ANALYTIC PROOF of the PRIME NUMBER THEOREM a Thesis Submitted to Kent State University in Partial Fulfillment of the Requirement

Home , Euler product, Riemann (crater), Von Mangoldt function

ANALYTIC PROOF OF THE PRIME NUMBER THEOREM

A thesis submitted

To Kent State University in partial

Fulfillment of the requirements for the

Degree of Master of Science

Maha Mosaad Alghamdi

May, 2019

Except for previously published materials

Thesis written by

Maha Mosaad Alghamdi

B.S., Dammam University, 2011

M.S., Kent State University, 2019

Approved by

Gang Yu , Advisor

Andrew Tonge , Chair, Department of Mathematical Sciences

James L. Blank , Dean, College of Arts and Sciences

TABLE OF CONTENTS

TABLE OF CONTENTS ...... III

LIST OF FIGURES ...... V

ACKNOWLEDGMENTS...... VI

CHAPTERS

1. INTRODUCTION ...... 1

2. SUMMATION FORMULAS...... 6

2.1 Abelian Summation Formula ...... 6

2.2 Euler-Maclaurin Summation Formula ...... 8

3. DIRICHLET SERIES AND EULER PRODUCTS ...... 10

3.1 The Half-Plane of Absolute Convergence of a Dirichlet Series ...... 11

3.2 The Function Defined by a Dirichlet Series ...... 12

3.3 Multiplication of the Dirichlet Series ...... 13

3.4 Euler Product ...... 16

4. BASIC PROPERTIES OF ζ(s) ...... 17

4.1 Definition and Basic Properties of the Euler Gamma Function ...... 17

4.2 Stirlig’s Formula ...... 19

4.3 Analytic Continuation of ζ(s) ...... 20

4.4 A Zero-Free Region of ζ(s) ...... 30

5. PRIME NUMBER THEOREM ...... 34

5.1 Truncated Form of Perron’s Formula ...... 36

5.2 Proof of the Prime Number Theorem ...... 40

5.3 An Explicit Formula of ...... 42

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BIBLIOGRAPHY ...... 46

LIST OF FIGURES

Figure 5.1. Rectangular contour with vertices (푏 ± 푖푇, −푈 ± 푖푇) oriented

counterclockwise ...... 36

Figure 5.2. Rectangular contour with vertices (푏 ± 푖푇, −푈 ± 푖푇) oriented

clockwise ...... 37

Figure 5.3. The integral 퐽 around the contour defined by Γ where 퐽 is defined by

1 휁′ 푥푠 퐽 = ∫ (− (푠)) 푑푠 ...... 41 2휋푖 Γ 휁 푠

Figure 5.4. The integral 퐽 around the contour Γ where 퐽 is defined by

1 휁′ 푥푠 퐽 = ∫ (− (푠)) 푑푠 ...... 43 2휋푖 Γ 휁 푠

ACKNOWLEDGMENTS

I owe my thanks and appreciation to my thesis advisor, Dr. Gang Yu, for his patience, advice, guidance, and assistance throughout the time of my thesis research. He was always available and willing to help me and explain any difficulties I faced. I am really proud to have a professor like him who is full of knowledge and experience, which makes someone learn a lot from him on both the scientific and the personal sides.

I wish to express my sincere thanks to both Dr. Ulrike Vorhauer and Dr. Morley

Davidson for agreeing to be the members of my defense committee. Also, I am thankful for our graduate coordinator, Dr. Artem Zvavitch, for his help and guidance.

I would love to thank Kent State University and especially the Department of

Mathematical Sciences, for providing me an opportunity to study for a master’s degree in pure mathematics. I also would like to thank Imam Abdulrahman Bin Faisal

University for its support and for giving me the opportunity to complete the master’s degree. Thanks as well to officials at the Saudi Arabian Cultural Mission in the United

States for their efforts and cooperation while I was studying for my master’s degree and for their support of my family.

I would like to thank for the blessing of my life my dear mother for her patience and for letting me find a way forward. Thank you, my dear mother, for sowing the ambition in my heart until what I am now. Thank you, my dear mother, for your support and sincere prayers, which makes me very happy. Thank you, for your patience and waiting for me in the days and years until I return home.

Thank you, my brothers and sisters, for your support, encouragement, and motivation, and especially for your trust and belief in me.

I could not achieve success easily without the support of my beloved husband and my daughters, Malak, Jumanh, Sadeem, my wonderful child Abdullah, and my little angel

Ward. A ton of thanks to my husband and children for their sacrifices and patience as I worked to achieve my goal. They supported and helped me unconditionally in my success. They share the most accurate details who rejoice to my joy and sadness for my sorrow.

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CHAPTER 1

INTRODUCTION

A prime number is a positive integer that has exactly two positive integer divisors:

1 and itself. In other words, a number is prime if it is greater than 1 and cannot be written as a product of two natural numbers that are both less than it.

The most important property of prime numbers is the Fundamental Theorem of

Arithmetic (FTA), which describes the relationship between prime numbers and natural numbers.

Every natural number n can be uniquely written as

푎1 푎푘 푛 = 푝1 … 푝푘 , where 푎1, … , 푎푘 are natural numbers, and 푝1 < 푝2 < ⋯ < 푝푘 are prime numbers.

This theorem shows that prime numbers are the multiplicative building blocks of natural numbers.

Based on the FTA, there is an infinite number of prime numbers. Proof of the infinitude of prime numbers based on the FTA has a long history. Euclid was the first to do so in roughly 300 BC. In the 18th century, Leonhard Euler offered a different proof, though it, too, was based on the FTA. Euler proved a factorization formula, which said that for any fixed real number s > 1, the following is true when the product is taken over all prime numbers:

1 1 −1 ∑∞ = ∏ (1 − ) (1.1) 푛=1 푛푠 푝 푝푠

This formula is essentially the analytic equivalence of the FTA. Euler’s formula represents an important discovery because it does much more than merely the infinitude of prime numbers. The formula also relates prime numbers to the more familiar natural numbers.

While a simple formula does not exist to distinguish prime numbers from composite numbers, the quantitative distribution of prime numbers within the natural numbers becomes extremely important. For a positive real number 푥, let

휋(푥) = ∑ 1, 푝≤푥 where 푝 runs over prime numbers up to 푥. Thus, 휋(푥) serves as the “counting function” of prime numbers. While the unboundedness of 휋(푥) had long been known, a natural question is: Does 휋(푥) have an asymptotic formula as 푥 → ∞? The Prime Number

Theorem (PNT), which serves as the focus of this thesis, gives an affirmative answer to the question posed.

In 1798, based on the existing numerical evidence, Adrien-Marie Legendre made the following very accurate conjecture:

푥 휋(푥)~ as 푥 → ∞. (1.2) 푙표푔 푥−1.08366

In 1849, Johann Carl Friedrich Gauss wrote about his own findings in a letter to

German astronomer Johann Franz Encke, saying in 1792 or 1793, when he was still a boy, that he had noticed that the average density of prime numbers up to 푥 should be

1 . In other words, log 푥

푥 휋(푥)~ as 푥 → ∞. (1.3) log 푥

As 푥 → ∞, Equation 1.2 and Equation 1.3 are equivalent; however, Equation 1.3 is the weakest form of the PNT and often is referred to as the PNT without an error term.

For the next few years, the PNT attracted the attention of mathematicians but remained unproved. In 1852 or so, Pafnuty Lvovich Chebyshev identified the existence of two positive constants: 퐶1 and 퐶2, such that

푥 푥 퐶 < 휋(푥) < 퐶 (푥 > 2) 1 log 푥 2 log 푥

Chebyshev also introduced two functions, which now are referred to as

Chebyshev’s functions:

휃(푥) = ∑ log 푝 푝≤푥 and

휓(푥) = ∑ Λ(푛), 푛≤푥 where Λ(푛) is the von Mangoldt function and is defined, for every integer 푛 ≥ 1, by

log푝 , if 푛 = 푝푚 for some prime 푝 and some 푚 ≥ 1, Λ(푛) = { 0 , otherwise.

The PNT Equation (1.3) is equivalent to

휓(푥)~ 푥 as 푥 → ∞ or 휃(푥) ~ 푥 as 푥 → ∞. (1.4)

In 1859, Georg Friedrich Bernhard Riemann published a paper about number theory in which he investigated Euler’s Identity Equation (1.1). Unlike Euler, Riemann considered the function defined on the left side of Equation (1.1) to be a complex variable function

1 휁(푠) = ∑∞ (Re s >1). (1.5) 푛=1 푛푠

That function is now known as the Riemann zeta-function. It can be analytically extended to a meromorphic function in the whole complex plane with a simple pole at s

= 1. Riemann continued to study the function and found, for example, that the distribution of the prime numbers is closely related to the distribution of the zeros of

휁(푠). (Some of Riemann’s findings were not tested as rigorously and later were corrected by Jacques Salomon Hadamard and von Mangoldt).

In 1896, after withstanding attacks for nearly a century, the PNT was eventually proved, independently and nearly simultaneously, by French mathematicians

Hadamard and Charles Jean de la Vallee Poussin. The key step in their proof is to show that

휁(1 + 푖푡) ≠ 0 , −∞ < 푡 < ∞.

In 1900, de la Vallee Poussin proved the PNT with an error term:

휋(푥) = 퐿푖(푥) + 푂 (푥 푒−푐√log푥) 푎푠 푥 → ∞, (1.6)

푥 푑푡 where for 푥 ≥ 2, 퐿푖(푥): = ∫ or, equivalently, 2 log 푡

휓(푥) = 푥 + 푂 (푥 푒−푐√log푥) 푎푠 푥 → ∞. (1.7)

This thesis focuses on the classic analytic proof of Equation 1.7, with subsequent chapters structured as follows:

• In chapter 2, we introduce two summation formulas often used in analytic

number theory: the Abelian summation and the Euler-Maclaurin summation. As

a more general form of the Abelian summation, the Euler-Maclaurin summation

plays a key role when 휁(푠) in Equation (1.5) is extended analytically to the whole

complex plane in chapter 4.

• In chapter 3, we review some basic ideas of the Dirichlet series, such as the half

plane of absolute convergence of the Dirichlet series, the function defined by a

Dirichlet series, and multiplication of the Dirichlet series. Also, we will discuss

the Euler product, which essentially is the analytic version of the FTA.

• In chapter 4, we study various analytic properties of 휁(푠). The chapter begins

with a review of some of the basic properties of Euler’s gamma function before

delving into a discussion about the analytic continuation of the Riemann zeta-

function to the entire complex plane, its function equation, and some results of

the zeros—in particular, the result of the de la Vallee Poussin theory on the

existence of a zero-free region of 휁(푠).

• In chapter 5, we prove the PNT by using the complex integration method. The

properties of 휁(푠), especially those covered in chapter 4, lead to the asymptotic

formula

휓(푥) = 푥 + 푂 (푥 푒−푐√log푥),

which is equivalent to the PNT in the form

휋(푥) = 퐿푖(푥) + 푂 (푥 푒−푐√log푥).

CHAPTER 2

SUMMATION FORMULAS

In analytic number theory, it often requires the estimation or an asymptotic formula for sums in the form of

∑ ( ) ( ) 푥1<푛≤푥2 푎 푛 푓 푛 , (2.1) where 푥1 < 푥2 are non-negative real numbers. Let 푎(푛) be a sequence of complex numbers and 푓(푥) be a complex-valued function defined as [0, +∞), which usually is sufficiently smooth.

In this chapter, we review two often-used summation formulas to deal with such sums: the Abelian summation and the Euler-Maclaurin summation.

2.1 Abelian Summation Formula

A proof of the Abelian summation formula is presented below. The formula deals with the partial sums of arithmetic functions in the form of Equation (2.1) with 푓(푛) differentiable and partial sums of 푎(푛) known. The formula is beneficial in number theory, especially when an asymptotic main term well approximates the partial sums of

푎(푛).

Theorem 1. Assume that 푎(푛), 푛 = 1,2, … is a complex sequence with the partial sum

퐴(푥) = ∑ 푎(푛) . 푛≤푥

Assume also that 0 ≤ 푥1 < 푥2 are real numbers, and let 푓(푥) have a continuous derivative on [푥1, 푥2]. Therefore,

푥2 ′ ∑ 푎(푛)푓(푛) = 퐴(푥2)푓(푥2) − 퐴(푥1)푓(푥1) − ∫ 퐴(푥)푓 (푥) 푑푥. 푥1 푥1<푛≤푥2

Proof. Using the Riemann Stieltjes Integration,

푥2 ∑ 푎(푛)푓(푛) = ∫ 푓(푥)푑퐴(푥). 푥1 푥1<푛≤푥2

Then, using integration by parts, the result is

푥2 ∑ 푎(푛)푓(푛) = 퐴(푥2)푓(푥2) − 퐴(푥1)푓(푥1) − ∫ 퐴(푥)푑푓(푥) 푥1 푥1<푛≤푥2

푥2 ′ = 퐴(푥2)푓(푥2) − 퐴(푥1)푓(푥1) − ∫ 퐴(푥)푓 (푥) 푑푥, 푥1 which proves the theorem. 

Theorem 1 apparently implies that, for 푥 > 1,

푥 ∑ 푎(푛)푓(푛) = 퐴(푥)푓(푥) − 퐴(푥)푓′(푥)푑푥. 푛≤푥 ∫1

When 퐴(푥) is well approximated asymptotically, let 퐴(푥) = 푀(푥) + 푟(푥), where 푀(푥) is the “main term,” which has a continuous derivative, and 푟(푥) is the “remainder term,” or

“error term.” It follows, then, that Theorem 1 is equivalent to

푥2 푥2 푥2 ∑ 푎(푛)푓(푛) = ∫ 푓(푥)푑퐴(푥) = ∫ 푓(푥)푑푀(푥) + ∫ 푓(푥)푑푟(푥) 푥1 푥1 푥1 푥1<푛≤푥2

푥2 = ∫ 푀′(푥) 푓(푥)푑푥 + 푅, 푥1 where

푥2 푥2 ′ 푅 = ∫ 푓(푥)푑푟(푥) = 푟(푥2)푓(푥2) − 푟(푥1)푓(푥1) − ∫ 푟(푥)푓 (푥) 푑푥, 푥1 푥1

which, presumably, gives the error term in most applications.

2.2 Euler-Maclaurin Summation Formula

In Theorem 1, assume 푎(푛) ≡ 1. Then, 퐴(푥) = [푥]. Let 퐴(푥) = 푀(푥) + 푟(푥) with

푀(푥) = 푥, then 푟(푥) = [푥] − 푥 = −{푥}, which implies that

푥2 푥2 푥2 ∑푥 <푛≤푥 푓(푛) = ∫ 푓(푥)푑(푥) + ([푥] − 푥)푓(푥)| − ∫ ([푥] − 푥)푓′(푥)푑푥. 1 2 푥1 푥1 푥1

If letting

1 1 푀(푥) = 푥 − 푎푛푑 푟(푥) = [푥] − 푥 + = −푏 (푥), 2 2 1 then,

푥2 푥2 푥2 ′ ∑푥 <푛≤푥 푓(푛) = ∫ 푓(푥)푑푥 − 푏1(푥)푓(푥)| + ∫ 푏1(푥)푓 (푥)푑푥. (2.2) 1 2 푥1 푥1 푥1

This is the simplest version of the Euler-Maclaurin summation formula. Before giving the general form of the formula, we generalize the function 푏1(푥).

Definition 1. Function 푏푙(푥) is the periodic function with period 1, such that

푏푙(푥) = ∫ 푏푙−1(푥)푑푥 , 푎푛푑 ∫ 푏푙(푥)푑푥 = 0. 0

Theorem 2 (Euler-Maclaurin Summation). Assume that 푓(푥) has a continuous 푙푡ℎ derivate on [푥1, 푥2], where 푙푁. Then,

푙 푥2 푗 (푗−1) 푥2 ∑ 푓(푛) = ∫ 푓(푥)푑푥 + ∑(−1) 푏푗(푥)푓 (푥)| 푥1 푥1 푥1<푛≤푥2 푗=1 푥2 푙+1 (푙) + (−1) ∫ 푏푙(푥)푓 (푥)푑푥. 푥1

Proof. Starting from Equation (2.2), we get

푥2 푥2 ∑ 푓(푛) = ∫ 푓(푥)푑푥 − 푏 (푥)푓(푥)|푥2 + ∫ 푏 (푥)푓′(푥)푑푥 1 푥1 1 푥1<푛≤푥2 푥1 푥1

푥2 푥2 = ∫ 푓(푥)푑푥 − 푏 (푥)푓(푥)|푥2 + 푏 (푥)푓′(푥)|푥2 − ∫ 푏 (푥)푓′′(푥)푑푥 1 푥1 2 푥1 2 푥1 푥1

푥2 푥2 = ∫ 푓(푥)푑푥 − 푏 (푥)푓(푥)|푥2 + 푏 (푥)푓′(푥)|푥2 − 푏 (푥)푓′′(푥)|푥2 + ∫ 푏 (푥)푓′′′(푥)푑푥 1 푥1 2 푥1 3 푥1 3 푥1 푥1 =….

푥2 푙 푥2 = ∫ 푓(푥)푑푥 + ∑(−1)푗푏 (푥)푓(푗−1)(푥)|푥2 + (−1)푙+1 ∫ 푏 (푥) 푓(푙)(푥)푑푥, 푗 푥1 푙 푥1 푗=1 푥1 which proves the theorem. 

CHAPTER 3

DIRICHLET SERIES AND EULER PRODUCTS

In this chapter, we review some basic properties of Dirichlet series and give some examples. Most of the content for this chapter comes from Tom Apostol’s book [1].

In 1737, Euler proved Euclid’s theorem on the existence of an infinite number of prime numbers by showing that the series ∑ 푝−1 diverges when extended over all prime numbers. Euler reached his conclusion by noting that the zeta function 휁(s), given by

1 휁(푠) = ∑∞ (3.1) 푛=1 푛푠 for real number 푠 > 1, tends to ∞ as s →1. One hundred years later, Dirichlet proved his theorem on prime numbers in arithmetic progressions when he studied the series

휒(푛) 퐿(푠, 휒) = ∑∞ , (3.2) 푛=1 푛푠 where 휒 is a Dirichlet character, and s > 1. The series in Equations (3.1) and (3.2) are examples of the Dirichlet series in the form

푓(푛) ∑∞ , (3.3) 푛=1 푛푠 where 푓(푛) is an arithmetic function. These are called Dirichlet series with coefficients

푓(푛). The series has become one of the most useful tools in analytic number theory.

To study the Dirichlet series further, we follow Riemann and let 푠 be a complex variable in the equation 푠 = 휎 + 푖푡, where 휎 and 푡 are real numbers. Then, 푛푠 =

푒푠 log푛 = 푒(σ + 푖푡)log푛 = 푛σ푒푖푡 log푛, which shows that |푛s| = 푛σ because |푛푖휃| = 1 for real

휃. The set of points defined by 푠 = 휎 + 푖푡 such that 휎 > 푎 is known as a half- plane. In the following section, we will show that each Dirichlet series has two half- planes: 휎 > 휎푐, where the series converges, and 휎 > 휎푎, where the series converges absolutely, and that in the convergence half-plane, the series represents an analytic function of the complex variable 푠.

3.1 The Half-Plane of Absolute Convergence of a Dirichlet Series

First note that if 휎 ≥ 푎, then |푛s| = 푛휎 ≥ 푛푎, and

푓(푛) |푓(푛)| | | ≤ . 푛s 푛푎

푓(푛) Therefore, if a Dirichlet series expressed by ∑ converges absolutely for 푠 = 푎 + 푖푏, 푛푠 then it also converges absolutely for all 푠, where 휎 ≥ 푎. These results lead to the following theorem:

푓(푛) Theorem 3. If the series ∑ | | does not converge or diverge for all 푠, then there 푛s exists a real number 휎푎, known as the abscissa of absolute convergence, where the

푓(푛) series ∑ converges absolutely if 휎 > 휎 but does not converge absolutely if 휎 < 푛푠 푎

푓(푛) 휎 . Note that if ∑ | | converges everywhere, then 휎 = – ∞. If that same series 푎 푛s 푎 converges nowhere, then 휎푎 = +∞.

We will consider two examples:

∞ −푠 • Example 1. The Dirichlet series ∑푛=1 푛 converges absolutely if 휎 > 1;

however, when 푠 = 1, the series diverges, with the result that 휎푎 = 1. The sum

of the series is denoted by 휁(푠) and is known as the Riemann zeta-function.

푓(푛) • Example 2. If 푓 is bounded, such that |푓(푛)| ≤ 푀 for all 푛 ≥ 1, then ∑ 푛푠

converges absolutely for 휎 > 1, which means that 휎푎 ≤ 1. More specifically, if

휒(푛) 휒 is a Dirichlet character, then the 퐿 -series 퐿(푠, 휒) = ∑ converges absolutely 푛푠

for 휎 > 1.

3.2 The Function Defined by a Dirichlet Series

푓(푛) To understand the function defined by a Dirichlet series, first, assume that ∑ 푛푠 converges absolutely for 휎 > 휎푎. Then, let 퐹(푠) represent the sum function

푓(푛) 퐹(푠) = ∑∞ for 휎 > 휎 . (3.4) 푛=1 푛푠 푎

Now, it is necessary to derive some properties of 퐹(푠).

Lemma 1. If 푁 ≥ 1 and 휎 ≥ 푐 > 휎푎, then

∞ ∞ 푓(푛) |푓(푛)| |∑ | ≤ 푁−(휎−푐) ∑ . 푛푠 푛푐 푛=푁 푛=푁

Proof. Lemma 1 is true since

∞ ∞ ∞ 푓(푛) |푓(푛)| |푓(푛)| |∑ | ≤ ∑ = ∑ 푛푠 푛휎 푛푐푛(휎−푐) 푛=푁 푛=푁 푛=푁

|푓(푛)| ≤ 푁−(휎−푐) ∑∞ .  푛=푁 푛푐

Theorem 4. If 퐹(푠) is given in Equation (3.4), then

lim 퐹(휎 + 푖푡) = 푓(1) 휎→+∞ is uniform for – ∞ < 푡 < +∞.

푓(푛) Proof. Since 퐹(푠) = 푓(1) + ∑∞ , we only want to show that the second term 푛=2 푛푠 tends toward 0 as 휎 → +∞. Choose 푐 > 휎푎. Then, for 휎 ≥ 푐, Lemma 1 implies

푓(푛) |푓(푛)| 퐴 |∑∞ | ≤ 2−(휎−푐) ∑∞ = , 푛=2 푛푠 푛=2 푛푐 2휎

퐴 where 퐴 is independent of 휎 and 푡. Since → 0 as 휎 → +∞, Theorem 4 is true.  2휎

3.3 Multiplication of the Dirichlet Series

Multiplication of the Dirichlet series involves relating products of the Dirichlet series with the Dirichlet convolution of their coefficients. Theorem 5 states why this can be done.

Theorem 5. Given two functions 퐹(푠) and 퐺(푠) represented by the Dirichlet series

∞ 푓(푛) 퐹(푠) = ∑ for 휎 > 푎 푛푠 푛=1 and

∞ 푔(푛) 퐺(푠) = ∑ for 휎 > 푏, 푛푠 푛=1 then,

ℎ(푛) 퐹(푠)퐺(푠) = ∑∞ for 휎 > max {푎, 푏} (3.5) 푛=1 푛푠 where ℎ = 푓 ∗ 푔, the Dirichlet convolution of 푓 and 푔, defined by

푛 ℎ(푛) = ∑ 푓(푑)푔 ( ) . 푑 푑|푛

훼(푛) On the other hand, if 퐹(푠)퐺(푠) = ∑ for all 푠 in a sequence {푠 } with 휎 → 푛푠 푘 푘

+∞ 푎푠 푘 → +∞, then 훼 = 푓 ∗ 푔.

Proof. For any 푠 where both series converge absolutely, we have

∞ ∞ ∞ ∞ 푓(푛) 푔(푚) 푓(푛)푔(푚) 퐹(푠)퐺(푠) = ∑ ∑ = ∑ ∑ . 푛푠 푚푠 (푚푛)푠 푛=1 푚=1 푛=1 푚=1

Because the series converges absolutely, it is possible to multiply the two series and then rearrange the terms in any way desired without changing the sum. For example, collect the terms for which 푚푛 is constant, such as 푚푛 = 푘. The possible values of 푘 are 1, 2, …. Therefore,

∞ ∞ ℎ(푘) 퐹(푠)퐺(푠) = ∑ ( ∑ 푓(푛)푔(푚)) 푘−푠 = ∑ , 푘푠 푘=1 푚푛=푘 푘=1 where ℎ(푘) = ∑푚푛=푘 푓(푛)푔(푚) = (푓 ∗ 푔)(푘). This proves the first assertion. To show the second assertion, use the uniqueness theorem. 

We will consider some examples:

1 휇(푛) • Example 1. Series ∑ and ∑ both converge absolutely for 휎 > 1. Taking 푛푠 푛푠

1 푓(푛) = 1 and 푔(푛) = 휇(푛) in Equation (3.5), ℎ(푛) = [ ]. Therefore, 푛

휇(푛) 휁(푠) ∑ = 1 if 휎 > 1. 푛푠

The resulting equation shows that 휁(푠) ≠ 0 for 휎 > 1 and that

휇(푛) 1 ∑∞ = if 휎 > 1.  푛=1 푛푠 휁(푠)

• Example 2. For a more general scenario, assume that 푓(1) ≠ 0, and let 푔 =

1 푓−1 be the Dirichlet inverse of 푓, 푖. 푒 푓 ∗ 푔(푛) = [ ]. Then, in any half-plane 푛

푓(푛) 푔(푛) where the series 퐹(푠) = ∑ and 퐺(푠) = ∑ both converge absolutely, 푛푠 푛푠

1 퐹(푠) ≠ 0 and 퐺(푠) = .  퐹(푠)

• Example 3. Start with 푓(푛) = 1 and 푔(푛) = 휑(푛), Euler’s totient. Since 휑(푛) ≤

휑(푛) 푛, the series ∑ converges absolutely for 휎 > 2. Also, ℎ(푛) = ∑ 휑(푑) = 푛푠 푑|푛

푛. Therefore, Equation (3.5) leads to

∞ ∞ 휑(푛) 푛 휁(푠) ∑ = ∑ = 휁(푠 − 1) if 휎 > 2. 푛푠 푛푠 푛=1 푛=1

Therefore,

∞ 휑(푛) 휁(푠 − 1) ∑ = if 휎 > 2.  푛푠 휁(푠) 푛=1

훼 훼 • Example 4. Take 푓(푛) = 1 and 푔(푛) = 푛 . Then, ℎ(푛) = ∑푑|푛 푑 = 휎훼(푛), and

Equation (3.5) gives us

휎 (푛) 휁(푠)휁(푠 − 훼) = ∑∞ 훼 if 휎 > 푚푎푥{1, 1 + 푅푒(훼)}.  푛=1 푛푠

• Example 5. Let Λ(푛) be the von Mangoldt function. Note that

푛 Λ(푛) = ∑ 휇(푑) log ( ). 푑|푛 푑

Then,

1 푓 (푠) = 푓 (푠)푓 (푠) = 푓 (푠). Λ 휇 푙표푔 휁(푠) log

When 푅푒 푠 > 1,

∞ ∞ ∞ ′ log푛 1 ′ 1 푓 (푠) = ∑ = − ∑ ( ) = − (∑ ) = −휁′(푠). log 푛푠 푛푠 푛푠 푛=1 푛=1 푛=1

Therefore,

Λ(푛) 휁′ 푓 (푠) = ∑∞ = − (푠).  Λ 푛=1 푛푠 휁

3.4 Euler Product

The next theorem, discovered and postulated by Euler in 1737, sometimes is referred to as the analytic version of the Fundamental Theorem of Arithmetic.

Theorem 6. Let 푓 be a multiplicative arithmetical function where the series 훴푓(푛) converges absolutely. Then, the sum of the series can be written as the absolutely convergent infinite product

∞ 2 ∑푛=1 푓(푛) = ∏푝{1 + 푓(푝) + 푓(푝 ) + ⋯ } (3.6) extended over all prime numbers. If 푓 is completely multiplicative, then the product simplifies to

1 ∑∞ 푓(푛) = ∏ . (3.7) 푛=1 푝 1−푓(푝)

Note that in each case, the product is referred to as the Euler product of the series.

We will consider the following example:

Example. By taking 푓(푛) = 1, 휇(푛), 휑(푛), and 휎훼(푛), respectively, we obtain the following Euler products:

∞ 1 1 휁(푠) = ∑ = Π if 휎 > 1. 푛푠 푝 1 − 푝−푠 푛=1

∞ 1 휇(푛) = ∑ = Π (1 − 푝−푠) if 휎 > 1. 휁(푠) 푛푠 푝 푛=1

∞ 휁(푠 − 1) 휑(푛) 1 − 푝−푠 = ∑ = Π if 휎 > 2. 휁(푠) 푛푠 푝 1 − 푝1−푠 푛=1

휎 (푛) 1 휁(푠)휁(푠 − 훼) = ∑∞ 훼 = Π if 휎 > max {1,1 + 푅푒(훼)}.  푛=1 푛푠 푝 (1−푝−푠)(1−푝훼−푠)

CHAPTER 4

BASIC PROPERTIES OF 휻(풔)

Euler’s gamma function, denoted by (s), can be defined in several ways. The

Weierstrass definition (which essentially belongs to Gauss) ( see [4], for example) will be used here.

4.1 Definition and Basic Properties of the Euler Gamma Function

Let (푠) be the Euler gamma function defined by the equation

1 푠 −푠 = 푠푒훾푠 ∏+∞ (1 + ) 푒 ⁄푛, (4.1) Γ(푠) 푛=1 푛 where  is the Euler constant defined as

푁 1 훾 ≔ lim (∑푛=1 − log푁). 푁⟶∞ 푛

It follows from the definition of the Euler gamma function that the RHS of Equation

(4.1) defines a holomorphic function with simple zeros at s = 0, –1, –2, …. Moreover,

1 is an entire function of order at most one. Therefore, (s) is meromorphic and has (푠) simple poles at s = 0, –1, –2, …, and (s) is zero-free.

Proposition 1. We have

1 푠 1 −푠 • = 푠 ∏+∞ (1 + ) (1 + ) . [Euler’s formula] (4.2) (s) 푛=1 푛 푛

1 푠(푠+1)….(푠+푁−1) • = lim . [Euler-Gauss formula] (4.3) (s) 푁⟶+∞ (푁−1)!푁푠

• If 푠 ≠ 0, then (s+1) = s(s). [function equation] (4.4)

푁 1 Proof. For the proof, it is known that 훾 = lim ( ∑푛=1 − lo푔푁 ). Therefore, 푁⟶+∞ 푛 ∞ 푁 푁−1 + −푠 1 1 1 푠 푒훾푠 = lim exp {푠 ∑ − 푠 ∑ log (1 + )} = ∏ (1 + ) 푒 ⁄푛 푁⟶+∞ 푛 푛 푛 푛=1 푛=1 푛=1 and, from the definition of the function (s), we get

1 푠 1 −푠 = 푠 ∏+∞ (1 + ) (1 + ) , (s) 푛=1 푛 푛 and thus Equation (4.2) is proved. Note that

푠+푛 1+푛 −푠 1 푠+푛 ∏푁−1 ( ) ( ) = ∏푁−1 . 푛=1 푛 푛 푁푠 푛=1 푛

Thus, from Equation (4.2), we get

푁−1 푁−1 1 푠 + 푛 1 + 푛 −푠 1 푠 + 푛 = 푠 lim ∏ ( ) ( ) = 푠 lim ∏ (s) 푁⟶+∞ 푛 푛 푁⟶+∞ 푁푠 푛 푛=1 푛=1

푠(푠 + 1) … . (푠 + 푁 − 1) 1 = lim ∙ , 푁⟶+∞ ( 푁 − 1)! 푁푠 which proves Equation (4.3).

If 푠 ≠ 0 , then from Equation (4.3), we get

1 (푠 + 1) … . (푠 + 푁) = lim (s + 1) 푁⟶+∞ (푁 − 1)! 푁푠+1

1 푠(푠 + 1) … . (푠 + 푁 − 1) 푠 + 푁 1 = lim ∙ = , 푠 푁⟶+∞ (푁 − 1)! 푁푠 푁 s(s) and this yields Equation (4.4). 

Corollary 1. For every natural number m, (푚 + 1) = 푚!.

Next, we state some useful propositions without giving a proof.

Proposition 2.

• [integral formula] If 푅푒 푠 > 0, then

+∞ Γ(s) = ∫ 푒−푡푡s−1푑푡. 0

• [addition formula] For any number 푠 not an integer, then

1 sin (푠) = . (푠)(1−푠) 

• [Legendre formula] For ∀ 푛 ∈ ℕ, then

1 1  ( ) (2푠) = 22푠−1(푠) (푠 + ). 2 2

4.2 Stirlig’s Formula

It is essential to know the behavior of (푠) as |푠| → ∞.

Theorem 7 (Stirlig’s Formula). If − 휋 + 훿 < arg 푠 < 휋 − 훿 , and 훿 > 0, then

1 1 1 log (푠) = (푠 − ) log 푠 − 푠 + log 2휋 + 푂 ( ), 2 2 |푠| where the constant in the symbol 푂 depends only on 훿. Moreover, in −휋 + 훿 < arg 푠 <

1 휋 − 훿, has no zeros. Therefore, by the definition of (푠), it follows that (푠)

+∞ 1 푠 푠 log = log 푠 + 훾푠 + ∑ (log (1 + ) − ). (s) 푛 푛 푛=1

Corollary 2.

′ 1 1 1 • − (푠) = +  + ∑+∞ ( − ) follows directly from the definition of the Euler  푠 푛=1 푛+푠 푛

gamma function.

′ 1 • (푠) = log푠 + 푂 ( ) if − 휋 + 훿 < arg푠 < 휋 − 훿.  |푠|

−휋|푡| 1 푖휋 1 휎− | | 휆 (휎− ) • (σ + it) = (1 + 푂(|푡|−1)√2휋 푒 2 |푡| 2 푒푖푡(log 푡 −1)푒 2 2 ,

1 if 푡 ≥ 1 where 휆 = { −1 if 푡 ≤ −1.

푛 푛 • As 푛 → +∞, 푛! ~√2휋푛 ( ) . 푒

1 1 1 • log 푛! = log Γ(푛 + 1) = (푛 + ) log(푛 + 1) − (푛 + 1) + log 2휋 + 푂 ( ) 2 2 푛

1 1 1 1 1 푛+2 = (푛 + ) log(푛) − 푛 + log 2휋 + 푂 ( ) + log (1 + ) − 1. 2 2 푛 푛

4.3 Analytic Continuation of 휻(풔)

In this section, we follow [4], [5], [6] and [7] in providing the definition of theta function and Theorems that are related to Riemann zeta function.

If 푅푒 푠 > 1, then the Riemann zeta-function 휁(푠) is defined by

+∞ 1 휁(푠) = ∑ . 푛푠 푛=1

It follows that ζ(s) is an analytic function in the half-plane 푅푒 푠 > 1. Note that for 푥 ≥ 1, then

+∞ 1 1 1 휁(푠) = ∑ = ∑ + ∑ 푛푠 푛푠 푛푠 푛=1 푛≤푥 푛>푥

푙 푁 (푗−1) 푁 푁 (푙) 1 푑푡 푗 1 푙+1 1 = ∑ + lim [∫ + ∑(−1) 푏푗(푡) ( ) | + (−1) ∫ 푏푙(푡) ( ) 푑푡]. 푛푠 푁→+∞ 푡푠 푡푠 푡푠 푛≤푥 푥 푗=1 푥 푥

Therefore, if 푅푒 푠 > 1, then

푙 1 푥1−푠 푏 (푥) 푏 (푥) 휁(푠) = ∑ + + 1 + ∑ 푠(푠 + 1) … (푠 + 푗 − 2) . 푗 푛푠 푠 − 1 푥푠 푥푠+푗−1 푛≤푥 푗=2 +∞ 푏 (푡) − 푠(푠 + 1) … (푠 + 푙 − 1) ∫ 푙 푑푡. 푡푠+푙 푥

When 푅푒 푠 > −푙 + 1, the RHS is analytic. Therefore, 휁(푠) has been analytically extended to the half-plane 푅푒 푠 > −푙 + 1 in this way. Let 푙 ⟶ +∞ to get the analytic extension of

휁(푠) to the whole complex plane ℂ.

Corollary 3. 휁(푠) is a meromorphic function with a simple pole at 푠 = 1 with residue

Next, we give the functional equation of 휁(푠) after stating a property of the classic theta-function.

Definition 2. For 푅푒 푠 > 0, the theta function 휃(푠) is defined by

+∞ −푛2휋푠 θ(푠) = ∑푛=−∞ 푒 .

Theta function satisfies a functional equation. In particular, when 푠 is a positive number, the functional equation takes the following form.

Lemma 2. For real number 푥 > 0, we have

1 휃 ( ) = 푥 θ(푥). 푥 √

Theorem 8 (Functional Equation of Riemann zeta Function). For this theorem,

−푠 푠 −(1−푠) 1−푠 휋 2 Γ ( ) 휁(푠) = 휋 2 Γ ( ) 휁(1 − 푠). (4.5) 2 2

Proof. When 푅푒 푠 > 0, we have

+∞ 푠 푠 Γ ( ) = ∫ 푒−푡푡2−1푑푡. 2 0

Substituting 푡 = 푛2휋푥, we get +∞ 푠 2 푠 Γ ( ) = ∫ 푒−푛 휋푥(푛2휋푥)2−1푑(푛2휋푥) 2 0

푠 +∞ 2 −1 −푛 휋푥( 2 )2 2 =∫0 푒 푛 휋푥 푛 휋 푑푥

푠 푠 +∞ 2 −1 푠 2 −푛 휋푥 2 = 푛 휋 ∫0 푒 푥 푑푥.

Therefore,

+∞ −푠 푠 2 푠 휋 2 Γ ( ) 푛−푠 = ∫ 푒−푛 휋푥 푥2−1 푑푥. 2 0

Assume 푅푒 푠 > 1. By summing over 푛, we have

−푠 푠 푠 +∞ 2 −1 휋 2 Γ ( ) 휁(푠) = ∑+∞ ∫ 푒−푛 휋푥 푥2 푑푥. 2 푛=1 0

Since 푥 > 0, by the Fubini Theorem, we can interchange the order of summation and integration. Therefore, for 푠 ∈ ℂ with 푅푒 푠 > 1,

−푠 푠 푠 +∞ −1 휋 2 훤 ( ) 휁(푠) = ∫ 휔(푥) 푥2 푑푥, (4.6) 2 0

+∞ −푛2휋푥 −휋푥 where 휔(푥) = ∑푛=1 푒 ≪ 푒 for 푥 > 0. By definition 2 and Lemma 2 ,

+∞ −푛2휋푥 θ(x) = ∑푛=−∞ 푒 = 1 + 2 휔(푥).

The objective is to replace the right-hand side of Equation (4.6) with something that converges for every 푠 ∈ ℂ. It is clear that for 푠 ∈ ℂ with 푅푒 푠 < 0, there are

∞ 1 problems if 푥 ↓ 0. To overcome these problems, we split the integral ∫0 into ∫0 +

∞ 1 ∞ −1 ∫1 and transform ∫0 into an integral ∫1 through the exchange of 푥 = 푥. By Lemma 2, we have

1 2휔 ( ) + 1 = 푥 (2 휔(푥) + 1), 푥 √ which gives 1 1 1 1 1 휔 ( ) = 푥2휔(푥) + 푥2 − . 푥 2 2

Therefore,

−푠 푠 푠 푠 +∞ −1 +∞ 1 1− 휋 2 Γ ( ) 휁(푠) = ∫ 휔(푥) 푥2 푑푥 − ∫ 휔 ( ) 푥 2푑푥−1 2 1 1 푥

+∞ +∞ 푠 1 1 1 1 푠 = ∫ 휔(푥) 푥2−1푑푥 + ∫ (푥2휔(푥) + 푥2 − ) 푥1−2 푥−2푑푥 2 2 1 1 −(1+푠) 푠 푠 −(1+푠) +∞ 1 − −1 +∞ −1 = ∫ (푥 2 − 푥 2 ) 푑푥 + ∫ 휔(푥) (푥2 + 푥 2 ) 푑푥. 1 2 1

Because 푅푒 푠 > 1, the first integral is equal to

1 2 −(푠−1) 2 −푠 +∞ 1 2 2 1 1 1 [− 푥 2 + 푥 2 ] = [ − ] = − = . 2 푠 − 1 푠 1 2 푠 − 1 푠 푠 − 1 푠 푠(푠 − 1)

Therefore,

−푠 푠 −(1+푠) 푠 1 +∞ −1 휋 2 Γ ( ) 휁(푠) = + ∫ 휔(푥) (푥2 + 푥 2 ) 푑푥. (4.7) 2 푠(푠−1) 1

Since 휔(푥) ≪ 푒−휋푥 as 푥 ⟶ +∞, It follows from Equation (4.7) that the right-hand side of Equation (4.6) is an analytic function for any s≠ 0,1, and keeps the same with the interchange of 푠 and 1 − 푠; that is,

−푠 푠 −(1−푠) 1−푠 휋 2 Γ ( ) 휁(푠) = 휋 2 Γ ( ) 휁(1 − 푠).  2 2

Note; form Equation (4.7), let

−푠 푠 휉(푠) = 푠(푠 − 1)휋 2 Γ ( ) 휁(푠), 2 then

+∞ 푠 −(1+푠) 휉(푠) = 1 + 푠(푠 − 1) ∫ 휔(푥) (푥2−1 + 푥 2 ) 푑푥. 1 This directly yields

Corollary 4. The function

1 −푠 푠 휉(푠) = 푠(푠 − 1)휋 2 Γ ( ) 휁(푠) 2 2 is entire, and

휉(푠) = 휉(1 − 푠).

푠 From Theorem 8, it follows that since Γ ( ) has simple poles at 푠 = 2

−2, −4, −6, … , −2푛, …, then 휁(푠) = 0 (i.e., has simple zeros) at these points. At 푠 = 0 the

푠 zeta function is not equal to 0 because the simple pole of Γ ( ) at 푠 = 0 cancels the 2 simple pole of 휁(푠) at 푠 = 1. Moreover, the zeros of 휁(푠) at 푠 = −2, −4 , −6, … are all simple because

1+2푛 1 + 2푛 휋+푛Γ(−푛)휁(−2푛) = 휋− 2 Γ ( ) 휁(1 + 2푛) ∈ ℝ ≠ 0. 2

In addition to the trivial zeros, the zeta function has infinitely many nontrivial zeros, all distributing in the strip 0 ≤ 푅푒 푠 ≤ 1.

Theorem 9. Let 휌푛 be the zeros of 휉(푠) such that 0 ≤ 푅푒 휌푛 ≤ 1, where 휉(푠) is an entire function of order one that has infinitely many zeros 휌푛. The zeros of 휉(푠) are the

1 1 nontrivial zeros of 휁(푠). Then, the series ∑ diverges and ∑ 1+휀 converges for any |휌푛| |휌푛|

휀 > 0.

1 1 Proof. For 푅푒 푠 > 1, 푠 = 휎 + 푖푡. Since = ∏ (1 − ) , 휁(푠) 푝 푝푠

1 −1 휁(2휎) 1 −1 (1 + ) = > 0 |휁(푠)| = |∏푝(1 − 푠) | ≥ ∏푝 푝푅푒 푠 휁(휎) . 푝

Therefore, when 푅푒 푠 > 1, the zeta function has no zero, thus 휉(푠) ≠ 0 for 푅푒 푠 > 1.

It follows from the functional equation that 휉(푠) ≠ 0 for 푅푒 푠 < 0. While all trivial zeros

푠 of 휁(푠) (푠 = −2, −4, … ) are canceled by the poles of Γ ( ), the zeros of 휉(푠) are exactly 2 the non-trivial zeros of 휁(푠). Therefore, all these zeros are distributed in the strip 0 ≤

푅푒 푠 ≤ 1.

Now let us determine the order of 휉(푠). We just need to do this for 푅푒 푠 > 0 because of the functional equation. Note that, when 푅푒 푠 > 0,

s +∞ 휁(푠) = − s ∫ 푡−푠−1{푡} 푑푡. 푠−1 1

We thus have 휁(푠) ≪ |푠| when 푅푒 푠 > 0. Moreover, since |Γ(푠)| ≤ 푒푐|푠|log|푠| as |푠| ⟶ +

∞, we have

|휉(푠)| ≪ 푒푐′|푠| log|푠|, the order of 휉(푠) thus is at most 1. On the other hand, on the real axis log Γ(푠) ~ 푠 log푠 as 푠 ⟶ +∞, thus the order of 휉(푠) equals 1. Besides, since log|휉(푠)| ≫ |푠|log|푠|, from

1 the classic result about entire functions, ∑ diverges where 휌푛 are the zeros of |휌푛|

휉(푠). Consequently, 휉(푠) has infinitely many zeros; that is, 휁(푠) has infinitely many non-

1 trivial zeros. The series ∑ 1+휀 converges for any 휀 > 0.  |휌푛|

Corollary 5. We have

푠 푠 퐴+퐵푠 +∞ 휌 휉(푠) = 푒 ∏푛=1 (1 − ) 푒 푛. 휌푛

Proof. Proof of Corollary 5 is clear because 휉(푠) is an entire function of order 1. 

Corollary 6. The nontrivial zeros of the zeta function are distributed symmetrically

1 with respect to the lines 푅푒 푠 = and 퐼푚 푠 = 0. 2

Proof. The proof comes directly from the functional equation of Riemann zeta function

휉(푠) = 휉(1 − 푠) 푎푛푑 ̅휉̅(̅̅푠̅̅) = 휉(푠̅). 

Theorem 10. We have the formula

+∞ +∞ 휁′ 1 1 1 1 1 (푠) = − + ∑ ( + ) + ∑ ( − ) + 퐵0 , 휁 푠 − 1 푠 − 휌푛 휌푛 푠 + 2푛 2푛 푛=1 푛=1 where 퐵0 is an absolute constant, and 휌푛 are the nontrivial zeros of 휁(푠).

Proof. Take the logarithmic differentiation of the left and right sides of Corollary 5.

We get

휉′ +∞ 1 1 (푠) = 퐵 + ∑푛=1 ( + ). 휉 푠−휌푛 휌푛

We know from Corollary 4 that

−푠 푠 휉(푠) = 푠(푠 − 1)휋 ⁄2Γ ( ) 휁(푠). (4.8) 2

Take the logarithmic differentiation of the left and right sides of Equation (4.8).

Therefore,

휁 휉 1 1 1 1 Γ′ 푠 ′ (푠) = ′ (푠) − − + log휋 − ( ). 휁 휉 푠 푠−1 2 2 Γ 2

We know from Corollary 2 that

Γ′ 1 1 1 − (푠) = + 훾 + ∑+∞ ( − ). Γ 푠 푛=1 푛+푠 푛

Therefore,

1 Γ′ s 1 1 1 − ( ) = + 훾 + ∑+∞ ( − ). 2 Γ 2 푠 푛=1 2푛+푠 2푛

The proof is complete. 

Theorem 11. Let 휌푛 = 훽푛 + 푖훾푛, 푛 = 1,2, … be the nontrivial zeros of 휁(푠), and let

푇 ≥ 2. Then,

+∞ 1 ∑푛=1 2 ≤ 푐 log 푇. 1+(푇−훾푛)

Proof. Let s = 2 + 푖푇 in Theorem 10. Then,

+∞ +∞ 휁′ −1 1 1 1 1 (2 + 푖푇) = + ∑ ( + ) + ∑ ( − ) + 퐵0. 휁 (2 + 푖푇) − 1 2 + 푖푇 − 휌푛 휌푛 2 + 푖푇 + 2푛 2푛 푛=1 푛=1

Note that

+∞ 1 1 1 1 2 + 푖푇 |∑ ( − ) | ≪ ∑ ( + ) + ∑ | | 2 + 푖푇 + 2푛 2푛 푛 푛 2푛(2푛 + 2 + 푖푇) 푛=1 푛≤푇 푛>푇

푇 ≪ log 푇 + ∑ ≪ log 푇 + 1 ≪ log 푇. (4.9) 푛>푇 푛2

휁′ Λ(푛) Since | (푠)| = |∑+∞ | < 푐 and, according to Theorem 10, we have 휁 푛=1 푛2+푖푇 1

+∞ +∞ 휁′ 1 1 1 1 1 −푅푒 (푠) = 푅푒 ( − 퐵0 − ∑ ( − ) ) − 푅푒 ∑ ( + ) 휁 푠 − 1 푠 + 2푛 2푛 푠 − 휌푛 휌푛 푛=1 푛=1

+∞ 1 1 ≤ 퐶2 log 푇 − 푅푒 ∑ ( + ). 푠 − 휌푛 휌푛 푛=1

It follows that

+∞ 1 1 푅푒 ∑푛=1 ( + ) ≤ 퐶3 log 푇 (4.10) 푠−휌푛 휌푛

for some absolute constant 퐶3 > 0.

Note that

1 훽푛 푅푒 = 2 2 ≥ 0. (4.11) 휌푛 훽푛 +훾푛 Since 0 ≤ 훽푛 ≤ 1, we have

1 1 푅푒 = 푅푒 푠 − 휌푛 (2 − 훽푛) + 푖(푇 − 훾푛)

1 1 (2−훽푛) (2−훽푛) 2 = 2 2 ≥ 2 ≥ 2 . (4.12) (2−훽푛) +(푇−훾푛) 1+(푇−훾푛) 1+(푇−훾푛)

Therefore, from Equations (4.10), (4.11), and (4.12), we get

+∞ 1 ∑푛=1 ( 2) ≤ 퐶 log 푇.  1+(푇−훾푛)

Corollary 7. The number of zeros for 휌푛 in the zeta function that satisfy

푇 ≤ |퐼푚 휌푛| ≤ 푇 + 1 is at most 퐶′푙표푔푇.

Proof. According to Theorem 11,

+∞ 1 ∑ 2 ≤ 퐶 log 푇. 1 + (푇 − 훾푛) 푛=1

Thus

1 ∑푇≤퐼푚휌 ≤푇+1 2 ≤ 퐶 log 푇. (4.13) 푛 1+(푇−훾푛)

Note that, when 푇 ≤ 훾푛 ≤ 푇 + 1, we have

1 1 2 ≥ . 1 + (푇 − 훾푛) 2

푇ℎ푢푠

# { 휌푛, 푇 ≤ 퐼푚휌푛 ≤ 푇 + 1} ≤ 2 퐶 푙표푔푇.

The corollary then follows from this and corollary 6. 

Corollary 8. For 푇 ≥ 2, we have

1 ∑|푇−훾 |>1 2 = 푂(log 푇). 푛 |푇−훾푛|

2 Proof. Note that if |푇 − 훾푛| > 1, then |푇 − 훾푛| > 1. Therefore,

2 2 1 + |푇 − 훾푛| < 2|푇 − 훾푛| , which implies that

1 1 2 > 2. 1+|푇−훾푛| 2|푇−훾푛|

Therefore,

1 1 1 > . 1+(푇−훾)2 2 (푇−훾)2

The proof is complete. 

Corollary 9. For 푠 = 휎 + 푖푡, where −1 ≤ 휎 ≤ 2 and |푡| ≥ 2, we have

휁′ 1 (푠) = ∑|푡−훾 |≤1 + 푂(log|푡|), 휁 푛 푠−휌푛 where 휌푛 are the nontrivial zeros of 휁(푠), 푛 = 1,2, …, and 훾푛 = Im 휌푛.

Proof. Since Equation (4.9) holds for all 푠 = 휎 + 푖푡, with |푡| ≥ 2 and −1 ≤ 휎 ≤ 2,

휁′ +∞ 1 1 (푠) = ∑푛=1 ( + ) + 푂(log |푡|). (4.14) 휁 푠−휌푛 휌푛

From Equation (4.14), we subtract the same equation using 푠 = 2 + 푖푡, and we get

+∞ 휁′ 휁′ 1 1 (푠) − (2 + 푖푡) = ∑ ( − ) + 푂(log |푡|). 휁 휁 푠 − 휌 2 + 푖푡 − 휌 푛=1 푛 푛 Thus, +∞ 휁′ 1 1 (푠) = ∑ ( − ) + 푂(log |푡|). 휁 푠 − 휌푛 2 + 푖푡 − 휌푛 푛=1

휁′ 1 1 1 1 (푠) = ∑ ( − ) + ∑ ( − ) + 푂(log |푡|). 휁 푠 − 휌푛 2 + 푖푡 − 휌푛 푠 − 휌푛 2 + 푖푡 − 휌푛 |훾푛−푡|≤1 |훾푛−푡|>1

Now, if |훾푛 − 푡| > 1, then

1 1 2−휎 3 | − | ≤ 2 ≤ 2. 휎+푖푡−휌푛 2+푖푡−휌푛 (훾푛−푡) (훾푛−푡)

According to Corollary 8,

1 1 3 | ∑ ( − )| ≤ ∑ 2 ≪ log|푡|. 푠 − 휌푛 2 + 푖푡 − 휌푛 (훾푛−푡) |훾푛−푡|>1 |훾푛−푡|>1

Therefore,

휁′ 1 1 (푠) = ∑ ( − ) + 푂(log |푡|). 휁 푠 − 휌푛 2 + 푖푡 − 휌푛 |훾푛−푡|≤1

1 1 = ∑ − ∑ + 푂(log |푡|). 푠 − 휌푛 2 + 푖푡 − 휌푛 |훾푛−푡|≤1 |훾푛−푡|≤1

From Corollary 7,

1 ∑ ≪ log|푡|. 2 + 푖푡 − 휌푛 |훾푛−푡|≤1

Therefore,

휁′ 1 (푠) = ∑ + 푂(log|푡| + #{휌푛 ; 푡 − 1 ≤ 훾푛 ≤ 푡 + 1}) 휁 푠 − 휌푛 |푡−훾푛|≤1

1 = ∑|푡−훾 |≤1 + 푂(log |푡|).  푛 푠−휌푛 4.4 A Zero-Free Region of 휻(풔)

Theorem 12 (de la Vallee-Poussin). There exists an absolute constant 퐶 > 0 such that 휁(푠) ≠ 0 in the domain

푐 푅푒 푠 = 휎 ≥ 1 − . log(|푡|+2)

Proof. First, claim that for 휎, 푡 ∈ 푅 with 휎 > 1. We have

휁′ 휁′ 휁′ 3 {− (휎)} + 4 {−푅푒 (휎 + 푖푡)} + {−푅푒 (휎 + 2푖푡)} ≥ 0. (4.15) 휁 휁 휁

To prove the claim, we first note that for any real number θ,

3 + 4 cos휃 + cos(2휃) = 2(1 + cos (휃))2 ≥ 0. (4.16)

For 푠 = 휎 + 푖푡 with Re s = 휎 > 1,

+∞ +∞ 휁′ Λ(푛) – (푠) = ∑ = ∑ Λ(푛)푛−휎푒−푖푡 log푛. 휁 푛푠 푛=1 푛=1

Therefore,

+∞ 휁′ – 푅푒 (푠) = ∑ Λ(푛)푛−휎 cos(푡 log푛). 휁 푛=1

It follows that from Equation (4.16),

휁′ 휁′ 휁′ 3 {− (휎)} + 4 {−푅푒 (휎 + 푖푡)} + {−푅푒 (휎 + 2푖푡)} 휁 휁 휁 +∞ = ∑(Λ(푛)푛−휎(3 + 4 cos(푡 log 푛) + cos (2푡 log 푛)) ≥ 0. 푛=1

Thus, the claim is proved.

Next, note that 휁(푠) has a pole at 푠 = 1, and thus there exists a positive constant 훾0 such that 휁(푠) ≠ 0 in the domain |푠 − 1| ≤ 훾0. Let 휌푛 = 훽푛 + 푖훾푛 be a zero of 휁(푠) such that |훾푛| > 훾0 and 0≤ 훽푛 ≤1. We want to prove that 훽푛 and 훾푛 satisfy the inequality given in the theorem. From Theorem 10 and Corollary 7, we obtain for 푠 = 휎 > 1,

휁′ 1 − (휎) < + 퐵 , (4.17) 휁 휎−1 1 for some constant 퐵1 such that 퐵1 > 0. In addition, from Theorem 10, we get for s = 휎 +

푖푡 with 1 < 휎 ≤ 2 and |푡| > 훾0

+∞ +∞ 휁′ 1 1 1 1 1 −푅푒 (푠) = 푅푒 − 푅푒 { ∑ ( + )} + 푅푒 {∑ ( − )} − 퐵0 휁 푠 − 1 푠 − 휌푘 휌푘 2푘 푠 + 2푘 푘=1 푘=1 for some constant 퐵0 . Note that

+∞ 1 1 |∑ ( − )| ≤ 퐴 log(|푡| + 2). 2푘 푠 + 2푘 푘=1

Therefore,

휁′ +∞ 1 1 −푅푒 (푠) < 퐴 log(|푡| + 2) − 푅푒 ∑푘=1 ( + ), 휁 푠−휌푘 휌푘 where 퐴 > 0 is an absolute constant. Since 0 ≤ 훽푘 ≤1,

1 휎 − 훽푘 푅푒 = 2 > 0. 푠 − 휌푘 |푠 − 휌푘|

Moreover,

1 훽 훽 푅푒 = 푘 = 푘 ≥ 0. 2 2 2 휌푘 훽푘 + 훾푘 |휌푘|

Therefore,

′ 휁 휎−훽푛 −푅푒 (휎 + 푖푡) < 퐴1 log(|푡| + 2) − 2. (4.18) 휁 |푠−휌푛|

It follows from this inequality that

휁 −푅푒 ′ (휎 + 2푖푡) < 퐴 log(2|푡| + 2). (4.19) 휁 2

Putting Equations (4.17), (4.18), and (4.19) into the claim, we get

3 4(휎−훽푛) − 2 2 + 퐴 log(|푡| + 2) ≥ 0. (4.20) 휎−1 (휎−훽푛) +(푡−훾푛)

The inequality in Equation (4.20) holds for any 푡 such that |푡| > 훾0 and 1 < 휎 ≤ 2,

1 where 퐴 > 0 is an absolute constant. Next, let 푡 = 훾푛 and 휎 = 1 + . 2퐴 log(|훾푛|+2)

Therefore, Equation (4.20) gives

4 3 ≤ + 퐴 log(|훾푛| + 2) = 7 퐴 log(|훾푛| + 2) 휎 − 훽푛 휎 − 1 and then

4 휎 − 훽푛 ≥ , 7 퐴 log(|훾푛|+2) which implies that

1 훽푛 ≤ 1 − , 14 퐴 log(|훾푛|+2)

and the proof is complete. 

Corollary 10. Let 푐 > 0 be the absolute constant in Theorem 12, and let 푇 ≥ 2. Then, in the domain

푐 휎 ≥ 1 − , 2 ≤ |푡| ≤ 푇, (4.21) 2 log(푇+2) we have

휁 | ′ (푠)| ≪ log2푇. 휁

Proof. From Corollary 9,

휁′ 1 (푠) = ∑ + 푂(log|푡|). 휁 푠 − 휌푛 |푡−훾푛|≤1

For 푠 = 휎 + 푖푡 in the region in Equation (4.21),

휁′ 1 | (푠)| ≤ ∑ + 푂(log 푇). 휁 |휎 − 훽푛 + 푖(푡 − 훾푛)| |푡−훾푛|≤1

푐 푐 Since 휎 ≥ 1 − and 훽 ≤ 1 − , then 2 log(푇+2) 푛 log(푇+2)

휁′ 2 | (푠)| ≤ log(푇 + 2) ∑ 1 + 푂(log 푇) ≪ log2푇, 휁 푐 |푡−훾푛|≤1 which is what we needed to prove. 

CHAPTER 5

PRIME NUMBER THEOREM

Let 휋(푥) be the number of primes up to 푥 such that

휋(푥) ≔ |{푝 ≤ 푥: 푝 is prime}|.

For 푥 > 2, let the logarithm integral be defined as

푥 푑푡 푥 푥 2푥 퐿푖(푥) ≔ ∫ = + + + ⋯. log푡 log 푥 log2푥 log3푥 2

The weakest form of the Prime Number Theorem states that

푥 푥 휋(푥) = + 표 ( ) 푎푠 푥 ⟶ +∞. log 푥 log 푥

Let

휋(푥) = 퐿푖(푥) + 푅(푥), (5.1) a stronger form of PNT gives an error term, the 푅(푥) in (5.1), much smaller than 퐿푖(푥).

In the thesis, we want to prove the stronger form with

푅(푥) ≪ 푥푒−푐′√log 푥 for some absolute constant 푐′ > 0. Instead of studying 휋(푥) directly, it is more convenient to study

휓(푥) ≔ ∑ Λ(푛), 푛≤푥 where Λ(푛) is the Von Mongoldt function. Note that if 휋(푥) has an asymptotic formula,

Equation (5.1), with 푅(푥) being small, then

휓(푥) = ∑ Λ(푛) = ∑ log 푝 + ∑ log 푝 푛≤푥 푝≤푥 푝푘≤푥 푘≥2 푥 = ∫ log 푡 푑 휋(푡) + 푂(√푥 log 푥) 1 푥 푥 푅(푡) = ∫ log 푡 푑 퐿푖(푡) + 푅(푥) log 푥 − ∫ 푑푡 + 푂(√푥 log 푥) 푡 1 1

= 푥 + 푂(√푥 log 푥 + log 푥 푚푎푥푡≤푥|푅(푡)|).

Conversely, if

휓(푥) = 푥 + Δ(푥) for Δ(푥) = 표(푥) 푎푠 푥 ⟶ +∞, then

푥 log 푝 1 휋(푥) = ∑ = ∫ 푑 ∑ log 푝 log 푝 log 푡 푝≤푥 2− 푝≤푡

푥 1 = ∫ 푑(휓(푡) + 푂(√푡 log 푡)) log 푡 2−

푥 1 =∫ 푑(푡 + Δ(푡) + 푂(√푡 log 푡)) 2− log 푡

1 = 퐿푖(푥) + 푂 ( 푥 + 푚푎푥 |Δ(푡)|).  √ log 푥 푡≤푥

Lemma 3. The asymptotic formula, Equation (5.1), with

푅(푥) ≪ 푥푒−푐√log 푥⁄log 푥 is equivalent to

휓(푥) = 푥 + 푂 (푥푒−푐√log 푥). (5.2)

In the rest of the chapter, we will be devoted to proving Equation (5.2). Some of the content comes from [4], [5] and [8].

5.1 Truncated Form of Perron’s Formula

Lemma 4. Assume that 푏 > 1 and that 푇 ≥ 2. Then,

푎푏 1 + 푂 ( ) , 푖푓 푎 > 1 1 푏+푖푇 푎푠 푇| log 푎| ∫ 푑푠 = { . (5.3) 2휋푖 푏−푖푇 푠 푎푏 푂 ( ) , 푖푓 0 < 푎 < 1 푇| log 푎|

Proof. We will consider the two cases separately. In the case of 푎 > 1, let 푈 be a large parameter, and consider a rectangular contour with vertices (푏 ± 푖푇, −푈 ± 푖푇) oriented counterclockwise as shown in Figure 5.1.

I iT

-U 0 1 b

III -iT

Figure 5.1. Rectangular contour with vertices (푏 ± 푖푇, −푈 ± 푖푇) oriented counterclockwise.

According to Cauchy’s residue theorem, we have

1 푏+푖푇 푎푠 (∫ + ∫ + ∫ + ∫ ) = 푅푒 푠 = 1. 2휋푖 푏−푖푇 퐼 퐼퐼 퐼퐼퐼 푠=0 푠

It is enough to prove that when 푎 > 1,

푎푏 ∫ + ∫ + ∫ ≪ . 퐼 퐼퐼 퐼퐼퐼 푇| log 푎|

Note that the integrals along 퐼 and 퐼퐼퐼 are equal in absolute value. If 푎 > 1, then

푎푠 푏 푎휎푑휎 푎푏 |∫ 푑푠 | ≤ ∫ ≤ . 퐼 푠 −푈 √푇2+휎2 푇 log 푎

푎푠 The same bound holds for ∫ 푑푠, and 퐼퐼퐼 푠

푎푠 푇 푎−푢 |∫ 푑푠 | ≤ ∫ 푑푡 ⟶ 0 푎푠 푈 ⟶ +∞. 2 2 퐼퐼 푠 −푇 √푈 + 푡

Therefore, the result holds for 푎 > 1.

In the case 0 < 푎 < 1, let 푈 > 푏 and consider the rectangular contour with vertices

(푏 ± 푖푇, 푈 ± 푖푇) oriented clockwise as shown in Figure 5.2.

iT I

0 b U

-iT III

Figure 5.2. Rectangular contour with vertices (푏 ± 푖푇, 푈 ± 푖푇) oriented clockwise.

According to Cauchy’s residue theorem, we have

1 푏+푖푇 (∫ + ∫ + ∫ + ∫ ) = 0. 2휋푖 푏−푖푇 퐼 퐼퐼 퐼퐼퐼

It suffices to show that, when 0 < 푎 < 1,

푎푏 ∫ + ∫ + ∫ ≪ . 퐼 퐼퐼 퐼퐼퐼 푇| log 푎|

Note that

푈 푎푠 푎휎푑휎 푎푏 |∫ 푑푠 | ≤ ∫ ≤ 2 2 퐼 푠 √푇 + 휎 푇|log 푎| 푏

푎푠 and that the same bound holds for ∫ 푑푠. Also, 퐼퐼퐼 푠

푎푠 푇 푎푢 |∫ 푑푠 | ≤ ∫ 푑푡 ⟶ 0 푎푠 푈 ⟶ +∞. 퐼퐼 푠 −푇 √푈2+푡2

Thus, the result holds for 0 < 푎 < 1. 

Next, let 푓(푠) be the Dirichlet series for an arithmetic function 푎푛, where

+∞ 푎 푓(푠) = ∑ 푛 , 푛푠 푛=1 and 푅푒 푠 is sufficiently large.

Theorem 13 (Truncated Perron’s Formula). Suppose that the series 푓(푠) converges absolutely for 휎 > 1 and that |푎푛| ≤ 퐴(푛), where A(n) > 0 is a monotonically increasing function and that

+∞ |푎 | ∑ 푛 = 푂((휎 − 1)−훼), 훼 > 0 푛휎 푛=1

1 as 휎 ⟶ 1 + 0. Suppose also that for any 푏 ≥ 푏 > 1, 푇 ≥ 1 and 푥 = 푁 + , where N∈ ℕ, 0 2 then we have

1 푏+푖푇 푥푠 푥푏 푥퐴(2푥)log푥 Φ(푥) = ∑ 푎 = ∫ 푓(푠) 푑푠 + 푂 ( ) + 푂 ( ), 푛≤푥 푛 2휋푖 푏−푖푇 푠 푇(푏−1)훼 푇 where the constants in the 푂 symbols depend only on 푏0.

1 푥 Proof. Since 푥 = 푁 + , then ≠ 1 for ∀ 푛 ∈ 푁. The series 푓(푠) converges absolutely 2 푛 for 푠 = 푏 + 푖푡. Integrating term by term and using Lemma 4, we get

+∞ 1 푏+푖푇 푥푠 1 푏+푖푇 (푥/푛)푠 ∫ 푓(푠) 푑푠 = ∑ 푎 ( ∫ 푑푠) 2휋푖 푠 푛 2휋푖 푠 푏−푖푇 푛=1 푏−푖푇

= ∑ 푎푛 + 푅, 푛≤푥 where

푥 푏 푥 −1 푅 = 푂 (∑+∞ |푎 | ( ) 푇−1 |log | ). 푛=1 푛 푛 푛

Next, we split the error term into two parts. The first part contains the summands for

푥 1 푥 which ≤ 표푟 ≥ 2. For these terms, 푛 2 푛

푥 |log | ≥ log2. 푥

Under the conditions of Theorem 13,

+∞ |푎 | 1 ∑ 푛 = 푂 ( ) . 푛푏 (푏 − 1)훼 푛=1

Therefore,

1 푥 For the terms with < < 2, 2 푛

푥 푏 푥 −1 ∑ |푎 | ( ) 푇−1 |log | 푛 푛 푛 1 2푥<푛<2푥

푏 −1 푁+1 푏 −1 푥 −1 푥 푥 −1 푥 = ∑ |푎푛| ( ) 푇 |log | + ∑ |푎푛| ( ) 푇 |log | 푛 푛 푛=푁−1 푛 푛 1 2푥<푛<2푥 푛≠푁−1,푁,푁+1

푁−1 1 −1 2푥 −1 퐴(2푥) 푁 + 푢 푥퐴(2푥) ≪ { ∫ (log 2) 푑푢 + ∫ (log ) 푑푢} + 푇 푢 1 푇 푥 푁+1 푁 + 2 2

푥퐴(2푥) log 푥 ≪ , 푇 where the last estimate follows from the fact that

푁−1 1 −1 2푥 −1 푁 + 푢 ∫ (log 2) 푑푢 + ∫ (log ) 푑푢 ≪ 푥log푥. 푢 1 푥 푁+1 푁 + 2 2  Now we are ready to prove the Prime Number Theorem.

5.2 Proof of the Prime Number Theorem

Theorem 14 (de la Vallee-Poussin). There exists an absolute constant 푐 > 0 such that

휓(푥) = ∑ Λ(푛) = 푥 + 푂 (푥푒−푐√log 푥). 푛≤푥

Proof. For 푹풆 풔 > ퟏ, we have

휁′ Λ(푛) − (푠) = ∑+∞ . 휁 푛=1 푛푠

1 Without loss of generality, suppose that 푥 = 푁 + ≥ 100. Note that 2

휁′ 1 |− (푠)| ≪ as 휎 → 1 + 0. 휁 휎−1

1 Now, in Theorem 13 we can take 훼 = 1 , 푏 = 1 + , 퐴(푛) = log푛, and 푇 = 푒√log 푥. log 푥

Then

1 푏+푖푇 휁′ 푥푠 푥log2 푥 휓(푥) = ∑ 훬(푛) = ∫ (− (푠)) 푑푠 + 푂 ( ). (5.4) 푛≤푥 2휋푖 푏−푖푇 휁 푠 푇

From Theorem 12 and Corollary 10 in chapter 4, there exists an absolute constant 푐1 >

푐 0 such that 휁(푠) has no zero in the domain 푅푒 푠 = 휎 ≥ 휎 = 1 − 1 and |푡| ≤ 푇. 1 2 log(푇+2)

Moreover,

휁′ (푠) ≪ log2푇 휁 for 푠 = 휎1 + 푖푡, −T ≤ t ≤ T and 푠 = 휎 + 푖푇, 휎1 ≤ 휎 ≤ 푏 . Now consider the integral 퐽 defined by 1 휁′ 푥푠 퐽 = ∫ (− (푠)) 푑푠, 2휋푖 Γ 휁 푠 where the contour Γ is as shown in Figure 5.3.

Figure 5.3. The integral 퐽 around the contour defined by Γ where 퐽 is 1 휁′ 푥푠 defined by 퐽 = ∫ (− (푠)) 푑푠. 2휋푖 Γ 휁 푠

휁′ 푥푠 Inside the contour, the integrand (− (푠)) has a pole of order 1 with residue equal to 휁 푠

푥. Therefore,

1 푏+푖푇 휁′ 푥푠 ∫ (− (푠)) 푑푠 = 푥 − 푅, (5.5) 2휋푖 푏−푖푇 휁 푠 where 푅 is the sum of the integrals along I, II, and III of Γ. These integrals need to be estimated. Integrals I and III are equal in absolute value and can be estimated as follows:

1 푏+푖푇 휁′ 푥푠 푏 휁′ 푥휎 푥 log2푇 ∫ + ∫ ≪ | ∫ (− (푠)) 푑푠| ≪ ∫ | (휎 + 푖푇)| 푑휎 ≪ . (5.6) 퐼 퐼퐼퐼 2휋푖 휎1+푖푇 휁 푠 휎1 휁 푇 푇

Similarly,

1 휎1+푖푇 휁′ 푥푠 푇 휁′ 푥휎1+푖푡 ∫ = ∫ (− (푠)) 푑푠 ≪ ∫ | (휎 + 푖푡) | 푑푡, 2휋푖 휁 푠 휁 1 휎 + 푖푡 퐼퐼 휎1−푖푇 −푇 1 which is bounded by

1 푑푡 푇 푑푡 ≪ 푥휎1log2푇 (∫ + ∫ ) ≪ 푥휎1log3푇. (5.7) 0 휎1 1 푡

푐 From Equations (5.4), (5.5), (5.6), and (5.7) and because 휎 = 1 − 1 and T = 1 2log (푇+2)

푒√log 푥, we have

휓(푥) = 푥 + 푂 (푥푒−푐√log 푥) for some absolute constant 푐 > 0. 

5.3 An Explicit Formula of 흍(풙)

The method of complex integration allows us to get a very important explicit formula that connects 휓(푥) with the zeroes of 휁(푠).

Theorem 15. Let 2 ≤ 푇 ≤ 푥. Then,

푥휌 푥log2푥 휓(푥) = ∑ Λ(푛) = 푥 − ∑ + 푂 ( ), 푛≤푥 |퐼푚휌|≤푇 휌 푇 where the 휌 are the nontrivial zeros of 휁(푠).

Proof. According to Theorem 10 in chapter 4,

휁′ 1 +∞ 1 1 +∞ 1 1 − (푠) = − ∑푛=1 ( + ) − ∑푛=1 ( − ) − 퐵0, (5.8) 휁 푠−1 푠−휌푛 휌푛 푠+2푛 2푛 where the 휌푛 are the nontrivial zeros of 휁(푠). Just as in the proof of Theorem 14, with

1 푏 = 1 + , log 푥

1 푏+푖푇 휁′ 푥푠 푥log2푥 휓(푥) = ∫ 1 (− (푠)) 푑푠 + 푂 ( ), (5.9) 2휋푖 푏−푖푇1 휁 푠 푇

1 where T≤ 푇 ≤ 푇 + 1 and 푇 is chosen so that the line 퐼푚 푠 = 푇 is ≫ away from 1 1 1 log 푇 any zero of 휁(s). This choice of 푇1is always possible because 휁(s) has 푂(logT) zeroes in the strip 푇 ≤ 퐼푚 푠 ≤ 푇 + 1. Next, consider the integral 퐽 around the contour Γ (see

Figure 5.4), where 퐽 is defined by

1 휁′ 푥푠 퐽 = ∫ (− (푠)) 푑푠. 2휋푖 Γ 휁 푠

iT1 Γ

-1/2 0 1 b

-iT1 III

Figure 5.4. The integral 퐽 around the contour Γ where 퐽 is defined by 1 휁′ 푥푠 퐽 = ∫ (− (푠)) 푑푠. 2휋푖 Γ 휁 푠

By Cauchy’s theorem and Equation (5.8), we have

푥휌 휁′ 퐽 = 푥 − ∑ − (0). (5.10) |퐼푚 휌|≤푇1 휌 휁

The next task is to estimate the integrals over the sides 퐼, 퐼퐼, and 퐼퐼퐼 of The integrals over sides 퐼 and 퐼퐼퐼 are equal in absolute value and can be estimated as follows:

푏+푖푇1 푠 푏 1 휁′ 푥 1 휁′ 휎 ∫ ≪ | ∫ (− (푠)) 푑푠| ≪ ∫ | (휎 + 푖푇1)| 푥 푑휎 2휋푖 1 휁 푠 푇1 1 휁 퐼+퐼퐼퐼 −2+푖푇1 −2

푥 푏 휁′ 1 ( ) ≪ ∫− | 휎 + 푖푇1 | 푑휎 (5.11) 푇1 2 휁 and that

1 − +푖푇 ′ 푠 1 1 2 1 휁 푥 − 푇1 휁′ 1 푑푡 ∫ ≪ | ∫ 1 (− (푠)) 푑푠| ≪ 푥 2 ∫ | (− + 푖푡)| 1 . (5.12) 퐼퐼 2휋푖 − −푖푇 휁 푠 −푇1 휁 2 +|푡| 2 1 2

By Corollary 9 in chapter 4, we have,

휁′ ( ) ∑ 1 ( | | ) 휎 + 푖푡 = |푡−훾푛|≤1 + 푂 log 푡 + 2 . (5.13) 휁 (휎−훽푛)+푖(푡−훾푛)

Note that when 푠 = 휎 + 푖푡 ∈ 퐼, 퐼퐼, or 퐼퐼퐼, we have

1 | | ≪ log 푇, (휎 − 훽푛) + 푖(푡 − 훾푛) and thus

휁′ (휎 + 푖푡) ≪ ∑ log푇 + log 푇 ≪ log2 푇. 휁 |푇1−훾푛|≤1

Hence, from Equation (5.11),

푥log2푇 푥log2푥 ∫ ≪ ≪ , (5.14) 퐼+퐼퐼퐼 푇1 푇 and, from Equation (5.12),

2 2 3 2 log 푇 푇1 푑푡 log 푇 푇1 푑푡 log T 푥log 푥 ∫ ≪ [∫ 1 ] ≪ [∫ 1 ] ≪ ≪ . (5.15) 퐼퐼 √푥 −푇1 +|푡| √푥 0 +|푡| √푥 푇 2 2

Therefore, Theorem 15 follows from Equations (5.9), (5.10), (5.14), and (5.15).

Corollary 11. Assuming the Riemann hypothesis,

1 휓(푥) = 푥 + 푂(푥2 log2푥).

Proof. Recall that for 푇 ≥ 2, 휁(s) has 푂(log 푇) zeroes in the strip

푇 ≤ |퐼푚 푠| ≤ 푇 + 1. It is easy, then, to see that for 푇 ≥ 2, there are 푂(푇 log 푇) nontrivial zeroes of 휁(s) that satisfy |퐼푚 휌| ≤ 푇. Under the Riemann hypothesis, every nontrivial

1 zero 휌 of the Riemann zeta-function satisfies 푅푒 휌 = . Therefore, from Theorem 15, 2 take 푇 = 푥. The result for large 푥 is

푥휌 휓(푥) = 푥 − ∑ + 푂(log2푥). (5.16) |퐼푚휌|≤푥 휌

Note that 푥휌 1 1 | ∑ | ≤ 푥 ∑ ≪ 푥 ∑ √ √ 1 휌 |휌| + |훾| |퐼푚 휌|≤푥 |퐼푚 휌|≤푥 |훾|≤푥 2 훾=퐼푚 휌

푥 1 ≪ √푥 ∫ 1 푑 ∑훾≤푡 1. 0 +푡 2

Using integration by part, we get

푥 (∑ 1) 푥 (∑ 1) ≪ 푥 { 훾≤푡 | + 훾≤푡 푑푡}. √ 1 ∫0 1 2 ( +푡) ( +푡) 2 0 2

When 푡 ≥ 2, we have ∑|퐼푚 휌|≤푡 1 ≪ 푡 log푡. Thus, the above quantity is

푥 푡 log 푡 ≪ √푥 {log 푥 + 1 + ∫ 2 푑푡} 2 푡

≪ √푥 log2푥 which, along with Equation (5.16), yields Corollary 12. 

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