝑥) is ∞

{𝑀𝑓}(𝑠) = 𝜑(𝑠) = ∫ 𝑓(𝑥)𝑥 𝑠−1 𝑑𝑥 0

where 𝑠 is a complex variable. The Mellin inversion theorem states that the function 𝑓(𝑥) can be retrieved from its Mellin transform as follows {𝑀−1 𝜑}(𝑥) = 𝑓(𝑥) =

1 𝑎+𝑖∞ −𝑠 ∫ 𝑥 𝜑(𝑠)𝑑𝑠 2𝜋𝑖 𝑎−𝑖∞

The Zeta Function and Counting the Prime Numbers

131

Where 𝑎 is an arbitrary number, which means the integral is from −∞ to +∞ over any vertical line in the complex plane. Now we can close the bracket and go back to equation (7) ∞ log 𝜁(𝑠) = ∫ 𝐽(𝑥)𝑥 −𝑠−1 𝑑𝑥 𝑠 0

(𝜎 > 1 )

We notice the right-hand side is similar to the Mellin transform except that 𝑠 has a negative sign, we can solve this problem easily by changing the variable to 𝑧 = −𝑠 and 𝑑𝑧 = −𝑑𝑠 ∞ log 𝜁(−𝑧) = ∫ 𝐽(𝑥)𝑥 𝑧−1 𝑑𝑥 −𝑧 0

(𝜎 > 1 )

And now we can use Mellin inverse transform and extract 𝐽(𝑥) 1 𝑎+𝑖∞ log 𝜁(−𝑧) −𝑧 𝐽(𝑥) = ∫ 𝑥 𝑑𝑧 2𝜋𝑖 𝑎−𝑖∞ −𝑧 We can go back to the original 𝑠 variable, and keeping in mind that when 𝑧 goes from 𝑎 − 𝑖∞ to 𝑎 + 𝑖∞, 𝑠 goes from−𝑎 + 𝑖∞ to −𝑎 − 𝑖∞

𝐽(𝑥) = −

1 −𝑎−𝑖∞ log 𝜁(𝑠) 𝑠 ∫ 𝑥 𝑑𝑠 2𝜋𝑖 −𝑎+𝑖∞ 𝑠

And the negative sign will disappear if we reverse the direction of the integral

𝐽(𝑥) =

1 −𝑎+𝑖∞ log 𝜁(𝑠) 𝑠 ∫ 𝑥 𝑑𝑠 2𝜋𝑖 −𝑎−𝑖∞ 𝑠

132

Naji Arwashan And since 𝑎 is arbitrary number then we can replace −𝑎 by 𝑎 𝑎+𝑖∞ log 𝜁(𝑠)

1

𝐽(𝑥) = 2𝜋𝑖 ∫𝑎−𝑖∞

𝑠

𝑥 𝑠 𝑑𝑠

(8)

And voila we found an analytic expression for the function 𝐽(𝑥). Certainly, it needs more simplification, but the main goal is achieved by finding an analytical way to express the prime counting number. One final thing we should mention with exclamation is that again Riemann paper proceeded Mellin’s work by 40 years!

The Calculation of 𝐋𝐨𝐠 𝜻(𝒔) The next step is to find an expression for log 𝜁(𝑠). Here we need to make use of the function 𝜉(𝑠) that we introduced in Chapter 4 (Section 4.9), as follows 𝜉(𝑠) =

𝑠(𝑠 − 1) −𝑠 𝑠 𝜋 2 Γ ( ) ζ(𝑠) 2 2

And we found that it has the functional equation of 𝜉(𝑠) = 𝜉(1 − 𝑠) At this point we will make a minor alteration to 𝜉(𝑠). We know from the functional equation of gamma (Equation 1, Chapter 4) that 𝑠

𝑠

𝑠

𝑠

2

𝑠

Γ (2 + 1) = Γ (2) 2, and then replacing Γ (2) = 𝑠 Γ (2 + 1) in 𝜉(𝑠) gives 𝑠

𝑠

𝜉(𝑠) = Γ (2 + 1) (𝑠 − 1)𝜋 −2 ζ(𝑠)

(9)

The Zeta Function and Counting the Prime Numbers

133

Riemann assumed that it was possible to factor 𝜉(𝑠) in terms of its roots as follows: 𝑠 𝜉(𝑠) = 𝑓(𝑠) ∏ (1 − ) 𝜌 𝜌

Where 𝜌 are the roots of 𝜉(𝑠) [recall that we found them in Chapter 5 (Section 5.7) to be equal to the nontrivial roots of zeta]; and 𝑓(𝑥) is a non-vanishing function, and he showed that 𝑓(𝑥) must be constant. The constant value of 𝑓(𝑠) = 𝑓(0) can be found to be equal to 𝜉(0) by replacing 𝑠 by 0 in the above equation 0 𝜉(0) = 𝑓(0) ∏ (1 − ) = 𝑓(0) 𝜌 𝜌

And therefore: 𝑠

𝜉(𝑠) = 𝜉(0) ∏𝜌 (1 − 𝜌)

(10)

The last equation implies that 𝜉(𝑠) is like a polynomial of infinite degree. The proof of the validity of the factorization was shown by Hadamard in 1893, some 34 years after Riemann published his paper. Now equating 𝜉(𝑠) from (9) and (10) gives: 𝑠 𝑠 𝑠 𝜉(0) ∏ (1 − ) = Γ ( + 1) (𝑠 − 1)𝜋 −2 ζ(𝑠) 𝜌 2

𝜌

And by taking the log of both sides and rearranging we obtain: 𝑠 𝑠 𝑠 log ζ(𝑠) = log 𝜉(0) + ∑ log (1 − ) − log Γ ( + 1) + log π − log (s − 1) 𝜌 2 2 𝜌

134

Naji Arwashan

Replacing log ζ(𝑠) from the last expression into (8) is a very intricate mathematical operation that leads to the Riemann explicit formula for 𝐽(𝑥) ∞

𝐽(𝑥) = 𝑙𝑖(𝑥) − ∑𝜌 𝑙𝑖(𝑥 𝜌 ) − log(2) + ∫𝑥

𝑑𝜏 𝜏(𝜏2 −1) log(𝜏)

(11)

The very first observation one can have about this formula is that how 𝐽(𝑥), which is a real number, can be calculated from a formula with complex numbers? The answer to that, it is true that the roots of zeta are complex numbers but as we saw in Chapter 5 (Section 5.6) the conjugate of every zero is a zero, and it can be shown that when a pair of zeros is replaced in 𝐽(𝑥)their imaginary parts will cancel out and we will be left only with a real value for 𝐽(𝑥). This point will be addressed again and demonstrated in the next chapter (Section 7.5). Also, in next chapter will show a different mathematical approach to obtaining the same explicit formula in (11).

6.5. NUMERICAL CALCULATION Let us list one more time the two equations defining the prime counting function ∞

𝐽(𝑥) = 𝑙𝑖(𝑥) − ∑ 𝑙𝑖(𝑥

𝜌)

− log(2) + ∫ 𝑥

𝜌

and ∞

𝜋(𝑥) = ∑ 𝑛=1

1 𝜇(𝑛) 𝐽 (𝑥 𝑛 ) 𝑛

𝑑𝜏 𝜏(𝜏 2 − 1) log(𝜏)

The Zeta Function and Counting the Prime Numbers

135

Riemann approximated the 𝐽(𝑥) function by its first term 𝑙𝑖(𝑥), and obtained a new approximation for 𝜋(𝑥) as follows: ∞

𝜋(𝑥) ≈ ∑ 𝑛=1

1 𝜇(𝑛) 𝑙𝑖 (𝑥 𝑛 ) 𝑛

He called the new estimate function 𝑅(𝑥) ∞

𝑅(𝑥) = ∑ 𝑛=1

1 𝜇(𝑛) 𝑙𝑖 (𝑥 𝑛 ) 𝑛

And in this case the exact value of 𝜋(𝑥) becomes ∞

𝜋(𝑥) = 𝑅(𝑥) + ∑ 𝑛=1

1 𝜇(𝑛) 𝑓 (𝑥 𝑛 ) 𝑛

Where 𝑓(𝑥) is 𝐽(𝑥) without the first term of 𝑙𝑖(𝑥) ∞

𝑓(𝑥) = − ∑ 𝑙𝑖(𝑥 𝜌 ) − log(2) + ∫ 𝜌

𝑥

𝜏(𝜏 2

𝑑𝜏 − 1) log(𝜏)

That is normally the approach taken by sources like “Prime Numbers and Computer Methods for Factorization” [9] and “Wolfram Demonstration Project” [3] where they numerically calculate 𝑅(𝑥) and then calculate the correction coming from the zeros of zeta through the function 𝑓(𝑥). However here in this book we will use different approach, and we will focus instead on the calculation of 𝐽(𝑥), because once 𝐽(𝑥) is found the calculation of 𝜋(𝑥) becomes a simple matter. Let us start by a numerical example and attempt to calculate 𝐽(20). The first term is 𝑙𝑖(20), the third one is log(2) and the forth

136 ∞

∫𝑥

Naji Arwashan 𝑑𝜏

𝜏(𝜏2 −1) log(𝜏)

all deal with real number and their calculation is simple.

The challenge arises from the second term ∑𝜌 𝑙𝑖(𝑥 𝜌 ) where complex logarithm is involved. That second term is a sum over all non-trivial zeros of zeta. Well, let us start by the first zero 𝜌1 = 0.5 + 𝑖14.134725142 we begin by calculating 200.5+𝑖14.134725142 200.5+𝑖14.134725142 = 200.5 20𝑖14.134725142 = 200.5 𝑒 𝑖(log 20)14.134725142 = 4.472136𝑒 𝑖42.343853 = −0.302303 − i4.46191 Then next step is to calculate 𝑙𝑖(−0.302303 − i4.46191) −0.302303−i4.46191

𝑙𝑖(−0.302303 − i4.46191) =

∫ 0

Figure 6-8. The contour of the integral li(200.5+i14.134725142).

𝑑𝑠 log𝑠

The Zeta Function and Counting the Prime Numbers

137

𝑠 is a complex number 𝑠 = 𝜎 + 𝑖𝑡 and the integral is a complex 1

integral, the integrand log 𝑠 has only one pole at s=1, then and as we saw in Chapter2 (Section 2.10) the integral is independent of the path from 0 to (−0.302303 − i4.46191) as long we avoid the pole, so let us calculate the integral following a direct linear path as shown in Figure 6-8. Recall the formula of contour integral from Chapter 2 (equation 1) 𝜏𝑛

∫ 𝑓(𝑠)𝑑𝑠 = ∫ 𝑓(𝛾(𝜏)) 𝛾 ′ (𝜏)𝑑𝜏 𝜏0

𝛾

(note we switch to use the variable 𝜏 in order to avoid any confusion with the variable 𝑡 which is reserved for the imaginary part of s) In our case, the equation 𝛾(𝜏) of the linear path from 0 to (−0.302303 − i4.46191) is −4.46191

𝛾(𝜏) = 𝜏 + 𝑖 −0.302303 𝜏 = 𝜏 + 𝑖14.759728𝜏

(where 𝜏 varies from

0 to -0.302303) And 𝛾′(𝜏) = 1 + 𝑖14.759728 Applying to our integral we have: −0.302303−i4.46191

∫ 0

𝑑𝑠 = log𝑠

−0.302303

∫ 0

1 + 𝑖14.759728 𝑑𝜏 log (𝜏 + 𝑖14.759728𝜏)

The denominator of the integrand is a complex logarithm which we found in Chapter 2 (Section 2.8) to be equal to log 𝑠 = log |𝑠| + 𝑖(𝜃 + 2𝜋𝑘)

138

Naji Arwashan So, in our case |𝑠| = √𝜏 2 + 14.759728𝜏 2

And 𝑠 moves on a straight line with argument equal to -1.6384 (see Figure 6-8), and therefore we have log(𝜏 + 𝑖14.759728𝜏) = log (√𝜏 2 + 14.759728𝜏 2 ) + 𝑖(−1.6384 + 2𝜋𝑘) And the integral becomes −0.302303

∫ 0

1 + 𝑖14.759728 log (√𝜏 2 + (14.759728𝜏)2 ) + 𝑖(−1.638445 + 2𝜋𝑘)

𝑑𝜏

Which can be simplified into real and imaginary parts as follows: −0.302303

∫

log (√𝜏 2 + (14.759728𝜏)2 ) + 14.759728(42.343853) [log (√𝜏 2

0

−0.302303

+𝑖

∫ 0

+

2 (14.759728𝜏)2 )]

+ (−1.638445 +

𝑑𝜏

2𝜋𝑘)2

14.759728log (√𝜏 2 + (14.759728𝜏)2 ) − 42.343853 2

[log(√𝜏 2 + 14.759728𝜏 2 )] + (−1.638445 + 2𝜋𝑘)2

𝑑𝜏

And now we have two integrals with real numbers that can be evaluated numerically with ease. But before we do that, we still have to decide what value to assign to 𝑘 (or in other words what branch of the complex logarithm to use). Normally the logarithmic integral 𝑙𝑖 uses the principal branch (𝑘 = 0). And if do that here we obtain for the integral the numerical value of 1.99797-i0.772249. We can repeat the same

The Zeta Function and Counting the Prime Numbers

139

calculation for the second, third …. zero of zeta but when we add the outcome together we find that the sum is diverging! How could that have happened?! Well the answer is the choice of the principal branch (𝑘 = 0) is not the correct choice. It turned out that the selection of the 7th branch (k=7) that will lead to convergence. And the numerical value of the integral in this case of k=7 is 200.5+i14.134725142

∫ 0

𝑑𝑠 = − 0.105384 + 𝑖0.005895 log𝑠

Well what is the secret about the seventh branch? Why 𝑘 = 7 works? The answer is when 𝑘 = 7 the term (−1.638445 + 2 𝜋7) in the integrand becomes equal to 42.343853 and this is the argument of the end point of the contour of the integral as it came from the calculation of 200.5+𝑖14.134725142 (see Figure 6-8, and above calculation of 200.5+𝑖14.134725142 ). We can repeat the same steps for different values of 𝑥 and generate the following graphs. In the first one, we show the exact staircase function of 𝐽(𝑥), and the curve obtained for 𝐽(𝑥) from Riemann explicit

Figure 6-9. The exact 𝑱(𝒙) function versus the one obtained from Riemann explicit formula (with no zeros of zeta).

140

Naji Arwashan

formula without accounting for any of the zeros of zeta, and in the second graph (Figure 6-10) it is when 10 zeros are accounted for, and the third one (Figure 6-11) is similar but now we use 50 zeros instead of 10; and we can see how the forming of the rises becomes more noticeable. And finally, in Figure 6-12 we used 1000 zeros and see how the curve become amazingly like a staircase. It is impressive how the zeros of zeta know the location of the prime powers and are gradually carving rises for them in the curve of the function 𝑙𝑖(𝑥).

Figure 6-10. The exact 𝑱(𝒙) function versus the one obtained from Riemann explicit formula (with 10 zeros of zeta).

Figure 6-11. The exact 𝑱(𝒙) function versus the one obtained from Riemann explicit formula (with 50 zeros of zeta).

The Zeta Function and Counting the Prime Numbers

141

Figure 6-12. The 𝑱(𝒙) function as obtained from Riemann explicit formula (with 1000 zeros of zeta).

Finally, if we use the equation 1 1 1 1 1 1 1 1 𝜋(𝑥) = 𝐽(𝑥) − 𝐽 (𝑥 2 ) − 𝐽 (𝑥 3 ) − 𝐽 (𝑥 5 ) + 𝐽 (𝑥 6 ) …. 2 3 5 6

Figure 6-13. The exact 𝝅(𝒙) function versus the one obtained from Riemann explicit formula (with 50 zeros of zeta).

We can build 𝜋(𝑥) from 𝐽(𝑥). For instance, in Figure 6-13 we show the exact 𝜋(𝑥) versus the one obtained from the equation above where

142

Naji Arwashan

𝐽(𝑥) was calculated using 50 zeros of zeta. And when a large number of zeros (1000) is used to calculate 𝐽(𝑥), then 𝜋(𝑥) becomes very close to a staircase as shown in Figure 6-14. And we notice in that figure that the rises at 4, 8, 9, … (𝑝𝑛 : 𝑛 > 1) that we used to have in 𝐽(𝑥) disappear in 𝜋(𝑥) and leave a minor trace that will completely disappear with the use of increasing large number of zeros; and the only rises left in 𝜋(𝑥) are the ones at the primes.

Figure 6-14. The 𝜋(𝑥) function as obtained from Riemann explicit formula (with 1000 zeros of zeta).

The final question is how the numerical calculation was done by other sources including “Prime Obsession” [5] and “Wolfram Demonstration Project” [3]. Well the answer is in the next section.

6.6. THE EXPONENTIAL INTEGRAL The exponential integral is defined as follows: ∞

𝑒 −𝑡 𝐸𝑖(𝑥) = − ∫ 𝑑𝑡 𝑡 −𝑥

The Zeta Function and Counting the Prime Numbers

143

If we make a variable change of 𝑡 = −𝑦 and 𝑑𝑡 = −𝑑𝑦 then replacing gives us 𝑥

−∞

𝑒𝑦 𝑒𝑦 𝐸𝑖(𝑥) = − ∫ − 𝑑𝑦 = ∫ 𝑑𝑦 −𝑦 𝑦 𝑥

−∞

And therefore, we can write: 𝑥

∞

𝑒 −𝑡 𝑒𝑡 𝐸𝑖(𝑥) = − ∫ 𝑑𝑡 = ∫ 𝑑𝑡 𝑡 𝑡 −𝑥

−∞

Let us now see the connection to the logarithmic integral. 𝑙𝑖(𝑥) was defined as follows: 𝑥

𝑙𝑖(𝑥) = ∫ 0

𝑑𝑡 log𝑡

If we define the following variable change 𝑧 = log𝑡 ⇒ 𝑡 = 𝑒 𝑧 and 𝑑𝑡 = 𝑒 𝑧 𝑑𝑧 When 𝑡 goes to 0, 𝑧 goes to −∞ and when t goes to 𝑥, 𝑧 goes to log(𝑥) and now substituting in 𝑙𝑖(𝑥) gives us log𝑥

𝑙𝑖(𝑥) = ∫ −∞

𝑒𝑧 𝑑𝑧 𝑧

And therefore, we can conclude that 𝑙𝑖(𝑥) = 𝐸𝑖(log 𝑥)

144

Naji Arwashan

The advantage of 𝐸𝑖(𝑥) is that it can be calculated numerically by the following series: ∞

𝐸𝑖(𝑥) = 𝛾 + log(𝑥) + ∑ 𝑛=1

𝑥𝑛 𝑛𝑛!

Where 𝛾 is the Euler-Mascheroni constant and is equal to 0.57721 When one is dealing with real number, using the above series to calculate 𝐸𝑖(𝑥) and later to calculate 𝑙𝑖(𝑥) would be perfectly fine. However, if one is dealing with complex numbers then he should be very careful. To illustrate our warning let us use the case of our numerical example from the previous section. 𝑙𝑖(200.5+i14.134725142 ) = 𝐸𝑖[log(200.5+i14.134725142 )] = 𝐸𝑖[(0.5 + i14.134725142)log(20)] = −0.105384 + i3.147487 Note that the real part is correct and matches our calculation in the previous section where we obtained (−0.105384 + 𝑖0.005895). However, the imaginary part is wrong! But the good news is that zeta’s zeros come in pairs of conjugate numbers, and when the sum of 𝑙𝑖(𝑥 𝜌 ) is taken, the imaginary parts cancel out leaving the final result for the calculation of 𝐽(𝑥) correct even though the numerical approach that was followed is not perfectly sound.

Chapter 7

VON MANGOLDT FORMULA TO THE RESCUE Riemann explicit formula for the 𝐽(𝑥) function though was a breakthrough accomplishment but was mostly based on mathematical intuition and lacking formal proof; also, the presence of the logarithmic integral in the formula made it difficult to use and work with. Until help came 35 years later from the German mathematician Hans von Mangoldt.

7.1. CHEBYSHEV FUNCTIONS AND VON MANGOLDT EXPLICIT FORMULA The Russian mathematician Pafnuty Chebyshev (1821-1894) introduced two new prime counting functions, named after him. The first one 𝜗(𝑥) 𝜗(𝑥) = ∑ log 𝑝 𝑝≤𝑥

146

Naji Arwashan And the second one 𝜓(𝑥) 𝜓(𝑥) = ∑ log 𝑝 𝑝𝑛 ≤𝑥

Both functions are step function: the first one jumps every time 𝑥 hits a prime 𝑝 by a value of log 𝑝, whereas the second one jumps every time 𝑥 hits a prime 𝑝 or a prime power 𝑝𝑛 by a value of log 𝑝. And similar to the functions 𝜋(𝑥) and 𝐽(𝑥) Chebyshev’s functions assume the halfway value at each jump. Furthermore, Chebyshev was able to prove that the Prime Number Theorem is equivalent to either of his functions being asymptotic to 𝑥. In other words, PNT is true if lim

𝑥→∞

𝜗(𝑥) 𝑥

= 1 or lim

𝑥→∞

𝜓(𝑥) 𝑥

= 1.

In this chapter, we will focus our attention on the second function 𝜓(𝑥), and let us start first by establishing the relationship between 𝐽(𝑥) and 𝜓(𝑥). The function 𝐽(𝑥) was defined as a step function that jumps by 1⁄𝑛 when 𝑥 is equal to 𝑝𝑛 , therefore 𝑑𝐽 = lim ∆𝐽 is equal to zero everywhere except when 𝑥 is 𝑝𝑛 , 𝑑𝐽 is equal to ∆𝑥→0

1⁄𝑛. And also we know from the definition of 𝜓(𝑥) that 𝑑𝜓 = lim ∆𝜓 ∆𝑥→0

is zero everywhere except when 𝑥 is equal to 𝑝𝑛 it is equal to log 𝑝; but 1

log 𝑝 can be written as 𝑛 log 𝑝𝑛 , and therefore we can write: 1 1 log 𝑝𝑛 = log 𝑥 = 𝑑𝐽 log 𝑥 𝑛 𝑛 and 𝑑𝜓 = 𝑑𝐽 = 0

𝑑𝜓 = log 𝑝 =

when 𝑥 = 𝑝𝑛 when 𝑥 ≠ 𝑝𝑛

Consequently 𝑑𝜓(𝑥) = log 𝑥 𝑑𝐽(𝑥)

(1)

Von Mangoldt Formula to the Rescue

147

35 years after Riemann published his paper, the German mathematician Hans von Mangoldt studied the function 𝜓(𝑥) and brilliantly was able to find a relatively simple explicit formula for it that eventually paved the way for the proof of PNT. The explicit formula is shown below 1

1

𝜓(𝑥) = 𝑥 − log(2𝜋) − 2 log (1 − 𝑥 2) − ∑𝜌

𝑥𝜌

(2)

𝜌

It consists of the sum of four terms: the first one is simply 𝑥, the 1

1

2

𝑥2

second one (− log 2𝜋) is a constant, the third one − log (1 −

)

becomes negligible for large value of 𝑥, and the forth and the important one is the sum of

𝑥𝜌 𝜌

over all the non-trivial zeros of zeta. We will

present the proof of the above equation in two parts.

7.2. THE FIRST PART OF THE POOF OF VON MANGOLDT FORMULA (DERIVATION OF THE INTEGRAL) In this section we will focus on the derivation of the following equation for 𝜓(𝑥) 𝑎+𝑖∞

1 𝜁 ′ (𝑠) 𝑥 𝑠 ∫ − 𝑑𝑠 = 𝜓(𝑥) 2𝜋𝑖 𝜁(𝑠) 𝑠 𝑎−𝑖∞

H.M. Edwards suggested in his book [6] at page 50 that the technique to prove the above equation is similar to the one used for the explicit formula of 𝐽(𝑥) that we used and demonstrated in Chapter 6

148

Naji Arwashan

(Section 6.4). Well, let us put that suggestion to the test and let us start from equation (6) in Chapter 6. ∞

log 𝜁(𝑠) = ∫ 𝑥 −𝑠 𝑑𝐽

(𝜎 > 1 )

0

Before proceeding with the integral by parts for the right-hand side, we differentiate both side with respect to 𝑠 ∞ 𝜁 ′ (𝑠) = ∫ −𝑥 −𝑠 log 𝑥 𝑑𝐽 𝜁(𝑠) 0

(𝜎 > 1 )

We found earlier in (1) that 𝑑𝜓 = log 𝑥 𝑑𝐽 therefore, replacing gives us: 𝜁 ′ (𝑠) 𝜁(𝑠)

∞

= ∫0 −𝑥 −𝑠 𝑑𝜓

(𝜎 > 1 )

(3)

Now we integrate the right-hand side by parts (∫ 𝑢 𝑑𝑣 = 𝑢𝑣 − ∫ 𝑣 𝑑𝑢 where 𝑢 = −𝑥 −𝑠 and 𝑑𝑣 = 𝑑𝜓) ∞ 𝜁 ′ (𝑠) −𝑠−1 = [−𝑥 −𝑠 𝜓(𝑥)]∞ )𝜓(𝑥) 𝑑𝑥 0 − ∫ (𝑠𝑥 𝜁(𝑠) 0

(𝜎 > 1 )

From the definition of 𝜓(x), we know that 𝜓(𝑥) and all its derivatives are equal to zero when 𝑥 < 2 and therefore

𝜓(𝑥) 𝑥𝑠

= 0 when

𝑥 → 0 ; to justify this result we can use the same argument we made in Section 6.4 which is a repeat application of Hospital rule until the limit becomes zero. And when 𝑥 → ∞, −𝑥 −𝑠 𝜓(𝑥) → 0 because 𝜓(𝑥) 𝜓(𝑥) 𝜓(𝑥) = 𝜎+𝑖𝑡 = 𝜎 𝑠 𝑥 𝑥 𝑥 (cos𝑡 + 𝑖sin𝑡)

Von Mangoldt Formula to the Rescue

149

Since 𝜎 > 1 we can write 𝜎 = 1 + 𝜀 𝜓(𝑥) 𝜓(𝑥) 1 = 𝑠 𝜀 𝑥 𝑥 𝑥 (cos𝑡 + 𝑖sin𝑡) From PNT we know that 𝑙𝑖𝑚

𝑥→∞

𝜓(𝑥) 𝑥

= 1 [However at this point PNT

was not proved yet! And hence, this route though leads to the right result, its use is not justified. Still we will follow this approach till the end and then present an alternative technique]. And since clearly lim

1

𝑥→∞ 𝑥 𝜀

= 0 then the last equation becomes:

−1 𝜁 ′ (𝑠) 𝑠

𝜁(𝑠)

∞

= ∫0 𝜓(𝑥)𝑥 −𝑠−1 𝑑𝑥

(𝜎 > 1 )

(4)

And the same way we applied in Chapter 6 (Section 6.4) Millen inverse transform on ∞ log 𝜁(𝑠) = ∫ 𝐽(𝑥)𝑥 −𝑠−1 𝑑𝑥 𝑠 0

And were able to obtain

𝐽(𝑥) =

1 𝑎+𝑖∞ log 𝜁(𝑠) 𝑠 ∫ 𝑥 𝑑𝑠 2𝜋𝑖 𝑎−𝑖∞ 𝑠

Will do also the same here on equation (4) and obtain: 𝜓(𝑥) =

1

𝑎+𝑖∞

− ∫ 2𝜋𝑖 𝑎−𝑖∞

𝜁(𝑠)′ 𝑥 𝑠 𝜁(𝑠) 𝑠

𝑑𝑠

(5)

As mentioned above this technique required PNT to be true (which was not the case at the time of the derivation), and therefore will

150

Naji Arwashan

present in the following a different technique presented in Havil’s book [7] at page 200. We start this time from equation (4) of Chapter 6. ∞

log 𝜁(𝑠) = ∑(𝑝𝑛 )−𝑠 𝑝𝑛

1 𝑛

(𝜎 > 1 )

Differentiating both sides with respect to 𝑠 gives ∞

𝜁 ′ (𝑠) −𝑛 log 𝑝 1 =∑ 𝜁(𝑠) 𝑝𝑛𝑠 𝑛 𝑛

(𝜎 > 1 )

𝑝

Simplifying gives ∞

𝜁 ′ (𝑠) log 𝑝 − = ∑ 𝑛𝑠 𝜁(𝑠) 𝑝 𝑛

(𝜎 > 1 )

𝑝

Multiplying both sides by

𝑥𝑠

gives

𝑠 ∞

𝜁 ′ (𝑠) 𝑥 𝑠 𝑥 𝑠 log 𝑝 − = ∑ 𝑛𝑠 𝜁(𝑠) 𝑠 𝑠 𝑛 𝑝

(𝜎 > 1 )

𝑝

Integrating both side from 𝑎 − 𝑖∞ to 𝑎 + 𝑖∞ (where 𝑎 is an arbitrary constant greater than 0) and dividing them by 2𝜋𝑖 we get 𝑎+𝑖∞

𝑎+𝑖∞

∞

1 𝜁 ′ (𝑠) 𝑥 𝑠 1 𝑥𝑠 log 𝑝 ∫ − = ∫ ∑ 𝑛𝑠 𝑑𝑠 2𝜋𝑖 𝜁(𝑠) 𝑠 2𝜋𝑖 𝑠 𝑛 𝑝 𝑎−𝑖∞

𝑎−𝑖∞

𝑝

(𝜎 > 1 )

Von Mangoldt Formula to the Rescue

151

Rearranging we get 𝑎+𝑖∞

∞

𝑎+𝑖∞

1 𝜁 ′ (𝑠) 𝑥 𝑠 1 𝑥 𝑠1 ∫ − 𝑑𝑠 = ∑ log 𝑝 ∫ ( 𝑛 ) 𝑑𝑠 2𝜋𝑖 𝜁(𝑠) 𝑠 2𝜋𝑖 𝑝 𝑠 𝑛 𝑝

𝑎−𝑖∞

(𝜎 > 1 )

𝑎−𝑖∞

𝑥

Calling 𝑦 = 𝑝𝑛 we obtain 𝑎+𝑖∞

𝑎+𝑖∞

∞

1 𝜁 ′ (𝑠) 𝑥 𝑠 1 𝑦𝑠 ∫ − 𝑑𝑠 = ∑ log 𝑝 [ ∫ 𝑑𝑠] 2𝜋𝑖 𝜁(𝑠) 𝑠 2𝜋𝑖 𝑠 𝑛 𝑝

𝑎−𝑖∞

(𝜎 > 1 )

𝑎−𝑖∞

The integral in brackets is a well-known integral, and will show in Appendix D that it is equal to 𝑎+𝑖∞

1 𝑦𝑠 ∫ 𝑑𝑠 = { 2𝜋𝑖 𝑠 𝑎−𝑖∞

0, 1 , 2 1,

0 𝑥, 𝑦 is less than 1, and the integral is 0. So, we are left with the sum when 𝑝𝑛 < 𝑥 (in this case the integral is 1) and when 𝑝𝑛 = 𝑥 (in this case the integral is ½). Recall that 𝜓(𝑥) = ∑𝑝𝑛≤𝑥 log 𝑝 and therefore the righthand side of the last equation is equal to 𝜓(𝑥) [note that when 𝑝𝑛 = 𝑥 the ½ value we obtain from the integral is consistent with the definition of 𝜓(𝑥) where we said the function jumps by halfway value at every jump], and the last equation becomes 1

𝑎+𝑖∞

− ∫ 2𝜋𝑖 𝑎−𝑖∞

𝜁 ′ (𝑠) 𝑥 𝑠 𝜁(𝑠) 𝑠

𝑑𝑠 = 𝜓(𝑥)

(6)

152

Naji Arwashan

7.3. THE SECOND PART OF THE PROOF OF VON MANGOLDT FORMULA (APPLICATION OF THE RESIDUE THEOREM) The advantage of the integral in the last equation (6) is that it can be evaluated according to the residue theorem that we covered in Chapter 2 (Section 2.11) and we will state it here again: If the function 𝑔(𝑧) has 𝑛 poles in the domain D, then 𝑛

∮

𝑔(𝑧)𝑑𝑧 = 2𝜋𝑖 ∑ 𝑅𝑒𝑠(𝑔, 𝑧𝑘 )

𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝐷

𝑘=1

But hold on, shouldn’t the integral in the residue theorem be over a closed contour? Well the strategy here is to evaluate the integral in equation (6) over a closed box as shown in the Figure 7-1, then prove

Figure 7-1. The closed contour of the integral.

Von Mangoldt Formula to the Rescue

153

that the integral over the top, left, and bottom sides of the box go to zero when ℎ and 𝐾 go to infinity, and therefore we end up with the integral equal to all the residues in the infinite box times 2𝜋𝑖 and consequently 𝜓(𝑥) equal to the sum of all the residues of [−

𝜻′ (𝒔) 𝒙𝒔 𝜻(𝒔) 𝒔

].

An important point to mention here before we move to the next step and calculate the residues is that the calculation of the above integral over the infinite box wouldn’t have been possible if zeta was not defined in the whole C-plane, and therefore we see here the need and benefit of the extension of zeta that we covered in chapters 3 and 4.

The Calculation of the Residues of [− The integrand in the last equation [−

𝜻′ (𝒔) 𝒙𝒔 𝜻(𝒔) 𝒔

𝜁 ′ (𝑠) 𝑥 𝑠 𝜁(𝑠) 𝑠

]

] is not defined when

the denominator is equal to zero (or when 𝑠 or 𝜁(𝑠) is equal to 0) and also is not defined when 𝑠 is equal to 1 (because 𝜁(𝑠) is not defined when 𝑠 = 1). Let us calculate the residue when 𝑠 = 0 𝑅𝑒𝑠 [−

𝜁 ′ (𝑠) 𝑥 𝑠 𝜁 ′ (𝑠) 𝑥 𝑠 𝜁 ′ (0) , 0] = lim − (𝑠 − 0) = − 𝑠→0 𝜁(𝑠) 𝑠 𝜁(𝑠) 𝑠 𝜁(0)

In order to calculate

𝜁 ′ (0) 𝜁(0)

we need to start from the functional

equation of zeta (equation 10 in Chapter 4) 𝜋𝑠 𝜁(𝑠) = 2𝑠 𝜋 𝑠−1 sin ( ) Γ(1 − 𝑠)𝜁(1 − 𝑠) 2

154

Naji Arwashan We take the logarithm of both side 𝜋𝑠 log 𝜁(𝑠) = 𝑠log2 + (𝑠 − 1)log𝜋 + log sin ( ) + logΓ(1 − 𝑠) 2 + log 𝜁(1 − 𝑠) And we differentiate both side with respect to 𝑠 1

𝜋𝑠

cos ( 2 ) Γ′(1 − 𝑠) 𝜁′(1 − 𝑠) 𝜁′(𝑠) = log2 + log𝜋 + 2 − − 𝜋𝑠 𝜁(𝑠) Γ(1 − 𝑠) 𝜁(1 − 𝑠) sin ( ) 2

Note the negative sign of the last two terms of the right-hand side is explained by the fact that the variable in those two terms is (1 − 𝑠) and the derivative of (1 − 𝑠) with respect to 𝑠 is −1 Let us now take the limit when 𝑠 → 1 𝜁′(𝑠) Γ′(1 − 𝑠) 𝜁′(0) = log 2𝜋 + 0 − lim − 𝑠→1 𝜁(𝑠) 𝑠→1 Γ(1 − 𝑠) 𝜁(0)

lim

Rearranging gives 𝜁′(0) 𝜁(0)

= log2𝜋 − lim

Let us consider

𝜁 ′ (𝑠)

𝑠→1 𝜁(𝑠) 𝜁(𝑠) Γ(1−𝑠)

Γ′ (1−𝑠)

− lim Γ(1−𝑠) 𝑠→1

(7)

when 𝑠 → 1, we have both numerator and

denominator go to infinity and therefore we can apply Hospital rule 𝜁(𝑠) 𝜁′(𝑠) = lim 𝑠→1 Γ(1 − 𝑠) 𝑠→1 −Γ′(1 − 𝑠) lim𝜁′(𝑠) 𝑠→1 = lim − Γ′(1 − 𝑠)

lim

𝑠→1

Von Mangoldt Formula to the Rescue

155

Which leads to Γ′(1 − 𝑠) 𝜁′(𝑠) = − lim 𝑠→1 Γ(1 − 𝑠) 𝑠→1 𝜁(𝑠)

lim

Replacing in (7) gives us 𝜁′(0) = log2𝜋 𝜁(0) And consequently

𝑅𝑒𝑠 [−

𝜁′(𝑠) 𝑥 𝑠 , 0] = −log2𝜋 𝜁(𝑠) 𝑠

Let us now calculate the residue when s is equal to T (trivial zeros of zeta)

𝑅𝑒𝑠 [−

𝜁′(𝑠) 𝑥 𝑠 𝜁′(𝑠) 𝑥 𝑠 (𝑠 − 𝑇)] , 𝑇] = lim [− 𝑠→𝑇 𝜁(𝑠) 𝑠 𝜁(𝑠) 𝑠 (𝑠 − 𝑇) 𝑥𝑠 = −lim lim 𝜁′(𝑠) 𝑠→𝑇 𝜁(𝑠) 𝑠→𝑇 𝑠

We apply Hospital rule to the first limit and get:

𝑅𝑒𝑠 [−

𝜁′(𝑠) 𝑥 𝑠 1 𝑥𝑠 , 𝑇] = −lim lim 𝜁′(𝑠) 𝑠→𝑇 𝜁′(𝑠) 𝑠→𝑇 𝜁(𝑠) 𝑠 𝑠 𝑥𝑠 𝑥𝑇 = − lim = − 𝑠→𝑇 𝑠 𝑇

156

Naji Arwashan

The same way we can find the residue when s is equal to ρ (nontrivial zeros of zeta):

𝑅𝑒𝑠 [−

𝜁′(𝑠) 𝑥 𝑠 𝑥𝜌 , 𝜌] = − 𝜁(𝑠) 𝑠 𝜌

And finally, when s =1, will start by the definition of zeta in terms of eta 𝜁(𝑠) =

𝜂(𝑠) 1 − 21−𝑠

Taking the logarithm of both sides we get log𝜁(𝑠) = log 𝜂(𝑠) − log(1 − 21−𝑠 ) Differentiating both sides with respect to 𝑠: 𝜁 ′ (𝑠) 𝜂′ (𝑠) (1 − 21−𝑠 )′ = − 𝜁(𝑠) 𝜂(𝑠) (1 − 21−𝑠 ) 𝜂′ (𝑠) 21−𝑠 log 2 = − 𝜂(𝑠) 1 − 21−𝑠 𝜁 ′ (𝑠)

Let us calculate the limit of lim 𝜁(𝑠) (𝑠 − 1) 𝑠→1

𝜁 ′ (𝑠) 𝜂′ (𝑠) 21−𝑠 log 2 (𝑠 − 1) = lim [ − ] (𝑠 − 1) 𝑠→1 𝜁(𝑠) 𝑠→1 𝜂(𝑠) 1 − 21−𝑠 𝜂′ (𝑠) 21−𝑠 log 2 (𝑠 − 1) − lim (𝑠 − 1) = lim 𝑠→1 𝜂(𝑠) 𝑠→1 1 − 21−𝑠 (𝑠 − 1) = 0 − 20 log 2 lim 𝑠→1 1 − 21−𝑠

lim

Von Mangoldt Formula to the Rescue

157

Applying Hospital rule gives (𝑠 − 1)′ 𝜁 ′ (𝑠) (𝑠 − 1) = − log 2 lim 𝑠→1 𝜁(𝑠) 𝑠→1 (1 − 21−𝑠 )′ 1 = − log 2 lim 1−𝑠 = −1 𝑠→1 2 log 2

lim

And now we can easily calculate the residue of [−

𝜁 ′ (𝑠) 𝑥 𝑠 𝜁(𝑠) 𝑠

] when

𝑠=1 𝑅𝑒𝑠 [−

𝜁 ′ (𝑠) 𝑥 𝑠 𝜁 ′ (𝑠) 𝑥 𝑠 , 1] = lim [− (𝑠 − 1)] 𝑠→1 𝜁(𝑠) 𝑠 𝜁(𝑠) 𝑠 𝜁 ′ (𝑠) 𝑥𝑠 = −lim (𝑠 − 1)lim 𝑠→1 𝜁(𝑠) 𝑠→1 𝑠 1 𝑥 = −(−1) = 𝑥 1

And hence we have now

𝜓(𝑥) = 𝑥 − log(2𝜋) − ∑ 𝑇

𝑥𝑇 𝑥𝜌 −∑ 𝑇 𝜌 𝜌

But remember the trivial zeros of zeta are −2, −4, −6, … or 𝑇 = −2𝑛 and therefore the last equation becomes: 𝜓(𝑥) = 𝑥 − log(2𝜋) − ∑𝑛

𝑥 −2𝑛 −2𝑛

− ∑𝜌

𝑥𝜌 𝜌

𝑥 −2𝑛

For the sum of the trivial zeros ∑𝑛 2𝑛 as follows:

(8)

an expression can be found

158

Naji Arwashan

Let us consider the Maclaurin series of log(1 + 𝑦) (from Chapter2, Section 2.5) log(1 + 𝑦) = 𝑦 −

𝑦2 𝑦3 𝑦4 𝑦5 + − + −⋯ 2 3 4 5

1

Making 𝑦 = − 𝑥 2 and multiplying both sides by ½ yields: 1 1 1 1 1 1 1 log (1 − 2 ) = (− 2 − 4 − 6 − 8 − ⋯ ) 2 𝑥 2 𝑥 2𝑥 3𝑥 4𝑥 −2 −4 −6 −8 𝑥 𝑥 𝑥 𝑥 = + + + +⋯ −2 −4 −6 −8 𝑥 −2𝑛 =∑ −2𝑛 𝑛

And now equation (8) becomes: 1

1

𝜓(𝑥) = 𝑥 − log(2𝜋) − 2 log (1 − 𝑥 2) − ∑𝜌

𝑥𝜌 𝜌

(9)

With an explicit formula for 𝜓(𝑥) is found we can now easily calculate 𝑑𝐽(𝑥) =

𝑑𝜓(𝑥) log 𝑥

and then integrate from 0 to 𝑥 to obtain

𝐽(𝑥) and that leads us to find the two important terms of Riemann’s formula for 𝐽(𝑥) [namely 𝑙𝑖(𝑥) - ∑𝜌 𝑙𝑖(𝑥 𝜌 ), see equation (11) of Chapter 6]. But the other two terms of the formula [ ∞

− log(2) + ∫𝑥

𝑑𝜏 𝜏(𝜏2 −1) log(𝜏)

] cannot be derived this route. Von

Mangoldt followed a different approach [6], far more complicated, and was able to obtain and rigorously prove the full Riemann’s formula for 𝐽(𝑥). As one can tell the new explicit formula for 𝜓(𝑥) is much simpler than Riemann original explicit formula for 𝐽(𝑥), and pretty much used its place in most modern textbooks.

Von Mangoldt Formula to the Rescue

159

7.4. THE WAVE COMPOSITION OF 𝝍(𝒙) We found that 𝜓(𝑥) can be expressed in an elegant explicit formula as function of the roots of zeta. But these roots are complex numbers, and to understand their contribution to the explicit formula, and to be able to calculate 𝜓(𝑥) numerically we need to write 𝜌 in terms of their real and imaginary parts as it is well presented and discussed in [13]. First let us remember that the zeros of zeta come in pair, because as we have seen previously (Section 5.7 of Chapter 5) the conjugate of every zero is a zero itself. So, if 𝜌 = 𝜎 + 𝑖𝑡 is a zero of zeta then 𝜌̅ = 𝜎 − 𝑖𝑡 is also a zero, and the contribution of the pair toward 𝜓(𝑥) can be evaluated as follows: 𝑥 𝜌 𝑥 𝜌̅ 𝑥 𝜎+𝑖𝑡 𝑥 𝜎−𝑖𝑡 − = − − 𝜌 𝜌̅ 𝜎 + 𝑖𝑡 𝜎 − 𝑖𝑡 ̅ 𝜌 𝜌 𝜎 [cos 𝑥 𝑥 𝑥 ( 𝑡 log 𝑥) + 𝑖 sin(𝑡 log 𝑥 )] − − = − 𝜌 𝜌̅ 𝜎 + 𝑖𝑡 𝜎 [cos 𝑥 (𝑡 log 𝑥) − 𝑖 sin(t log 𝑥)] − 𝜎 − 𝑖𝑡 −

In order to eliminate the imaginary constant from the denominator of the two terms of the right-hand side, we multiply and divide each term by the conjugate of its denominator

−

𝑥 𝜌 𝑥 𝜌̅ 𝑥 𝜎 [cos (𝑡 log 𝑥) + 𝑖 sin(𝑡 log 𝑥) ](𝜎 − 𝑖𝑡) − =− (𝜎 + 𝑖𝑡) (𝜎 − 𝑖𝑡) 𝜌 𝜌̅ 𝜎 [cos(𝑡 𝑥 log 𝑥 ) − 𝑖 sin(𝑡 log 𝑥 )](𝜎 + 𝑖𝑡) − (𝜎 − 𝑖𝑡) (𝜎 + 𝑖𝑡)

160

Naji Arwashan Adding the two terms on the right-hand side gives us 𝑥 𝜌 𝑥 𝜌̅ −2𝑥 𝜎 [𝜎 cos(𝑡 log 𝑥 ) + 𝑡 sin(𝑡 log 𝑥 )] − − = 𝜌 𝜌̅ 𝜎2 + 𝑡2 Factoring out

−

−2𝑥 𝜎 √𝜎 2 +𝑡 2

we get

𝑥 𝜌 𝑥 𝜌̅ −2𝑥 𝜎 𝜎 − = cos(𝑡 log 𝑥 ) [ 𝜌 𝜌̅ √𝜎 2 + 𝑡 2 √𝜎 2 + 𝑡 2 t + sin(𝑡 log 𝑥)] √𝜎 2 + 𝑡 2 𝑡

Defining 𝛼 = arctan 𝜎 𝑥 𝜌 𝑥 𝜌̅ −2𝑥 𝜎 [cos 𝛼 cos(𝑡 log 𝑥 ) + sin𝛼 sin(𝑡 log 𝑥)] − − = 𝜌 𝜌̅ √𝜎 2 + 𝑡 2 We can simplify the expression in brackets and get 𝑥 𝜌 𝑥 𝜌̅ −2𝑥 𝜎 − − = cos (𝑡 log 𝑥 − 𝛼) 𝜌 𝜌̅ √𝜎 2 + 𝑡 2 We notice that, and as expected, the imaginary part has disappeared because 𝜓(𝑥) , like 𝜋(𝑥), is a real number and no room for imaginary part in it. And if we account for all the pairs of zeros then the explicit formula becomes: 1 1 𝜓(𝑥) = 𝑥 − log (1 − 2 ) − log 2𝜋 2 𝑥 ∞ 𝜎𝑘 2𝑥 −∑ cos (𝑡𝑘 log 𝑥 − 𝛼𝑘 ) √𝜎𝑘 2 + 𝑡𝑘 2 𝑘=1

Von Mangoldt Formula to the Rescue

161

As we found in the case of 𝐽(𝑥), if we don’t account for the zeros of zeta, we obtain from the new explicit formula the straight line shown in Figure 7-2. The line follows nicely the slope of the staircase of 𝜓(𝑥) but it doesn’t recognize any of the jumps, it is only when we account for the zeros that we start, almost mysteriously, seeing the jumps forming at the locations of 𝑝𝑛 , and with more zeros accounted for the jumps become more pronounced as shown in Figures 7-3 and 7-4 where we account for 10 and 50 zeros respectively. It is almost as if the roots of zeta know the locations of the primes on the grid of the natural numbers. It is certainly one of the marvelous mysteries of mathematics.

Figure 7-2. The actual 𝝍(𝒙) function versus the one obtained from Von Mangoldt explicit formula (with no zeros of zeta).

Figure 7-3. The actual 𝝍(𝒙) function versus the one obtained from Von Mangoldt explicit formula (with 10 zeros of zeta).

162

Naji Arwashan

Figure 7-4. The actual 𝝍(𝒙) function versus the one obtained from Von Mangoldt explicit formula (with 50 zeros of zeta).

And if we account for a really large number of zeros like 100,000 then 𝜓(𝑥) becomes clearly a staircase function as shown below.

Figure 7-5. The 𝝍(𝒙) function as obtained from von Mangoldt explicit formula (with 100,000 zeros of zeta).

Von Mangoldt Formula to the Rescue And as one would expect if we calculate the

𝑑𝜓 𝑑𝑥

163 with the same

number of zeta’s zeros (100,000), will see spikes at 𝑝𝑛 and zero everywhere else as shown in Figure 7-6.

𝒅𝝍

Figure 7-6. The function as obtained from Von Mangoldt explicit formula (with 𝒅𝒙 100,000 zeros of zeta).

Finally, let us examine the characteristics of the cosine wave curves composing the 𝜓(𝑥) function. For the zeta zero 𝑘, we can write 𝑌𝑘 (𝑥) =

2𝑥 𝜎𝑘 √𝜎𝑘 2 + 𝑡𝑘 2

cos (𝑡𝑘 log 𝑥 − 𝛼𝑘 )

First we notice that the amplitude of the wave is not constant but equal to

2𝑥 𝜎𝑘 √𝜎𝑘 2 +𝑡𝑘 2

and therefore it is going to grow with 𝑥. Second since

the argument of the cosine is not 𝑥 but log 𝑥 then the period of the wave 𝑇 will not be constant, and it can be calculated as follows: cos (𝑡𝑘 log 𝑥 − 𝛼𝑘 ) = 𝑐𝑜𝑠 [𝑡𝑘 log(𝑥 + 𝑇) − 𝛼𝑘 ] => 𝑡𝑘 log 𝑥 − 𝛼𝑘 = [𝑡𝑘 log(𝑥 + 𝑇) − 𝛼𝑘 ] + 2𝜋

164

Naji Arwashan Simplifying gives log(𝑥 + 𝑇) − log 𝑥 =

2𝜋 𝑡𝑘

𝑇 2𝜋 log (1 + ) = 𝑥 𝑡𝑘 𝑇 = 𝑥(𝑒 2𝜋/𝑡𝑘 − 1) So, the period also will grow with 𝑥. On the other hand, we can easily see that both the amplitude and the period get smaller with the increase of 𝑡𝑘 (or with the increase ranking of the zero). To illustrate that we show in Figures 7-7 and 7-8 the wave curve associated with the 1st zero (𝜎1 = 0.5, 𝑡1 = 14.134) and the 10th one (𝜎10 = 0.5, 𝑡10 = 49.774) respectively

Figure 7-7. The plot of the function 𝑌1 (𝑥) =

𝑥 𝜌1 𝜌1

+

𝜌1 𝑥̅̅̅̅

̅̅̅̅ 𝜌1

=

2𝑥 0.5 √0.52 +14.134 2

cos (14.13 4log 𝑥 − 1.57)

Von Mangoldt Formula to the Rescue

165

Figure 7-8. The plot of the function 10 𝑥 𝜌10 𝑥 𝜌̅̅̅̅̅ 2𝑥 0.5 𝑌10 (𝑥) = + = cos (49.774 log 𝑡 − 1.56) 𝜌10 𝜌10 ̅̅̅̅ √0.52 + 49.7742

7.5. PROVING OF THE PRIME NUMBER THEOREM We said in the beginning of this chapter that Chebyshev found a new version of PNT equivalent to lim

𝑥→∞

𝜓(𝑥) 𝑥

= 1, and since now we have

an explicit formula for 𝜓(𝑥) (equation 8) then we can write 1

1

𝑥 − log(2𝜋) − 2 log (1 − 𝑥 2) − ∑𝜌

𝑥𝜌

𝜓(𝑥) 𝜌 = 𝑥 𝑥 𝜓(𝑥) log(2𝜋) 1 1 𝑥 𝜌−1 =1 − − log (1 − 2 ) − ∑ 𝑥 𝑥 2𝑥 𝑥 𝜌 𝜌

When 𝑥 → ∞, it is easy to see that the second and third terms go to zero and as a result to prove PNT we need to prove that ∑𝜌 lim 𝑥 𝜌−1 = 𝑥→∞

0. Let us replace 𝜌 = 𝜎 + 𝑖𝑡 𝑥 𝜌−1 = 𝑥 𝜎−1+𝑖𝑡 = 𝑥 𝜎−1 [cos 𝑡 log 𝑥 + 𝑖 sin 𝑡 log 𝑥 ]

166

Naji Arwashan

We know 0 ≤ 𝜎 ≤ 1, if all the nontrivial roots of zeta (𝜌) have 𝜎 < 1 then 𝑥 𝜎−1 will go to zero when 𝑥 goes to infinity, taking with it to zero the product 𝑥 𝜎−1 [cos 𝑡 log 𝑥 + 𝑖 sin 𝑡 log 𝑥 ] and it can be proved that in this case the infinite sum of zeros (since we are taking the sum over all the roots of zeta) will end to zero as well, and consequently PNT is true. However if one of the roots has 𝜎 = 1 then 𝑥 𝜎−1 becomes equal to 1, and the product 𝑥 𝜎−1 [cos 𝑡 log 𝑥 + 𝑖 sin 𝑡 log 𝑥 ] will not vanish when 𝑥 goes to ∞, and consequently PNT becomes false. And therefore, the proof of PNT becomes equivalent to proving the absence of any zeta’s root on the line of 𝜎 = 1. The proof came in year 1896 by independently two mathematicians Jacques Hadamard and Charles Jean de la Vallee-Poussin. And was a breakthrough accomplishment in the history of math. For several years after that, it was believed that relying on the zeta’s zeros and complex analysis is the only way to prove PNT, and that no elementary proof can be found (elementary in the sense that it doesn’t rely on complex analysis). Until 1948 when the Norwegian mathematician Atle Selberg proved by elementary means the following inequality 𝑥 |𝜗(𝑥) log(𝑥) + ∑ log(𝑝) 𝜗 ( ) − 2𝑥 log(𝑥)| ≤ 𝑀𝑥 𝑝 𝑝≤𝑥

Where 𝜗(𝑥) is the first Chebyshev function introduced in Section 7.1, and 𝑀 is some constant. Few months later, the prolific and famous mathematician Paul Erdos used the inequality above as starting point to come up with an elementary proof of PNT. In parallel Selberg used his newly found inequality to come up with his own elementary proof of PNT. The dispute, misunderstanding and miscommunication between the two mathematicians was an unfortunate episode in the history books of mathematics of the twentieth century.

Von Mangoldt Formula to the Rescue

167

With PNT proved, we know that the relative error between 𝑙𝑖(𝑥) and 𝜋(𝑥) goes to zero in infinity. But what do we know about the direction of that error? Will 𝑙𝑖(𝑥) continue larger than 𝜋(𝑥)? In 1914 John Littlewood showed that there will be infinite number of crossing between the two curves. And in 1933 his student Stanly Skewes showed, assuming Riemann Hypothesis, that the first crossing should 1034

take place before 1010 . Later, in 1955, Skews showed, without Riemann Hypothesis, that 𝜋(𝑥) must exceed 𝑙𝑖(𝑥) for some 𝑥 smaller 10964

than 1010 . By now this bound has been improved enormously. We know now that the two functions cross somewhere near 1.397 × 10316 , but we don’t know if this is the first crossing.

7.6. A BRIEF WORD ABOUT THE SIGNIFICANCE OF RIEMANN HYPOTHESIS It is very important here to clarify a common misunderstanding and to explain that Riemann explicit formula (and von Mangoldt’s one as well) predicts the distribution of primes regardless of the location of the zeros of zeta, or in other words regardless whether the R.H. is true or not. That is actually what Riemann himself alluded to in his paper. He calculated the first four zeros and found them with 𝜎 = 1⁄2 but couldn’t prove that all the zeros are like that, and wrote in his paper: “It is certainly preferable to find a mathematical proof for this statement: however, after a few unsuccessful attempts I have given up for now, because it did not seem necessary for the present work”. And the present work Riemann refers to is nothing else than his derivation of an explicit formula for 𝐽(𝑥). On the other hand, in 1901 the Swedish mathematician Helge von Koch [14] was able to prove that the R.H. is equivalent to 𝜋(𝑥) = 𝑙𝑖(𝑥) + 𝑂(√𝑥 log 𝑥)

168

Naji Arwashan The big O notation above means the absolute difference (error)

between 𝑙𝑖(𝑥) and 𝜋(𝑥) is always less than 𝐶 √𝑥 log 𝑥 (where C is some constant). And this is really significant because it means if the R.H. is true then 𝑙𝑖(𝑥) is not only asymptotic equivalent to 𝜋(𝑥) but also a very good estimate of it, and the gap between the two will stay below some constant C times √𝑥 log 𝑥. In addition, one should recognize that numerous theorems and PhD theses started and built on the assumption that the Riemann Hypothesis is true, and therefore proving it will lead to the proof of a significantly large amount of mathematical work and results at once. And what if the Riemann Hypothesis turned to be false? Well that would certainly be a tragic surprise! The Fields Medal recipient, Enrico Bombieri wrote in the Official Problem Description of the Riemann Hypothesis on the website of the Clay Institute “the failure of Riemann Hypothesis would create havoc in the distribution of prime numbers”. Finally, to highlight one last time the prominence of the Riemann Hypothesis in the world of mathematics will close by a famous quote from the celebrated German mathematician David Hilbert “If I were to awaken after having slept for a thousand years, my first question would be: Has the Riemann Hypothesis been proven?”

APPENDIX A: FOURIER TRANSFORM Fourier Transform is a powerful mathematical technique astoundingly used in different areas of mathematics. In particular it is heavily used in the derivation of the functional equation of the Jacobi theta function (see Appendix B) which is a key element in the derivation of the functional equation of zeta. We designate this appendix to introduce and explain informally and briefly the important and key aspects of Fourier Transform.

A.1. DEFINITION OF FOURIER TRANSFORM Fourier transform has important applications in mechanics and physics, but mathematically speaking, if we have a complex function 𝑠(𝑡), its Fourier transform 𝔉(𝑠(𝑡)) is defined as follows: ∞

𝔉(𝑠(𝑡)) = 𝑠̂ (𝜔) = ∫−∞ 𝑠(𝑡) 𝑒 −𝑖2𝜋𝜔𝑡 𝑑𝑡

(1)

𝑠(𝑡) can be retrieved from its Fourier transform by the inverse transform as follows:

170

Naji Arwashan First multiply both sides of (1) by 𝑒 𝑖2𝜋𝜔𝜏 ∞

𝑠̂ (𝜔)𝑒

𝑖2𝜋𝜔𝜏

=𝑒

𝑖2𝜋𝜔𝜏

∫ 𝑠(𝑡) 𝑒 −𝑖2𝜋𝜔𝑡 𝑑𝑡 −∞

Next, take the integral with respect to 𝜔 from −∞ to ∞ ∞

∞

∞

∫ 𝑠̂ (𝜔) 𝑒 𝑖2𝜋𝜔𝜏 𝑑𝜔 = ∫ [𝑒 𝑖2𝜋𝜔𝜏 ∫ 𝑠(𝑡) 𝑒 −𝑖2𝜋𝜔𝑡 𝑑𝑡] 𝑑𝜔 −∞

−∞ ∞

−∞ ∞

= ∫ ∫ 𝑠(𝑡) 𝑒 −𝑖2𝜋𝜔𝑡 𝑑𝑡 𝑒 𝑖2𝜋𝜔𝜏 𝑑𝜔 −∞ −∞

Combining the exponents on the right-hand side gives ∞

∞

∞

∫−∞ 𝑠̂ (𝜔) 𝑒 𝑖2𝜋𝜔𝜏 𝑑𝜔 = ∫−∞ ∫−∞ 𝑠(𝑡) 𝑒 𝑖2𝜋𝜔(𝜏−𝑡) 𝑑𝑡𝑑𝜔

(2)

Fourier Inversion Theorem states that ∞

∞

𝑠(𝜏) = ∫−∞ ∫−∞ 𝑠(𝑡) 𝑒 𝑖2𝜋𝜔(𝜏−𝑡) 𝑑𝑡𝑑𝜔

(3)

Thus, replacing (3) in (2) we can retrieve the function 𝑠 from its Fourier transform 𝑠̂ ∞

𝑠(𝜏) = ∫−∞ 𝑠̂ (𝜔) 𝑒 𝑖2𝜋𝜔𝜏 𝑑𝜔

(4)

Note that the real and imaginary parts of a complex function can be seen as the x and y coordinates of a point in the c-plane; and when the complex function has a single variable, and if we consider this variable as representing time then the complex function becomes the motion of that point in the c-plane. For example the zeta function is a complex function of two variables (𝜎, 𝑡), but if we fix 𝜎 to 0.5 for instance, and if we consider the variable 𝑡 to be representing time, then 𝜁(0.5 +

Appendix A

171

𝑖𝑡) becomes the trajectory of a point moving in the complex plane as shown below (the motion can be seen in the YouTube video: “The Riemann Hypothesis, Explained”, the Quanta magazine, January 2021, com/watch?v=zlm1aajH6gY ). Please note the same graph was presented in Chapter 5 (Figure 5https://www.youtube.com/watch?v=zlm1aajH6gY). Please note the 10) as the zeta mapping of the critical line. same graph was presented in Chapter 5 (Figure 5-10) as the zeta mapping of the critical line.

Figure A-1. The plot of the function 𝜻(𝟎. 𝟓 + 𝒊𝒕) seen as a motion of a point in the complex plane.

As it will become more evident in the next two sections Fourier transform 𝑠̂ (𝜔) really represents the contribution of different frequencies in the motion of a point moving in the complex plane and is often called the spectrum of 𝑠. To illustrate that idea let us consider a point that has a simple circular motion around the origin with a radius equal to 1, and a constant frequency of 𝜔0 , its motion can be described by the function

𝑠(𝑡) = cos(2𝜋𝜔0 𝑡) + 𝑖 sin(2𝜋𝜔0 𝑡) = 𝑒 𝑖2𝜋𝜔0𝑡 and if we take Fourier transform of the above function, we find it to be equal to zero everywhere except when 𝜔 = 𝜔0 it goes to infinity. Such

172

Naji Arwashan

function is called the Dirac delta function and is designated by 𝛿(𝜔 − 𝜔0 )

A.2. FOURIER SERIES If the trajectory of a point moving in the complex plane closes on itself to one loop, and keeps circulating in that loop over and over in a periodic motion of a period 𝑇 and frequency of 𝜔0 = 1/𝑇, then the French mathematician Jean-Baptiste Joseph Fourier proved that the motion of that point can be described by the circulation of an infinite number of circles rotating at the frequencies of 𝜔𝑘 = 𝑘𝜔0 , and having radii 𝐶𝑘 (called Fourier’s coefficient), and the infinite sum is called Fourier series:

Figure A-2. The periodic motion of a point in the x-y plane. ∞

𝑠(𝑡) = ∑ 𝐶𝑘 cos(2𝜋𝜔1 𝑡) + 𝑖𝐶𝑘 sin(2𝜋𝜔1 𝑡) 𝑘=−∞ ∞

𝑠(𝑡) = ∑ 𝐶𝑘 𝑒 𝑖2𝜋𝑘𝜔0𝑡 𝑘=−∞

Appendix A

173

To find the coefficient 𝐶𝑘 , we will select a given value for 𝑘 and let 1

us call it 𝑗, and we will multiply both sides by 𝑇 𝑒 −𝑖2𝜋𝑗𝜔0𝑡 ∞

1 −𝑖2𝜋𝑗𝜔 𝑡 1 −𝑖2𝜋𝑗𝜔 𝑡 0 𝑠(𝑡) = 0 𝑒 𝑒 ∑ 𝐶𝑘 𝑒 𝑖2𝜋𝑘𝜔0𝑡 𝑇 𝑇 𝑘=−∞

Then integrate both sides from 𝑡 = 0 to 𝑇 as follows: 𝑇

𝑇

𝑡=0

𝑡=0

∞

1 1 ∫ 𝑒 −𝑖2𝜋𝑗𝜔0 𝑡 𝑠(𝑡)𝑑𝑡 = ∫ [ 𝑒 −𝑖2𝜋𝑗𝜔0𝑡 ∑ 𝐶𝑘 𝑒 𝑖2𝜋𝑘𝜔0𝑡 ] 𝑑𝑡 𝑇 𝑇 𝑘=−∞

∞

=

𝑇

𝐶𝑘 ∑ ∫ 𝑒 −𝑖2𝜋𝑗𝜔0𝑡 𝑒 𝑖2𝜋𝑘𝜔0𝑡 𝑑𝑡 𝑇 𝑘=−∞ 𝑡=0

Combining the exponents on the right-hand side gives: 𝑇

𝑇

∞

1 𝐶𝑘 ∫ 𝑒 −𝑖2𝜋𝑗𝜔0𝑡 𝑠(𝑡)𝑑𝑡 = ∑ ∫ 𝑒 𝑖2𝜋(𝑘−𝑗)𝜔0𝑡 𝑑𝑡 𝑇 𝑇 𝑡=0

𝑘=−∞ 𝑡=0

Let us examine the integral on the right-hand side. The variable 𝑘 is an integer that varies from −∞ to ∞. When 𝑘 ≠ 𝑗 we have 𝑇

∫𝑡=0 𝑒 𝑖2𝜋(𝑘−𝑗)𝜔0𝑡 𝑑𝑡= 𝑇

𝑇

= ∫ cos 2𝜋(𝑘 − 𝑗)𝜔0 𝑡 𝑑𝑡 + 𝑖 ∫ sin 2𝜋(𝑘 − 𝑗)𝜔0 𝑡 𝑑𝑡 𝑡=0

𝑡=0

The period of the cosine in the last equation is

1 (𝑘−𝑗)𝜔0

𝑇

= (𝑘−𝑗) and

therefore we are integrating the cosine over a whole number of its period and thus the integral in this case is zero. The same can be said

174

Naji Arwashan

about the sine, and therefore the right-hand side is zero when 𝑘 ≠ 𝑗. When 𝑘 = 𝑗 the integral in the right-hand side becomes 𝑇

∫ 1 𝑑𝑡 = 𝑇 𝑡=0

And therefore, Fourier’s coefficients 𝐶𝑘 (radii of rotating circles) are 𝐶𝑘 =

1 𝑇 ∫ 𝑠(𝑡) 𝑒 −𝑖2𝜋𝑘𝜔0𝑡 𝑑𝑡 𝑇 𝑡=0

Note that the first and the slowest circle has the largest radius, and the coefficients (radii) become smaller and smaller with the higher frequencies, and often the first few circles provide a good practical approximation of the motion.

A.3. RELATIONSHIP BETWEEN FOURIER TRANSFORM AND SERIES Suppose we have a complex function 𝑓(𝑡) defined over the interval 𝑇 and equal to zero outside the interval, as the one shown in Figure A-3

Figure A-3. The real and imaginary parts of the function f(t).

Appendix A

175

𝑓(𝑡) is clearly not a periodic function but let us construct another function 𝑔(𝑡) to be periodic with period 𝑇, where 𝑔(𝑡) is identical to 𝑓(𝑡) between 0 and 𝑇 as shown below

Figure A-4. The real part of the function g(t).

Figure A-5. The imaginary part of the function g(t).

Since 𝑔(𝑡) is periodic function then we can write it in terms of Fourier series as follows 𝑖2𝜋𝑘𝜔0 𝑡 𝑔(𝑡) = ∑∞ 𝑘=−∞ 𝐶𝑘 𝑒

(5)

With the coefficient 𝐶𝑘 equal to 𝐶𝑘 =

1 𝑇 ∫ 𝑔(𝑡) 𝑒 −𝑖2𝜋𝑘𝜔0𝑡 𝑑𝑡 𝑇 𝑡=0

But as we said 𝑓(𝑡) and 𝑔(𝑡) are identical between 0 and 𝑇 then we can replace 𝑔(𝑡) by 𝑓(𝑡) in the last equation to obtain

176

Naji Arwashan 𝐶𝑘 =

1 𝑇 ∫ 𝑓(𝑡) 𝑒 −𝑖2𝜋𝑘𝜔0𝑡 𝑑𝑡 𝑇 𝑡=0

And since 𝑓(𝑡) is zero outside the interval 𝑇, then we can extend the integral to±∞, and will also replace the frequency 𝜔0 by its value of 1⁄𝑇 𝐶𝑘 =

𝑘 1 ∞ ∫ 𝑓(𝑡) 𝑒 −𝑖2𝜋𝑇𝑡 𝑑𝑡 𝑇 −∞

But note now that the integral is nothing else than Fourier transform 𝑘

of 𝑓(𝑡) evaluated at the value of 𝜔=𝑇 and therefore we can write 𝐶𝑘 =

1 𝑘 𝑓̂ ( ) 𝑇 𝑇

And consequently, replacing in (5) gives ∞

𝑘 1 𝑘 𝑔(𝑡) = ∑ [ 𝑓̂ ( )] 𝑒 𝑖2𝜋(𝑇)𝑡 𝑇 𝑇

𝑘=−∞

And if we limit 𝑡 only to the domain of [0, 𝑇], then we can replace 𝑔(𝑡) by 𝑓(𝑡) and write ∞

𝑓(𝑡) = ∑ 𝑘=−∞

1 𝑘 𝑖2𝜋(𝑘)𝑡 𝑇 𝑓̂ ( ) 𝑒 𝑇 𝑇

0≤𝑡≤𝑇

Now let us treat 𝑡 as constant, and consider the new variable to be, 𝜔, (the argument of 𝑓̂). In this case the sum is evaluated at values of 𝜔

Appendix A 1

2

3

177 1

equal to ± 𝑇 , ± 𝑇 , ± 𝑇 , … and notice that 𝑇 is ∆𝜔 and with this we can rewrite the last sum as ∞

𝑓(𝑡) = ∑ 𝑓̂(𝜔) 𝑒 𝑖2𝜋(𝜔)𝑡 ∆𝜔 𝑘=−∞

∶𝜔=

𝑘 𝑇

We recognize the last sum to be Riemann sum, and informally 1

speaking when 𝑇 goes to infinity, ∆𝜔 = 𝑇 goes to zero and the sum becomes the following integral: ∞

𝑓(𝑡) = ∫ 𝑓̂(𝜔)𝑒 𝑖2𝜋(𝜔)𝑡 𝑑𝜔 −∞

And this is the inverse Fourier Transform!

APPENDIX B: THE FUNCTIONAL EQUATION OF THE JACOBI THETA FUNCTION B.1. INTRODUCTION Jacobi Theta function (named after Carl Gustav Jacob Jacobi) is a function of the form: ∞

𝜗(𝑧, 𝜏) = ∑ 𝑒 (𝜋𝑖𝑛

2 𝜏+2𝜋𝑖𝑛𝑧)

𝑛=−∞

Where z and 𝜏 are both complex variables, and 𝐼𝑚(𝜏) > 0. When 𝑧 = 0 and 𝜏 = 𝑖𝑢 then the Jacobi theta function becomes −𝜋𝑛 𝜙(𝑢) = ∑∞ 𝑛=−∞ 𝑒

2𝑢

(1)

Which is the function we encountered in the derivation of the functional equation of the zeta function (Chapter 4, Section 4.4), and we made use of its own functional equation that takes the form of 𝜙(𝑢) =

1 √𝑢

𝜙(1/𝑢)

180

Naji Arwashan

The proof of the above functional equation depends on two formulae: the Poisson Summation formula, and the Fourier transform of 𝑓(𝑥) = 𝑒 −𝜋𝑥

2𝑎

which we will present in the next two sections.

B.2. POISSON SUMMATION FORMULA If 𝐹 is a smooth function (indefinitely differentiable) and vanishes at ∞, then Poisson Summation Formula is the equality between the following two series ∞ ̂ ∑∞ 𝑛=−∞ 𝐹(𝑛) = ∑𝑚=−∞ 𝐹 (𝑚)

(2)

Where 𝐹̂ is the Fourier transform of 𝐹. To prove (2) we start by defining the function 𝐺(𝑣) = ∑∞ 𝑛=−∞ 𝐹(𝑛 + 𝑣)

(3)

And 𝐺(𝑣 + 1) is equal to ∞

∞

𝐺(𝑣 + 1) = ∑ 𝐹(𝑛 + (𝑣 + 1)) = ∑ 𝐹((𝑛 + 1) + 𝑣) 𝑛=−∞

𝑛=−∞

And since n varies from -∞ to ∞ then adding 1 to it doesn’t make any difference and therefore 𝐺(𝑣 + 1) = 𝐺(𝑣), which means 𝐺(𝑣) is a periodic function with a period of 1. And since it is periodic, we can represent it in terms of Fourier series as follows: ∞

𝐺(𝑣) = ∑ 𝐶𝑚 𝑒 2𝜋𝑖𝑚𝑣 𝑚=−∞

Appendix B

181

Where 𝐶𝑚 is Fourier coefficient and is equal to 1

𝐶𝑚 = ∫ 𝐺(𝑣)𝑒 −2𝜋𝑖𝑚𝑣 𝑑𝑣 0

Replacing 𝐺(𝑣) by its value from (3) we get 1

∞

𝐶𝑚 = ∫ ∑ 𝐹(𝑛 + 𝑣) 𝑒 −2𝜋𝑖𝑚𝑣 𝑑𝑣 0 𝑛=−∞ ∞ 1

= ∑ ∫ 𝐹(𝑛 + 𝑣)𝑒 −2𝜋𝑖𝑚𝑣 𝑑𝑣 𝑛=−∞ 0

Let us make the following variable change: 𝑛 + 𝑣 = 𝑥 with 𝑑𝑣 = 𝑑𝑥; and for the limit of the integral, when 𝑣 = 0, 𝑥 is equal to 𝑛, and when 𝑣 = 1, 𝑥 is equal to 𝑛 + 1 ∞

𝑛+1

𝐶𝑚 = ∑ ∫

𝐹(𝑥)𝑒 −2𝜋𝑖𝑚(𝑥−𝑛) 𝑑𝑥

𝑛=−∞ 𝑛

Separating 𝑛 from 𝑥 in the exponent gives ∞

𝐶𝑚 = ∫ 𝐹(𝑥)𝑒 −2𝜋𝑖𝑚(𝑥) 𝑒 −2𝜋𝑖𝑚(−𝑛) 𝑑𝑥 −∞

But note that 𝑒 −2𝜋𝑖𝑚(−𝑛) = 𝑒 2𝜋𝑖𝑚𝑛 = cos(2𝜋𝑚𝑛) + 𝑖sin(2𝜋𝑚𝑛) = 1

182

Naji Arwashan And therefore ∞

𝐶𝑚 = ∫ 𝐹(𝑥)𝑒 −2𝜋𝑖𝑚𝑥 𝑑𝑥 −∞

The last expression is by definition Fourier transform of 𝐹, and therefore 𝐶𝑚 = 𝐹̂ (𝑚), and as a result ∞

∞

𝐺(𝑣) = ∑ 𝐹(𝑛 + 𝑣) = ∑ 𝐹̂ (𝑚)𝑒 2𝜋𝑖𝑚𝑣 𝑛=−∞

𝑚=−∞

The last equation is true for any 𝑣 including 𝑣 = 0, and hence we have Poisson Summation ∞

∞

∑ 𝐹(𝑛) = ∑ 𝐹̂ (𝑚) 𝑛=−∞

𝑚=−∞

B.3. FOURIER TRANSFORM OF 𝒇(𝒙) = 𝒆−𝝅𝒙

𝟐𝒂

In this section we will prove that 𝔉(𝑒

−𝜋𝑥 2 𝑎

)=

1 √𝑎

𝑒

−𝜋

𝑢2 𝑎

To prove this transform we will start first by proving that the 2

Fourier transform of 𝑔(𝑥) = 𝑒 −𝜋𝑥 is equal to itself, or in other words, 2

𝑔̂(𝑢) = 𝑒 −𝜋𝑢 . So, we need to show that 2

∞

2

𝔉(𝑒 −𝜋𝑥 ) = ∫ 𝑒 −𝜋𝑥 𝑒 −2𝜋𝑖𝑥𝑢 𝑑𝑥 = 𝑒 −𝜋𝑢 −∞

2

Appendix B

183

2

Dividing both sides by 𝑒 −𝜋𝑢 gives 2

∞

2

𝑒 𝜋𝑢 ∫ 𝑒 −𝜋𝑥 𝑒 −2𝜋𝑖𝑥𝑢 𝑑𝑥 = 1 −∞ 2

𝑒 𝜋𝑢 is independent of 𝑥 and therefore we can bring it as a constant inside the integral ∞

2

2

∫ 𝑒 −𝜋𝑥 𝑒 −2𝜋𝑖𝑥𝑢 𝑒 𝜋𝑢 𝑑𝑥 = 1 −∞

Combining all the exponents of 𝑒 gives ∞

∫ 𝑒 −𝜋(𝑥

2 +2𝜋𝑖𝑥𝑢+𝑖𝑢2 )

𝑑𝑥 = 1

−∞

Realizing that the exponent in parenthesis is a perfect square gives ∞

2

∫ 𝑒 −𝜋(𝑥+𝑖𝑢) 𝑑𝑥 = 1 −∞

To prove the last equality, First let us consider the special case +∞

2

when u=0, and let us call 𝐼 = ∫−∞ 𝑒 −𝜋𝑥 𝑑𝑥 (note that I is greater than 0) +∞

𝐼2 = ∫ 𝑒

+∞ −𝜋𝑥 2

−∞ +∞ +∞

2

𝑑𝑥 ∫ 𝑒 −𝜋𝑦 𝑑𝑦 −∞ 2

2

𝐼2 = ∫ ∫ 𝑒 −𝜋𝑥 𝑒 −𝜋𝑦 𝑑𝑥 𝑑𝑦 −∞ −∞

184

Naji Arwashan +∞ +∞

𝐼2 = ∫ ∫ 𝑒 −𝜋(𝑥

2 +𝑦 2 )

𝑑𝑥 𝑑𝑦

−∞ −∞

Using polar coordinates: 𝑥 = 𝑟cos𝜃 , 𝑦 = 𝑟sin𝜃

𝑥2 + 𝑦2 = 𝑟2 And 𝜕(𝑥, 𝑦) | 𝑑𝜃𝑑𝑟 𝜕(𝜃, 𝑟) 𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦 =| − | 𝑑𝜃𝑑𝑟 𝜕𝜃 𝜕𝑟 𝜕𝑟 𝜕𝜃 = |−𝑟𝑠𝑖𝑛𝜃𝑠𝑖𝑛𝜃 − 𝑐𝑜𝑠𝜃𝑟𝑐𝑜𝑠𝜃|𝑑𝜃𝑑𝑟 = 𝑟𝑑𝜃𝑑𝑟 where 𝑟 > 0

𝑑𝑥𝑑𝑦 = |

And when 𝑥 and 𝑦 go from -∞ to ∞, 𝑟 and 𝜃 go from 0 to ∞ and from 0 to 2𝜋 respectively 2𝜋 +∞ 2

𝐼 =∫ ∫ 𝑒 0

2𝜋 −𝜋𝑟 2

+∞ 2

𝑟𝑑𝜃𝑑𝑟 = ∫ 𝑑𝜃 ∫ 𝑒 −𝜋𝑟 𝑟𝑑𝑟

0

0

0

1 −𝜋𝑟2 ∞ 1 = [𝜃]2𝜋 𝑒 [ ] = [2𝜋 − 0] [0 − ]=1 0 −2𝜋 −2𝜋 0 Since 𝐼 > 0, we conclude that 𝐼 = 1 For the general case when 𝑢 can take any value (not only zero). Let us take the derivative with respect to 𝑢 for the integral ∞

2

∫−∞ 𝑒 −𝜋(𝑥+𝑖𝑢) 𝑑𝑥 𝜕 ∞ −𝜋(𝑥+𝑖𝑢)2 ∫ 𝑒 𝜕𝑢 −∞

∞

𝜕

2

𝑑𝑥 = ∫−∞ (𝜕𝑢 𝑒 −𝜋(𝑥+𝑖𝑢) ) 𝑑𝑥

(4)

Appendix B 𝜕

Note that

𝜕𝑢 𝜕

And that

𝜕𝑥

Therefore

185

2

𝑒 −𝜋(𝑥+𝑖𝑢) = −𝑖2𝜋(𝑥 + 𝑖𝑢)𝑒 −𝜋(𝑥+𝑖𝑢) 2

𝑒 −𝜋(𝑥+𝑖𝑢) = −2𝜋(𝑥 + 𝑖𝑢)𝑒 −𝜋(𝑥+𝑖𝑢)

𝜕

2

𝜕

𝑒 −𝜋(𝑥+𝑖𝑢) = 𝑖 𝜕𝑥 𝑒 −𝜋(𝑥+𝑖𝑢) 𝜕𝑢

2

2

2

Replacing in (4) gives ∞ 𝜕 ∞ −𝜋(𝑥+𝑖𝑢)2 𝜕 −𝜋(𝑥+𝑖𝑢)2 ∫ 𝑒 𝑑𝑥 = 𝑖 ∫ 𝑒 𝑑𝑥 𝜕𝑢 −∞ −∞ 𝜕𝑥 2

∞

= 𝑖[𝑒 −𝜋(𝑥+𝑖𝑢) ]−∞ = 𝑖[0 − 0] = 0 ∞

2

So if the derivative of ∫−∞ 𝑒 −𝜋(𝑥+𝑖𝑢) 𝑑𝑥 with respect to 𝑢 is zero ∞

2

that means ∫−∞ 𝑒 −𝜋(𝑥+𝑖𝑢) 𝑑𝑥 is independent of 𝑢, and its value is the same for any value of 𝑢. And we already found its value when 𝑢 is zero to be 1 and therefore ∞

2

∫ 𝑒 −𝜋(𝑥+𝑖𝑢) 𝑑𝑥 = 1 −∞

With this we established that

2

𝔉(𝑒 −𝜋𝑥 ) = 𝑒 −𝜋𝑢

2

And now we can go back to our original aim of finding 𝔉(𝑒 which can be written as 2

2

𝔉(𝑒 −𝜋𝑥 𝑎 ) = 𝔉 (𝑒 −𝜋(𝑥√𝑎) )

(5) −𝜋𝑥 2 𝑎

),

186

Naji Arwashan Expanding Fourier transform on the right-hand side gives ∞

2

2

𝔉(𝑒 −𝜋𝑥 𝑎 ) = ∫ 𝑒 −𝜋(𝑥√𝑎) 𝑒 −2𝜋𝑖𝑥𝑢 𝑑𝑥 −∞

We make the variable change of 𝑣 = 𝑥 √𝑎 and 𝑑𝑥 = 𝑑𝑣⁄√𝑎 ∞

2

2

𝔉(𝑒 −𝜋𝑥 𝑎 ) = ∫ 𝑒 −𝜋𝑣 𝑒

−2𝜋𝑖

𝑣 𝑢 √𝑎

𝑑𝑣 √𝑎

−∞

Rearranging we get 1

2

𝔉(𝑒 −𝜋𝑥 𝑎 ) =

Calling t=

√𝑎

∞

2

∫ 𝑒 −𝜋𝑣 𝑒

−2𝜋𝑖𝑣

𝑢 √𝑎

𝑑𝑣

−∞

𝑢

√𝑎

2

𝔉(𝑒 −𝜋𝑥 𝑎 ) =

1 √𝑎

∞

2

∫ 𝑒 −𝜋𝑣 𝑒 −2𝜋𝑖𝑣𝑡 𝑑𝑣 −∞ 2

We recognize that the integral is Fourier transform of (𝑒 −𝜋𝑣 ), therefore 2

𝔉(𝑒 −𝜋𝑥 𝑎 ) =

1

2

√𝑎

𝔉(𝑒 −𝜋𝑣 )

2

2

From (5) we know that 𝔉(𝑒 −𝜋𝑣 ) = 𝑒 −𝜋𝑡 and replacing it in the last equation gives 2

𝔉(𝑒 −𝜋𝑥 𝑎 ) =

1 √𝑎

𝑒 −𝜋𝑡

2

Appendix B

187

Switching back to the 𝑢 variable where we defined earlier 𝑡 =

𝔉(𝑒

−𝜋𝑥 2 𝑎

)=

1 √𝑎

𝑒

−𝜋(

𝑢 √𝑎

𝑢 2 ) √𝑎

And with that we achieved our goal for this section 2

𝔉(𝑒 −𝜋𝑥 𝑎 ) =

1

𝑒 −𝜋 𝑎

𝑢2 𝑎

(6)

√

B.4. THE FUNCTIONAL EQUATION OF THE JACOBI THETA FUNCTION Now we are ready to derive the functional equation of 𝜙(𝑎) where 𝜙(𝑎) is defined in (1) as ∞

𝜙(𝑎) = ∑ 𝑒 −𝜋𝑛

2𝑎

𝑛=−∞

Applying Poisson Summation Formula: ∞

𝜙(𝑎) = ∑ 𝑒

∞ −𝜋𝑛2 𝑎

2

= ∑ 𝔉( 𝑒 −𝜋𝑛 𝑎 )

𝑛=−∞

𝑛=−∞

2

Replacing 𝔉( 𝑒 −𝜋𝑛 𝑎 ) from (6) gives ∞

∑ 𝑒 𝑛=−∞

∞ −𝜋𝑛2 𝑎

= ∑ 𝑚=−∞

1 √𝑎

𝑒

−𝜋

𝑚2 𝑎

188

Naji Arwashan

=

1

∞

∑ 𝑒 −𝜋 𝑚

√𝑎 𝑚=−∞

And hence the functional equation of 𝜙(𝑎) 𝜙(𝑎) =

1 √𝑎

𝜙(√𝑎)

21 𝑎

APPENDIX C: THE FUNCTIONAL EQUATIONS OF THE GAMMA FUNCTION (LEGENDRE’S DUPLICATION AND EULER’S REFLECTION) The proof of the equivalent forms of the Zeta functional equation presented in Chapter 4 (Section 4.7) depended on couple of functional equations of the Gamma function namely: Legendre Duplication Formula and Euler Reflection Formula. We will show and prove both of them in this appendix.

C.1. THE BETA FUNCTION AND THE PROOF OF LEGENDRE DUPLICATION FORMULA Legendre Duplication formula is the following

Γ(2𝑠) =

22𝑠−1

1 Γ(𝑠)Γ (𝑠 + ) 2 √𝜋

190

Naji Arwashan

To prove it we need to introduce first the beta function 𝐵(𝑥, 𝑦) as below 1

𝐵(𝑥, 𝑦) = ∫0 𝑡 𝑥−1 (1 − 𝑡)𝑦−1 𝑑𝑡

(1)

Where 𝑥 and 𝑦 are both complex variable with 𝑅𝑒 (𝑥) > 0 and 𝑅𝑒 (𝑦) > 0 The beta function has the following relation with the gamma function

𝐵(𝑥, 𝑦) =

Γ(𝑥)Γ(𝑦)

(2)

Γ(𝑥+𝑦)

And the proof goes as follows: We saw in Chapter 4 that the Gamma function is defined as ∞

Γ(𝑥) = ∫ 𝑢 𝑥−1 𝑒 −𝑢 𝑑𝑢 0

So, we can write ∞

∞

Γ(𝑥)Γ(𝑦) = ∫ 𝑢 𝑥−1 𝑒 −𝑢 𝑑𝑢 ∫ 𝑣 𝑦−1 𝑒 −𝑣 𝑑𝑣 𝑢=0 ∞

𝑣=0 ∞

= ∫ ∫ 𝑒 −𝑢−𝑣 𝑢 𝑥−1 𝑣 𝑦−1 𝑑𝑢𝑑𝑣 𝑢=0 𝑣=0

We change the variables 𝑢 and 𝑣 as follows: 𝑢 = 𝑧𝑡 and 𝑣 = 𝑧 − 𝑧𝑡 = 𝑧(1 − 𝑡). When 𝑡 = 0, obviously 𝑢 is 0, and when 𝑡 goes to 1 and 𝑧 goes to ∞, 𝑢 goes to ∞. Now let us examine 𝑣, when 𝑡 = 1, 𝑣 = 0 and when 𝑡 goes to 0, and 𝑧 goes to ∞, 𝑣 goes to ∞ . Therefore, with the

Appendix C

191

variables 𝑡 and 𝑧, the integral will be over the new domain of 0 < 𝑡 < 1 and 0 < 𝑧 < ∞ as shown below.

Figure C-1. The domain of the integral of the function Γ(𝑥)Γ(𝑦) in the u-v plane (left), and in the t-z plane (right).

Replacing 𝑢 and 𝑣 by the new variables 𝑡 and 𝑧 gives ∞

1

Γ(𝑥)Γ(𝑦) = ∫ ∫ 𝑒 −𝑧𝑡−(𝑧−𝑧𝑡) (𝑧𝑡)𝑥−1 (𝑧 − 𝑧𝑡)𝑦−1 |𝐽(𝑡, 𝑧)|𝑑𝑡𝑑𝑧 𝑧=0 𝑡=0

Where 𝜕(𝑢, 𝑣) 𝜕𝑢 𝜕𝑣 𝜕𝑢 𝜕𝑣 − |=| | 𝜕(𝑡, 𝑧) 𝜕𝑡 𝜕𝑧 𝜕𝑧 𝜕𝑡 = |𝑧(1 − 𝑡) − 𝑡(𝑧)| = |𝑧| = 𝑧 where 𝑧 > 0

|𝐽(𝑡, 𝑧)| = |

Substituting and simplifying gives ∞

1

Γ(𝑥)Γ(𝑦) = ∫ ∫ 𝑒 −𝑧 𝑧 𝑥+𝑦−1 𝑡 𝑥−1 (1 − 𝑡)𝑦−1 𝑑𝑡𝑑𝑧 𝑧=0 𝑡=0

192

Naji Arwashan Separating the two integrals gives ∞

1 −𝑧 𝑥+𝑦−1

Γ(𝑥)Γ(𝑦) = ∫ 𝑒 𝑧 𝑧=0

𝑑𝑧 ∫ 𝑡 𝑥−1 (1 − 𝑡)𝑦−1 𝑑𝑡 𝑡=0

The first integral is the definition of Γ(𝑥 + 𝑦) and the second one is the definition of 𝐵(𝑥 + 𝑦) and therefore we have equation (2) Γ(𝑥)Γ(𝑦) = Γ(𝑥 + 𝑦)𝐵(𝑥 + 𝑦) Now let us go back to the definition of beta function and consider the special case when 𝑥 = 𝑦 = 𝑠 1

𝐵(𝑠, 𝑠) = ∫ 𝑡 𝑠−1 (1 − 𝑡)𝑠−1 𝑑𝑡 0

Let us make the variable change of 𝑡 = (1 + 𝑝)⁄2 and 𝑑𝑡 = 𝑑𝑝⁄2 ; when 𝑡 is 0 𝑝 is − 1, and when 𝑡 is 1 𝑝 is 1; and now 𝐵(𝑠, 𝑠) becomes equal to 1 + 𝑝 𝑠−1 1 + 𝑝 𝑠−1 𝑑𝑝 𝐵(𝑠, 𝑠) = ∫ ( ) (1 − ) 2 2 2 −1 1

Simplifying gives 𝐵(𝑠, 𝑠) =

1 1 1 ∫ (1 + 𝑝)𝑠−1 (1 − 𝑝)𝑠−1 𝑑𝑝 2𝑠−1 2𝑠−1 2 −1 1

Clearly (1 + 𝑝)(1 − 𝑝) = (1 − 𝑝2 ) and that gives 1 1 𝐵(𝑠, 𝑠) = 22−2𝑠 ∫ (1 − 𝑝2 )𝑠−1 𝑑𝑝 2 −1

Appendix C

193

But (1 − 𝑝2 ) = (1 − (−𝑝)2 ), and therefore the integral from -1 to 1 is equal to 2 times the integral from 0 to 1. 1 1 𝐵(𝑠, 𝑠) = 22−2𝑠 [2 ∫ (1 − 𝑝2 )𝑠−1 𝑑𝑝] 2 0

Simplifying gives 1

𝐵(𝑠, 𝑠) = 22−2𝑠 ∫ (1 − 𝑝2 )𝑠−1 𝑑𝑝 0 1

1

Changing the variable again to 𝑝 = √𝑞, 𝑑𝑝 = 2 𝑞 −2 𝑑𝑞 and the limits of the integral stay the same because when 𝑝 is 0, 𝑞 is 0; and when 𝑝 equal 1, 𝑞 is equal 1, and the integral becomes 1 1 1 𝐵(𝑠, 𝑠) = 22−2𝑠 ∫ (1 − 𝑞)𝑠−1 ( 𝑞 −2 𝑑𝑞) 2 0

Simplifying gives 1

1

𝐵(𝑠, 𝑠) = 21−2𝑠 ∫ 𝑞 −2 (1 − 𝑞)𝑠−1 𝑑𝑞 0 1

But the integral now is nothing else than the definition of 𝐵 (2 , 𝑠) and therefore we have 1 𝐵(𝑠, 𝑠) = 21−2𝑠 𝐵 ( , 𝑠) 2 Now replacing beta function by its value in terms of the gamma function from (2) we get

194

Naji Arwashan 1

Γ ( ) Γ(𝑠) Γ(𝑠)Γ(𝑠) = 21−2𝑠 21 Γ(𝑠 + 𝑠) Γ ( + 𝑠) 2

Cancelling Γ(𝑠) from both sides, and isolating Γ(2𝑠) gives: Γ(2𝑠) =

22𝑠−1 1 2

Γ( )

1

Γ(𝑠)Γ (𝑠 + 2)

(3) 1

There are many ways to calculate Γ ( ) , the easiest one is to use 2 Euler Reflection Formula (next section). [We should note here that the proof we present next for the Euler Reflection Formula depends on the above equation (3) but doesn’t need the final format of Legendre Duplication Formula, therefore it is ok at this point to use the reflection 1

formula to calculate 𝛤 (2) ]. Euler Reflection formula states that Γ(𝑠)Γ (1 − 𝑠) =

Replacing s by

𝜋 sin (𝜋𝑠)

1 2

1 1 𝜋 Γ ( ) Γ (1 − ) = 𝜋 2 2 sin ( ) 2

Which yields: 1 Γ ( ) = √𝜋 2

Appendix C

195

1

And finally replacing Γ ( ) by its numerical value in (3) gives us 2 Legendre Duplication Formula

Γ(2𝑠) =

22𝑠−1

1 Γ(𝑠)Γ (𝑠 + ) 2 √𝜋

C.2. PROOF OF THE EULER REFLECTION FORMULA Euler Reflection Formula is: Γ(𝑠)Γ (1 − 𝑠) =

𝜋 sin (𝜋𝑠)

There are multiple ways of proving this formula, will present here the proof from the classic book “the Gamma Function” by E. Artin [1]. We start by defining the function: 𝜑(𝑠) = Γ(𝑠)Γ (1 − 𝑠) sin(𝜋𝑠)

(4)

And our goal is to prove that this function is constant and equal to 𝜋. First let us calculate 𝜑(𝑠 + 1) 𝜑(𝑠 + 1) = Γ(𝑠 + 1)Γ (1 − (𝑠 + 1)) sin (𝜋(𝑠 + 1)) Simplifying gives 𝜑(𝑠 + 1) = Γ(𝑠 + 1)Γ (−𝑠)[−sin (𝜋𝑠)]

(5)

Recall the functional equation of gamma from Chapter 4, Section 4.1

196

Naji Arwashan Γ (𝑠 + 1) = 𝑠Γ (𝑠)

(6)

(6) is valid for any value of s including -s Γ (−𝑠 + 1) = (−𝑠)Γ (−𝑠) Isolating Γ (−𝑠) we get Γ (−𝑠) = −

Γ (1−𝑠)

(7)

𝑠

Replacing Γ (𝑠 + 1) and Γ (−𝑠) from (6) and (7) into (5) gives 𝜑(𝑠 + 1) = [𝑠Γ (𝑠)] [−

Γ (1 − 𝑠) ] [−sin (𝜋𝑠)] 𝑠

Simplifying gives 𝜑(𝑠 + 1) = Γ (𝑠) Γ (1 − 𝑠) sin(𝜋𝑠) And the right-hand side is nothing else than 𝜑(𝑠) and therefore, 𝜑(𝑠 + 1) = 𝜑(𝑠)

(8)

The last equation means 𝜑(𝑠) is a periodic function with a period of 1. For the next step we will use equation (3) from the proof of Legendre Duplication formula (previous Section). The equation was

Γ(2𝑠) =

22𝑠−1

1 Γ(𝑠)Γ (𝑠 + ) 2 Γ (2) 1

Appendix C 1

197 1

Γ ( ) is a constant; and if we call 𝑏 = 2Γ ( ) and replace 2 2 accordingly in the previous equation we get 1 𝑏2−2𝑠 Γ(2𝑠)𝑏 = Γ(𝑠)Γ (𝑠 + ) 2 Rewriting the last equation with 𝑠⁄2 as variable gives 𝑠

𝑠+1

𝑏2−𝑠 Γ(𝑠) = Γ ( ) Γ ( 2

2

)

(9)

And rewriting the last equation with (1 − s ) as variable gives 1−𝑠

𝑏2𝑠−1 Γ(1 − s ) = Γ (

2

𝑠

) Γ (1 − 2)

(10)

Now we can go back to the function 𝜑(𝑠), and calculate 𝑠

𝑠+1

𝜑 (2) 𝜑 (

2

) as follows:

𝑠 𝑠+1 𝜑( )𝜑( )= 2 2 𝑠 𝑠 𝑠 𝑠+1 𝑠+1 𝑠+1 [Γ ( ) Γ (1 − ) sin (𝜋 )] [Γ ( ) Γ (1 − ) sin (𝜋 )] 2 2 2 2 2 2

Simplifying gives 𝑠 𝑠+1 𝜑( )𝜑( )= 2 2 𝑠 𝑠 𝜋𝑠 𝑠+1 1−𝑠 𝜋𝑠 Γ ( ) Γ (1 − ) sin ( ) Γ ( )Γ ( ) cos ( ) 2 2 2 2 2 2

198

Naji Arwashan 𝑠

𝑠+1

Γ( )Γ( 2 replaced 𝜋𝑠

2

1−𝑠

) can be replaced from (9), and Γ (

from

(10);

and

we

know

from

2

𝑠

) Γ (1 − 2) can be

trigonometry

𝜋𝑠

sin ( 2 ) cos ( 2 ) is equal to 1/2 sin(𝜋𝑠). Replacing we obtain 𝑠 𝑠+1 1 𝜑( )𝜑( ) = [𝑏2−𝑠 Γ(𝑠)][𝑏2𝑠−1 Γ(1 − s )] sin (𝜋𝑠) 2 2 2 Simplifying gives: 𝑠 𝑠+1 𝑏2 𝜑( )𝜑( )= Γ(𝑠) Γ(1 − s )sin (𝜋𝑠) 2 2 4 Let us rename the constant

𝑏2 4

to 𝑐

𝑠 𝑠+1 𝜑( )𝜑( ) = 𝑐 𝜑(𝑠) 2 2 And let us take the natural logarithm of both sides 𝑠 𝑠+1 log 𝜑 ( ) + log 𝜑 ( ) = log 𝑐 + log 𝜑(𝑠) 2 2 Let us call 𝑢 = 𝑠⁄2 and 𝑣 = (𝑠 + 1)⁄2 log 𝜑(𝑢) + log 𝜑(𝑣) = log 𝑐 + log 𝜑(𝑠) Let us differentiate both side with respect to s: 𝑑[log 𝜑(𝑢)] 𝑑𝑢 𝑑[log 𝜑(𝑣)] 𝑑𝑣 𝑑[log 𝜑(𝑠)] + = 0+ 𝑑𝑢 𝑑𝑠 𝑑𝑢 𝑑𝑠 𝑑𝑠

that

Appendix C But from the definition of 𝑢 and 𝑣 we have

199 𝑑𝑢 𝑑𝑠

1

= 2 and

𝑑𝑣 𝑑𝑠

1

=2

replacing gives 1 𝑑[log 𝜑(𝑢)] 𝑑[log 𝜑(𝑣)] 𝑑[log 𝜑(𝑠)] + { }= 2 𝑑𝑢 𝑑𝑣 𝑑𝑠 Let us differentiate one more time with respect to 𝑠: 1 𝑑 2 [log 𝜑(𝑢)] 𝑑𝑢 𝑑 2 [log 𝜑(𝑣)] 𝑑𝑣 𝑑 2 [log 𝜑(𝑠)] + = { } 2 𝑑𝑢2 𝑑𝑠 𝑑𝑣 2 𝑑𝑠 𝑑𝑠 2 𝑑𝑢

Again replacing 𝑑𝑠 =

𝑑𝑣 𝑑𝑠

1

= 2 gives

1 𝑑 2 [log 𝜑(𝑢)] 𝑑 2 [log 𝜑(𝑣)] 𝑑 2 [log 𝜑(𝑠)] + = { } 4 𝑑𝑢2 𝑑𝑣 2 𝑑𝑠 2 If we call 𝑔 =

𝑑 2 [log 𝜑(𝑠)] 𝑑𝑠2

then we get:

1 𝑠 𝑠+1 )] = 𝑔(𝑠) [𝑔 ( ) + 𝑔 ( 4 2 2 Since we found earlier in equation 8 that 𝜑(𝑠) is periodic, then log 𝜑(𝑠) and 𝑔(𝑠) are periodic also. And if 𝑔(𝑠) is periodic then it is bounded by a value 𝑀, and |𝑔(𝑠)| ≤ 𝑀 The last equation can be written as follows: 1 𝑠 1 𝑠+1 𝑔(𝑠) = [𝑔 ( )] + [𝑔 ( )] 4 2 4 2 1 𝑠 1 𝑠+1 |𝑔(𝑠)| ≤ |𝑔 ( )| + |𝑔 ( )| 4 2 4 2

200

Naji Arwashan 1 1 𝑀+ 𝑀 4 4 1 |𝑔(𝑠)| ≤ 𝑀 2

|𝑔(𝑠)| ≤

We can repeat using the last equation infinite times, every time cutting the bound M by a half, leading in the end to 𝑔(𝑠) being equal to zero. But 𝑔(𝑠) is the second derivative of log 𝜑(𝑠) and since it is null then log 𝜑(𝑠) is linear. But we know also that log 𝜑(𝑠) is periodic, therefore it must be constant. And if log 𝜑(𝑠) is constant then 𝜑(𝑠) is constant as well. Now it is left for us to find the constant value of 𝜑(𝑠). We started by defining 𝜑(𝑠) as: 𝜑(𝑠) = Γ(𝑠)Γ (1 − 𝑠) sin (𝜋𝑠) From the functional equation of the gamma function, we can write Γ(𝑠) =

Γ(𝑠 + 1) 𝑠

From the Maclaurin series of sine function from Chapter 2 (Section 2.5) we can write

sin(𝜋𝑠) = 𝜋𝑠 −

(𝜋𝑠)3 (𝜋𝑠)5 + −⋯ 3! 5!

Replacing Γ(𝑠) and sin(𝜋𝑠) in 𝜑(𝑠) we obtain:

𝜑(𝑠) =

(𝜋𝑠)3 (𝜋𝑠)5 Γ(𝑠 + 1) Γ (1 − 𝑠) (𝜋𝑠 − + − ⋯) 𝑠 3! 5!

= Γ(𝑠 + 1)Γ (1 − 𝑠) (𝜋 −

𝜋3𝑠2 𝜋5𝑠4 + −⋯) 3! 3!

Appendix C And when s = 0 𝜑(0) = Γ(1)Γ (1) (𝜋 − 0 + 0 − ⋯ ) = 𝜋 And therefore, the constant value of 𝜑(𝑠) is 𝜋 𝜑(𝑠) = Γ(𝑠)Γ (1 − 𝑠) sin(𝜋𝑠) = 𝜋 And finally, Euler Reflection Formula is Γ(𝑠)Γ (1 − 𝑠) =

𝜋 sin(𝜋𝑠)

201

APPENDIX D: CALCULATION OF THE 𝒂+𝒊∞ 𝒚𝒔 INTEGRAL 𝟐𝝅𝒊 ∫𝒂−𝒊∞ 𝒔 𝟏

𝒅𝒔 (𝒚 > 𝟎, 𝒂 > 𝟎)

The value of the above integral plays an important role in the derivation of von Mangoldt formula (Chapter 7) and therefore we decided in this appendix to show its calculation as it is outlined in [6]. There are three cases for this integral with three different outcomes: 0 < 𝑦 < 1, 𝑦 = 1, and 𝑦 > 1

The Case when 𝟎 < 𝒚 < 𝟏 Let us evaluate the integral over the closed contour shown in Figure D-1. The integrand has one pole when 𝑠= 0, and since the pole is outside the domain of interest then the integral according to equation (6) from Chapter 2, Section 2.10 is equal to 0 𝑎+𝑖ℎ

𝐾+𝑖ℎ

𝐾−𝑖ℎ

𝑎−𝑖ℎ

𝑦𝑠 𝑦𝑠 𝑦𝑠 𝑦𝑠 ∫ 𝑑𝑠 + ∫ 𝑑𝑠 + ∫ 𝑑𝑠 + ∫ 𝑑𝑠 = 0 𝑠 𝑠 𝑠 𝑠

𝑎−𝑖ℎ

𝑎+𝑖ℎ

𝐾+𝑖ℎ

𝐾−𝑖ℎ

204

Naji Arwashan

Figure D-1. The contour of the integral when 0 < 𝑦 < 1.

Isolating

1

𝑎+𝑖ℎ 𝑦 𝑠

∫ 2𝜋𝑖 𝑎−𝑖ℎ

𝑠

𝑑𝑠 and reversing the integral direction for the

last two integrals by multiplying them by (-) we get 𝑎+𝑖ℎ 𝑦 𝑠

∫𝑎−𝑖ℎ

𝑠

𝐾+𝑖ℎ 𝑦 𝑠

𝑑𝑠 = − ∫𝑎+𝑖ℎ

𝑠

𝐾+𝑖ℎ 𝑦 𝑠

𝑑𝑠 + ∫𝐾−𝑖ℎ

𝑠

𝐾−𝑖ℎ 𝑦 𝑠

𝑑𝑠 + ∫𝑎−𝑖ℎ

𝑠

𝑑𝑠

(1)

Let us examine the second integral on the right-hand side, 𝑠 = 𝜎 + 𝑖𝑡 and it goes from 𝐾 − 𝑖ℎ to 𝐾 + 𝑖ℎ therefore we can write it as 𝑠 = 𝐾 + 𝑖𝑡 with 𝑡 varies from −ℎ to ℎ, and the integrand

𝑦 𝐾+𝑖𝑡 𝐾+𝑖𝑡

becomes

equal to 𝑦 𝐾+𝑖𝑡 𝑦 𝐾 (cos 𝐾 log 𝑦 + 𝑖 sin 𝐾 log 𝑦) = 𝐾 + 𝑖𝑡 𝐾 + 𝑖𝑡 It is easy to see that the magnitude of the numerator is less than 𝑦 𝐾 and the magnitude of the denominator is greater than 𝐾, and therefore 𝑦 𝐾+𝑖𝑡 𝑦𝐾 | |≤ 𝐾 + 𝑖𝑡 𝐾

Appendix D

205

And applying ML inequality (4) from Chapter 2 Section 2.10, we get 𝐾+𝑖ℎ

𝑦𝑠 𝑦𝐾 | ∫ 𝑑𝑠 | ≤ 2ℎ 𝑠 𝐾 𝐾−𝑖ℎ

Now let us shift our attention to the first integral, it is from 𝑎 = 𝑖ℎ to 𝑘 + 𝑖ℎ, therefore the path is defined by 𝛾(𝜎) = 𝜎 + 𝑖ℎ and 𝛾′(𝜎) = 1. Using the complex integral formula (equation 3, Chapter 2), we get: 𝐾+𝑖ℎ

𝐾

𝑦𝑠 𝑦 𝜎+𝑖ℎ ∫ 𝑑𝑠 = ∫ (1)𝑑𝜎 𝑠 𝜎 + 𝑖ℎ

𝑎+𝑖ℎ

𝑎

Since |𝜎 + 𝑖ℎ| ≥ ℎ then we can write 𝐾

𝐾

𝑦 𝜎+𝑖ℎ 𝑦 𝜎+𝑖ℎ |∫ 𝑑𝜎 | ≤ |∫ 𝑑𝜎 | 𝜎 + 𝑖ℎ ℎ 𝑎

𝑎

And since |𝑦 𝜎+𝑖ℎ | ≥ 𝑦 𝜎 we can write 𝐾

𝐾

|𝑦 𝐾 − 𝑦 𝑎 | 𝑦 𝜎+𝑖ℎ 𝑦𝜎 𝑦𝜎 𝐾 |∫ 𝑑𝜎 | ≤ |∫ 𝑑𝜎 | = |[ ] |= 𝜎 + 𝑖ℎ ℎ ℎ log 𝑦 𝑎 ℎ|log 𝑦| 𝑎

𝑎

And therefore, 𝐾+𝑖ℎ

|𝑦 𝐾 − 𝑦 𝑎 | 𝑦𝑠 | ∫ 𝑑𝑠 | ≤ 𝑠 ℎ|log 𝑦| 𝑎+𝑖ℎ

206

Naji Arwashan The same result can be found for the third integral on the right

𝐾−𝑖ℎ 𝑦 𝑠 ∫𝑎−𝑖ℎ 𝑠

𝑑𝑠 . And now replacing everything in (1) we get 𝑎+𝑖ℎ

|𝑦 𝐾 − 𝑦 𝑎 | 𝑦𝑠 𝑦𝐾 | ∫ 𝑑𝑠 | ≤ 2ℎ + 2 𝑠 𝐾 ℎ|log 𝑦| 𝑎−𝑖ℎ

In the current case 𝑦 is between 0 < 𝑦 < 1, and therefore when 𝐾 → ∞, 𝑦 𝐾 → 0 and the above bound becomes 𝑎+𝑖ℎ

| ∫ 𝑎−𝑖ℎ

𝑦𝑠 𝑦𝑎 𝑑𝑠 | ≤ 2 𝑠 ℎ|log 𝑦|

And when ℎ → ∞ 𝑎+𝑖∞

| ∫ 𝑎−𝑖∞

𝑦𝑠 𝑑𝑠 | ≤ 0 𝑠

And therefore, 𝑎+𝑖∞

∫ 𝑎−𝑖∞

𝑦𝑠 𝑑𝑠 = 0 𝑠

And consequently 𝑎+𝑖∞

1 𝑦𝑠 ∫ 𝑑𝑠 = 0 2𝜋𝑖 𝑠 𝑎−𝑖∞

Appendix D

207

The Case When 𝒚 = 𝟏 In this case the integral becomes 𝑎+𝑖∞

𝑎+𝑖ℎ

𝑦𝑠 𝑑𝑠 1 ∫ 𝑑𝑠 = lim ∫ = lim [log 𝑠]𝑎+𝑖ℎ 𝑎−𝑖ℎ ℎ→∞ 𝑠 𝑠 2𝜋𝑖 ℎ→∞

𝑎−𝑖∞

𝑎−𝑖ℎ

= lim [log(𝑎 + 𝑖ℎ) − log(𝑎 − 𝑖ℎ)] ℎ→∞

From the definition of complex logarithm, we have 𝑎+𝑖∞

∫ 𝑎−𝑖∞

𝑦𝑠 ℎ 𝑑𝑠 = lim {[log √𝑎2 + ℎ2 + 𝑖 (arctan )] ℎ→∞ 𝑠 𝑎 − [log √𝑎2 + (−ℎ)2 + 𝑖 (arctan

−ℎ )]} 𝑎

Simplifying gives 𝑎+𝑖∞

1 𝑦𝑠 1 ℎ −ℎ ∫ 𝑑𝑠 = lim {[𝑖 (arctan )] − [𝑖 (arctan )]} 2𝜋𝑖 𝑠 2𝜋𝑖 ℎ→∞ 𝑎 𝑎 𝑎−𝑖∞

Accounting for the principal branch of the logarithm only, when ℎ

ℎ → ∞, (arctan 𝑎) → 𝑎+𝑖∞

𝜋 2

𝑦𝑠 𝜋 𝜋 ∫ 𝑑𝑠 = {[𝑖 ( )] − [𝑖 (− )]} = 𝑖𝜋 𝑠 2 2

𝑎−𝑖∞

208

Naji Arwashan And finally 𝑎+𝑖∞

1 𝑦𝑠 1 ∫ 𝑑𝑠 = 2𝜋𝑖 𝑠 2 𝑎−𝑖∞

The Last Case Is When 𝒚 > 𝟏 The approach in this case is similar to the first case but this time we will choose the closed contour to include the pole as shown below,

Figure D-2. The contour of the integral when 𝑦 > 1.

And now according to the residue theorem equation (6) from Chapter2, the integral over the closed boundary of the above domain (going counterclockwise) becomes equal to the value of the residue (which can be easily found to be 𝑦 0 = 1) times 2𝜋𝑖. Therefore, we can write

Appendix D 𝑎+𝑖ℎ 𝑦 𝑠

∫𝑎−𝑖ℎ

𝑠

−𝐾+𝑖ℎ 𝑦 𝑠

𝑑𝑠 + ∫𝑎+𝑖ℎ

𝑠

−𝐾−𝑖ℎ 𝑦 𝑠

𝑑𝑠 + ∫−𝐾+𝑖ℎ

𝑠

209 𝑎−𝑖ℎ 𝑦 𝑠

𝑑𝑠 + ∫−𝐾−𝑖ℎ

𝑠

𝑑𝑠 = 2𝜋𝑖

(2)

Rearranging, and reversing the direction of the last two integrals on the right by multiplying them by (-) 𝑎+𝑖ℎ

−𝐾+𝑖ℎ

𝑦𝑠 ∫ 𝑑𝑠 − 2𝜋𝑖 = − ∫ 𝑠

𝑎−𝑖ℎ

𝑎+𝑖ℎ

𝑦𝑠 𝑑𝑠 + 𝑠

−𝐾+𝑖ℎ

∫ −𝐾−𝑖ℎ

−𝐾−𝑖ℎ

𝑦𝑠 𝑑𝑠 + ∫ 𝑠

𝑎−𝑖ℎ

𝑦𝑠 𝑑𝑠 𝑠

Following steps similar to the first case; we start by the second integral, the variable 𝑠 is equal to 𝑠 = −𝐾 + 𝑖𝑡 with t varies from −ℎ to ℎ, and the integrand

𝑦 −𝐾+𝑖𝑡 −𝐾+𝑖𝑡

is equal to

𝑦 −𝐾+𝑖𝑡 𝑦 −𝐾 (cos 𝐾 log 𝑦 + 𝑖 sin 𝐾 log 𝑦) = −𝐾 + 𝑖𝑡 −𝐾 + 𝑖𝑡 It is easy to see that the magnitude of the numerator is less than 𝑦 −𝐾 and the magnitude of the denominator is greater than |−𝐾|, and therefore

|

𝑦 −𝐾+𝑖𝑡 𝑦 −𝐾 |≤ −𝐾 + 𝑖𝑡 𝐾

And applying ML inequality (4) from Chapter 2 Section 2.10, we get −𝐾+𝑖ℎ

| ∫ −𝐾−𝑖ℎ

𝑦𝑠 𝑦 −𝐾 𝑑𝑠 | ≤ 2ℎ 𝑠 𝐾

210

Naji Arwashan

For the first integral, 𝑠 goes from (𝑎 + 𝑖ℎ) to (−𝐾 + 𝑖ℎ), therefore the path is defined by 𝛾(𝜎) = 𝜎 + 𝑖ℎ and 𝛾′(𝜎) = 1. Using the complex integral formula, we get: −𝐾+𝑖ℎ

∫ 𝑎+𝑖ℎ

−𝐾

𝑦𝑠 𝑦 𝜎+𝑖ℎ 𝑑𝑠 = ∫ (1)𝑑𝜎 𝑠 𝜎 + 𝑖ℎ 𝑎

Since |𝜎 + 𝑖ℎ| ≥ ℎ then we can write −𝐾

−𝐾

𝑎

𝑎

𝑦 𝜎+𝑖ℎ 𝑦 𝜎+𝑖ℎ |∫ 𝑑𝜎 | ≤ |∫ 𝑑𝜎 | 𝜎 + 𝑖ℎ ℎ And since |𝑦 𝜎+𝑖ℎ | ≥ 𝑦 𝜎 we can write −𝐾

−𝐾

𝑎

𝑎

|𝑦 −𝐾 − 𝑦 𝑎 | 𝑦 𝜎+𝑖ℎ 𝑦𝜎 𝑦 𝜎 −𝐾 |∫ 𝑑𝜎 | ≤ |∫ 𝑑𝜎 | = |[ ] |= 𝜎 + 𝑖ℎ ℎ ℎ log 𝑦 𝑎 ℎ log 𝑦 And therefore, −𝐾+𝑖ℎ

| ∫ 𝑎+𝑖ℎ

|𝑦 −𝐾 − 𝑦 𝑎 | 𝑦𝑠 𝑑𝑠 | ≤ 𝑠 ℎ log 𝑦

The same result can be found for the third integral on the right −𝐾−𝑖ℎ 𝑦 𝑠 ∫𝑎−𝑖ℎ 𝑠

𝑑𝑠

And now replacing everything in (2) gives 𝑎+𝑖ℎ

| ∫ 𝑎−𝑖ℎ

|𝑦 −𝐾 − 𝑦 𝑎 | 𝑦𝑠 𝑦 −𝐾 𝑑𝑠 − 2𝜋𝑖 | ≤ 2ℎ + 2 𝑠 𝐾 ℎ log 𝑦

Appendix D Letting 𝐾 → ∞, (since 𝑦 > 1) then 𝑦 −𝐾 → 0 𝑎+𝑖ℎ

| ∫ 𝑎−𝑖ℎ

𝑦𝑠 𝑦𝑎 𝑑𝑠 − 2𝜋𝑖 | ≤ 2 𝑠 ℎ log 𝑦

And letting ℎ → ∞ 𝑎+𝑖∞

| ∫ 𝑎−𝑖∞

𝑦𝑠 𝑑𝑠 − 2𝜋𝑖 | ≤ 0 𝑠

And therefore, 𝑎+𝑖∞

∫ 𝑎−𝑖∞

𝑦𝑠 𝑑𝑠 = 2𝜋𝑖 𝑠

Or 𝑎+𝑖∞

1 𝑦𝑠 ∫ 𝑑𝑠 = 1 2𝜋𝑖 𝑠 𝑎−𝑖∞

211

REFERENCES [1] [2] [3]

[4] [5] [6] [7] [8] [9]

Artin, Emil. 1964. The Gamma Function. New York: Holt Rinehart and Winston. Apostol, Tom. 1976. Introduction to Analytic Number Theory. New York: Springler-Verlag. Baillie, Robert. 2011. “How the Zeros of the Zeta Function Predict the Distribution of Primes.” Wolfram Demonstrations Project. Bonfert-Taylor Petra. 2007. Analysis of a Complex Kind. Wesleyan University, Cosmo Learning. Derbyshire, John. 2004. Prime Obsession. New York: Penguin Group. Edwards, H.M. 1974. Riemann’s Zeta Function. New York: Academic Press. Havil, Julian. 2003. Gamma, Exploring Euler’s Constant. Princeton: Princeton University Press. Riemann, Bernard. 1859. On the Number of Prime Numbers less than a Given Quantity. Riesel, Hans. 1994. Prime Numbers and Computer Methods for Factorization, 2nd ed. New York: Springer.

214

Naji Arwashan

[10] Sondow, Jonathan. 2003. “Zeros of the Alternating Zeta Function on the Line R(s)=1.” arXiv math.NT/0209393v2 [11] Sondow, Jonathan. 2012. “A Simple Couterexample to Havil’s Reformation of the Riemann Hypothesis.” Elemente der Mathematik 67. [12] Titchmarsh, E.C. 1986. The Theory of the Riemann ZetaFunction, 2nd ed. Oxford: Clarendon Press. [13] Van der Veen, Roland and Van de Craats, Jan. 2015. The Riemann Hypothesis, A million Dollar Problem. Washington: MAA Press. [14] Von Koch, Heldge. 1901. Sur la distribution des nombres premiers [On the distribution of prime numbers]. Acta Mathematica, volume 24 (1901), pages 159-182. [15] Zagier, Don. 1977. “The First 50 Million Prime Numbers.” The Mathematical Intelligencer.

INDEX A absolute convergence of a series, 2, 30, 126 absolute difference, 168 absolute value, 22, 23, 30, 31, 33, 47, 53, 60 absolute value of the summand, 53, 60 absolutely convergent series, 32 alteration to ξ(s), 132 alternate series test, 3 alternating harmonic series, 32 alternating series, 31, 58 alternating series test, 58 alternating zeta series, 57 amplitude of the wave, 163 analytic, 19, 20, 29, 35, 38, 46, 48, 49, 54, 55, 62, 132, 213 analytic continuation, 62 analytic function, 29, 46 analytic solution, 19 analyticity, 35 antiderivative, 40, 46 approximation, 112, 113, 135, 174 argument of a complex number, 22 asymptotically equivalent, 113, 115 Atle Selberg, 166

August Ferdinand Mobius, 118

B Basel problem, 15 Bernard Riemann, ix, 1, 2, 32, 67, 73 beta function, 190, 192, 193 branch of the complex logarithm, 138 branches of logarithm, 35 Brook Taylor, 26

C Carl Fredric Gauss, 112 Carl Gustav Jacob Jacobi, 179 cartesian plane, 21 Cauchy residue formula, 52 Cauchy-Riemann equations, 38 Charles Jean de la Vallee-Poussin, 166 Chebyshev functions, 145 Clay institute, ix, x, 93, 168 complex analysis, ix, 39, 115, 125, 127, 166 complex differentiation, 35 complex function, 24, 25, 34, 35, 38, 44, 104, 169, 170, 174

216

Index

complex integral, 42, 43, 48, 137, 205, 210 complex integration, 39 complex logarithm, 34, 35, 136, 137, 207 complex mapping, 24 complex number, 1, 17, 18, 21, 22, 23, 26, 33, 34, 35, 39, 42, 53, 55, 58, 59, 62, 63, 73, 134, 137, 144, 159 complex plane, 36, 39, 42, 62, 75, 84, 91, 131, 171, 172 complex variable, 24, 130, 179, 190 conditional convergence, 31 conjugate, 17, 21, 95, 106, 134, 144, 159 contour integral formula, 44, 45 contour integrals, 42 convergence of the eta series, 58 convergence tests of a series, 2 coprime, 15 critical line, 1, 93, 94, 104, 105, 106, 171 cubic equation, 19

E. Landau, 63 Enrico Bombieri, 168 equivalent form of the functional equation, 87 estimation Lemma, 47 eta function, 57, 77, 104 eta series, vii, 57, 58, 60, 64 Euclid, 110 Euler formula, 26, 29, 55 Euler product formula, 7, 9, 10, 15, 91, 125 Euler reflection formula, 85, 87, 88, 90, 189, 194, 195, 201 Euler-Mascheroni constant, 144 explicit formula, 15, 115, 125, 134, 147, 158, 159, 160, 161, 162, 163, 165, 167 exponential integral, 142 extension of zeta, vii, 57, 73, 153

D

factor ξ(s), 133 factorial, 75 factorial function, 75 Fourier inversion theorem, 170 Fourier series, 172, 175, 180 Fourier transform, 169, 170, 171, 174, 176, 177, 180, 182, 186 frequency, 171, 172, 176 function R(x), 135 function li(x), 140 function ξ(s), 89, 90, 107, 132 function ψ(x), 146, 147 function ξ(s), 89, 107 functional equation, vii, 73, 74, 77, 78, 80, 84, 85, 86, 87, 89, 90, 92, 93, 107, 108, 132, 153, 169, 179, 180, 187, 188, 189, 195, 200 functional equation of ϕ(a), 187, 188 functional equations of the Gamma function, 189 fundamental theorem of arithmetic, 7, 109

D.V. Widder, 63 David Hilbert, ix, 168 density of primes, 114 derivative, 11, 12, 13, 36, 37, 38, 41, 43, 70, 114, 154, 184, 185, 200 derivative of zeta, 11, 12 differentiation of a real function, 35 Dirac delta function, 172 direct comparison test, 2 Dirichlet series, 58 discriminant, 18 distribution of prime numbers, x, 115, 168, 214 distribution of primes, 111, 115, 125, 167 Don Zagier, 111

E E. Artin, 195

F

Index fundamental theorem of calculus, 27, 39, 40, 45

G gamma function, 73, 74, 75, 86, 87, 93, 190, 193, 200 gap between two primes, 111 geometric representation, 21 geometric series, 3, 4, 7, 8, 13, 14, 76, 78 grid of the natural numbers, 161 Guillaume de l’Hôpitale, 68

H H.M. Edwards, 117, 125, 147 Hadamard, 133 halfway value, 146, 151 Hans von Mangoldt, 145, 147 harmonic series, 4, 5, 32 Havil, 99, 150, 213, 214 Helge von Koch, 167 Hjalmar Mellin, 130 hospital rule, 68, 69, 70, 71, 114, 130, 148, 154, 155, 157

I imaginary axis, 33, 37 imaginary constant, 17, 20, 21, 159 indefinite integral, 40 indefinitely differentiable, 180 infinitely differentiable, 29, 35, 38 infinity, 4, 31, 33, 39, 43, 67, 68, 69, 70, 92, 112, 113, 153, 154, 166, 167, 171, 177 integral by parts, 148 integrand, 137, 139, 153, 203, 204, 209 internet, 86 inventor of complex numbers, 20

217 J

J. Sondow, 58, 64 Jacobi Theta function, 179 Jacques Hadamard, 166 Jean-Baptiste Joseph Fourier, 172 John Derbyshire, 117, 125 John Littlewood, 167

L Legendre duplication formula, 87, 189, 194, 195, 196 length of the curve, 48 Leonhard Euler, 7 logarithmic integral, 114, 138, 143, 145

M Maclaurin series, 29, 30, 31, 32, 125, 158, 200 magnitude, 22, 25, 106, 204, 209 magnitude of the denominator, 204, 209 magnitude of the numerator, 204, 209 maximum magnitude, 47 mean value theorem, 41 Mellin inverse transform, 131 Mellin inversion theorem, 130 Mellin transform, 130, 131 ML inequality, 47, 60, 205, 209 Mobius function, 12, 118, 119, 123, 124 Mobius inversion formula, 118, 122 modulus, 22, 33, 34 motion of a point, 171 multi-valued function, 35

N natural logarithm, 13, 35, 125, 198

218

Index

natural numbers, 1, 4, 7, 10, 15, 76, 77, 79, 109, 110, 111, 129 negative even integers, 93 Nicole Oresme, 5 nontrivial zeros, 1, 107 non-trivial zeros of zeta, 93, 108, 136, 147, 156 norm of the partition, 128 number of prime numbers, ix, 1, 213 number theory, ix, 11, 12, 109, 115, 118, 127

prime number theorem, ix, 109, 113, 115, 146, 165 prime numbers, 7, 8, 10, 109, 111, 125 prime obsession, 117 primitive, 46 principal argument, 22, 35 principal branch, 35, 138, 207

Q quadratic equation, 18 Quanta magazine, 171

O O notation, 168 offset logarithmic integral, 113

P Pafnuty Chebyshev, 145 pair of zeros, 134 parabola, 18 partial derivatives, 38, 54, 55 path of the integral, 42 Paul Erdos, 166 period, 100, 102, 163, 164, 172, 173, 175, 180, 196 period of the wave, 163 periodic motion, 172 Peter Gustav Lejeune Dirichlet, 58, 113 Peter Lejeune-Dirichlet, 32 plane of complex numbers, 73, 87 Poisson summation formula, 180, 187 polar coordinates, 26, 33, 34, 184 pole, 52, 71, 86, 87, 90, 108, 137, 203, 208 polynomial of infinite degree, 133 positive even integer, 93 prime counting function, 111, 115, 117, 134, 145 prime factors, 12, 118, 119

R Rafael Bombelli, 20 Ramanujan, 86 randomness, 111 ratio test, 2, 4, 33 real and imaginary parts, 17, 42, 53, 55, 63, 95, 96, 105, 138, 159, 170, 174 real axis, 33 real number(s), vii, 1, 3, 4, 6, 17, 21, 22, 39, 41, 42, 46, 53, 62, 73, 92, 134, 136, 138, 144, 160 reciprocal of zeta, 11 relative error, 112, 167 residue, 48, 52, 71, 86, 152, 153, 155, 156, 157, 208 residue of zeta, 71, 86 Riemann explicit formula, 125, 134, 139, 140, 141, 142, 145, 167 Riemann hypothesis, vii, ix, x, 26, 91, 93, 96, 99, 105, 109, 167, 168, 171, 214 Riemann sum, 67, 128, 177 Riemann-Stieltjes sum, 127, 128 roots of the zeta function, vii, 91

Index S smooth function, 29, 180 special values of zeta, 15 spectrum, 171 staircase, 110, 111, 139, 140, 142, 161, 162 Stanly Skewes, 167 step function, 129, 146 Stieltjes, 128, 129 summand, 31, 53, 54 symmetrical, 90 symmetry of the zeros, 99, 106

uniqueness, 110 unsolved problem in math, 99

V vectors, 22 von Mangoldt explicit formula, 145, 162

W wave composition of ψ(x), 159 Wolfram Demonstration Project, 135, 142

T Taylor expansion, 26, 28, 29, 35 Taylor series, 27, 29, 35 trajectory of a point, 171, 172 trivial zeros of zeta, 92, 93, 155, 157 twin primes, 112

U unique prime-factorization theorem, 109 unique product of primes, 7, 109

219

Y youtube videos, 86

Z zeros of the zeta function, 109, 115 zeta function, vii, 1, 7, 53, 54, 55, 57, 61, 73, 75, 76, 78, 85, 86, 87, 91, 99, 109, 117, 124, 170, 179, 189, 213, 214 zeta series, 4, 7, 15, 53, 57, 58, 60, 61, 85, 92