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AKADEMIN FÖR TEKNIK OCH MILJÖ Avdelningen för elektronik,
matematik och naturvetenskap
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
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Omslagsbild: Figuren visar grafen av [Table[Prime[𝑛] (𝑛Log[𝑛])⁄
, {𝑛, 2, 5000}]]. Figur: Daud Nawaz, 2018.
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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.
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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 .
15
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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
16
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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.
19
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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
20
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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
24
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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
27
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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) .
28
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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