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The Dirichlet Series To The Riemann Hypothesis

The zeta-function and primes

Daud Nawaz

2018

Examensarbete, Grundnivå (kandidatexamen), 15 hp Matematik

Fristående kurs

Handledare: Johan Björklund Examinator: Rolf Källström

Omslagsbild: Figuren visar grafen av [Table[Prime[𝑛] (𝑛Log[𝑛])⁄ , {𝑛, 2, 5000}]]. Figur: Daud Nawaz, 2018.

Daud Nawaz

Abstract

is paper examines the Riemann zeta-function and its relation to the prime distribution.In this work, I present the journey from the Dirichlet series to the Riemann hypothesis.Furthermore, I discuss the prime counting function, the Riemann prime counting functionand the Riemann explicit function for distribution of primes. is paper explains that thenon-trivial zeros of the zeta-function are the key to understand the prime distribution.

Contents1 Introduction 1

2 Dirichlet series and its properties 32.1 Dirichlet series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.2 e half plane of absolute convergence of a Dirichlet series . . . . . . . . . . . . 42.3 e half plane of convergence of a Dirichlet series . . . . . . . . . . . . . . . . . . 52.4 Power series and Dirichlet series . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.5 Analytic properties of Dirichlet series . . . . . . . . . . . . . . . . . . . . . . . . . 6

3 Euler product and zeta-function ζ(s) 83.1 Leonhard Euler . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83.2 Euler product and zeta function . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83.3 e Gamma function and its properties . . . . . . . . . . . . . . . . . . . . . . . . 103.4 Prime counting function and zeta-function . . . . . . . . . . . . . . . . . . . . . . 11

4 Zeta-function ζ(s) and its properties 144.1 Functional equation of Riemann zeta-function . . . . . . . . . . . . . . . . . . . . 144.2 Properties of zeta-function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

4.2.1 Zeta-function for negative integers ζ(−n) . . . . . . . . . . . . . . . . . . 154.2.2 Zeta-function evaluated at even integers ζ(2n) . . . . . . . . . . . . . . . 17

4.3 Zero free region for zeta-function ζ(s) . . . . . . . . . . . . . . . . . . . . . . . . 18

5 e Riemann hypothesis and No(T ) 215.1 e Riemann hypothesis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215.2 Zero density estimate, see [7] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215.3 e critical zeros of ζ(s) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

5.3.1 A lower bound for No(T ), see [7] . . . . . . . . . . . . . . . . . . . . . . . 245.3.2 A positive proportion of No(T ) . . . . . . . . . . . . . . . . . . . . . . . . 26

6 Primes distribution and ζ(s) zeros 276.1 Riemann’s prime counting function J(x) , see [4] . . . . . . . . . . . . . . . . . . 27

6.1.1 e Golden key . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 296.2 e explicit formula of Riemann, see [4] . . . . . . . . . . . . . . . . . . . . . . . 306.3 e Riemann hypothesis in math and physics . . . . . . . . . . . . . . . . . . . . 33

References 35

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1 IntroductionHuman beings have been interested in numbers regardless of region and time. We have usedngers, stones, sticks and othermethods for communicating in numbers. Interest in the propertiesof numbers has been great and it still has the potential of extensive research for mathematiciansas well as for the layman. e study of properties of numbers is called the number theory. Innumber theory, some problems are easy to formulate as well as to prove. While, some problemsmay seem of an easy nature but we have not been able to give a mathematical proof. It wasdiscovered early that dierent numbers have dierent properties and they behave dierently.For example, even numbers can be divided by two while prime numbers cannot be divided intosmaller numbers. Prime numbers have great importance in mathematics and in this work weshall discuss the Riemann zeta-function which is related to the prime numbers. e Riemannzeta-function is a specic case of the Dirichlet series. Before Riemann, Euler; who was a Swissmathematician, worked with the zeta-function. He applied calculus methods to solve the numbertheory problems. What Riemann did was that he expanded the properties of the zeta-functionby applying analytical methods. Riemann is considered the founder of analytical number theory.In 1859 he suggested in his paper that there is a relation between zeros of the zeta-function andprimes distribution. Further, in his paper, he suggested that all zeros to the zeta-function lie ona critical line, which is called the Riemann hypothesis (RH). e RH has great importance in theprime number theory. A proof to the RH would help mathematicians to solve many problemsrelated to the prime numbers.

is paper will explain the development of the zeta-function from Dirichlet series to the Riemannzeta-function. During the journey, we shall go through the Euler product and the functionalequation of the Riemann zeta-function. We shall nish this work by going through the explicitformula of Riemann which shows a connection between zeros of the zeta-function and primesdistribution. is paper consists of ve parts which are as follows:

• Dirichlet series and its properties

• Euler product and zeta-function

• Zeta-function and its properties

• e Riemann hypothesis and critical zeros of the zeta-function

• Prime distribution and zeros of the zeta-function

e purpose of this work is to give an introduction of the zeta-function to the reader by explainingthe background of the zeta-function and how its properties have been expanded by Riemann.Further on, a clear relation between the zeta-function and prime numbers will be discussed.

Certainly, this paper is not able to show everything related to the zeta-function, the Riemannhypothesis or the analytical number theory. is work has its restrictions in the form of timeand my prior knowledge of the analytic number theory. ese limits do not allow me to providea deeper study than this which would require more time. Hopefully, this paper will give a shortand understandable introduction to the eld and evoke some interest in the reader to study the

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subject further. is paper assumes that the reader has previous knowledge in complex analysisand calculus. See for instance Fundamentals of Complex Analysis wrien by Edward B. Sa andArthur David Snider.

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2 Dirichlet series and its propertiesis sectionwill present Dirichlet series and the relation betweenDirichlet series and the zeta-function.At the end of this section, the analytical properties of Dirichlet series will be explained. epurpose of this section is to provide a background to the zeta-function.

2.1 Dirichlet seriesDirichlet series are a special case of analytical functions and even though Dirichlet series are notas important as power series when it comes to analysis. Dirichlet series has a great importancein applications of complex analysis to the number theory [8].All series of the form

∑∞n=1

f(n)ns

are called Dirichlet series with coecient f(n) where f(n)is an arithmetic function. Such series are very important in the analytic number theory [2].Dirichlet series were dened for 1 [8] but Riemann changed the concept. G. F. B. Riemannwas a German mathematician who lived 1826 to 1866. He is generally regarded as the founderof the analytic number theory. Even though his father had other hopes for him, he becamea mathematician because of his shyness which was an obstacle to become a minister. He isrecognized as the most brilliant mathematical mind of the nineteenth century. He started with afunction called the Zeta-function, which has great importance in the analytic number theory. Weshall discuss the function later [10]. Riemann let the s be the complex variable so that s = σ+ itwhere σ, t � R and further we have that

|ns| = |nσ+it| = |nσnit| = |nσ.eit logn| (1)

and we know that |eiθ| = 1 for θ � R =⇒ |ns| = nσ. By log we mean the natural logarithm.We shall show that a Dirichlet series converges for half plane σ > σc and converges absolutelyfor another half plane where σ > σa . Half plane is dened as the points such that σ > awhere s = σ + it [2]. Before we start to discuss Dirichlet series some examples of Dirichletseries are as follows:

Example : ζ(s) . For 1 , ζ(s) is dened by a Dirichlet series of the form

ζ(s) =∞∑n=1

1

ns

and ζ(s) is analytical function for 1 [8].Example : For 1 we have

−ζ′(s)

ζ(s)=

∞∑n=1

Λ(n)

ns

where Λ(n) is Mangoldt’s function,

Λ(n) =

{log p if n = pk , p is a prime number and k ≥ 10 if n 6= pk

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e values of Λ(n) for rst ten positive integers are 0, log(2), log(3), log(2), log(5), 0, log(7),log(2), log(3) and 0. e Mangoldt’s function shows a relation between the zeta-function andprime numbers which we shall discuss in more detail later [8].Example : For 1 we have

1

ζ(s)=

∞∑n=1

µ(n)

ns

where µ(n) is the Möbius function. e Möbius function also shows a relation between thezeta-function and primes just as Mangoldt’s function [8].

2.2 e half plane of absolute convergence of a Dirichlet seriesAs it is explained earlier that for σ ≥ a , we have |ns| = nσ ≥ na therefore we get thefollowing inequality ∣∣∣∣f(n)ns

∣∣∣∣ ≤ |f(n)|nais inequality means that if Dirichlet series

∑∞n=1

f(n)ns

converges absolutely for s = a + bithen an absolute converges for all such s that σ ≥ a is also sure [2]. is result leads us tothe following theorem.

eorem 1. [2] Suppose the series∑|f(n)n−s| does not converge for all s or diverge for all s .

en there exists a real number σa , called the abscissa of absolute convergence, such that the series∑f(n)n−s converges absolutely if σ > σa but does not converge absolutely if σ < σa .

Before we give a proof we need to learn two facts. e rst one is that if a Dirichlet seriesconverges for s = s0 then it converges absolutely for s0 + a where a > 1. Let us say a seriesconverges for s0 then the series must converge absolutely for s0 + a because |ns0+a| = nσ0+a. Bythat we get | f(n)

ns0+a| = |f(n)

ns0| · 1

na≤ 1

nafor large values of n since f(n)

ns0tends to zero. which shows

us that the series converges absolutely for s0 + a if series converges for s0 where a > 1.e second fact is that if a Dirichlet series does not diverge or converge for all 1 and has a pole when s = 1 [2]. e

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zeta-function behave like 1s−1 near s = 1 which makes it clear that if s = 1 + it where t 6= 0 then

the zeta-function is analytic and the series is convergent [4].Example :

∑nnn−s is a series which diverges for all s so σa = +∞ as stated above [2].

Example :∑n−nn−s is a series which converges absolutely for all s that‘s why σa = −∞

[2].

2.3 e half plane of convergence of a Dirichlet serieseorem 2. [2] If the series

∑f(n)n−s converges for s = σo + ito then it also converges for

all s with σ > σo . If it diverges for s = σo + ito then it diverges for all σ < σo .

Proof. e second statement follows from the rst. To prove the rst statement choose any swith σ > σo . en lemma 1 shows that∣∣∣∣ ∑

a σo we have∣∣∣∣ ∑a 0 and if we suppose so = 0 = σo then for σ > 0 we have∣∣∣∣ ∑a 0 because |∑

n≤x(−1)n| ≤ 1 .

eorem 3. [2] If the series∑f(n)n−s does not converge everywhere or diverge everywhere, then

there exists a real number σc called the abscissa of convergence, such that the series converges forall s in the half plane σ > σc and diverges for all s in the half plane σ < σc .

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Proof. We argue as in the proof of theorem 1 taking σc to be the least upper bound of all σ forwhich

∑f(n)n−s diverges.

As we did for absolute convergence, we dene σc = −∞ if the series converges everywhereand σc = +∞ if the series diverges everywhere [2].

2.4 Power series and Dirichlet seriesWe can nd a resemblance between power series and Dirichlet series. Power series has a disk ofconvergence where the inside of the disk is a domain of absolute convergence. While Dirichletseries has a half plane of convergence where the domain of absolute convergence may be theproper subset of a domain of convergence. A power series represents an analytic function insidethe disk of convergence while a Dirichlet series represents an analytic function in the half-planeof convergence [2].

2.5 Analytic properties of Dirichlet serieseorem 4. [2] e sum function F (s) =

∑f(n)n−s of a Dirichlet series is analytic in its half

plane of convergence σ > σc and its derivative F ′(s) is represented in this half plane by theDirichlet series

F ′(s) = −∞∑n=1

f(n) log n

ns(2)

obtained by dierentiating term by term.

Dierentiating of the sum function term by term is possible because aDirichlet series convergesuniformly on every compact subset which lies interior to the half-plane of convergence. Further,we have to note that the abscissa of the convergence and absolute convergence for equation 2is same as for the F (s).By the help of eorem 4 we can obtain K − th derivative of F (s) and it is represented by

FK(s) = (−1)K∞∑n=1

f(n)(log n)K

nsfor σ > σc

A real example can be the zeta function for σ > 1

ζ(s) =∞∑n=1

1

ns

ζ ′(s) = −∞∑n=1

log(n)

ns

e fact that a Dirichlet series converges uniformly on every compact subset which lies interiorto the half-plane of convergence and the next coming Lemma prove the eorem 4. e lemmastates that

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Lemma 2. [2] Let {fn} be a sequence of functions that are analytic on an open subset s of thecomplex plane and assume that {fn} converges uniformly on every compact subset of s to a limitfunction f . en f is analytic on s and the sequence of derivatives {f ′n} converges uniformlyon every compact subset of s to the derivative f ′ .

Proof. Since fn is analytic on s we have Cauchy integral formula

fn(a) =1

2πi

∫∂D

fn(z)

z − adz

where D is any compact disk in s , ∂D is positively oriented boundary and a is any interiorpoint of D . Because of uniform convergence we can pass to the limit under the integral signand obtain

f(a) =1

2πi

∫∂D

f(z)

z − adz

which implies that f is analytic inside D . For the derivatives we have

f ′n(a) =1

2πi

∫∂D

fn(z)

(z − a)2dz

andf ′(a) =

1

2πi

∫∂D

f(z)

(z − a)2dz

from which it follows easily that f ′n(a) → f ′(a) uniformly on every compact subset of s asn→ ∞ .

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3 Euler product and zeta-function ζ(s)is section shows a direct relation between prime numbers and the zeta-function in the form ofthe Euler product. Further on, the section presents the gamma-function and a relation betweenthe prime counting function and the zeta-function. Knowledge about the gamma-function isnecessary to understand the functional equation of the zeta-function which will be presented inthe next section.

3.1 Leonhard EulerLeonhard Eulerwas a Swissmathematicianwho lived 1707-1783. Hewas a remarkablemathematicianand served almost all branches of mathematics. His work has been prized ten times by the FrenchAcademy. Euler has been a part of the academy in St. Petersburg and the academy in Berlin. Helost eyesight in one eye early as a young man but this did not stop him from being creative. Eulerwas the one who started to apply calculus methods to the number theory. He gave a productwhich relates prime numbers to the zeta-function; a special case of Dirichlet series [10].

3.2 Euler product and zeta functionWe described earlier the zeta-function as a Dirichlet series, the same function can be representedby the Euler productwhich gives us a direct relation between the zeta-function and prime numbers.

eorem 5. [9] Euler product: For s = σ + it , σ > 1 the following holds

ζ(s) =∏p

(1

1− 1ps

)where the product on the right is taken over all prime numbers p.

Proof. Let x ≥ 2 we dene the function ζx(s) as

ζx(s) = Π(p ≤ x)(

1

1− 1ps

)en with each of the factor on the right, as Re(s) ≥ 1 we can always rewrite the term as ageometric series : 1

1− 1p−s

=∑∞

k=01pks

where each geometric series converges absolutely. Hencewe could multiply term by term and replace with the new equation.

ζ(s) = Π

( ∞∑k=0

1

pks

)=

∞∑k1=0

...........∞∑kj=0

1

(pm11 .........pmjj )

s(3)

when 2 = p1 < p2 < .......... pj as pj represents all prime numbers up to x . By Usingfundamentals theoremof arithmetic, we know that for any positive integer n is uniquely determinedby

n = pm11 · pm22 · · · pmjj

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when m1 · · ·mj are nonnegative integers and p1 · · · pj are prime numbers. Consequently wecan take the right side of equation 3 as the form

∑n≤X

1

ns+

′∑n>X

1

ns

where the ” ′ ” stands for summation over those natural numbers n > X whose prime divisorare all ≤ X .We can give an upper bound of this sum by∣∣∣∣∑

n>X

1

ns

∣∣∣∣ ≤ ∑n>X

1

nσ<∑n>X

1

nσ≤ 1

Xσ+

∫ ∞X

du

uσ≤ σ

σ − 1X1−σ (4)

e right formula gives an upper bound of the sum. Now combining the denitions of ζX(s)with equation 3 and 4 we obtain the relation between ζX(s) and the zeta-function as

ζX(s) = Π

(1− 1

ps

)−1=∑n≤X

1

ns+O

(σ

σ − 1X1−σ

)

Here O(

σσ−1X

1−σ)

represents the upper bound. As we take the limit as X → ∞ . We have

X1−σ → 0 since σ > 1 . Hence the upper bound of the dierence vanished and we proved∞∑n=1

1

ns= Π

(1

1− 1ps

)

eorem 5 was discovered by Euler in 1737. Sometime it is referred to as an analytic versionof the Fundamental eorem of Arithmetic.In 1737 Euler showed that there exists an innite number of primes because

∑primes

1p

= ∞ .We can elaborate it, we know from the beginning of this paper that

∞∑n=1

1

ns→ ∞ as s→ 1+

and we are also familiar with that∞∑n=1

1

ns=∏p

(1

1− 1ps

)We mentioned earlier that the zeta-function is not dened for s = 1 but if we approach s = 1from right side of real-axis thenwe get the limit is equal to∞. Which leads us to the contradictionbecause If therewere nitelymany primes then

∑∞n=1

1nwould converge. Hence by contradiction

there are innitely many primes [2].e same result that there are innitely many primes was already proven by Euclid before Euler.

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eorem 6. [5] let 2, 3, 5, . . . , p be the aggregate of primes up to p , and let

q = 2 · 3 · 5 · . . . · p+ 1

en q is not divisible by any of the numbers 2, 3, 5, . . . , p . It is therefore either prime, or divisibleby a prime between p and q . In either case there is a prime greater than p , which proves thetheorem.

Euclid way to prove is very important because by using the same method one can prove thatthere are innitely many primes of the form 4n + 3 , 6n + 5 and 8n + 5 [5]. As far as ζ(s)is concerned, the Euler product is of more interest.

3.3 e Gamma function and its propertiesIn the functional equation of ζ(s) , gamma function/Euler gamma-function appears, thats whywe should have some understanding of the gamma function. Gamma function Γ(s) is representedin integral form as

Γ(s) =

∫ ∞0

xs−1 e−x dx

e Γ(s) is analytic everywhere in the s-plane except the points s = 0,−1,−2, ........ whereit has poles and there, poles have residue of (−1)

n

n!at s = −n. Γ(s) can be represented as a limit

function also

Γ(s) = limn→∞

nsn!

s(s+ 1)......(s+ n)s 6= 0,−1,−2,−3, .........

Γ(s) is never zero and Γ(s) satises the two conditions for all s except s = 0,−1,−2,−3, . . .

(1) Γ(s+ 1) = s Γ(s)

(2) Γ(s) Γ(1− s) = πsin πs

Further can be noted that Γ(n + 1) = n! if n is a positive integer [2]. ese properties ofgamma-function can be illustrated by the following gure where it is clearly shown that Γ(s) isnever zero for s > 0 and Γ(s) has poles at −n where n is a positive integer.

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Figure 1: e gamma-function Γ(s) ploed for −6 ≤ s ≤ 5

3.4 Prime counting function and zeta-functionBefore describing a relation between Riemann zeta-function and the prime counting function π(x) ,we have to introduce π(x) . e prime counting function is dened as

π(x) = The number of Primes p satisfying 2 ≤ p ≤ x

π(x) is a function which counts primes smaller or equal to x . e graph of π(x) is asfollowing

Figure 2: e prime counting function ploed for 0 ≤ x ≤ 50

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It may seem that primes are distributed irregularly but by analyzing carefully it can be seenthat distribution of primes is quite regular.

x π(x) xlog x

π(x)x

log x

10 4 4.3 0.93102 25 21.7 1.15103 168 144.9 1.16104 1,229 1,086 1.11105 9,592 8,686 1.10106 78,498 72,464 1.08107 664,579 621,118 1.07108 5,761,455 5,434,780 1.06109 50,847,534 48,309,180 1.051010 455,052,512 434,294,482 1.048

By analyzing the prime table as shown above, but for x ≤ 106 Gauss and Legendre proposedthat the quotient

π(x)x

log x

is close to 1 as x → ∞ , but they could not prove it. Later on, in 1851 Chebyshev, a Russianmathematician proved that if there is such limit then it must be 1. He could neither prove theGauss and Legendre conjecture. In 1859 Riemann used an analytical approach to the problem byusing Euler’s formula

ζ(s) =∞∑n=1

1

ns.

Riemann studied the problem for s as a complex variable where s > 1 . Riemann also sketcheda relation between ζ(s) and primes distribution. Hewas not able to solve the problem completelyand there is still much to prove and disprove, however by manipulating analytical tools in 1896Hardmard and de la Vallée Poussin proved that

limx→∞

π(x) log x

x= 1 .

at is called the prime number theorem [2]. e purpose of the explanation above is to show thatπ(x) is an important part of the number theory [3]. What we are going to do in this sub-sectionis to show a relation between π(x) and ζ(s) by using the Euler product.

eorem 7. [3] e relation between the zeta-function and the prime counting function is thefollowing

log ζ(s) = s

∫ ∞2

π(x)

x(xs − 1)dx

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Proof. [3]

ζ(s) =∏P

(1− 1

ps

)−1

log ζ(s) = log∏P

(1− 1

ps

)−1=

= −∑P

log

(1− 1

ps

)(5)

From equation 5 to equation 6, we jump from the prime numbers to the natural numbers. We needto clarify here that the right hand side of 5 and 6 are exactly the same. Recall the prime countingfunction, whenever we choose n in equation 6 that is not a prime numberwe get π(n)−π(n−1) =0. Furthermore, whenever n is a prime number we have π(n) − π(n − 1) = 1. Which meansthat we are still summing over the primes but by using the natural numbers. We use the naturalnumbers because we are not fully aware of the prime distribution.

log ζ(s) = −∞∑n=2

{π(n)− π(n− 1)} log(

1− 1ns

)= (6)

= −∞∑n=2

π(n)

{log

(1− 1

ns

)− log

(1− 1

(n+ 1)s

)}=

=∞∑n=2

π(n)

∫ n+1n

s

x(xs − 1)dx =

= s

∫ ∞2

π(x)

x(xs − 1)dx (7)

It is very important to study π(x) and it has been intensively studied. Equation 7 givespossibilities to study the analytical character of log ζ(s) and then reach out to π(x) by inverseintegral transform [3]. We shall discuss deeper about the relation between the prime countingfunction and the zeta-function in the last section.

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4 Zeta-function ζ(s) and its propertiesis section is about the functional equation of Riemann zeta-functionwhich gives an insight intothe zeros of the zeta-function. is section has its focus on the zeros of the zeta-function and somespecic values of the zeta-function. Zeros for negative integers and values of the zeta-functionfor positive even integers will be explained here. e last subsection of this section discusses thezero-free region where the zeta-function has no zeros.

4.1 Functional equation of Riemann zeta-functione dierence between Euler’s zeta function and Riemann’s zeta function is that the former wasdened only for s > 1 while the laer is dened over the whole s-plane with a simple poleat s = 1 . e Riemann zeta function has some zeros ζ(−2n) = 0 which are called trivialzeros. is should be more clear when we look at the functional equation. Riemann extendedholomorphically the zeta-function from the half-plane 1 to thewhole complex plane

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using this to replace the product Γ(1− s) sin(πs2

)in equation ( 9) we obtain

π−s2 Γ

(s

2

)ζ(s) = π−

(1−s)2 Γ

(1− s

2

)ζ(1− s)

In other words, the functional equation takes the form

φ (s) = φ ( 1− s )

whereφ (s) = π−

s2 Γ

(s

2

)ζ(s)

e function φ(s) has simple poles at s = 0 and s = 1 . Following Riemann, we multiplyφ(s) by s(s−1)

2to remove the poles and dene

ξ(s) =1

2s(s− 1) φ(s)

ξ is an entire function of s and satises the functional equation

ξ(s) = ξ(1− s)

A more interesting form is when s = 12

+ it and that leads to ξ(

12

+ it

)= ξ

(12− it

).

When Re(s) = 12, it has very interesting status in the Riemann hypothesis and is called the

critical line which we shall discuss later [2].

4.2 Properties of zeta-functionWe already know that ζ(s) has trivial zero for −2n where n is a positive integer. ζ(s) isanalytic over whole s − plane with a simple pole at s = 1 . Before we go to zeros of ζ(s)in general we are going to study special cases of ζ(s) where s = −n and s = 2n , n ispositive integer [2]. One thing we need to recall is that

Bn =∞∑k=0

[nk

]Bk if n ≥ 2

and Bn is called Bernoulli number. ere are given some values of Bernoulli number at the endof this paper, which we are going to use later to calculate the Basel problem; the Basel problemasks for the exact summation of the innite series

∑∞n=1

1n2.

4.2.1 Zeta-function for negative integers ζ(−n)

eorem 8. [2] If n ≥ 0 we have

ζ(−n) = −(Bn+1n+ 1

)Also, if n ≥ 1 we have ζ(−2n) = 0 , B2n+1 = 0 .

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Proof. Proof of ζ(−n) = −(Bn+1n+1

)follows directly from the fact that

I(s, a) =1

2πi

∫c

zs−1eaz

1− ezdz = Resz=0

(zs−1eaz

1− ez

)and

ζ(s, a) = Γ(1− s) I(s, a)

If s = −n then we have

I(−n, a) = Resz=0(z−n−1 eaz

1− ez

)I(−n, a) = −Resz=0

(z−n−2

zeaz

ez − 1

)I(−n, a) = −Resz=0

(z−n−2

∞∑m=0

Bm(a)

m!zm)

I(−n, a) = −Bn+1(a)(n+ 1)!

In our case a = 1 which gives us

ζ(−n) = −Bn+1n+ 1

If we draw a graph for the zeta-function we shall see the same result as shown above that thezeta-function is zero for negative even integers. ese are the same pointswhere the gamma-functionhas its poles. e graph of the zeta-function is following

Figure 3: e zeta-function ploed for −10 ≤ σ ≤ 0

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4.2.2 Zeta-function evaluated at even integers ζ(2n)

eorem 9. [2] If k is a positive integer then we have

ζ(2k) = (−1)k+1 (2π)2kB2k

2(2k)!

Proof. We take s = 2k in the functional equation for ζ(s) to obtain

ζ(1− 2k) = 2(2π)−2k Γ(2k) cos(πk) ζ(2k)

or−B2k

2k= 2(2π)−2k (2k − 1)! (−1)k ζ(2k)

which implies

ζ(2k) = (−1)k+1 (2π)2kB2k

2(2k)!

It is important to note here that there is no simple formula to calculate ζ(2k+1) not even fora special case, for example, ζ(3) . Following examples demonstrate the application ofeorem 9.e rst example is well known as the Basel problem which Euler already solved without usingeorem 9.

Example:

ζ(2) =π2

6=

∞∑n=1

n−2

ζ(4) =π4

90=

∞∑n=1

n−4

We know that B2 = 16 and B4 = −130

. According to theorem 9

ζ(2k) =(−1)k+1 (2π)2k B2k

2(2k)!

ζ(2) =(−1)2(2π)2

2(2)!.

1

6=

4π2

4.

1

6=

π2

6=

∞∑n=1

n−2

ζ(4) =(−1)3(2π)4

2(4)!

(− 1

30

)= −π

4

3

(− 1

30

)=

π4

90∑∞n=1 n

−4 = π4

90and we can carry on, if we want to.

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4.3 Zero free region for zeta-function ζ(s)First of all, we shall investigate the line when σ = 1 . According to the following theorem

eorem 10. [2] ζ(1 + it) 6= 0 for every real t .

eorem 10 means that the zeta-function for s = 1 + it where t is every real t, is never zero.e proof of eorem 10 is drawn from the following Lemma.

Lemma 3. [2] For σ > 1 we have

ζ3(σ) | ζ(σ + it) |4 | ζ(σ + 2it) | ≥ 1

Proof. For proof of eorem 10 we need only to consider t 6= 0 and rewrite the above equationin the form

{(σ − 1) ζ(σ)}3∣∣∣∣ζ(σ + it)σ − 1

∣∣∣∣4 | ζ(σ + 2it) | ≥ 1σ − 1 (10)is is valid if σ > 1 . Now let σ → 1+ in equation 10. e rst factor approaches 1 sinceζ(s) has residue 1 at the pole s = 1 . e third factor tends to | ζ(1 + 2it) | . If ζ(1 + it) wasequal to zero, the middle factor could be wrien as∣∣∣∣ζ(σ + it)− ζ(1 + it)σ − 1

∣∣∣∣4 → | ζ ′(1 + it) |4 as σ → 1+erefore if for some t 6= 0 we had ζ(1 + it) = 0 the le member of equation 10 wouldapproach the limit

| ζ ′(1 + it) |4 | ζ(1 + 2it) | as σ → 1+

But the right member tends to innity as σ → 1+ and this gives a contradiction.

It was already shown that the zeta-function has no zeros for σ > 1 . Based on theorem 10, itcan be concluded that ζ(s) 6= 0 if σ ≥ 1 and by analyzing the functional equation of ζ(s)

ζ(s) = 2(2π)1−s Γ(1− s) sin(πs

2

)ζ(1− s)

we come to know that ζ(s) has zeros at s = −2n which are called as trivial zeros. e trivial

zeros result from the vanishing of sin(πs2

). e equation also shows that the trivial zeros are

the only zeros for σ ≤ 0 . Until now, we have reached the conclusion that ζ(s) 6= 0 forσ ≥ 1 and ζ(s) 6= 0 for σ ≤ 0 except the trivial zeros [2]. e rst statement that thezeta-function is not zero for σ ≥ 1 can be illustrated by gure 4

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Figure 4: e zeta-function ploed for 0.5 ≤ σ ≤ 25

Another way to look into it, is through the relation

1

|ζ(s)|=

∣∣∣∣∏p

(1− 1

ps

)∣∣∣∣ ≤ ∞∑n=1

1

nσ< 1 +

∫ ∞1

du

uσ=

σ

σ − 1

which means|ζ(s)| > σ − 1

σ> 0

e relation conrms what we have discussed earlier that ζ(s) has no zeros in the half planewhere 1 and neither in the half plane where 0 we have

(1− 21−s) ζ(s) =∞∑n=1

(−1)n−1

ns(11)

is implies that ζ(s) < 0 if s is real and 0 < s < 1 .

Proof. First assume that σ > 1 . en we have

(1− 21−s) ζ(s) =∞∑n=1

1

ns− 2

∞∑n=1

1

(2n)s

(1− 21−s) ζ(s) = (1 + 2−s + 3−s + ..........) − 2 (2−s + 4−s + 6−s + ..........)

(1− 21−s) ζ(s) = 1− 2−s + 3−s − 4−s + 5−s − 6−s + ..........

Which proves equation (11) for σ > 1 . However if σ > 0 the series on the right convergesso equation (11) also holds for σ > 0 by analytic continuation. when s is real the series inequation (11)) is an alternating series with a positive sum. If 0 < s < 1 the factor (1− 21−s)is negative hence ζ(s) is also negative.

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Now, we come to a nal statement that except of the trivial zeros, all zeros lie in the stripe0 < s < 1 but not on the x-axis [2] [8]. Furthermore if ρ is a zero then ρ is also a zeroto ζ(s) . Which shows that the zeros are symmetrical with respect to the real axis around thecritical line σ = 1

2[8].

Figure 5: e critical stripe and the critical line

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5 e Riemann hypothesis and No(T )is section starts with the famous Riemann hypothesis. Aer that, the zero density conjectureis discussed which elaborates the concentration of the zeta-function’s zeros around the criticalline. at is followed by an analysis of the ratio of critical zeros, where boundaries for the criticalzeros are represented.

5.1 e Riemann hypothesisIt is proved that all nontrivial zeros lie in the critical strip but Riemann stated in his memoir onπ(x) published in 1859 that all nontrivial zeros of the zeta function lie on the line σ = 1

2. at is

known as the Riemann hypothesis (RH). Riemann did not prove the hypothesis. e hypothesishas caught the aention of mathematicians and even though much is in favor of the Riemannhypothesis, it is still unsolved [8].

e RH has been an obsession of mathematicians in the twentieth century. e RH is one ofthe old open problems of math, with the resolution of other old open problems, obsession forthe RH has become greater. Two of the old open problems which have been solved are; theFour-Color eorem and the Fermat’s Last eorem. e Four-Color eorem was originated in1852 and resolved in 1976 while the Fermat’s Lasteoremwas originated in 1637 and resolved in1994. David Hilbert, one of the greatest mathematician of his time stated the RH as exceedinglyimportant. Director of the Institute for Advanced Study in Princeton and a former professor ofmathematics at Harvard University, he said that one of the three most challenging and interestingproblems of mathematics is the RH. Importance of the RH in our time can be guessed by the priceof one million dollars oered by the Clay Institute for a proof or disproof of the RH [4].

e Riemann hypothesis has great importance for the analytic number theory because non-trivialzeros of the zeta-function have a connection with prime distribution [8]. We shall study therelation between non-trivial zeros of the zeta-function and prime distribution deeper in the nextsection.

5.2 Zero density estimate, see [7]We are going to represent some statements in this subsection without proof because of thecomplexity of the subject. e reader may see the references. Since we don’t know whetherthe Riemann hypothesis is true or false, we consider a region to analyze ζ(s) zeros. We areaware that there exists no zero at the line s = 1 . A hypothesis which states that the further weare from the critical line σ = 1

2, the considered region must contain a smaller amount of zeros

is called zero density hypothesis. e region which is needed in practice to analyze the densityof zeros is a rectangle

R(α, T ) = { s = σ + it ; σ ≥ α , | t | ≤ T }

for 12≤ α ≤ 1 and T ≥ 3 . Let N(α, T ) denote as number of zeros of the Riemann

zeta-function and zeros are given by ρ = β + iγ where

β ≥ α |γ| ≤ T

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If we assume that RH is true then N(α, T ) = 0 for α > 12

and T ≥ 3 . But the wellknown zero free region is represented by

N(α, T ) = 0 if α ≥ 1− c (log T )−23 (log log T )

13

Where c is an absolute positive constant. When 0 ≤

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is bounded by O(T (log T )5−4A) for A > 1 . Two years aer A.E Ingham in 1942 A. Selbergproved

N(α, T ) 1

2

A. Selberg’s estimate is valid for 12< α ≤ 1. If φ(T ) is a positive function which tends to innity

as T tends to innity then almost all zeros of the zeta-function lie in the domain∣∣∣∣ β − 12∣∣∣∣ < φ(T )log T if φ(T )→∞ as T →∞

Aer A.Selberg, M.N. Huxley gave a beer result for the density conjecture near the line s = 1 .e result was given by the following theorem

eorem 12. [7] For any α > 56and T ≥ 2 we have

N(α, T )

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Figure 6: e rst few critical zeros of zeta-function

5.3.1 A lower bound for No(T ), see [7]

Innitely many zeros of ζ(s) lie on the critical line was proven by G.H. Hardy in 1914 and lateron he and J.E. Lilewood proved another theorem.

eorem 13. G.H. Hardy and J.E. Lilewood proved that for a large T

No(T ) >> T (16)

Proof. We let

f(u) =

g

(12

+ iu

)∣∣∣∣g(12 + iu)∣∣∣∣ ζ

(1

2+ iu

)

Where g(s) = π− s2 Γ(s2

). By the functional equation ζ(s) g(s) = g(1−s) ζ(1−s) it shows

that f(u) is real and even. We consider two integrals

I(t) =

∫ t+4t

f(u) du

L(t) =

∫ t+4t

| f(u) | du

Where 4 is xed positive number. Clearly | I(t) | ≤ L(t) . If

| I(t) | < L(t) (17)

then f(u) must change sign in the interval (t, t+4) , hence f(u) must have a zero in (t, t+

4) , so has ζ(

12

+ iu

)because g(s) does not vanish anywhere. erefore our problem reduces

to showing that equation (17) occurs quite oen. To this end we are going to prove a lower boundfor I(t) and upper bound for L(t) on average over the segment [T, 2T ]. To prove equation (16)we need two lemmas which state

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Lemma 4. Let 4 ≥ 1 . ere exists a function K(t) , which also depends on 4 , such that

L(t) ≥ 4 − K(t)∫ 2TT

| K(t) |2 dt

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5.3.2 A positive proportion of No(T )

A. Selberg showed that a positive proportion of ζ(s) zeros lie on the critical line. He stated that

No(T ) >> T log T

B. Conrey went further and gave a percentage on them. He showed that if T is suciently largethen

No(T ) >2

5N(T ) (18)

which shows that if T is large enough then the amount of the critical zeros is always more thenforty percent of all zeros of the zeta-function. e result does not deny the RH because it onlyshows the lower boundary of the critical zeros. is means that the ratio of the critical zeroscould be much higher, such that all zeros are the critical zeros [7]. We need to wait for such aremarkable result to come.

A proof of equation 18 needs a proper knowledge in Dirichlet polynomial which will lead to anew area of discussion. at’s why we avoid it. e proof is available in Analytic Number eoryby H. Iwaniec and E. Kowalski if the reader is interested [7].

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6 Primes distribution and ζ(s) zerosAer all the eort we have put into studying the zeta-function and its properties. We are nearto the essence of Riemann 1859 paper or what Derbyshire (2003) calls the Golden key. In thissection, we shall study that the Riemann’s prime-counting function has some resemblance to theprime counting function π(x) , although they are not the same. Representation of Riemann’sprime-counting function is followed by a discussion of the explicit formula of Riemann. at willshow us a direct relation between nontrivial zeros of the zeta-function and primes distribution.At the very end, we shall discuss the importance of the Riemann hypothesis.

6.1 Riemann’s prime counting function J(x) , see [4]Wehave earlier in this paper talked about the prime counting function π(x) . e prime countingfunction is a step function for example, the prime counting function has same value 4 for x =10, 10.1, 10.2 . . . . e function keeps the same value and then jumps suddenly to 5 at x =11, 11.1, 11.2 . . . until x = 13 .We are going to introduce a function which is alike the prime counting function in the sensethat both are step functions. Both are nite sums and both count the prime numbers but it isnecessary to mention that they are not exactly the same. is function was originally denoted asa f function by Riemann but we refer to it as J as following Harold Edwards because nowadaysf is commonly used for any function. J(x) is dened as follows

J(x) = π(x) +1

2π (√x) +

1

3π ( 3√x) +

1

4π ( 4√x) +

1

5π ( 5√x) + . . . . . .

Where x is any positive number and π is the prime counting function. J(x) is a nite sumand it can be seen by the denition of the function. No maer what x we chose, aer sometime we shall end up with a x smaller than 2 . Because the prime counting function is zero forx < 2 , it bounds the J(x) to a nite sum. For example, if we take x = 100 by the seventh rootand all roots aer that we shall get π(x) = 0 as shown in the following equation (19)

J(100) = π(100) +1

2π(10) +

1

3π(4.64 . . .) +

1

4π(3.16 . . .) +

1

5π(2.51 . . .) +

1

6π(2.15 . . .) +

(19)0 + 0 . . .

which in the terms of primes, means

J(100) = 25 +

(1

2× 4

)+

(1

3× 2

)+

(1

4× 2

)+

(1

5× 1

)+

(1

6× 1

)and that sums ups to 28.5333 . . . . As it is said earlier, for any nonnegative x , J(x) is a nitesum. Why is J(x) important? Well, we have already dened J(x) in terms of π(x) . Beforegoing to the next stage we need to show π(x) in the terms of J(x) . Riemann used the Möbiusinversion to express π(x) in the terms of J(x) . We are not going to discuss theMöbius inversionhere because of the lack of space. e reader is referred to [2] for studies of the Möbius inversion,if necessary. Anyhow, by applying the inversion Riemann got the following expression

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π(x) = J(x) − 12J (√x) − 1

3J ( 3√x) − 1

5J ( 5√x) +

1

6J ( 6√x) − 1

7J ( 7√x) +

1

10J ( 10√x) − · · ·

(20)By analyzing the above expression (20), we can notice that some terms are missing (the fourth,eighth and ninth) and the sign is alternating. at leads us to the next expression (21) which hasthe Möbius function (see appendix) in it. e expression, we get is

π(x) =∑n

µ(n)

nJ( n√x) (21)

Now, we have shown π(x) in the terms of J(x) . By doing that Riemann founded a way toexpress J(x) in terms of ζ(x) . at is a wonderful thing in mathematics. Let us put it to thetest and account π(100) by using the new expression.

π(100) = J(100) − 12J(10) − 1

3J(4.64 . . .) − 1

5J(2.51 . . .) +

1

6J(2.15 . . .) − 0 + 0 − · · ·

(22)

= 288

15−(

1

2× 51

3

)−(

1

3× 21

2

)−(

1

5× 1)

+

(1

6× 1)

= 288

15− 22

3− 5

6− 1

5+

1

6

We notice that π(x) is also a nite sum because J(x) is zero for x < 2 . e result of theequation (22) is 25 and that is magical. Because it gives us the exact number of primes up to 100 .e dierence between J(x) and π(x) can be shown by the following gure

Now we shall discuss the Golden key which shows a direct connection between J(x) and ζ(s) .

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6.1.1 e Golden key

We have shown earlier that ζ(s) can be wrien in terms of the product of only primes. eEuler’s product

ζ(s) =∏p

(1

1− 1ps

)Next equation is just another way of expressing the Euler’s product

ζ(s) =1

1− 12s

× 11− 1

3s

× 11− 1

5s

× 11− 1

7s

× 11− 1

11s

× 11− 1

13s

× · · · · · · (23)

If we take log of both sides of equation (23). en by manipulating the log property such thatlog(a · b) = log(a) + log(b) gives us the following

log ζ(s) = log

(1

1− 12s

)+ log

(1

1− 13s

)+ log

(1

1− 15s

)+ log

(1

1− 17s

)+ log

(1

1− 111s

)+· · ·

since log 1a

= − log(a) , we have

− log(

1− 12s

)− log

(1− 1

3s

)− log

(1− 1

5s

)− log

(1− 1

7s

)− log

(1− 1

11s

)−· · · (24)

By using Taylor expansion for log(1− x) , we can expand every log term in equation (24) intoan innite series. We are allowed to do that because x is between −1 and 1 , so long as s ispositive. Since all conditions are fullled, we have

1

2s+

(1

2× 1

22s

)+

(1

3× 1

23s

)+

(1

4× 1

24s

)+

(1

5× 1

25s

)+

(1

6× 1

26s

)+ · · ·

+1

3s+

(1

2× 1

32s

)+

(1

3× 1

33s

)+

(1

4× 1

34s

)+

(1

5× 1

35s

)+

(1

6× 1

36s

)+ · · ·

+1

5s+

(1

2× 1

52s

)+

(1

3× 1

53s

)+

(1

4× 1

54s

)+

(1

5× 1

55s

)+

(1

6× 1

56s

)+ · · ·

+1

7s+

(1

2× 1

72s

)+

(1

3× 1

73s

)+

(1

4× 1

74s

)+

(1

5× 1

75s

)+

(1

6× 1

76s

)+ · · ·

+1

11s+

(1

2× 1

112s

)+

(1

3× 1

113s

)+

(1

4× 1

114s

)+

(1

5× 1

115s

)+

(1

6× 1

116s

)+ · · ·

We pick one term from the second series above for example 12× 1

32s. By following Riemann, one

can show that

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1

2× 1

32s=

1

2× s×

∫ ∞32

x−s−1 dx (25)

A deeper study of Riemann’s method is made in Derbyshire (2003). Expression (25) above is justa way to come closer to J(x) in terms of ζ(s) . By expressing (25), Riemann showed that J(x)could be shown as a continuous sum of integrals.

e idea behind this is to show that every stripe in J(x) can be represented by using primes.For example, stripe-2 where jump for J(x) is 1

2can be represented by∫ ∞

2

1× x−s−1 dx +∫ ∞22

1

2× x−s−1 dx +

∫ ∞23

1

3× x−s−1 dx + · · · · · · · · ·

Stripe-3 in J(x) can be expressed by using prime number 3 and then 5 and so on. Withoutgoing to deep into this, idea is that an innite sum of an area under the step function J(x) canbe expressed in terms of an innite sum of integrals. Which eventually leads us to the Golden key

log ζ(s) = s ×∫ ∞0

J(x)x−s−1dx

or1

slog ζ(s) =

∫ ∞0

J(x)x−s−1dx

is is a very important result because π(x) and J(x) belong to the number theory and ζ(s)belongs to calculus and analysis. By showing a relation between J(x) and ζ(s) , Riemanncreated a bridge between two dierent elds of mathematics. In other words a great opportunityfor math and mathematicians [4]. e Golden key leads us to the heart of Riemann’s paper. atis called the explicit formula of the Riemann and we shall discuss that now.

6.2 e explicit formula of Riemann, see [4]Now, we have come to the heart of Riemann’s 1859 paper. Earlier, we have shown ζ(s) in termsof J(x) and now its time to express J(x) in terms of ζ(s) . e expression which shows J(x)in terms of ζ(s) is called the explicit formula of Riemann and is stated as

J(x) = Li(x) −∑ρ

Li(xρ) − log 2 +∫ ∞x

dt

t(t2 − 1) log t(26)

e expression (26) is the main result of Riemann’s paper. at is a very important result becauseit gives us the opportunity to knowmore about primes distribution by studying the zeta-function.We divide the explicit formula into four terms and study them separately. We start with the easyterms the third and fourth and then continue to the rst and the second.

e third term log(2) is not that complicated and simply equals 0.693147180 . . . . e fourthterm is the area under the curve 1

t(t2−1) log(t) for t from x to ∞ . e Riemann prime countingfunction is zero for x < 2 that’s why we are interested in the area of the curve from x = 2to ∞ . e maximum value for the fourth term is 0.14001010 . . . . at gives us, the value of

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both the third and the fourth term together is in between −0.6931 to −0.5531 , which is notsignicant for the prime counting function, because the prime counting function is about millionsand trillions and such a small value is insignicant.

e rst term Li(x) is called the principle term. Li(x) = li(x)− li(2), where li(x) is dened asthe area under the curve 1

log(t)from 0 to x . In other words

Li(x) = li(x)− li(2) =∫ x2

dt

log t

By subtracting li(2), we get rid of the singularity of li(x) at x = 1. We are allowed to do thatbecause J(x) = 0 for 0 ≤ x < 2.

Figure 7: e graph of li(x) ploed 0 ≤ x ≤ 40

Li(x) is asymptotic to the prime counting function, π(x) ∼ Li(x) . We presented earlier theprime number theorem, π(x) ∼ Li(x) is an improved form of the prime number theorem.Figure 8 shows that Li(x) approximates π(x) beer than x

log(x). Now, back to J(x). e rst term

Figure 8: An approximation of the prime counting function π(x) by Li(x) and xlog x

Li(x) is not a big problem either because one can obtain values for the rst term from math tablebooks or soware likeMathematica orMaple. e most important term in the explicit formula isthe second term

∑ρ Li(x

ρ) .

e second term is referred to as a periodic term and is central to the explicit formula. If we

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explain the second term, in other words, we can say that we add up Li(xρ) over roots ρ , whereρ are nontrivial zeros of the zeta-function. e question is, how the nontrivial zeros turned upin the J(x) ? Remember the Golden key

1

slog ζ(s) =

∫ ∞0

J(x)x−s−1dx

We know from algebra and polynomial that one function can be expressed in a form of its zeros.e problem here is that all polynomials are entire functions but the zeta-function is not, becauseit has a singularity at s = 1 . Riemann during the inversion of the Golden key, in order to expressJ(x) in terms of ζ(s) succeeded to transform the zeta-function into a slightly dierent entirefunction. All zeros of the new function are the nontrivial zeros of the zeta-function. Now, thenew function is entire and we can write it in form of its zeros. All trivial zeros of the zeta-functionvanished under the process of transformation. e new function in terms of the nontrivial zerosis the second term

∑ρ Li(x

ρ) . J(x) is a real function, and when we sum Li(xρ) over ρ all theIm(s) of ρ vanishes. at’s why

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Figure 10: An approximation of the prime counting function by the Riemann prime countingfunction by using the rst 300 nontrivial zeros ρ of the zeta-function

Two graphs 9 and 10 show an approximation of the prime counting function by the Riemannprime counting function. Both graphs show us that the higher number of nontrivial zeros usedin summation of the second term

∑ρ Li(x

ρ) give a more precise approximation of the primecounting function. e point that the Riemann hypothesis is important for the prime numbertheory and primes distribution should be more clear now. A clear conclusion is that more weknow about the nontrivial zeros of the zeta-function, the more precise approximation can beobtained. In other words, we can obtain minimum of J(x) − π(x) by knowing more aboutnontrivial zeros of the zeta-function.

6.3 e Riemann hypothesis in math and physicse simplest innite Dirichlet series, the zeta-function is fundamental in the theory of the primenumbers. It is and has been very important to study the properties of ζ(s) , specically zerosof ζ(s) for the analytic number theory [5]. As the Riemann zeta-function is important for thenumber theory, a solution to the Riemann hypothesis would be essential for several problemsrelated prime distribution. ere are several questions about the prime numbers which would beimpacted by a solution to the Riemann hypothesis [10].

For example, are there innitely many pair primes (p, p+ 2) ? Are there innitely many primetriplets of the type (p, p + 2, p + 6) and (p, p + 4, p + 6) ? ese ideas have been discussedthrough the history of mathematics but they are still to be proven [5]. An answer to the RHwill be inuential for such simple problems. Every even number ≥ 6 is the sum of two oddprimes and every odd number ≥ 9 is the sum of three odd primes. e last two statementswere conjectured by Goldbach in 1742 and Euler agreed to the truth of these conjectures butthey were unable to give a proof. A solution to the RH might be helpful to prove or disprove theconjecture [10]. We can conclude that ζ(s) is of great importance in the number theory.

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A solution to the RH is not only concerned with prime distribution, there are several otherproblems which could be solved if we were fully aware of ζ(s) properties. For example, the wellknown problem of Dirichlet series divisor in the number theory which solution needs knowledgeabout the behavior of |ζ(s)| . ζ(s) is even used in quantum theory to study Feynman’s ”pathintegral”. at is because ζ(s) signals are connected with the law of distribution of the primenumbers in a natural series [8].

Further, it seems that there is a relation between nontrivial zeros of the zeta-function and eigenvaluesof Hermitian matrices. e wonder here is, how eigenvalues of Hermitian matrices are related toprime distribution? Because nontrivial zeros of the zeta-function are related to prime distributionwhile eigenvalues of randomHermitianmatrices turn up in the result of inquiries into the behaviorof system of subatomic particles under the law of quantum mechanics. Whether true or false, theRH will have an importance in quantum physics [4].

ere are some equivalent functions to ζ(s) likeDavenport-Heilbronn function f(s) and Epstein’szeta-function ζQ(s) which fulll the functional equation of Riemann zeta-function ζ(s) butfails the RH. Because they have zeros outside the critical line. e resemblance between ζ(s)and two functions mentioned above has an end because they do not have the Euler product.Which leads us to an inability to make nal decisions about the RH by studying such equivalentfunctions [8]. Finally, it is not already decided that the RH must be true. It may be false but sucha groundbreaking result is worth waiting for[4].

AcknowledgmentsI would like to express my deep sense of gratitude to my supervisor Johan Björklund for hisinspiring suggestions. I am deeply indebted to him for givingme a chance to study this subject andproviding guidance throughout this work. I acknowledge with thanks, the assistance providedby my younger brother, Zikriya Nawaz, who helped me to understand Latex.Finally, I would like to thank my parents who paid for my early education and my wife forsupporting me on this journey.

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References[1] A. R. Adams and C. Essex. Calculus. PEARSON, 2013.

[2] Tom M. Apostol. Introduction to Analytic Number eory. Springer-Verlag, 1976.

[3] Titchmarsh E. C. THE THEORY OF THE RIEMANN ZETA-FUNCTION. CLARENDON PRESS.OXFORD, 1986.

[4] J Derbyshire. Prime Obsession. Joseph Henry Press Washington, D.C., 2003.

[5] H.G. Hardy and M. E. Wright. An Introduction to the eory of Numbers. CLARENDONPRESS. OXFORD, 1979.

[6] K. Ireland and M. Rosen. A Classical Introduction to Modern Number eory. Springer, 1990.

[7] H. Iwaniec and E. Kowalski. Analytic Numbereory. American Mathematical Society, 2004.

[8] A. A. Karatsuba. COMPLEX ANALYSIS in NUMBER THEORY. CRC Press, 1995.

[9] A. A. Karatsuba and M. S. Voronin. e Riemann Zeta-Function. Walter de Gruyter. Berlin.New York 1992, 1992.

[10] O. Ore. Number eory and Its History. DOVER PUBLICATION, INC., New York, 1948.

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Appendix

Arithmetic FunctionA real- or complex-valued function dened on the positive integers is called an arithmetic functionor a number-theoretic functions. Several arithmetic functions have an important role in the studyof prime distribution. An example of such function is the Möbius function µ(n) . µ(n) is denedas follows

µ(1) = 1

if n > 1 , write n = pa11 . . . . . . pakk . en

µ(n) = (−1)k if a1 = a2 = . . . . . . = ak = 1

µ(n) = 0 otherwise.Note that µ(n) = 0 if and only if n has a square factor > 1. e values of µ(n) for rst sevenpositive integers are 1,−1,−1, 0,−1, 1 and− 1 [2].

Bernoulli Numberse sequence number B0, B1, B2, . . . . . . . e Bernoulli numbers are dened inductively asfollows. B0 = 1 and if B1, B2, . . . . . . , Bm−1 are already determined then Bm is dened by

(m + 1) Bm = −m−1∑k=0

(m+ 1k

)Bk

which gives us a combination of linear equations

1 + 2B1 = 0

1 + 3B1 + 3B2 = 0

1 + 4B1 + 6B2 + 4B3 = 0

1 + 5B1 + 10B2 + 10B3 + 5B4 = 0

e solution to this equation system gives us that B1 = −12 , B2 =16, B3 = 0, B4 =

− 130, B5 = 0, B6 =

142, . . . . . . etc. Bernoulli numbers for odd integers > 1 vanishes and

non-zero Bernoulli numbers alternate in sign [6].

Big-O NotationWe write f(x) = O (u(x)) as x → a , provided that

|f(x)| ≤ k |u(x)|

hold for some constant k on some open interval containing x = a . Similarly, f(x) = g(x) +O (u(x)) as x → a if f(x) − g(x) = O (u(x)) as x → a , that is, if

| f(x) − g(x) | ≤ k |u(x)| near a.

For example, sinx = O(x) as x → 0 because | sinx| ≤ |x| near 0 [1].

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Cauchy InequalityAccording to the Cauchy inequality, square of a series of a product uv where u, v ∈ R is equalto or smaller than the product of series square of both variables in rst product. Mathematicallystated, the Cauchy inequality means( n∑

i=1

uivi

)2≤( n∑

i=1

u2i

) ( n∑i=1

v2i

)where the inequality holds if and only if ui

vi= k where k ∈ R+ for all 1 ≤ i ≤ n which

have uivi 6= 0 [8].

Dirichlet polynomialA Dirichlet polynomial is a nite Dirichlet series

D(s) =∑

1≤n≤N

ann−s

with complex coecients an [7].

Real part of a complex number