An introduction to the Riemann hypothesis

Author:Alexander Bielikabielik@kth.se

Supervisor:Par Kurlberg

SA104X – Degree Project in Engineering PhysicsRoyal Institute of Technology (KTH)

Department of Mathematics

September 13, 2014

Abstract

This paper exhibits the intertwinement between the prime numbers and the zeros of the Riemann zeta function,drawing upon existing literature by Davenport, Ahlfors, et al.

We begin with the meromorphic continuation of the Riemann zeta function ζ and the gamma function Γ. We thenderive a functional equation that relates these functions and formulate the Riemann hypothesis.

We move on to the topic of finite-ordered functions and their Hadamard products. We show that the xi functionξ is of finite order, whence we obtain many useful properties. We then use these properties to find a zero-free regionfor ζ in the critical strip. We also determine the vertical distribution of the non-trivial zeros.

We finally use Perron’s formula to derive von Mangoldt’s explicit formula, which is an approximation of the Cheby-shev function ψ. Using this approximation, we prove the prime number theorem and conclude with an implication ofthe Riemann hypothesis.

Contents

Introduction 2

1 The statement of the Riemann hypothesis 31.1 The Riemann zeta function ζ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 The gamma function Γ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.3 The functional equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.4 The critical strip . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2 Zeros in the critical strip 102.1 Functions of finite order . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102.2 The Hadamard product for functions of order 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.3 Proving that ξ has order at most 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.4 A zero-free region for ζ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162.5 The number of zeros in a rectangle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

3 The distribution of prime numbers 233.1 Perron’s formula for Dirichlet series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233.2 An approximation of the Chebyshev function ψ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253.3 von Mangoldt’s explicit formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273.4 Proving the explicit formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283.5 The prime number theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303.6 The smallest possible error term . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

Appendices 33

A Additional proofs 34A.1 The Weierstrass form of Γ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34A.2 Stirling’s formula for ln Γ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35A.3 Jensen’s formula for holomorphic functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37A.4 The uniqueness theorem for Dirichlet series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

Bibliography 39

1

Introduction

In 1859, the German mathematician Georg Friedrich Bernhard Riemann proposed a hypothesis [Riemann, 1859,pp. 1-9] about prime numbers that would later bear his name, the Riemann hypothesis. The prime numbers donot appear to follow any obvious pattern. However, Riemann observed a close relation between the behavior of anelaborate function, the so-called Riemann zeta function, and the frequency of prime numbers. Riemann calculateda few zeros of this function, and quickly noted that the interesting ones lay on a certain vertical straight line in thecomplex plane. Riemann subsequently conjectured that all non-trivial zeros lie on this line. Today, over 1013 zerosare known [Gourdon, 2004, pp. 19-25], and all of them agree with the hypothesis.

The Riemann hypothesis has important implications for the distribution of prime numbers and is strongly con-nected to the prime number theorem, which gives a good approximation of the density of prime numbers. In particular,the Riemann hypothesis gives a precise answer to how good the approximation given by the prime number theorem is.In a sense, the Riemann hypothesis conveys the idea that the prime numbers are distributed as regularly as possible.This regularity would tell a great deal about the average behavior of prime numbers in the long run.

In today’s society, the importance of prime numbers has increased rapidly, especially with the advance of in-formation technology and cryptography. However, the importance of the Riemann hypothesis goes far beyond itsconsequences for the distribution of prime numbers. It has been shown that hundreds of statements in number theoryfollow from it [Gowers et al., 2008, p. 715]. With this background, it might not be a surprise that some mathemati-cians consider the hypothesis to be the most important problem in pure mathematics, but it remains unresolved fornow. The Riemann hypothesis is one of the seven Millennium Prize Problems that were stated by the Clay Mathe-matics Institute in 2000, carrying a million dollar prize for a correct solution. It is also part of the eighth problem inDavid Hilbert’s list of unsolved problems.

Other than the Riemann zeta function and its zeroes, there is currently no known approach to establish the distri-bution of prime numbers with desired precision. A disproof of the Riemann hypothesis would reveal a lot about howdisordered the primes numbers really are. Enrico Bombieri, a prominent number theorist, remarked that ”the failureof the Riemann hypothesis would create havoc in the distribution of prime numbers” [Havil, 2003, p. 205]. In 1770,Euler argued more pessimistically that ”mathematicians have tried in vain to discover some order in the sequence ofprime numbers but we have every reason to believe that there are some mysteries which the human mind will neverpenetrate” [Gowers et al., 2008, p. 348].

2

1. The statement of the Riemannhypothesis

1.1 The Riemann zeta function ζ

For complex numbers s with real part greater than 1, we define the Riemann zeta function by the absolutelyconvergent series

ζ(s) :=

∞∑n=1

1

ns= 1 +

1

2s+

1

3s+ · · · . (1.1)

By the Weierstrass M-test, we find that the convergence is uniform in the region Re(s) ≥ 1 + δ for any δ > 0. Wenow show that ζ is holomorphic within Re(s) > 1. To this end, we consider the sequence of holomorphic functionsdefined by

fi(s) :=

i∑n=1

1

ns. (1.2)

Since {fi}∞i=1 converges uniformly to ζ on any compact subset of Re(s) > 1, it follows that the function is holomorphicthere.

In 1737, Euler deduced that ∏p∈P

1

1− p−s=∏p∈P

∞∑n=0

(1

ps

)n=∏p∈P

(1 +

1

ps+

1

p2s+ · · ·

)= 1 +

∑p∈P

1

ps+∑p,q∈P

1

psqs+ · · ·

= 1 +1

2s+

1

3s+ · · · =

∞∑n=1

1

ns= ζ(s) (1.3)

for any integer s > 1, where P denotes the set of prime numbers, though his argument can be extended to any complexnumber s with Re(s) > 1. In the above derivation, we first used the formula for a geometric series. Then, we rewrotethe product and used the fundamental theorem of arithmetic, which states that each positive integer equals exactlyone product of prime powers.

This useful relation is usually called the Euler product formula. Some writers describe it as ”the Golden Key”[Derbyshire, 2004, p. 105]. Taking the logarithm, we find that∣∣∣∣∣∣ln

∣∣∣∣∣∣∏p∈P

1

1− p−s

∣∣∣∣∣∣∣∣∣∣∣∣ =

∣∣∣∣∣∣∑p∈P

ln

∣∣∣∣ 1

1− p−s

∣∣∣∣∣∣∣∣∣∣

=

∣∣∣∣∣∣∑p∈P

ln

∣∣∣∣ ps

ps − 1

∣∣∣∣∣∣∣∣∣∣

=

∣∣∣∣∣∣∑p∈P

ln |ps| − ln |ps − 1|

∣∣∣∣∣∣=

∣∣∣∣∣∣∑p∈P

ˆ |ps||ps−1|

dx

x

∣∣∣∣∣∣ ≤∑p∈P

ˆ |ps||ps|−1

dx

x<∑p∈P

1

|ps| − 1, (1.4)

3

which clearly converges for Re(s) > 1. It follows that the product on the left-hand side of (1.3) converges. Hence,the formula (1.3) allows us to conclude that ζ has no zeros in the region Re(s) > 1, as each factor in the convergentproduct is different from zero.

By partial summation, we can extend the domain of the function to Re(s) > 0:

ζ(s) =

∞∑n=1

n−s

=

∞∑n=1

n(n−s − (n+ 1)−s

)= s

∞∑n=1

n

ˆ n+1

n

x−s−1 dx

= s

ˆ ∞1

bxcx−s−1 dx

= s

ˆ ∞1

x−s dx− sˆ ∞

1

{x}x−s−1 dx

=s

s− 1− sˆ ∞

1

{x}x−s−1 dx. (1.5)

Here, the symbols bxc and {x} = x− bxc stand for the integral and fractional parts of x, respectively.

Since |{x}| ≤ 1, we see that the integral on the right of (1.5) converges absolutely in this extended domain.Furthermore, the convergence is uniform in the region Re(s) ≥ δ for any δ > 0. It follows that this new function is ameromorphic continuation of ζ. We observe that its only pole in this domain is a simple pole at s = 1 with residue

Res(ζ, 1) = lims→1

(s− 1)ζ(s) = 1. (1.6)

We also observe that both terms in (1.5) are real and negative for real numbers 0 < s < 1, so ζ(s) < 0 on this line.Consequently, there are no zeros in this interval.

For completeness, we mention another way to meromorphically continue ζ to Re(s) > 0. For Re(s) > 1, we observethat (

1− 2

2s

)ζ(s) =

∞∑n=1

1

ns−∞∑n=1

2

(2n)s=

∞∑n=1

(−1)n−1

ns=: η(s), (1.7)

whence ζ(s) = (1− 21−s)−1η(s). Here, η is the Dirichlet eta function, which can be shown to converge for Re(s) > 0.While the right-hand side has singularities at s = 1+k 2πi

ln 2 , where k ∈ Z, only the singularity at s = 1 is non-removable.

1.2 The gamma function Γ

The gamma function is an extension of the factorial function to complex numbers. For numbers s with positivereal part, it is defined by the convergent integral

Γ(s) :=

ˆ ∞0

ts−1e−t dt. (1.8)

Let us show that this function is holomorphic. This time, we use Morera’s theorem. Let C be any closed curve withinRe(s) > 0. Then ‰

C

Γ(s) ds =

‰C

ˆ ∞0

ts−1e−t dtds =

ˆ ∞0

e−t(‰

C

ts−1 ds

)dt (1.9)

by Fubini’s theorem. We now observe that the inside integral is 0 by the Cauchy–Goursat theorem. It follows fromMorera’s theorem that Γ is holomorphic in this domain.

Using integration by parts, we find that Γ satisfies the following functional equation:

Γ(s+ 1) =

ˆ ∞0

tse−t dt =[−tse−t

]∞t=0

+ s

ˆ ∞0

ts−1e−t dt = sΓ(s). (1.10)

We also observe that

Γ(1) =

ˆ ∞0

e−t dt = 1, (1.11)

4

so it follows thatΓ(n+ 1) = n! (1.12)

for positive integers n. In this sense, the gamma function is an extension of the factorial function.

The functional equation (1.10) enables us to obtain the meromorphic continuation of Γ by a step-by-step procedure.For Re(s) > −1, we define Γ1 by

Γ1(s) :=1

sΓ(s+ 1), (1.13)

so that Γ1(s) = Γ(s) for Re(s) > 0. We note that Γ1 is holomorphic for Re(s) > −1, except for the simple pole ats = 0. We now let k be any positive integer and define Γk by

Γk(s) :=1

s(s+ 1) . . . (s+ k − 1)Γ(s+ k) (1.14)

for Re(s) > −k. We similarly find that Γk is holomorphic in this domain, except for the simple poles at the non-positiveintegers from 0 to 1− k with residue

Res(Γ,−k) = lims→−k

(s+ k)Γ(s)

= lims→−k

(s+ k)Γ(s+ k)

s(s+ 1) . . . (s+ k − 1)

= lims→−k

Γ(s+ k + 1)

s(s+ 1) . . . (s+ k − 1)

=Γ(1)

(−k)(−k + 1) . . . (−1)

=(−1)k

k!. (1.15)

We also note that Γk(s) = Γ(s) for Re(s) > 0. Letting k → ∞, we obtain the meromorphic continuation of Γ to thewhole complex plane with simple poles at the non-positive integers. From now on, this infinitely extended functionwill be denoted by Γ; cf. [Ireland and Rosen, 2010, pp. 261-262].

We now wish to derive some useful formulae. For this purpose, we use the Weierstrass form of Γ (see appendixA.1 for proof; cf. [Ahlfors, 1966, p. 198]), which is valid for any complex number:

Γ(s) =e−γs

s

∞∏n=1

(1 +

s

n

)−1

es/n. (1.16)

We see that Γ is nowhere zero, since each factor in the convergent product is different from zero. It follows that thereciprocal of Γ is entire. We can also confirm that there are simple poles at the non-positive integers.

Comparing the Weierstrass products (cf. [Ahlfors, 1966, p. 195]) for the entire functions

1

Γ (s)= seγs

∞∏n= 1

(1 +

s

n

)e−s/n (1.17)

andsin (πs) = πs

∏n 6=0

(1− s

n

)es/n, (1.18)

we find that1

−sΓ (−s) Γ (s)=

sin (πs)

π. (1.19)

Using the functional equation Γ(1 + s) = sΓ(s), it follows that

Γ (1− s) Γ (s) = −sΓ (−s) Γ (s) =π

sin (πs), (1.20)

for non-integers. We call this Euler’s reflection formula. If we set s = 12 , we get

Γ

(1

2

)=√π. (1.21)

This is perhaps the best-known value of Γ at a non-integer argument, and will be used shortly in the next derivation.

5

By the Weierstrass form (1.17), it follows that

d

ds

(Γ′(s)

Γ(s)

)=

∞∑n=0

1

(s+ n)2. (1.22)

Furthermore, we note that Γ(s)Γ(s+ 12 ) and Γ(2s) have the same poles. Taking the second derivative of the logarithm

of Γ(s)Γ(s+ 12 ), we get

d

ds

(Γ′(s)

Γ(s)

)+

d

ds

(Γ′(s+ 1

2

)Γ(s+ 1

2

) ) =

∞∑n=0

1

(s+ n)2+

∞∑n=0

1(s+ n+ 1

2

)2= 4

[ ∞∑n=0

1

(2s+ 2n)2+

∞∑n=0

1

(2s+ 2n+ 1)2

]

= 4

∞∑m=0

1

(2s+m)2

= 2d

ds

(Γ′(2s)

Γ(2s)

), (1.23)

where we used the chain rule in the last step. Integrating both sides twice and then taking the exponential, we obtainthe relation

Γ(s)Γ

(s+

1

2

)= eas+bΓ(2s), (1.24)

where a and b are undetermined constants. Taking s = 12 and s = 1, we get

Γ

(1

2

)Γ(1) = ea/2+bΓ(1) (1.25)

and

Γ(1)Γ

(1 +

1

2

)= ea+bΓ(2), (1.26)

where Γ(1) = Γ(2) = 1, Γ( 12 ) =

√π and Γ(1 + 1

2 ) = 12Γ( 1

2 ) = 12

√π. Taking the exponential, we are led to the system

of equations {12a+ b = 1

2 lnπ,

a+ b = 12 lnπ − ln 2.

(1.27)

It follows that a = −2 ln 2 and b = 12 lnπ + ln 2. Insertion into the original relation gives

Γ(s)Γ

(s+

1

2

)= 21−2s

√πΓ(2s), (1.28)

which is known as Legendre’s duplication formula. Note that the formula is not valid at the non-positive integersand half-integers.

By substituting s with 1−s2 in Euler’s reflection formula (1.20), we get

Γ

(s+ 1

2

)Γ

(1− s

2

)=

π

cos(πs2

) . (1.29)

By substituting s with s2 in Legendre’s duplication formula (1.28), we get

Γ(s

2

)Γ

(s+ 1

2

)= 21−s√πΓ(s). (1.30)

If we take the quotient of the two recently derived formulae (1.29) and (1.30), we obtain a new functional equation:

Γ(s

2

)/Γ

(1− s

2

)= 21−sπ−1/2 cos

(πs2

)Γ(s). (1.31)

We finally mention Stirling’s formula for Γ, which can be stated as follows:

Γ(s) =√

2πe−sss−1/2(1 +O(|s|−1)

). (1.32)

This approximation is valid for large |s| such that | arg s| < π. Taking the logarithm of both sides, we can rewrite theapproximation as

ln Γ(s) =

(s− 1

2

)ln s− s+

1

2ln 2π +O(|s|−1), (1.33)

which is valid under the same conditions (see appendix A.2 for proof; cf. [Davenport, 2000, p. 73]).

6

1.3 The functional equation

We first introduce the Jacobi theta function θ, which is defined by

θ(s) :=

∞∑n=−∞

e−πn2s (1.34)

on the right half-plane Re(s) > 0, where it is holomorphic. We shall find a functional equation for θ, and then

transform it into a functional equation for ζ. Recall that e−αx2

is a fixed point of the Fourier transform with

Fx[e−αx

2]

(ξ) =

√π

αe−(πξ)2/α (1.35)

for complex numbers α with Re(α) > 0, where Fx denotes the Fourier transform with respect to the variable x.Setting α = πs, we get

Fx[e−πx

2s]

(ξ) =1√se−πξ

2/s. (1.36)

Thus, we obtain the following functional equation by the Poisson summation formula:

θ(s) =

∞∑n=−∞

e−πn2s =

∞∑k=−∞

1√se−πk

2/s =1√sθ

(1

s

). (1.37)

We now define the helper function ω by

ω(s) :=∞∑n=1

e−πn2s =

θ(s)− 1

2. (1.38)

It follows that

ω

(1

s

)=θ( 1s )− 1

2=

√sθ(s)− 1

2=

√s

2(2ω(s) + 1)− 1

2= −1

2+

1

2

√s+√sω(s). (1.39)

Let Mx denote the Mellin transform with respect to the variable x. By the definition of Γ in (1.8), we have

Mx

[e−πn

2x]

(s) =

ˆ ∞0

xs−1e−πn2x dx =

{t := πn2x

}= (πn2)−s

ˆ ∞0

ts−1e−t dt = (πn2)−sΓ(s) (1.40)

for Re(s) > 0, whence

Mx

[e−πn

2x] (s

2

)=

ˆ ∞0

xs/2−1e−πn2x dx = π−s/2Γ

(s2

)n−s. (1.41)

By the series representation (1.1) of ζ, it follows that

π−s/2Γ(s

2

)ζ(s) =

∞∑n=1

π−s/2Γ(s

2

)n−s =

∞∑n=1

ˆ ∞0

xs/2−1e−πn2x dx =

ˆ ∞0

xs/2−1ω(x) dx =Mx [ω(x)](s

2

)(1.42)

for Re(s) > 1, where we changed the order of summation in accordance with Fubini’s theorem.

We are now ready to use the symmetry of ω. We split the integral on the right-hand side of (1.42) into two pieces,one from 0 to 1, where we substitute x by 1

x , and the other from 1 to ∞:

Mx [ω(x)](s

2

)=

ˆ 1

0

xs/2−1ω(x) dx+

ˆ ∞1

xs/2−1ω(x) dx

=

ˆ ∞1

x−s/2−1ω

(1

x

)dx+

ˆ ∞1

xs/2−1ω(x) dx. (1.43)

Next, using the functional equation for ω, we find that

ˆ ∞1

x−s/2−1ω

(1

x

)dx =

ˆ ∞1

x−s/2−1

(−1

2+

1

2

√x+√xω(x)

)dx

= −1

s+

1

s− 1+

ˆ ∞1

x−(s+1)/2ω(x) dx, (1.44)

and so

π−s/2Γ(s

2

)ζ(s) =Mx [ω(x)]

(s2

)= − 1

s (1− s)+

ˆ ∞1

(xs/2−1 + x−(s+1)/2

)ω(x) dx, (1.45)

7

which is what we wanted to show.

We note that the integral on the right of (1.45) is absolutely convergent for any s, and converges uniformly in anybounded part of the plane, because

ω(x) = O(e−πx). (1.46)

It follows that the above expression gives the meromorphic continuation of ζ to the whole complex plane, so we mayuse it to define the function for all remaining non-zero complex numbers. A little more care is needed for s = 0.

Since the right-hand side of (1.45) is invariant under the substitution s 7→ 1− s, we find that

π−s/2Γ(s

2

)ζ(s) = π−(1−s)/2Γ

(1− s

2

)ζ(1− s). (1.47)

This, together with the functional equation for Γ in (1.31), shows that ζ satisfies

ζ(1− s) = 2(2π)−s cos(πs

2

)Γ(s)ζ(s), (1.48)

which is known as the Riemann functional equation. Substituting 1− s for s yields

ζ(s) = 2sπs−1 sin(πs

2

)Γ(1− s)ζ(1− s). (1.49)

Using continuity at s = 0, we obtain the value at this point by taking the limit as s → 0+ along any path in theregion Re(s) > 0, where the integral representation (1.5) of ζ is valid:

ζ(0) = lims→0

2sπs−1 sin(πs

2

)Γ(1− s)ζ(1− s)

= lims→0

(2π)s

π

( ∞∑n=0

(−1)n

(2n+ 1)!

(πs2

)2n+1)

Γ(1− s)(

1− s(1− s)− 1

− (1− s)ˆ ∞

1

{x}x−(1−s)−1 dx

)

= lims→0

(2π)s

π

(πs2

)( ∞∑n=0

(−1)n

(2n+ 1)!

(πs2

)2n)

Γ(1− s)(−1 + s

s− (1− s)

ˆ ∞1

{x}xs−2 dx

)

= lims→0

(2π)s

2

(1 +

∞∑n=1

(−1)n

(2n+ 1)!

(πs2

)2n)

Γ(1− s)(−1 + s− s(1− s)

ˆ ∞1

{x}xs−2 dx

)=

1

2· 1 · Γ(1) · (−1)

= −1

2. (1.50)

The Riemann functional equation also has a symmetric version. For Re(s) > 0, let

ξ(s) :=1

2π−s/2s(s− 1)Γ

(s2

)ζ(s) (1.51)

be the xi function, sometimes called Landau’s xi function. It follows immediately from the definition that

ξ(s) = ξ(1− s), (1.52)

and we use the above relation to define the function for Re(s) ≤ 0.

1.4 The critical strip

It is time to reach some conclusions. In the sine version (1.49) of the Riemann functional equation, we let s = −2nfor any positive integer n and note that sin

(πs2

)= 0. Since Γ(1− s) and ζ(1− s) are finite at these points, it follows

that the Riemann zeta function has simple zeros at the negative even integers; these are known as the trivial zeros.However, as we shall see, there are other values for which the function is zero; these are called the non-trivial zeros.

Let us try to locate the non-trivial zeros. By the definition, we have ζ(s) = ζ(s), so the zeros are symmetric aboutthe real axis. This, together with the Riemann functional equation, shows that the zeros are also symmetric aboutthe line Re(s) = 1

2 . It follows that there are no non-trivial zeros for Re(s) < 0, since there are no for Re(s) > 1.We conclude that every non-trivial zero must satisfy 0 ≤ Re(s) ≤ 1. For this reason, we call this region the criticalstrip. The Riemann hypothesis asserts that all zeros in the critical strip lie on the line Re(s) = 1

2 , which we namethe critical line. Figure 1.1 illustrates our recent conclusions.

8

(a) The Riemann zeta function ζ in the complex plane. Thewhite spot at s = 1 represents the function’s pole. The blackspots at the negative even integers represent the trivial zeros.The black spots on the critical line Re(s) = 1

2represent the

non-trivial zeros.

(b) The gamma function Γ in the complex plane. The whitespots at the non-positive integers represent the poles of thefunction. Note the lack of zeros.

(c) The xi function ξ in the complex plane. Note the symmetryand the lack of poles.

(d) The first few zeros of ζ on the critical line highlighted byplotting

∣∣ζ ( 12

+ it)∣∣ for −50 ≤ t ≤ 50.

Figure 1.1: Visualization of some complex-valued functions using the domain coloring technique. The magnitudeof the output is represented by the brightness, where black represents zero and white represents infinity, while theargument of the output is represented by the hue, where red represents zero.

We note that the Riemann hypothesis does not explicitly say anything about the multiplicity of the zeros. However,it has been shown (cf. [Bui and Heath-Brown, 2013, pp. 1-10]) that the hypothesis implies that at least 19

27 of thezeros are simple. Thus far, all zeros that have been located are simple.

9

2. Zeros in the critical strip

2.1 Functions of finite order

We say that an entire function f is of finite order if there exist real numbers C and α such that

|f(z)| ≤ C exp(|z|α) (2.1)

as |z| → ∞. It follows from Picard’s theorem that α > 0 if f is non-constant. We call the infinum of all numbersα for which this inequality holds for some number C the function order of f . We first show that a finite-orderedfunction f with no zeros must be of the form eg, where g is a polynomial. Moreover, we also show that the degree ofg must be equal to the order of f , so it is always an integer.

Since f is an entire function with no zeros, we know that its logarithmic derivative f ′

f is itself entire. Furthermore,it is known that any antiderivative of an entire function is entire, from which it follows that the single-valued functiong(z) = ln f(z) is entire. We now show that g is a polynomial.

We define h(z) := g(z)− g(0), so that h(0) = 0, and let

M(R) := sup|z|≤2R

Re(h(z)), (2.2)

where R > 0. We observe that M(R) ≥ 0, since h(0) = 0, and consider the holomorphic function

φ(z) :=h(z)

2M(R)− h(z)(2.3)

in the closed disk of radius 2R centered at the origin. We find that |φ(z)| ≤ 1, because |h(z)| ≤ |2M(R) − h(z)| bythe definition of M . Since φ(0) = 0, it follows that

ψ(z) :=2Rφ(z)

z(2.4)

is holomorphic in the disk. Furthermore, |ψ(z)| ≤ 1 on the boundary |z| = 2R. By the maximum modulus principle,the same is true in the disk. But then

|φ(z)| =∣∣∣ z2R

ψ(z)∣∣∣ ≤ 1

2(2.5)

for |z| ≤ R. By the definition of φ, it follows that 2|h(z)| ≤ |2M(R)−h(z)| ≤ 2M(R)+ |h(z)|, so that |h(z)| ≤ 2M(R)in the closed origin-centered disk of radius R.

Let 2R > α√

lnC. By the definition in (2.1), we note that the function g satisfies

Re(g(z)) = ln |f(z)| ≤ lnC + |z|α < 2(2R)α. (2.6)

in the closed disk of radius 2R. By the above inequality, we get

M(R) = sup|z|≤2R

Re(h(z)) = sup|z|≤2R

Re(g(z))−Re(g(0)) < 2(2R)α + 2(2R)α = 22+αRα. (2.7)

It follows that |h(z)| ≤ 2M(R) < 23+αRα for |z| ≤ R. Hence, there exists a constant k such that

|g(z)| = |g(0) + h(z)| < |g(0)|+ 23+αRα < k|z|α (2.8)

on the circle |z| = R. Note that the constant k does not depend on R as long as |z| is sufficiently large to restrain theterm |g(0)|.

10

By the entirety of g, we know that the function is equal to its Maclaurin series in the whole plane, so we can write

g(z) =

∞∑n=0

g(n)(0)

n!zn. (2.9)

Since |g(z)| < k|z|α on the circle |z| = R, Cauchy’s estimate gives

|g(n)(0)| < n!k|z|α

Rn(2.10)

for all |z| = R [Stewart and Tall, 1983, p. 184]. Letting R → ∞, we can deduce that |g(n)(0)| = 0 for n > α. Itfollows that g is a polynomial of degree at most α, which is the order of f . Since α was defined as the infimum of allnumbers that satisfy (2.1), we find that the degree of g must indeed be α.

We are now interested in obtaining the bound on the number of zeros of a finite-ordered function f in an opendisk of radius R about the origin. Suppose that f is an entire function of order α <∞, and denote its zeros by {zk}in non-decreasing order of |zk|, where multiple zeros are repeated as appropriate.

Suppose that {zk}nk=1 are the zeros of f in the open disk |z| < R, and that there are no zeros on the boundary|z| = R. For convenience, also assume that f(0) 6= 0. Then, Jensen’s formula (see appendix A.3 for proof; cf.[Ahlfors, 1966, p. 206]) states that

1

2π

ˆ 2π

0

ln |f(Reiθ)|dθ − ln |f(0)| =n∑k=1

ln

(R

|zk|

)= ln

Rn

|z1| . . . |zn|, (2.11)

from where we see that the modulus of f(Reiθ) depends on the moduli of the zeros in the disk. This enables us toprove that the zeros of a finite-ordered function cannot be too dense.

Let n(r) denote the number of zeros in the open disk |z| < r. Then, we can write the right-hand side as

lnRn

|z1| . . . |zn|= ln

|z2||z1|

+ 2 ln|z3||z2|

+ · · ·+ n lnR

|zn|=

ˆ R

0

n(r)

rdr, (2.12)

so that Jensen’s formula becomes

1

2π

ˆ 2π

0

ln |f(Reiθ)|dθ − ln |f(0)| =ˆ R

0

n(r)

rdr. (2.13)

Choose a number ε > 0. By the definition of function order, we know that

ln |f(Reiθ)| ≤ lnC +Rα < Rα+ε (2.14)

for all sufficiently large R. From here it follows that

n(R) ln 2 = n(R)

ˆ 2R

R

dr

r≤ˆ 2R

R

n(r)

rdr ≤

ˆ 2R

0

n(r)

rdr < (2R)α+ε − ln |f(0)| < 2(2R)α+ε (2.15)

by Jensen’s formula. But this means that n(R) = O(Rα+ε), which is what we wanted to show. It also follows thatthe sum

∞∑n=1

|zn|−β =

ˆ ∞0

dn(r)

rβ=

[n(r)

rβ

]∞0

+ β

ˆ ∞0

n(r)

rβ+1dr = β

ˆ ∞|z1|

n(r)

rβ+1dr � β

ˆ ∞|z1|

rα+ε−β−1 dr <∞ (2.16)

converges for β > α+ ε. We note that the sum converges for any β > α if we choose ε < β − α.

2.2 The Hadamard product for functions of order 1

We are ready to represent a finite-ordered function f by a product involving its zeroes. Henceforward, to avoidunnecessary details, we shall concern ourselves only with functions of order α = 1.

We previously observed that the sum∑|zn|−β converges for any β > α. By taking β = 2 and recalling the

Maclaurin series for the exponential function,

ez =

∞∑n=0

zn

n!= 1 + z +O(z2), (2.17)

11

we find that the product

P (z) :=

∞∏n=1

(1− z

zn

)ez/zn (2.18)

converges absolutely for all z. Now, if we take the logarithm of the product and recall that

ln(1− z) = −∞∑n=1

zn

n= −z +O(z2) (2.19)

for |z| < 1, then it follows that each term in the obtained sum satisfies

ln

(1− z

zn

)+

z

zn�∣∣∣∣ zzn∣∣∣∣2 � |zn|−2 (2.20)

uniformly in the disk |z| ≤ R, except possibly for the first n(R + δ) terms, where R > 0 and δ > 0 are arbitrary. Inconsequence, the convergence of the product (2.18) is uniform in any bounded region. Thus, P is an entire functionwith the same zeros as f , counting multiplicity, if we assume that f(0) 6= 0. Note that we make this assumption onlyfor the sake of simplicity; otherwise, we would multiply the product by the factor zm, where m is the multiplicity ofthe zero at z = 0. Hence, the function

F (z) :=f(z)

P (z)(2.21)

is entire and without zeros.

We now want to show that F is of order 1. Choose a number ε > 0. Since we know that f is of order 1, it wouldsuffice to prove that the estimate ∣∣∣∣ 1

P (z)

∣∣∣∣ = O(exp(|z|1+ε)) (2.22)

is valid as |z| → ∞ in order to establish that the bound

|F (z)| =∣∣∣∣ f(z)

P (z)

∣∣∣∣ = O(exp(|z|1+ε)) (2.23)

holds as |z| → ∞, and the statement would follow. Unfortunately, this is impossible to prove due to the zeros of P .Instead, our strategy will be to show that ∣∣∣∣ 1

P (z)

∣∣∣∣ = O(exp(R1+ε)) (2.24)

on some origin-centered circle |z| = r of radius r ∈ (R, 2R) for all sufficiently large R > 0. We then obtain the bound

|F (z)| =∣∣∣∣ f(z)

P (z)

∣∣∣∣ = O(exp(R1+ε)), (2.25)

which must also hold on the circle |z| = R by the maximum modulus principle, because F is entire.

Since the product P is absolutely convergent, we are free to change the order of multiplication. Let us writeP (z) = P1(z)P2(z)P3(z), where

P1(z) :=

∞∏k=1+n(2R)

(1− z

zk

)ez/zk , (2.26)

P2(z) :=

n(2R)∏k=1

ez/zk = exp

n(2R)∑k=1

z

zk

(2.27)

and

P3(z) :=

n(2R)∏k=1

(1− z

zk

). (2.28)

We now find upper bounds on the moduli of {ln |Pi(z)|}3i=1. These will yield both upper and lower bounds on

|Pi(z)|, because | ln |Pi(z)|| =∣∣∣∣ln ∣∣∣∣ 1

Pi(z)

∣∣∣∣∣∣∣∣. By our previous estimate (2.20), we know that the kth term of ln |P1(z)| is

O

(∣∣∣∣ zzk∣∣∣∣2)

. Using the same tricks of integration as in (2.16), we get

ln |P1(z)| � R2∞∑

k=1+n(2R)

|zk|−2 = R2

ˆ ∞2R

dn(r)

r2� R2

ˆ ∞2R

rε−2 dr = −R2 (2R)ε−1

ε− 1� R1+ε. (2.29)

12

We also note that the kth term of ln |P2(z)| is O(∣∣∣∣ zzk

∣∣∣∣), from where we conclude that

ln |P2(z)| � R

n(2R)∑k=1

|zk|−1 = R

ˆ 2R

|z1|

dn(r)

r� R

ˆ 2R

|z1|rε−1 dr =

R

ε((2R)ε − |z1|ε)� R1+ε. (2.30)

We now consider the last subproduct P3. This time, due to the zeros of P3, we must look for a bound that doesnot necessarily hold everywhere in the annulus R < |z| < 2R. Since

∑|zk|−2 converges, we know that the union

of intervals⋃

(|zk| − |zk|−2, |zk| + |zk|−2) is of finite length. It follows that for all sufficiently large R, there exists anumber R < r < 2R such that ||zk|− r| > |zk|−2 for all zk. With this in mind, we find that every factor of P3 satisfies∣∣∣∣1− z

zk

∣∣∣∣ =

∣∣∣∣zk − zzk

∣∣∣∣ ≥ ||zk| − |z|||zk|>|zk|−2

|zk|= |zk|−3 > (2R)−3 (2.31)

on the circle |z| = r, from which it follows that∣∣∣∣ 1

P3(z)

∣∣∣∣ =

n(2R)∏k=1

∣∣∣∣1− z

zk

∣∣∣∣−1

<((2R)3

)n(2R) �((2R)3

)R1+ε

= exp(3 ln(2R)R1+ε)� exp(R1+2ε), (2.32)

because the number of factors in the product is O(R1+ε) by Jensen’s formula.

Since the estimate for F holds for a sequence of values of R tending to infinity, we can conclude that F ≡ eg,where g is a polynomial of degree at most 1. We finally deduce that

f(z) = F (z)P (z) = eA+Bz∞∏n=1

(1− z

zn

)ez/zn , (2.33)

where A and B are some constants. This factorization is known as the Hadamard product for functions of order 1.

We finish with an observation. We know that the sum∑|zn|−1−ε converges for any ε > 0 by Jensen’s formula.

However, the sum∑|zn|−1 may or may not converge. Since |1 − z| ≤ 1 + |z| ≤ e|z| for any complex number z, we

find that |(1− z)ez| ≤ e2|z|. Hence, if the latter sum converges, then

|f(z)| =∣∣eA+Bz

∣∣ ∞∏n=1

∣∣∣∣(1− z

zn

)ez/zn

∣∣∣∣ ≤ e|A+Bz|∞∏n=1

e2|z/zn| = e|A+Bz| exp

(2|z|

∞∑n=1

|zn|−1

)< eC|z| (2.34)

for some real number C.

2.3 Proving that ξ has order at most 1

We shall apply our previous work on finite-ordered functions to ξ, which we defined in (1.51) as

ξ(s) =1

2π−s/2s(s− 1)Γ

(s2

)ζ(s). (2.35)

To that end, let us first prove that it has order at most 1. Hence, we wish to show that for any ε > 0, there exists areal number C such that

|ξ(s)| ≤ C exp(|s|1+ε) (2.36)

as |s| → ∞. By the symmetric version (1.52) of the functional equation, we know that ξ is an even function of s− 12 ,

so we only need to consider Re(s) ≥ 12 .

We now check that each factor of ξ satisfies the bound. For the trivial factors, it is evident that

1

2π−s/2s(s− 1) =

1

2e−(s/2) lnπ

(e2 ln s − eln s

)� exp(|s|+ ln |s|)� exp(|s|). (2.37)

For the Γ-factor, we can use Stirling’s formula, which we derived in (1.33) to be

ln Γ(s) =

(s− 1

2

)ln s− s+

1

2ln 2π +O(|s|−1) (2.38)

under the condition that |s| is large and | arg s| < π. Thanks to our assumption that Re(s) ≥ 12 , we do not have to

worry about the poles of Γ. Since the error term O(|s|−1) is small for large |s|, the largest contribution will be givenby the term

(s− 1

2

)ln s. Hence, we have

ln Γ(s

2

)� |s| ln |s|, (2.39)

13

from where it follows thatΓ(s

2

)� exp(|s| ln |s|). (2.40)

It remains to estimate the ζ-factor. We do this by using the representation we obtained in (1.5) by partial summation,

ζ(s) =s

s− 1− sˆ ∞

1

{x}x−s−1 dx, (2.41)

which is valid for Re(s) > 0. In the region of interest, Re(s) ≥ 12 , we see that the integral is bounded. We conclude

thatζ(s)� |s|. (2.42)

Together, these three bounds establish that

|ξ(s)| ≤ C exp(|s| ln |s|) (2.43)

as |s| → ∞ for some constant C. The knowledge that ln |s| � |s|ε for any ε > 0 completes the proof that ξ has orderat most 1.

We now show that ξ is entire, which is a prerequisite for our latest conclusion. From the representation we obtainedin (1.5) by partial summation, we know that the pole at s = 1 is the only singularity of ζ in the region Re(s) > 0.Therefore, let s ∈ C be such that Re(s) ≤ 0. It follows from the cosine version (1.48) of the Riemann functionalequation that there exists a complex number t with Re(t) ≥ 1 such that

ζ(s) = ζ(1− t) = 2(2π)−t cos

(πt

2

)Γ(t)ζ(t). (2.44)

We recall that the exponential and cosine functions are entire. It follows that the factors 2(2π)−t and cos(πt2

)have no poles. We also know that the factor Γ(t) has no poles when Re(t) ≥ 1. Thus, the only possible pole in thisregion is a simple pole at t = 1 from the factor ζ(t), but at this point we have

cos

(πt

2

)= cos

(π2

)= 0, (2.45)

so the pole is cancelled by the zero. We conclude that ζ has no other poles. It follows that ξ is entire, since the factors(s− 1) cancels the simple poles of Γ

(s2

)and ζ at s = 0 and s = 1, respectively.

To apply our previous results, we must also show that ξ(0) 6= 0, because we made that assumption for convenience.From the definition of ξ, we have the limit

ξ(1) =1

2π−1/2Γ

(1

2

)lims→1

(s− 1)ζ(s) =1

2, (2.46)

which we can plug into the function by continuity at s = 1. By the functional equation (1.52), we deduce thatξ(0) = ξ(1) = 1

2 6= 0. It now follows that

ξ(s) = eA+Bs∏ρ

(1− s

ρ

)es/ρ, (2.47)

where the ρ’s are the zeros of ξ listed with multiplicity. We shall use this product formula to obtain a partial-fractiondecomposition of the logarithmic derivative of ζ, which will be the basis for much of the later work.

By the series representation (1.1) of ζ, we can write

ζ(s) = 1 +

∞∑n=2

1

ns, (2.48)

so we see that ζ(s) → 1 as s → +∞ through real values. We further observe that ln Γ(s) ∼ s ln s by Stirling’sformula (1.33). It follows that ξ does not satisfy the inequality |ξ(s)| < eC|s| found in (2.34) for any constant C.We conclude that

∑|ρ|−1−ε converges for any ε > 0, but that

∑|ρ|−1 diverges. Consequently, ξ has infinitely many

zeros; otherwise, the latter sum would not diverge.

We also observe that the trivial zeros of the factor sζ(s) are cancelled by the poles of Γ( 12s), and that the zero

of the factor s − 1 is cancelled by the pole of ζ at s = 1. We have previously remarked that the factor 12π−s/2Γ

(s2

)has no zeros. Hence, the zeros of ξ are precisely the non-trivial zeros of ζ, and we have deduced that ζ has infinitely

14

many zeros in the critical strip. For curiosity, we note that ξ( 12 + it) is purely real.

Taking the logarithmic derivative of the infinite product (2.47), we get

ξ′(s)

ξ(s)= B +

∑ρ

(1

s− ρ+

1

ρ

). (2.49)

By the definition in (1.51), we have

ζ(s) =2

s(s− 1)πs/2Γ−1

(s2

)ξ(s) =

1

(s− 1)πs/2Γ−1

(s2

+ 1)ξ(s), (2.50)

so we also get a product formula for ζ. Note that we have shifted from Γ(s2

)to Γ

(s2 + 1

)in order to absorb the factor

1s , as ζ does not have a pole or zero at s = 0. As we promised some time ago, we take its logarithmic derivative andobtain the partial-fraction decomposition

ζ ′(s)

ζ(s)= B − 1

s− 1+

1

2lnπ − 1

2

Γ′(s2 + 1

)Γ(s2 + 1

) +∑ρ

(1

s− ρ+

1

ρ

). (2.51)

This representation will be useful to prove some interesting properties of ζ. It distinctly exhibits the pole at s = 1and the non-trivial zeros. Taking the logarithmic derivative of the Weierstrass product (1.17) for Γ, we get

− 1

2

Γ′(s2 + 1

)Γ(s2 + 1

) =1

2γ +

∞∑n=1

(1

s+ 2n− 1

2n

), (2.52)

which exhibits the trivial zeros. Here,

γ := limn→∞

(n∑k=1

1

k− lnn

)

= 1− limn→∞

(lnn−

n∑k=2

1

k

)

= 1− limn→∞

(n−1∑k=1

(ln(k + 1)− ln k)−n−1∑k=1

1

k + 1

)

= 1− limn→∞

n−1∑k=1

(ln(k + 1)− ln k − 1

k + 1

)

= 1− limn→∞

n−1∑k=1

(ln(k + 1)− ln k +

k

k + 1− 1

)

= 1− limn→∞

n−1∑k=1

[lnx+

k

x

]k+1

k

= 1− limn→∞

n−1∑k=1

ˆ k+1

k

x− kx2

dx

= 1− limn→∞

ˆ n

1

x− bxcx2

dx

= 1−ˆ ∞

1

x− bxcx2

dx

= 1−ˆ ∞

1

{x}x−2 dx (2.53)

is the Euler–Mascheroni constant.

We now determine the constants A and B for fun. By the product formula (2.47), we have ξ(0) = eA, but wepreviously noted that ξ(0) = 1

2 , so A = ln 12 . For B, we take the logarithmic derivative (2.49) and observe that

ξ′(0)

ξ(0)= B = −ξ

′(1)

ξ(1), (2.54)

where the latter equality follows from the functional equation (1.52). By the relation (2.50), we get

ξ′(s)

ξ(s)=ζ ′(s)

ζ(s)+

1

s− 1− 1

2lnπ +

1

2

Γ′(s2 + 1

)Γ(s2 + 1

) . (2.55)

15

Setting s = 1 in (2.52) and comparing with the series for ln 2, we obtain

− 1

2

Γ′(

32

)Γ(

32

) =1

2γ − 1 + ln 2, (2.56)

whence

B =1

2γ − 1 +

1

2ln(4π)− lim

s→1

(ζ ′(s)

ζ(s)+

1

s− 1

). (2.57)

By the integral representation (1.5) of ζ, we have

ζ(s) =s

s− 1− sI(s), (2.58)

where

I(s) :=

ˆ ∞1

{x}x−s−1 dx. (2.59)

Now, computation gives

lims→1

(ζ ′(s)

ζ(s)+

1

s− 1

)= 1− I(1) = γ (2.60)

by the definition of γ in (2.53). It follows that

B = −1

2γ − 1 +

1

2ln(4π) ≈ −0.023, (2.61)

There is an interesting interpretation of this result. Although the sum∑|ρ|−1 diverges, we note that

∑ρ−1

converges if we sum symmetrically in the sense that we take the terms from ρ and ρ together, because if ρ = α+ iβ,then

1

ρ+

1

ρ=

2α

α2 + β2≤ 2

|ρ|2, (2.62)

whose sum we know converges. It follows from the logarithmic derivative (2.49) and the functional equation (1.52)that

B +∑ρ

(1

1− s− ρ+

1

ρ

)= −B −

∑ρ

(1

s− ρ+

1

ρ

). (2.63)

We know that ρ is a zero if and only if 1− ρ is a zero, so the terms containing 1− s− ρ and s− ρ cancel. Hence,

B = −∑ρ

1

ρ= −2

∑β>0

α

α2 + β2. (2.64)

From this and the numerical value of B, we can deduce that |β| > 6 for all non-trivial zeros. Indeed, the smallest pairof zeros is s ≈ 1

2 ± i14.13 [Havil, 2003, p. 196].

2.4 A zero-free region for ζ

Let us write s = σ + it in the sections to come. We first show that ζ(s) 6= 0 on σ = 1. By the Euler productformula, derived in (1.3) to be

ζ(s) =∏p∈P

1

1− p−s, (2.65)

and the Maclaurin series (2.19) for the natural logarithm, we have

ln ζ(s) = −∑p∈P

ln(1− p−s) =∑p∈P

∞∑n=1

p−sn

n=∑p∈P

∞∑n=1

p−nσ

nexp(−itn ln p), (2.66)

for σ > 1, where we used that |p−s| < 1. It follows that

Re ln ζ(s) =∑p∈P

∞∑n=1

p−nσ

ncos(t ln(pn)). (2.67)

Now, we consider the identity3 + 4 cos θ + cos(2θ) = 2(1 + cos θ)2 ≥ 0. (2.68)

If we insert s = σ into (2.66) and s ∈ {σ + it, σ + 2it} into (2.67), we can apply the above identity to get

3 ln ζ(σ) + 4Re ln ζ(σ + it) + Re ln ζ(σ + 2it) ≥ 0, (2.69)

16

from which it follows by exponentiation that

ζ3(σ)|ζ4(σ + it)ζ(σ + 2it)| ≥ 1. (2.70)

This is valid for σ > 1. Let σ → 1+. Because of the pole at s = 1, we have

ζ(σ) ∼ 1

σ − 1. (2.71)

If s = 1 + it were a zero of ζ for some fixed t 6= 0, then there would exist a real number C such that

|ζ(σ + it)| < C(σ − 1). (2.72)

However, since ζ has no other poles, we know that ζ(σ + 2it) remains bounded in this limit, which is a contradictionto the inequality (2.70), because 4 > 3. We conclude that ζ(s) 6= 0 on σ = 1. By the functional equation (1.52), thesame is true on σ = 0.

We can actually do better than that – we can show that ζ(s) 6= 0 in a certain region to the left of σ = 1. This time,we choose to work with the logarithmic derivative of ζ rather than the logarithm of ζ in order to avoid difficultieswith meromorphic continuation to the left of σ = 1; this is so because the former function has only simple poles ats = 1 for σ ≥ 1 and at the non-trivial zeros of ζ for 1 > σ > 0, while the latter function has logarithmic singularitiesat s = 1 and at the non-trivial zeros of ζ.

We define the von Mangoldt function by

Λ(n) :=

{ln p if n = pm for some p ∈ P and m ∈ Z>0,

0 otherwise.(2.73)

We note that Λ(n) ≥ 0 for all n. We now derive a formula for σ > 1. If we take the logarithm of the Euler productformula (1.3), we get

ln ζ(s) = −∑p∈P

ln(1− p−s

). (2.74)

Now, by taking the derivative and identifying the infinite geometric series, we obtain

− ζ ′(s)

ζ(s)=∑p∈P

(ln p)p−s

1− p−s=∑p∈P

(ln p)

(1

1− p−s− 1

)=∑p∈P

(ln p)

∞∑n=1

p−sn =∑p∈P

∑m∈Z>0

ln p

pms=

∞∑n=1

Λ(n)

ns. (2.75)

This formula will prove very useful to our future work. It is also instrumental to observe that∑d|n

Λ(d) = lnn (2.76)

for σ > 1. To derive this result, we multiply the formula (2.75) by ζ(s) to get

∞∑n=1

1

ns

∑d|n

Λ(d) =

∞∑n=1

1

ns

∑km=n

Λ(m) =

∞∑k=1

∞∑m=1

Λ(m)

ksms=

( ∞∑n=1

1

ns

)( ∞∑n=1

Λ(n)

ns

)= ζ(s)

(−ζ′(s)

ζ(s)

)=

∞∑n=1

lnn

ns.

(2.77)Now, we recollect that Dirichlet coefficients are uniquely determined by the sum function (see appendix A.4 for proof;cf. [Apostol, 1976, pp. 226-227]), and so the desired identity follows from a comparison of the coefficients of the twoDirichlet series in (2.77).

By the formula (2.75), we observe that

− ζ ′(s)

ζ(s)=

∞∑n=1

Λ(n)

ns=

∞∑n=1

Λ(n)

nσexp(−it lnn). (2.78)

Taking the real part, we get

−Reζ ′(s)

ζ(s)=

∞∑n=1

Λ(n)

nscos(t lnn). (2.79)

Once again, we apply the identity (2.68). By the same token, we have

3

[−ζ′(σ)

ζ(σ)

]+ 4

[−Re

ζ ′(σ + it)

ζ(σ + it)

]+

[−Re

ζ ′(σ + 2it)

ζ(σ + 2it)

]≥ 0 (2.80)

17

for σ > 1. Let σ → 1+ at some fixed t 6= 0. For the first term in the above inequality, we know that there exists areal number C such that

− ζ ′(σ)

ζ(σ)<

1

σ − 1+ C, (2.81)

for 1 < σ ≤ 2, because ζ has a simple pole at s = 1 with residue 1. For the other two terms, we expect their behaviourto be influenced by the possible zeros at a height near to t or 2t just to the left of σ = 1. This observation is mademore precise by the partial-fraction decomposition of the logarithmic derivative of ζ, which we derived in (2.51) to be

− ζ ′(s)

ζ(s)=

1

s− 1−B − 1

2lnπ +

1

2

Γ′( 12s+ 1)

Γ( 12s+ 1)

−∑ρ

(1

s− ρ+

1

ρ

), (2.82)

where B is a constant determined by (2.61). Now, let 1 ≤ σ ≤ 2 and |t| ≥ 2 in the partial-fraction formula. Then,using the asymptotic series for the digamma function, which we find in appendix A.2 to be

Γ′(s)

Γ(s)= ln s− 1

2s+O(|s|−2), (2.83)

we conclude that the gamma term is O(ln |t|). It follows that

−Reζ ′(s)

ζ(s)< O(ln |t|)−

∑ρ

Re

(1

s− ρ+

1

ρ

). (2.84)

Write ρ = α+ iβ, where 0 < α < 1 in the critical strip. We note that

Re

(1

s− ρ

)=

σ − α|s− ρ|2

> 0 (2.85)

and

Re

(1

ρ

)=

α

|ρ|2> 0, (2.86)

so the sum over the non-trivial zeros has positive real part. Hence, if we wish, we may omit the sum from theright-hand side of the inequality (2.84). Putting s = σ + 2it in (2.84), we obtain

−Reζ ′(σ + 2it)

ζ(σ + 2it)< O(ln |t|). (2.87)

By (2.64), we know that |β| > 6. Thus, if we wish, we can choose t to coincide with β without leaving the region|t| ≥ 2. We may also pick out just the one term 1

s−ρ in the sum which corresponds to this zero, and leave out therest. Putting s = σ + it, we get

−Reζ ′(σ + it)

ζ(σ + it)< O(ln |t|)− 1

σ − α. (2.88)

Substituting these upper bounds in the inequality (2.80) gives

4

σ − α<

3

σ − 1+O(ln |t|). (2.89)

Write α = 1− δ, where 0 < δ ≤ 14 is near zero, and take σ = 1 + 4δ. This yields

1

20δ< O(ln |t|), (2.90)

which is equivalent to

δ � 1

ln |t|. (2.91)

By the definition of δ, it follows that there exists a positive number C such that

α < 1− C

ln |t|. (2.92)

Hence, we have proven that

σ ≥ 1− C

ln |t|(2.93)

defines a zero-free region for ζ for |t| ≥ 2. Since we know that |β| > 6, it is irrelevant to consider this statementoutside this region. Furthermore, it can be shown (cf. [Kadiri, 2005, p. 2]) that the inequality holds for C = 1

5.69693 .Figure 2.1 illustrates our conclusions.

18

Figure 2.1: The shaded area depicts a zero-free region for ζ(s) in the critical strip. The solid boundaries derive fromthe inequality C

ln |t| < α < 1− Cln |t| with C = 1

5.69693 , while the dotted boundaries derive from the bound |β| > 6. The

blue dots represent actual zeros, all of whom are consistent with the Riemann hypothesis.

2.5 The number of zeros in a rectangle

We already know that there are infinitely many zeros in the critical strip, but we want to find more informationabout their distribution. Let N(T ) denote the number of zeros of ζ in the rectangle {σ + it | 0 < σ < 1, 0 < t < T}.We are interested in an approximate formula for N . We notice that the number N(T ) is exactly half of what wewould call n(T ) in the context of the product formula for ξ. For the sake of simplicity, we choose to work with thefunction ξ instead of ζ due to its convenient functional equation; as we ascertained earlier, both functions have thesame zeros and poles in the open critical strip. To avoid complications, we also choose T so that there are no zeroson the line {σ + iT | 0 < σ < 1}, where it is to be understood that T is large.

We now introduce a helpful relation. Let f be a function meromorphic inside and on a closed contour C withoutzeros or poles on C. Then, the argument principle states that

‰C

f ′(z)

f(z)dz = 2πi(Nf − Pf ), (2.94)

where Nf and Pf denote the sum of the orders of all zeros and poles of f lying inside C, respectively. On the otherhand, we can interpret the contour integral on the left-hand side as

‰C

f ′(z)

f(z)dz =

‰C

(d

dzln f(z)

)dz =

‰C

d(Ln |f(z)|+ i arg f(z)) = i∆C arg f(z), (2.95)

where ∆C arg f(z) stands for the total variation of the argument of f as z travels around C. The last equality istrue because the single-valued function Ln |f(z)| has zero net excursion on any closed curve. Note that we capitalizedLn to point out that we deal with the principal value of the logarithm; we will use this convention when we want toemphasize single-valuedness. It follows that

2π(Nf − Pf ) = ∆C arg f(z). (2.96)

19

Figure 2.2: The boundary of the rectangle with vertices at 2, 2 + iT , −1 + iT and −1.

Let R be the positively oriented boundary of the rectangle with vertices at 2, 2 + iT , −1 + iT and −1, as depictedin Figure 2.2. By the argument principle (2.96), we have

2πN(T ) = ∆R arg ξ(s), (2.97)

because ξ has no poles anywhere and no zeros outside the critical strip. We also know that ξ is real and nowhere zeroon the real line, so there is no contribution to ∆R arg ξ(s) from the base of the rectangle. Furthermore, because ofthe functional equation (1.52) and the symmetric property ξ(s) = ξ(s), we have

ξ(σ + it) = ξ(1− σ − it) = ξ(1− σ + it). (2.98)

It follows that the variation of the argument of ξ for the path at the top-right corner (from 2 through 2+ iT to 12 + iT )

is the same as the path at the top-left corner (from 12 + iT through −1 + iT to −1). We denote the former path by

Q and conclude thatπN(T ) = ∆Q arg ξ(s). (2.99)

Recalling the definition in (1.51), we see that we can write

ξ(s) =1

2π−s/2s(s− 1)Γ

(s2

)ζ(s) = π−s/2(s− 1)Γ

(s2

+ 1)ζ(s), (2.100)

becauses

2Γ(s

2

)= Γ

(s2

+ 1). (2.101)

For curiosity, we observe that

ξ(2) = π−11Γ(2)ζ(2) =1

π

∞∑n=1

1

n2=π

6(2.102)

by Euler’s solution to the Basel problem. Seeing that the argument of a product is the sum of the arguments, we nowdecide to study the change in the argument of each factor of ξ as s moves along Q. For the first two factors, we have

∆Q arg π−s/2 = ∆Q arg exp(−s

2lnπ

)= ∆Q

(− t

2lnπ

)= −T

2lnπ (2.103)

and

∆Q arg(s− 1) = arg

(−1

2+ iT

)− arg 1 = π + arctan(−2T ) =

π

2+O(T−1). (2.104)

20

Using Stirling’s formula (1.33) for the Γ-factor, we get

∆Q arg Γ(s

2+ 1)

= ∆QIm ln Γ(s

2+ 1)

= Im ln Γ

(5

4+ i

T

2

)− Im ln Γ(2)

= Im

((3

4+ i

T

2

)ln

(5

4+ i

T

2

)− 5

4− iT

2+

1

2ln(2π) +O(T−1)

)=

3

4arg

(5

4+ i

T

2

)+T

2ln

∣∣∣∣54 + iT

2

∣∣∣∣− T

2+O(T−1)

=3π

8+T

2lnT

2− T

2+O(T−1). (2.105)

Collecting the variations, we find that

N(T ) =1

π∆Q

(arg π−s/2 + arg(s− 1) + arg Γ

(s2

+ 1)

+ arg ζ(s))

=T

2πln

T

2π− T

2π+

7

8+O(T−1) +

1

π∆Q arg ζ(s). (2.106)

The term 78 has a certain significance, but this is outside the scope of this paper.

Before we proceed with ∆Q arg ζ(s), let us show a simple lemma. We claim that∑ρ

1

1 + (T − β)2= O(lnT ), (2.107)

where ρ = α+ iβ runs through the non-trivial zeros in the critical strip. To prove this lemma, we first recall that

−Reζ ′(s)

ζ(s)< O(ln |t|)−

∑ρ

Re

(1

s− ρ+

1

ρ

)(2.108)

for 1 ≤ σ ≤ 2 and |t| ≥ 2; we derived this in (2.84) from the partial-fraction decomposition of the logarithmic derivativeof ζ. Putting s = 2 + iT in this inequality, we obtain∑

ρ

Re

(1

2 + iT − ρ+

1

ρ

)< O(lnT ), (2.109)

since the logarithmic derivative of ζ is uniformly bounded in this neighborhood, because by (2.75), we have∣∣∣∣−ζ ′(s)ζ(s)

∣∣∣∣ =

∣∣∣∣∣∞∑n=1

Λ(n)

ns

∣∣∣∣∣ <∞∑n=1

√n

|ns|=

∞∑n=1

1

nσ−1/2= ζ

(σ − 1

2

), (2.110)

which is obviously bounded for σ > 32 . Moreover, by the previous section, we know that each summand in the series

(2.109) has positive real part. Since 0 < α < 1, it is evident that

Re

(1

2 + iT − ρ

)=

2− α(2− α)2 + (T − β)2

≥ 1

4 + (T − β)2=

1

4

1

1 + 14 (T − β)2

>1

4

1

1 + (T − β)2, (2.111)

and the assertion in (2.107) of the lemma follows.

We observe two immediate consequences of (2.107). Firstly, it is immediate that |T − β| < 1 holds for at mostO(lnT ) non-trivial zeros. Secondly, we find that the sum over the non-trivial zeros with |T − β| ≥ 1 is also∑

ρ

1

(T − β)2=∑ρ

1 + (T − β)2

(T − β)2

1

1 + (T − β)2≤ 2

∑ρ

1

1 + (T − β)2= O(lnT ). (2.112)

We can deduce even more information from (2.107). Let −1 ≤ σ ≤ 2, and recall that we have chosen T so thatT 6= β for every non-trivial zero ρ. It follows from the partial-fraction decomposition (2.51) that

ζ ′(σ + iT )

ζ(σ + iT )− ζ ′(2 + iT )

ζ(2 + iT )=∑ρ

(1

σ + iT − ρ− 1

2 + iT − ρ

)+O(1). (2.113)

By the estimate (2.110), we haveζ ′(2 + iT )

ζ(2 + iT )= O(1). (2.114)

21

In the sum on the right-hand side of (2.113), we first consider the terms of the form 12+iT−ρ with |T − β| < 1. Since

|2 + iT − ρ| ≥ 1, it follows from the first immediate consequence of (2.107) that the sum of these terms is O(lnT ).As for the terms with |t− β| ≥ 1 of any form, we have∣∣∣∣ 1

σ + iT − ρ− 1

2 + iT − ρ

∣∣∣∣ =2− σ

|(σ + iT − ρ)(2 + iT − ρ)|≤ 3

|T − β|2, (2.115)

so by the second immediate consequence of (2.107), we find that the sum of these is also O(lnT ). It follows that

ζ ′(s)

ζ(s)=∑ρ

1

s− ρ+O(lnT ) (2.116)

for −1 ≤ σ ≤ 2, where the sum is limited to those non-trivial zeros for which |T − β| < 1.

We are now in a position to find an estimate for ∆Q arg ζ(s). Let us denote the vertical path from 2 to 2 + iT byV and the horizontal path from 2 + iT to 1

2 + iT by L. On the former path, we observe that Reζ(2 + it) > 0, because

ζ(2 + it) = 1 +

∞∑n=2

1

n2+it, (2.117)

where ∣∣∣∣∣∞∑n=2

1

n2+it

∣∣∣∣∣ ≤∞∑n=2

1

n2=π2

6− 1 < 1. (2.118)

Therefore, the variation along the vertical line σ = 2 contributes O(1). Recalling the argument principle, we find that

∆Q arg ζ(s) = ∆V arg ζ(s) + Im(i∆L arg ζ(s)) = O(1) +

ˆ 12 +iT

2+iT

Imζ ′(s)

ζ(s)ds. (2.119)

On the other hand, we also have ˆ 12 +iT

2+iT

Im1

s− ρds = ∆L arg(s− ρ). (2.120)

Since the above variation along L has at most absolute value π, and since the number of terms in the sum in (2.116)is O(lnT ) by the first immediate consequence of (2.107), we conclude that ∆Q arg ζ(s) = O(lnT ). We finally get theasymptotic relation

N(T ) =T

2πln

T

2π− T

2π+O(lnT ). (2.121)

Incidentally, if we enumerate the non-trivial zeros with β > 0 in order of increasing ordinate, we get

βn ∼ 2πn

lnn(2.122)

as n→∞; cf. [Titchmarsh, 1986, p. 214].

22

3. The distribution of prime numbers

3.1 Perron’s formula for Dirichlet series

In this section, we shall introduce the identity

δ(y) :=1

2πi

ˆ c+i∞

c−i∞

ys

sds =

0 if 0 < y < 1,12 if y = 1,

1 if y > 1,

(3.1)

where the not absolutely convergent integral is regarded as a principal value in the sense that

ˆ c+i∞

c−i∞

ys

sds := lim

T→∞

ˆ c+iT

c−iT

ys

sds (3.2)

and c > 0 is arbitrary. This identity, which is called Perron’s formula, allows one to pick up the terms with n ≤ xin any Dirichlet series. We note that the integral is independent of c (as long as c > 0), which is essential. Wealso observe that the integral can be interpreted as a unit step function that takes the average value at the point ofdiscontinuity. Figure 3.1 illustrates how we will carry out the integration.

Figure 3.1: To prove Perron’s formula, we first truncate the improper integral. We then consider three distinct cases.For y = 1, we compute the integral directly. For 0 < y < 1, we regard the integral as one side of a rectangle extendingto the right. For y > 1, we regard the integral as one side of a rectangle extending to the left, where we pick up thepole at s = 0 on the way. To prove Perron’s truncated formula, we also replace the vertical line by circular arcs.

Let us prove the formula (3.1). We begin with the special case y = 1, which is an easy exercise in integration:

1

2πi

ˆ c+iT

c−iT

1

sds =

1

2π

ˆ T

−T

dt

c+ it

=1

2π

ˆ T

0

(1

c+ it+

1

c− it

)dt

=1

2π

ˆ T

0

2c

c2 + t2dt

=1

2π

[2 arctan

(t

c

)]T0

=1

πarctan

(T

c

)−−−−→T→∞

1

2. (3.3)

23

We move on to the case 0 < y < 1. We want to shift the line of integration to the right, so the integrand becomessmall. By the residue theorem, we can write

ˆ c+iT

c−iT

ys

sds =

(ˆ d−iT

c−iT+

ˆ d+iT

d−iT−ˆ d+iT

c+iT

)ys

sds, (3.4)

where d > c > 0, because the integrand has no singularities to the right of the vertical line σ = c. We now estimatethe three integrals on the right-hand side. We find that the horizontal integrals satisfy∣∣∣∣∣

ˆ d±iT

c±iT

ys

sds

∣∣∣∣∣ ≤ˆ d±iT

c±iT

∣∣∣∣yss∣∣∣∣ |ds| ≤ ˆ d

c

yσ

Tdσ ≤ 1

T

ˆ ∞c

yσ dσ =1

T

[yσ

ln y

]∞c

=yc

T | ln y|, (3.5)

while the vertical integral satisfies∣∣∣∣∣ˆ d+iT

d−iT

ys

sds

∣∣∣∣∣ ≤ˆ d+iT

d−iT

∣∣∣∣yss∣∣∣∣ |ds| = ˆ T

−T

yd√d2 + t2

dt = yd[arsinh

(t

d

)]T−T

= 2yd arsinh

(T

d

). (3.6)

Letting d→∞, we find that this term vanishes, since yd → 0 for 0 < y < 1. It follows that∣∣∣∣∣ˆ c+iT

c−iT

ys

sds

∣∣∣∣∣ ≤ 2yc

T | ln y|. (3.7)

Now, letting T →∞, we get1

2πi

ˆ c+iT

c−iT

ys

sds −−−−→

T→∞0. (3.8)

We continue with the last case y > 1. This time, we want to shift the line of integration to the left, so the integrandbecomes small. By the residue theorem, we can write

ˆ c+iT

c−iT

ys

sds =

(−ˆ c−iT

−d−iT+

ˆ −d+iT

−d−iT+

ˆ c+iT

−d+iT

)ys

sds+ 2πi, (3.9)

where d > c > 0, because the integrand has a simple pole at s = 0 with residue

Res

(ys

s, 0

)= lims→0

sys

s= y0 = 1. (3.10)

By the same token, we find that the horizontal integrals satisfy∣∣∣∣∣ˆ c±iT

−d±iT

ys

sds

∣∣∣∣∣ ≤ yc

T | ln y|, (3.11)

while the vertical integral satisfies ∣∣∣∣∣ˆ −d+iT

−d−iT

ys

sds

∣∣∣∣∣ ≤ 2y−d arsinh

(T

d

). (3.12)

Since y−d → 0 as d→∞ for y > 1, it follows that

1

2πi

ˆ c+iT

c−iT

ys

sds −−−−→

T→∞1, (3.13)

and the identity follows.

In order to pick up the terms with n ≤ x in a Dirichlet series, we need to interchange an infinite sum with aconditionally convergent integral. Since this is not always permissible, we consider the truncated integral

δ(y, T ) :=1

2πi

ˆ c+iT

c−iT

ys

sds, (3.14)

which legitimizes the manipulation, but at the same time introduces an error term. We claim that

δ(y, T ) =

0 +O(yc min(1, T−1| ln y|−1)) if 0 < y < 1,12 +O(cT−1) if y = 1,

1 +O(yc min(1, T−1| ln y|−1)) if y > 1,

(3.15)

24

where c > 0 and T > 0. This formula is much more useful than (3.1), because it provides an explicit estimate for theerror term.

Let us now prove our assertion. We consider the special case y = 1 first. It follows from our previous computationin (3.3) that

δ(1, T ) =1

2πi

ˆ c+iT

c−iT

1

sds =

1

2π

ˆ T

0

2c

c2 + t2dt =

{u :=

t

c

}=

1

π

ˆ T/c

0

du

1 + u2=

1

2− 1

π

ˆ ∞T/c

du

1 + u2, (3.16)

where ˆ ∞T/c

du

1 + u2<

ˆ ∞T/c

du

u2=

[− 1

u

]∞T/c

=c

T. (3.17)

Hence,

δ(1, T ) =1

2+O

( cT

). (3.18)

We move on to the case 0 < y < 1. By (3.7), we have

|δ(y, T )| =

∣∣∣∣∣ 1

2πi

ˆ c+iT

c−iT

ys

sds

∣∣∣∣∣ ≤ 2yc

T | ln y|. (3.19)

It follows that

δ(y, T ) = O(

yc

T | ln y|

), (3.20)

which proves the first inequality. For the second inequality, we replace the extended rectangle used in (3.7) by acircular arc of radius R =

√c2 + T 2 centered at the origin (see Figure 3.1). On this arc, which we denote by C, we

have |ys| ≤ yc and |s| = R. It follows that

|δ(y, T )| ≤ 1

2πmaxs∈C

∣∣∣∣yss∣∣∣∣ `(C) =

1

2π

yc

RπR =

1

2yc (3.21)

by the estimation lemma (cf. [Saff and Snider, 2001, p. 170]), also known as the M-L inequality. Thus,

δ(y, T ) = O(yc). (3.22)

The last case y > 1 can be proven by similar arguments, the difference being that we use a rectangle or circular arcto the left and that we pick up the pole at s = 0. This concludes the proof.

3.2 An approximation of the Chebyshev function ψ

We finally have all the ingredients to derive an explicit formula for a certain arithmetic function that describes thedistribution of prime numbers. Let the Chebyshev function be defined by

ψ(x) :=∑n≤x

Λ(n) =∑pm≤x

ln p, (3.23)

which can also be interpreted as the logarithm of the least common multiple of all positive integers up to bxc. Tomitigate the jump discontinuities at the prime powers, we modify the function as follows:

ψ0(x) :=1

2

∑n≤x

Λ(n) +∑n<x

Λ(n)

=

{ψ(x)− 1

2Λ(x) if x = pm for some prime p and positive integer m,

ψ(x) otherwise.(3.24)

To extract information about the growth of this function, we consider the corresponding Dirichlet series

∞∑n=1

Λ(n)

ns= −ζ

′(s)

ζ(s), (3.25)

which converges absolutely for σ > 1 according to (2.75). We now use Perron’s truncated formula (3.15) to pick up

25

the terms with n ≤ x:

ψ0(x) =1

2

∑n≤x

Λ(n) +∑n<x

Λ(n)

=

∞∑n=1

Λ(n)δ(xn, T)

+R(x, T )

=

∞∑n=1

Λ(n)

(1

2πi

ˆ c+iT

c−iT

(x/n)s

sds

)+R(x, T )

=1

2πi

ˆ c+iT

c−iT

( ∞∑n=1

Λ(n)

ns

)xs

sds+R(x, T )

=1

2πi

ˆ c+iT

c−iT

[−ζ′(s)

ζ(s)

]xs

sds+R(x, T ), (3.26)

where R is the error term. Note that we allow x to be a prime power in this deduction. By the absolute convergenceof the Dirichlet series, we except (3.26) to hold for any c > 1.

Let us find an explicit expression for the error term R. By Perron’s truncated formula (3.15), we know that

R(x, T )�∞∑

n=1, n 6=x

Λ(n)(xn

)cmin

(1,

1

T | ln xn |

)+ Λ(x)

c

T. (3.27)

We observe that every variable and function that appears in the sum is non-negative. To avoid technicalities, we onlyconsider x > 1. We choose c = 1 + 1

ln x > 1 to ensure that xc = ex � x. We also recall that Λ(n) ≤ lnn. It followsthat

R(x, T )�∞∑

n=1, n6=x

Λ(n)x

ncmin

(1,

1

T | ln xn |

)+

lnx

T. (3.28)

We now subdivide all terms into three different groups. We first consider the terms with n ≤ 34x or n ≥ 4

3x. For theseterms, we have x

n ≥43 or x

n ≤34 , whence | ln x

n | ≥ | ln43 | > 0 has a positive lower bound. We also recollect that

ζ ′(σ)

ζ(σ)∼ −1

σ − 1(3.29)

near the simple pole at s = 1. It follows from (3.28) that the contribution of these terms is

� x

T

∞∑n=1

Λ(n)

nc=x

T

[−ζ′(c)

ζ(c)

]=x

T

[−ζ ′(1 + 1

ln x

)ζ(1 + 1

ln x

) ]� x

Tlnx. (3.30)

We now consider the terms with 34x < n < x. We note that this set vanishes if it does not contain a prime power.

Therefore, let us assume that there exists a greatest prime power q in this set. Since∣∣x−qx

∣∣ < ∣∣∣x/4x ∣∣∣ < 14 , we see from

the Maclaurin series (2.19) for the natural logarithm that

lnx

q= − ln

q

x= − ln

(1− x− q

x

)≥ x− q

x. (3.31)

Hence, by (3.28), this term contributes

� Λ(q)x

qcmin

(1,

x

T (x− q)

)� (lnx) min

(1,

x

T (x− q)

). (3.32)

For the other prime power terms in this set, we can write n = q−m for some integer 0 < m < 14x. By the same token,

lnx

n≥ ln

q

q −m= − ln

q −mq

= − ln

(1− m

q

)≥ m

q. (3.33)

It follows from (3.28) that their contribution to the sum is

�d 14x−1e∑m=1

Λ(q −m)x

(q −m)cmin

(1,

q

Tm

)�d 14x−1e∑m=1

(lnx)q

Tm� x ln2 x

T, (3.34)

26

sincen∑

m=1

1

m=

Γ′(n)

Γ(n)+ γ � lnn. (3.35)

The terms with x < n < 43x are dealt with similarly, the difference being that we consider the least prime power

in this set. For convenience, let us denote the distance from x to the nearest prime power other than itself by 〈x〉.Summing up the estimates, we conclude that the total error is

R(x, T )� x ln2 x

T+ (lnx) min

(1,

x

T 〈x〉

). (3.36)

It follows that

ψ0(x) =1

2πi

ˆ c+iT

c−iT

[−ζ′(s)

ζ(s)

]xs

sds+O

(x ln2 x

T+ (lnx) min

(1,

x

T 〈x〉

)), (3.37)

which will be our next starting point.

3.3 von Mangoldt’s explicit formula

Let us take a closer look at the formula

ψ0(x) =1

2πi

ˆ c+iT

c−iT

[−ζ′(s)

ζ(s)

]xs

sds+R(x, T ), (3.38)

where R is the error term (3.36) and x > 1. We want to compute the finite integral. To this end, we regard theintegral as one side of a rectangle extending to the left. The reason we want to move the vertical line of integrationaway to infinity on the left is because |xs| vanishes there.

As we move past the region of absolute convergence, we shall use the meromorphic continuation of ζ to estimate itslogarithmic derivative on the resulting contour and predict where its zeros and poles may be encountered. Indeed, themain obstacle in carrying out this integration is the difficulty to choose T so that the horizontal sides of the rectangleavoid the zeros in the critical strip.

Provided we are successful in the integration, we will we able to use the residue theorem to express the integralas the sum of the residues of the integrand at its poles. We recall that each singularity of the logarithmic derivativeof a meromorphic function is a simple pole, where the residue is simply the multiplicity of the corresponding zero, orminus the order of the corresponding pole, of the original function. We also note that the exponential function hasno poles.

We first observe that the simple pole of 1s at s = 0 contributes

Res

([−ζ′(s)

ζ(s)

]xs

s, 0

)= lims→0

s

[−ζ′(s)

ζ(s)

]xs

s= −ζ

′(0)

ζ(0)= − ln 2π, (3.39)

where the value of the constant follows from the partial-fraction decomposition (2.51) of the logarithmic derivative ofζ. We also see that each zero of ζ at s = ρ, whether trivial or not, contributes

Res

([−ζ′(s)

ζ(s)

]xs

s, ρ

)= lims→ρ

(s− ρ)

[−ζ′(s)

ζ(s)

]xs

s= −x

ρ

ρ. (3.40)

Considering the Taylor series

1

2ln(1− x−2) =

∞∑n=1

x−2n

−2n, (3.41)

we notice that the contribution from the trivial zeros may be written in a more compact way. We finally note thatthe simple pole of ζ at s = 1 contributes

Res

([−ζ′(s)

ζ(s)

]xs

s, 1

)= lims→1

(s− 1)

[−ζ′(s)

ζ(s)

]xs

s=x1

1= x, (3.42)

where we changed sign because it is a pole.

Summing everything up and letting T →∞, we arrive at the formula

ψ0(x) = x−∑ρ

xρ

ρ− 1

2ln(1− x−2)− ln 2π, (3.43)

27

which is called von Mangoldt’s explicit formula and illustrates how the zeros of the Riemann zeta function controlthe distribution of the prime numbers. Here, ρ runs over the non-trivial zeros and the not absolutely convergent seriesis to be understood as a symmetric sum in the sense that∑

ρ

xρ

ρ:= lim

T→∞

∑|Im(ρ)|<T

xρ

ρ. (3.44)

We note that the non-trivial zeros are not necessarily simple, so they may appear multiple times in the sum.

3.4 Proving the explicit formula

It is time to carry out the integration of

ψ0(x) =1

2πi

ˆ c+iT

c−iT

[−ζ′(s)

ζ(s)

]xs

sds+R(x, T ). (3.45)

Let us regard the vertical line of integration as one side of a rectangle extending to the left, as illustrated in Figure3.2. In this way, it follows from the residue theorem that

ψ0(x) =1

2πi

(−ˆ c−iT

−U−iT+

ˆ −U+iT

−U−iT+

ˆ c+iT

−U+iT

)[−ζ′(s)

ζ(s)

]xs

sds+x−

∑|β|<T

xρ

ρ−dU2 −1e∑n=1

x−2n

−2n− ln 2π+R(x, T ), (3.46)

where U > 0 and we write ρ = α+β. The contributions to the sum of the residues of the integrand at its poles insidethe rectangle follow from our observations in the previous section.

Figure 3.2: We regard the integral as one side of the rectangle with vertices at c−iT , c+iT , −U+iT and −U−iT . Wemake various estimates of the integrand on different parts of the contour. To prove von Mangoldt’s explicit formula,we show that all integrals vanish as U →∞ and T →∞. The dots show the different poles of the integrand.

We need to choose the values of U and T wisely. We see that if we take U to be a large odd integer, then the leftvertical side of the rectangle passes halfway between two trivial zeros, which turns out to be good choice. The choiceof T demands more care. By the first immediate consequence of (2.107), we know that the number of non-trivial zerosin the horizontal strip T − 1 < β < T + 1 is � lnT for any large T . It follows that there exists a zero-free substrip ofheight � 1

lnT . Thus, by varying T by at most 1, we can guarantee that

|T − β| � 1

lnT. (3.47)

for every non-trivial zero.

We recall that we deduced in (2.116) that

ζ ′(s)

ζ(s)=

∑|T−β|<1

1

s− ρ+O(lnT ) (3.48)

for s ∈ {σ+ iT | − 1 ≤ σ ≤ 2}. By the bound (3.47), we know that each term in the sum is O(lnT ). But the numberof terms in the sum is also O(lnT ), whence

ζ ′(s)

ζ(s)= O(ln2 T ) (3.49)

28

for −1 ≤ σ ≤ 2. We also recollect from (2.110) that∣∣∣∣−ζ ′(s)ζ(s)

∣∣∣∣ < ζ

(σ − 1

2

)≤ ζ

(3

2

), (3.50)

for σ ≥ 2, which is negligible in comparison to O(ln2 T ). Hence, the contribution from the horizontal integrals in thisrange is

ˆ c±iT

−1±iT

[−ζ′(s)

ζ(s)

]xs

sds� ln2 T

ˆ c±iT

−1±iT

∣∣∣∣xss∣∣∣∣ ds� ln2 T

ˆ c

−1

xσ

Tdσ � ln2 T

T

ˆ c

−∞xσ dσ =

ln2 T

T

[xσ

lnx

]c−∞� x ln2 T

T lnx,

(3.51)since c = 1 + 1

ln x .

It remains to estimate∣∣∣ ζ′(s)ζ(s)

∣∣∣ for −U ≤ σ ≤ −1. Fortunately, this is easy to do. Since we only consider σ ≤ −1,

we can write s = 1−w for some complex number w with Re(w) =: τ = 1− σ ≥ 2. We recall the unsymmetric version(1.48) of the functional equation, which is given by

ζ(s) = ζ(1− w) = 2(2π)−w cos(πw

2

)Γ(w)ζ(w). (3.52)

Taking the logarithmic derivative, we find that

ζ ′(s)

ζ(s)= −ζ

′(1− w)

ζ(1− w)= − ln 2π − π

2tan

(πw2

)+

Γ′(w)

Γ(w)+ζ ′(w)

ζ(w), (3.53)

where we used the product rule for differentiation. The first term is a constant that can be dropped. Since thefunction tan z has singularities at z = π

2 (1 + 2n), where n is any integer, we see that the second term is boundedif, say, |w − (1 + 2n)| ≥ 1

2 , which is equivalent to |(1 − w) − (−2n)| ≥ 12 . Hence, the second term is bounded when

s = 1 − w lies outside every circle of radius 12 that is centered at a trivial zero, which is the case with our choice of

U . We know that |w| = |1− s| ≤ 2|s| for σ ≤ −1, so the third term is O(ln |w|) = O(ln 2|s|). Finally, the last term isbounded by (3.50). It follows that

ζ ′(s)

ζ(s)= O(ln 2|s|), (3.54)

whence we see that the contribution from the horizontal integrals in this range is

ˆ −1±iT

−U±iT

[−ζ′(s)

ζ(s)

]xs

sds� ln 2T

ˆ −1±iT

−U±iT

∣∣∣∣xss∣∣∣∣ ds� lnT

ˆ −1

−U

xσ

Tdσ � lnT

T

ˆ −1

−∞xσ dσ =

lnT

T

[xσ

lnx

]−1

−∞=

lnT

Tx lnx,

(3.55)which is negligible in comparison to the contribution from (3.51). We also see that the contribution from the verticalintegral is ˆ −U+iT

−U−iT

[−ζ′(s)

ζ(s)

]xs

sds� ln 2U

ˆ −U+iT

−U−iT

∣∣∣∣xss∣∣∣∣ ds� lnU

ˆ T

−T

x−U

Udt� T lnU

UxU, (3.56)

which vanishes as U →∞.

Letting U →∞ and collecting the dominating error terms from (3.36) and (3.51), we get

ψ0(x) = x−∑|β|<T

xρ

ρ− 1

2ln(1− x−2)− ln 2π +R(x, T ), (3.57)

where

R(x, T )� x ln2(xT )

T+ (lnx) min

(1,

x

T 〈x〉

). (3.58)

Making T →∞ at constant x > 1, we see that R(x, T )→ 0, which proves the explicit formula (3.43). Note that theconvergence is only uniform on closed intervals that do not contain prime powers. This is because ψ0 has discontinu-ities at the prime powers.

When we derived the bound (3.47), we placed a restriction on T . We are now able to remove this restriction. Byvarying T by a bounded amount, we add or remove O(lnT ) terms to the sum

∑xρ

ρ . Since each term in the sum is

O( xT ), the total variation is O(x lnTT ), but this error is already covered by R.

29

3.5 The prime number theorem

Let us begin with some definitions. We define the prime-counting function as

π(x) := #{p ∈ P | p ≤ x}; (3.59)

that is, π(x) is the number of prime numbers less than or equal to x. We also define the offset logarithmic integralby

Li(x) :=

ˆ x

2

dt

ln t. (3.60)

In the year 1896, the prime number theorem, which asserts that

π(x) ∼ Li(x), (3.61)

was independently proven by Jacques Hadamard [Hadamard, 1896, pp. 199-220] and Charles-Jean de La ValleePoussin [de la Vallee Poussin, 1896, pp. 183-256]. Another weaker version of the theorem states that

π(x) ∼ x

lnx, (3.62)

but it can be shown that

Li(x) ∼ x

lnx

∞∑n=0

n!

(lnx)n(3.63)

is a better approximation. This is also the version we will prove. We shall conclude this paper by relating this theoremto the Riemann hypothesis, which significantly improves the error term in (3.61).

We first show thatψ(x) ∼ x, (3.64)

which is not unlike the prime number theorem in the sense that an arithmetic function is approximated by a simpleexpression. We recall from (3.57) that

ψ0(x) = x−∑|β|<T

xρ

ρ− 1

2ln(1− x−2)− ln 2π +R(x, T ), (3.65)

where

R(x, T )� x ln2(xT )

T+ (lnx) min

(1,

x

T 〈x〉

). (3.66)

We note that this formula is suggestive of the statement we wish to prove, still we need to find an estimate for thesum of the non-trivial zeros. To this end, we use the fact that the real part of a non-trivial zero cannot be arbitrarilyclose to 1. In particular, we proved in (2.92) that there exists a positive number C such that

α < 1− C

lnT. (3.67)

for every non-trivial zero with |β| < T . It follows that

|xρ| = xα < x1− ClnT = x exp

(−C lnx

lnT

). (3.68)

We now consider the sum∑

1|ρ| . By (2.64), we know that |β| > 6. Hence,

∑|β|<T

1

|ρ|=

∑6<|β|<T

1

|ρ|<

∑6<|β|<T

1

|β|= 2

∑6<β<T

1

β. (3.69)

Recall the zero-counting function N(t) defined as the number of zeros with 0 < β < t in the critical strip. Byintegration by parts, we have

2∑

6<β<T

1

β= 2

ˆ T

6

dN(t)

t=

2N(T )

T+ 2

ˆ T

6

N(t) dt

t2� lnT +

ˆ T

6

ln tdt

t= lnT +

[ln2 t

2

]T6

� ln2 T, (3.70)

since N(6) = 0 and N(t) = O(t ln t) by (2.121). It follows that∑|β|<T

xρ

ρ� x(ln2 T ) exp

(−C lnx

lnT

). (3.71)

30

Without loss of generality, we may assume that x is an integer that is not a prime power. In this case, we have〈x〉 ≥ 1, so that

R(x, T )� x ln2(xT )

T(3.72)

by the estimate (3.58). It follows that

ψ(x)− x� x ln2(xT )

T+ x(ln2 T ) exp

(−C lnx

lnT

). (3.73)

We choose T = e√

ln x, so that ln2 T = lnx. Insertion in (3.73) gives

ψ(x)− x� x(ln2 x) exp(−√

lnx)

+ x(lnx) exp(−C√

lnx). (3.74)

Let C1 = min(1, C)− 2ε for some ε > 0. Since lnx� eε√

ln x, we find that

ψ(x) = x+O(x exp

(−C1

√lnx))

, (3.75)

which is what we wanted to show.

We now prove the equivalent result for π. We note that integration by parts yields

Li(x) =

ˆ x

2

dt

ln t=

[t

ln t

]x2

−ˆ x

2

td

(1

ln t

)=

x

lnx− 2

ln 2+

ˆ x

2

dt

ln2 t=

x

lnx+

ˆ x

2

tdt

t ln2 t+O(1). (3.76)

On the other hand, if we let

π1(x) :=∑n≤x

Λ(n)

lnn, (3.77)

we find that

π1(x) =∑n≤x

Λ(n)

(1

lnx+

[− 1

ln t

]xn

)=

1

lnx

∑n≤x

Λ(n) +∑n≤x

Λ(n)

ˆ x

n

dt

t ln2 t=ψ(x)

lnx+

ˆ x

2

ψ(t) dt

t ln2 t. (3.78)

by the definition of ψ. The last step can be justified by

∑n≤x

Λ(n)

ˆ x

n

dt

t ln2 t=

∑2≤n<x

Λ(n)

ˆ x

n

dt

t ln2 t=

∑2≤m<x

ˆ m+1

m

∑n≤m

Λ(n)dt

t ln2 t

=

ˆ x

2

ψ(t) dt

t ln2 t. (3.79)

We further note that

π1(x) =∑n≤x

Λ(n)

lnn=∑pm≤x

ln p

m ln p=∑pm≤x

1

m. (3.80)

Since we sum over all prime powers less than or equal to x, we find that

π1(x) =

∞∑m=1

1

mπ( m√x) =

blb xc∑m=1

1

mπ( m√x) = π(x) +O

blb xc∑m=2

m√x

= π(x) +O(√x). (3.81)

Here, we used that π( m√x) = 0 for m

√x < 2 and π( m

√x) ≤ m

√x. Of course, it is to be understood that 1

√x = x and

lbx = log2 x. Thus, we have

π(x) =ψ(x)

lnx+

ˆ x

2

ψ(t) dt

t ln2 t+O(

√x). (3.82)

We now replace ψ in (3.82) by the approximation (3.75), obtaining

π(x)− Li(x)� x exp(−C1

√lnx)

+

(ˆ 4√x

2

+

ˆ x

4√x

)exp

(−C1

√ln t)

dt

� x exp(−C1

√lnx)

+

ˆ 4√x

2

1 dt+

ˆ x

4√x

exp

(−1

2C1

√lnx

)dt

� x exp(−C1

√lnx)

+ 4√x+ x exp

(−1

2C1

√lnx

)(3.83)

In the second step, we used the fact that√

ln t ≥ 12

√lnx for 4

√x ≤ t ≤ x. Note that the constant C1 is the same as

in (3.75). It follows that

π(x) = Li(x) +O(x exp

(−1

2C1

√lnx

)), (3.84)

which implies the prime number theorem. We also conclude that good estimates for ψ imply good estimates for π,and vice versa. Indeed, we were ultimately interested in the behaviour of π from the beginning, but it turns out thatdue to the Dirichlet series (2.75) for Λ, it is more convenient to study the behaviour of ψ.

31

3.6 The smallest possible error term

It is time to involve the Riemann hypothesis to the matter. Suppose that there exists a number 12 ≤ θ < 1 such

that α ≤ θ for all non-trivial zeros. Then, it follows that |xρ| ≤ xθ. Since we have already seen in (3.70) that the sum∑1|ρ| over zeros with |β| < T is O(ln2 T ), we find that

ψ(x)− x� x ln2(xT )

T+ xθ ln2 T (3.85)

by the explicit formula (3.57). Choosing T =√x, we get

ψ(x)− x�√x ln2 x+ xθ ln2 x� xθ ln2 x, (3.86)

whenceψ(x) = x+O(xθ ln2 x). (3.87)

By the relation (3.82) and the estimate (3.87), it follows that

π(x) =ψ(x)

lnx+

ˆ x

2

ψ(t) dt

t ln2 t+O(

√x)

=x

lnx+O(xθ lnx) +

ˆ x

2

dt

ln2 t+

ˆ x

2

O(tθ−1) dt+O(√x)

=x

lnx+O(xθ lnx) +

[Li(t)− t

ln t

]x2

+O(xθ) +O(√x)

= Li(x) +O(xθ lnx), (3.88)

since θ ≥ 12 . Note that the error term is lowered by a factor of lnx. In particular, the assumption of the Riemann

hypothesis implies thatπ(x) = Li(x) +O(

√x lnx), (3.89)

which is a significant improvement over (3.84).

We now prove the converse. Suppose that there exists a number 12 ≤ θ < 1 such that

ψ(x) = x+O(xθ). (3.90)

Recalling the formula (2.75) for the logarithmic derivative of ζ, we find that

− ζ ′(s)

ζ(s)=

∞∑n=1

Λ(n)

ns=

ˆ ∞1

x−s dψ(x) (3.91)

for σ > 1. Integrating by parts, we get

− ζ ′(s)

ζ(s)=[x−sψ(x)

]∞1−ˆ ∞

1

ψ(x) d(x−s) = s

ˆ ∞1

ψ(x)x−s−1 dx, (3.92)

where we used that ψ(1) = 0 and ψ(x) = O(x). Rearranging the integrand, we deduce that

− ζ ′(s)

ζ(s)= s

ˆ ∞1

(ψ(x)− x)x−s−1 dx+ s

ˆ ∞1

x−s dx = s

ˆ ∞1

(ψ(x)− x)x−s−1 dx+s

s− 1(3.93)

for σ > 1. But we assumed that ψ(x) − x � xθ, so the resulting integral extends to an holomorphic function forσ > θ. Thus, ζ cannot have zeros there, and hence we have α ≤ θ for all zeros. We also note that the same conclusionholds when we assume that π(x) = Li(x) +O(xθ). It follows that our best estimate (3.89) is valid if and only if theRiemann hypothesis holds true. Hence, the Riemann hypothesis is equivalent to the smallest possible error term inthe prime number theorem (3.61).

32

Appendices

33

A. Additional proofs

A.1 The Weierstrass form of Γ

Let us consider the product

g(s) :=1

s

∞∏n=1

(1 +

s

n

)−1

es/n. (A.1)

We first observe that

ln g(s) = − ln s+

∞∑n=1

(− ln

(1 +

s

n

)+s

n

)� − ln s+

∞∑n=1

( sn

)2

, (A.2)

so the product in (A.1) converges absolutely and uniformly in any compact subset of C \ {0,−1,−2, . . .}. We furthernote that g has simple poles at the non-positive integers and no zeros, just like Γ. It follows that Γ

g is an entire

function with no zeros. Thus, we can write Γg ≡ ef for some entire function f . We wish to show that f(s) = −γs,

where γ is the Euler–Mascheroni constant, for then

Γ(s) =e−γs

s

∞∏n=1

(1 +

s

n

)−1

es/n, (A.3)

which is the Weierstrass form of Γ.

We now observe that

g(s+ 1)

g(s)= limN→∞

s

s+ 1

N∏n=1

1 + sn

1 + s+1n

e1/n

= limN→∞

s

s+ 1

(N∏n=1

n+ s

n+ s+ 1

)exp

(N∑n=1

1

n

)

= limN→∞

s · N

N + s+ 1· exp

(− lnN +

N∑n=1

1

n

)= s · 1 · eγ (A.4)

for any s ∈ C\{0,−1,−2, . . .}. In the last step, we used the definition of γ in (2.53). It follows that g(s+1) = seγg(s).This looks similar to Γ(s+ 1) = sΓ(s), which we recall from (1.10).

We now introduce the entire function

q(s) :=Γ(s)

e−γsg(s)(A.5)

and note that

q(s+ 1) =Γ(s+ 1)

e−γ(s+1)g(s+ 1)=

sΓ(s)

e−γ(s+1)seγg(s)=

Γ(s)

e−γsg(s)= q(s), (A.6)

so q is periodic with period 1. We recollect that we want to show that f(s) = −γs, which is equivalent to q ≡ 1 bythe definition of f . We also observe that

lims→0

q(0) = lims→0

Γ(s)

e−γsg(s)=

lims→0

sΓ(s)

lims→0

se−γsg(s)=

1

1= 1 (A.7)

by the definition of g in (A.1) and the residues of Γ in (1.15).

34

Let s = σ + it be such that σ ≥ 1. We observe that∣∣∣∣g(σ + it)

g(σ)

∣∣∣∣ =|σ||σ + it|

∞∏n=1

∣∣1 + σn

∣∣∣∣1 + σ+itn

∣∣=

∞∏n=0

|n+ σ||n+ σ + it|

= exp

( ∞∑n=0

ln

√(n+ σ)2√

(n+ σ)2 + t2

)

= exp

(−1

2

∞∑n=0

ln

(1 +

t2

(n+ σ)2

)). (A.8)

We notice that the summand decreases with increasing n+ σ. It follows that

∞∑n=0

ln

(1 +

t2

(n+ σ)2

)≤ˆ ∞

0

ln

(1 +

t2

x2

)dx = |t|

ˆ ∞0

ln

(1 +

1

x2

)dx = 2|t|

ˆ ∞0

dx

1 + x2= π|t|. (A.9)

by integration by parts. We conclude that |g(σ + it)| ≥ |g(σ)|e−π|t|/2. We also note that

|Γ(σ + it)| =∣∣∣∣ˆ ∞

0

xσ+it−1e−x dx

∣∣∣∣ ≤ ˆ ∞0

∣∣xσ+it−1e−x∣∣ dx =

ˆ ∞0

xσ−1e−x dx = Γ(σ) (A.10)

by the integral definition of Γ in (1.8), which is valid for σ > 0. We find that

|q(σ + it)| =∣∣∣∣ Γ(σ + it)

e−γ(σ+it)g(σ + it)

∣∣∣∣ ≤ ∣∣∣∣ Γ(σ)

e−γσg(σ)e−π|t|/2

∣∣∣∣ = |q(σ)|eπ|t|/2 (A.11)

for σ ≥ 1.

Since q is periodic with period 1, we need only consider, say, σ ∈ [1, 2]. Hence, the inequality (A.11) is valideverywhere. Furthermore, setting C := sup

σ∈[1,2]

|q(σ)|, we get

|q(σ + it)| ≤ |q(σ)|eπ|t|/2 ≤ Ceπ|t|/2 < Ce2π|t|. (A.12)

Now, we consider the function

p(s) :=q(s− 1

4 )− q(0)

sin(2πs)− 1; (A.13)

it is entire, since the zeros of the denominator at s = 14 + n, where n ∈ Z, are cancelled by the zeros of the numerator

by the periodicity of q. We notice that p(s) = p(s+1). Moreover, for any real number B, we observe that p(σ+it)→ 0as |t| → ∞ at |σ| ≤ B, because |q(σ + it)| ≤ Ceπ|t|/2 and | sin(2πs)| ≥ 1

2

∣∣e2πt − e−2πt∣∣. Thus, p is a bounded entire

function. Hence, it follows from Liouville’s theorem that p is constant; that is, p ≡ A for some A ∈ C. But it must bethat A = 0, because otherwise q would have to be Θ(e2π|t|), which would contradict the inequality (A.11). Therefore,p ≡ 0, and so q ≡ 1, as desired.

A.2 Stirling’s formula for ln Γ

We shall make use of the previously deduced Weierstrass form in (A.3) to derive an approximation of ln Γ. By thedefinition of the Euler–Mascheroni constant γ in (2.53), we have

Γ(s) =e−γs

s

∞∏n=1

(1 +

s

n

)−1

es/n

= limN→∞

1

sexp

(−s

(N∑n=1

1

n− lnN

))exp

(s

N∑n=1

1

n

)N∏n=1

n

n+ s

= limN→∞

1

sNs

N∏n=1

n

n+ s. (A.14)

We consider the region |Arg s| < π, where the capitalization of Arg indicates that we use the principal branch of argwith values in (−π, π]. Since Γ is non-zero and holomorphic in this simply connected domain, which contains none

35

of the poles of Γ, we can define an holomorphic branch of ln Γ there by´ s

1Γ′(z)Γ(z) dz [Bak and Newman, 1996, p. 100],

which is real on the positive real axis. Taking this logarithm, we get

ln Γ(s) = limN→∞

(− ln s+ s lnN +

N∑n=1

lnn−N∑n=1

ln(n+ s)

)= limN→∞

(s lnN + lnN !−

N∑n=0

ln(n+ s)

). (A.15)

We recall the well-known Stirling’s formula for the factorial, which can be formulated as follows:

N ! =√

2πe−NNN+1/2(1 +O(N−1)

), (A.16)

where N is a large positive integer. Taking the logarithm, we get

lnN ! =

(N +

1

2

)lnN −N +

1

2ln 2π +O(N−1), (A.17)

which we can use to estimate the second term in (A.15). Insertion gives

ln Γ(s) = limN→∞

((s+N +

1

2

)lnN −N +

1

2ln 2π −

N∑n=0

ln(n+ s) +O(N−1)

). (A.18)

We now estimate the third term in (A.15). To this end, we use a neat identity. Consider any real function f ofdifferentiability class C2. Integration by parts twice yields

ˆ n+ 12

n− 12

f(x) dx =

(ˆ n

n− 12

+

ˆ n+ 12

n

)1 · f(x) dx

=

[(x− n+

1

2

)f(x)

]nn− 1

2

−ˆ n

n− 12

(x− n+

1

2

)f ′(x) dx

+

[(x− n− 1

2

)f(x)

]n+ 12

n

−ˆ n+ 1

2

n

(x− n− 1

2

)f ′(x) dx

= f(n)−ˆ n

n− 12

(x− n+

1

2

)f ′(x) dx−

ˆ n+ 12

n

(x− n− 1

2

)f ′(x) dx

= f(n) +1

2

(ˆ n

n− 12

(x− n+

1

2

)2

f ′′(x) dx+

ˆ n+ 12

n

(x− n− 1

2

)2

f ′′(x) dx

)

= f(n) +1

2

ˆ n+ 12

n− 12

⟨x+

1

2

⟩2

f ′′(x) dx (A.19)

where n is any integer and 〈x〉 is the distance from x to the nearest integer. Taking f(x) = ln(s+ x) and 0 ≤ n ≤ N ,it follows that

N∑n=0

ln(s+ n) =

ˆ N+ 12

− 12

ln(s+ x) dx− 1

2

ˆ N+ 12

− 12

〈x+ 12 〉

2

(s+ x)2dx, (A.20)

provided that |s| ≥ 1 (which we may assume in the limit |s| → ∞), so that s+x 6= 0 on the interval − 12 ≤ x ≤ N + 1

2 .Computing the first integral, we find that

ˆ N+ 12

− 12

ln(s+ x) dx = [(s+ x) ln(s+ x)− x]N+ 1

2

− 12

=

(s+N +

1

2

)ln

(s+N +

1

2

)−(s− 1

2

)ln

(s− 1

2

)−N − 1

=

(s+N +

1

2

)(lnN + ln

(1 +

1

N

(s+

1

2

)))−(s− 1

2

)ln

(s− 1

2

)−N − 1. (A.21)

We also observe that1

2

ˆ N+ 12

− 12

〈x+ 12 〉

2

(s+ x)2dx� |s|−1. (A.22)

36

Insertion of the above into the sum in (A.18) gives

ln Γ(s) = limN→∞

(−(s+N +

1

2

)ln

(1 +

1

N

(s+

1

2

))+

(s− 1

2

)ln

(s− 1

2

)+ 1 +

1

2ln 2π +O(N−1 + |s|−1)

)=

(s− 1

2

)ln

(s− 1

2

)− s+

1

2+

1

2ln 2π +O(|s|−1)

=

(s− 1

2

)(ln s− 1

2s+O(|s|−2)

)− s+

1

2+

1

2ln 2π +O(|s|−1)

=

(s− 1

2

)ln s− s+

1

2ln 2π +O(|s|−1) (A.23)

in the limit |s| → ∞, which is the approximation we wanted to prove for large |s| such that | arg s| < π; cf.[Whittaker and Watson, 1927, pp. 246-253].

Since ln Γ is holomorphic in the domain under consideration, it follows from Cauchy’s integral formula for thederivative that we can differentiate the asymptotic expansion for ln Γ term by term [Erdelyi, 1956, p. 21], obtaining

Γ′(s)

Γ(s)= ln s− 1

2s+O(|s|−2). (A.24)

We also note that the logarithmic derivative of Γ is often called the digamma function ψ in the literature, but alas wecannot use this notation because it conflicts with the Chebyshev function defined in (3.23).

A.3 Jensen’s formula for holomorphic functions

Suppose that f is holomorphic in a region which contains the closed disk |z| ≤ R, where R > 0. Let {zk}mk=1 be thezeros of f in the open disk |z| < R, arranged in order of increasing modulus and repeated according to multiplicity,and assume that there are no zeros on the boundary |z| = R.

We first consider the case f(0) 6= 0 and recall that Re ln z = ln |z|. We find that

Re

ˆ R

0

f ′(reiθ)eiθ

f(reiθ)dr = Re

ˆ R

0

(d

drln f(reiθ)

)dr = Re

(ln f(Reiθ)− ln f(0)

)= ln |f(Reiθ)| − ln |f(0)|, (A.25)

where we used the fundamental theorem of calculus in the region of holomorphicity of f . On the other hand, integratingover the angle θ, we have

1

2πRe

2πˆ

0

R

0

f ′(reiθ)eiθ

f(reiθ)dr dθ = Re

R

0

1

2πir

2πˆ

0

f ′(reiθ)

f(reiθ)ireiθ dθ dr = Re

R

0

1

2πir

‰

|z|=r

f ′(z)

f(z)dz dr = Re

R

0

n(r)

rdr

(A.26)by the argument principle, where n(r) denotes the number of zeros in the open disk |z| < r. Since n is real-valued,we may omit Re in front of the integral in (A.26). It follows that

ˆ R

0

n(r)

rdr =

1

2π

ˆ 2π

0

ln |f(Reiθ)|dθ − ln |f(0)|. (A.27)

Let 0 ≤ j ≤ m be an integer and write z0 = 0 and zm+1 = R for convenience. We observe that n is constant for|zj | < r < |zj+1|. We further note that n(r) = j when the interval |zj | < r < |zj+1| is non-empty. We get

ˆ R

0

n(r)

rdr =

m∑j=0

ˆ |zj+1|

|zj |

n(r)

rdr =

m∑j=1

ˆ |zj+1|

|zj |

j

rdr =

m∑j=1

j(ln |zj+1| − ln |zj |) = −m∑j=1

ln |zj |+m lnR (A.28)

We finally obtain

ln |f(0)| = −m∑j=1

ln

(R

|zj |

)+

1

2π

ˆ 2π

0

ln |f(Reiθ)|dθ, (A.29)

which is Jensen’s formula.

In case f(0) = 0, we can write f(z) = czh+ . . . , where h is the multiplicity of the zero at z = 0, since f is assumedto be holomorphic there. Applying (A.29) to

g(z) := f(z)

(R

z

)h, (A.30)

37

we find that

ln |c|+ h lnR = −m∑j=1

ln

(R

|zj |

)+

1

2π

ˆ 2π

0

ln |f(Reiθ)|dθ, (A.31)

because g(0) = cRh and |g(Reiθ)| = |f(Reiθ)|.

A.4 The uniqueness theorem for Dirichlet series

Suppose that

F (s) :=

∞∑n=1

f(n)

ns(A.32)

and

G(s) :=

∞∑n=1

g(n)

ns(A.33)

are two Dirichlet series that converge absolutely in the half-plane σ ≥ σ0, where σ0 is a real number and we writes = σ + it. We claim that if F ≡ G in this half-plane, then f(n) = g(n) for every positive integer n.

Now, we defineH(s) := F (s)−G(s) (A.34)

to be the Dirichlet series forh(n) := f(n)− g(n). (A.35)

By the absolute convergence of F and G in the half-plane σ ≥ σ0, we know that H is absolutely convergent there andH ≡ 0. We want to prove that h(n) = 0 for all n, so we assume the opposite.

Let m be the smallest positive integer such that h(m) 6= 0. Then

H(s) =

∞∑n=m

h(n)

ns=h(m)

ms+H1(s), (A.36)

where

H1(s) :=

∞∑n=m+1

h(n)

ns. (A.37)

Choosing s with σ ≥ σ0, we find thath(m) = −msH1(s), (A.38)

since H(s) = 0 in this half-plane. Now, we observe that

|H1(s)| =

∣∣∣∣∣∞∑

n=m+1

h(n)

ns

∣∣∣∣∣ ≤∞∑

n=m+1

|h(n)|nσ

=

∞∑n=m+1

|h(n)|nσ0nσ−σ0

≤ 1

(m+ 1)σ−σ0

∞∑n=m+1

|h(n)|nσ0

. (A.39)

It follows that

|h(m)| ≤ mσ|H1(s)| ≤ mσ

(m+ 1)σ−σ0

∞∑n=m+1

|h(n)|nσ0

=

(m

m+ 1

)σ(m+ 1)σ0

∞∑n=m+1

|h(n)|nσ0

=

(m

m+ 1

)σC, (A.40)

where

C := (m+ 1)σ0

∞∑n=m+1

|h(n)|nσ0

(A.41)

is a real number, independent of s, by the absolute convergence of H(σ0). Letting s → +∞ through real values, wefind that the right-hand side of (A.40) tends to 0, which contradicts our hypothesis that h(m) 6= 0, and the uniquenesstheorem follows.

38

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