An introduction to the Riemann hypothesis

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An introduction to the Riemann hypothesis

Author:

Alexander Bielik abielik@kth.se

Supervisor:

P¨ ar Kurlberg

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

Department of Mathematics

September 13, 2014

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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 then derive 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 function ξ is of finite order, whence we obtain many useful properties. We then use these properties to find a zero-free region for 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 of the Riemann hypothesis.

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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 do not appear to follow any obvious pattern. However, Riemann observed a close relation between the behavior of an elaborate function, the so-called Riemann zeta function, and the frequency of prime numbers. Riemann calculated a few zeros of this function, and quickly noted that the interesting ones lay on a certain vertical straight line in the complex plane. Riemann subsequently conjectured that all non-trivial zeros lie on this line. Today, over 10¹³ zeros are 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 its consequences for the distribution of prime numbers. It has been shown that hundreds of statements in number theory follow from it [Gowers et al., 2008, p. 715]. With this background, it might not be a surprise that some mathematicians consider the hypothesis to be the most important problem in pure mathematics, but it remains unresolved for now. 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 in David Hilbert’s list of unsolved problems.

Other than the Riemann zeta function and its zeroes, there is currently no known approach to establish the distribution of prime numbers with desired precision. A disproof of the Riemann hypothesis would reveal a lot about how disordered the primes numbers really are. Enrico Bombieri, a prominent number theorist, remarked that ”the failure of 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 of prime numbers but we have every reason to believe that there are some mysteries which the human mind will never penetrate” [Gowers et al., 2008, p. 348].

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1. The statement of the Riemann hypothesis

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 absolutely convergent series

ζ(s) :=

∞

X

n=1

1

n^s = 1 + 1 2^s + 1

3^s+ · · · . (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. We now show that ζ is holomorphic within Re(s) > 1. To this end, we consider the sequence of holomorphic functions defined by

f_i(s) :=

i

X

n=1

1

n^s. (1.2)

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

In 1737, Euler deduced that Y

p∈P

1

1 − p^−s =Y

p∈P

∞

X

n=0

1 p^s

ⁿ

=Y

p∈P

1 + 1

p^s + 1 p^2s+ · · ·

= 1 +X

p∈P

1 p^s + X

p,q∈P

1 p^sq^s+ · · ·

= 1 + 1 2^s+ 1

3^s+ · · · =

∞

X

n=1

1

n^s = 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 complex number s with Re(s) > 1. In the above derivation, we first used the formula for a geometric series. Then, we rewrote the product and used the fundamental theorem of arithmetic, which states that each positive integer equals exactly one 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

Y

p∈P

1 1 − p^−s

=

X

p∈P

ln

1 1 − p^−s

=

X

p∈P

ln

p^s p^s− 1

=

X

p∈P

ln |p^s| − ln |p^s− 1|

=

X

p∈P

ˆ |p^s|

|p^s−1|

dx x

≤X

p∈P

ˆ |p^s|

|p^s|−1

dx x <X

p∈P

1

|p^s| − 1, (1.4)

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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 convergent product is different from zero.

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

ζ(s) =

∞

X

n=1

n^−s

=

∞

X

n=1

n n^−s− (n + 1)^−s

= s

∞

X

n=1

n ˆ n+1

n

x^−s−1dx

= s ˆ ∞

1

bxcx^−s−1dx

= s ˆ ∞

1

x^−sdx − s ˆ ∞

1

{x}x^−s−1dx

= s

s − 1− s ˆ ∞

1

{x}x^−s−1dx. (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 a meromorphic continuation of ζ. We observe that its only pole in this domain is a simple pole at s = 1 with residue

Res(ζ, 1) = lim

s→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 observe that

1 − 2

2^s

ζ(s) =

∞

X

n=1

1 n^s−

∞

X

n=1

2 (2n)^s =

∞

X

n=1

(−1)ⁿ⁻¹

n^s =: η(s), (1.7)

whence ζ(s) = (1 − 2^1−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 positive real part, it is defined by the convergent integral

Γ(s) :=

ˆ ∞ 0

t^s−1e^−tdt. (1.8)

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

C

Γ(s) ds =

‰

C

ˆ ∞ 0

t^s−1e^−tdt ds = ˆ ∞

0

e^−t

‰

C

t^s−1ds

dt (1.9)

by Fubini’s theorem. We now observe that the inside integral is 0 by the Cauchy–Goursat theorem. It follows from Morera’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

t^se^−tdt =−t^se^−t∞ t=0+ s

ˆ ∞ 0

t^s−1e^−tdt = sΓ(s). (1.10) We also observe that

Γ(1) = ˆ ∞

0

e^−tdt = 1, (1.11)

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so it follows that

Γ(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 at s = 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 Γ_kis holomorphic in this domain, except for the simple poles at the non-positive integers from 0 to 1 − k with residue

Res(Γ, −k) = lim

s→−k(s + k)Γ(s)

= lim

s→−k

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

= lim

s→−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 the whole complex plane with simple poles at the non-positive integers. From now on, this infinitely extended function will 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 appendix A.1 for proof; cf. [Ahlfors, 1966, p. 198]), which is valid for any complex number:

Γ(s) = e^−γs s

∞

Y

n=1

1 + s n

⁻¹

e^s/n. (1.16)

We see that Γ is nowhere zero, since each factor in the convergent product is different from zero. It follows that the reciprocal 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) = se^γs

∞

Y

n = 1

1 + s

n

e^−s/n (1.17)

and

sin (πs) = πsY

n 6=0

1 − s n

e^s/n, (1.18)

we find that

1

−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.

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By the Weierstrass form (1.17), it follows that d ds

Γ⁰(s) Γ(s)

=

∞

X

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

Γ⁰(s) Γ(s)

+ d

ds

Γ⁰ s +¹₂ Γ s +¹₂

!

=

∞

X

n=0

1 (s + n)²+

∞

X

n=0

1 s + n +¹₂2

= 4

"_∞ X

n=0

1 (2s + 2n)² +

∞

X

n=0

1 (2s + 2n + 1)²

#

= 4

∞

X

m=0

1 (2s + m)²

= 2 d ds

Γ⁰(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 obtain the relation

Γ(s)Γ

s +1

2

= e^as+bΓ(2s), (1.24)

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

2

Γ(1) = e^a/2+bΓ(1) (1.25)

and

Γ(1)Γ

1 +1

2

= e^a+bΓ(2), (1.26)

where Γ(1) = Γ(2) = 1, Γ(¹₂) =√

π and Γ(1 + ¹₂) = ¹₂Γ(¹₂) =¹₂√

π. Taking the exponential, we are led to the system of equations

(₁

2a + b = ¹₂ln π,

a + b = ¹₂ln π − ln 2. (1.27)

It follows that a = −2 ln 2 and b =¹₂ln π + ln 2. Insertion into the original relation gives Γ(s)Γ

s +1

2

= 2^1−2s√

πΓ(2s), (1.28)

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

2

Γ s + 1 2

= 2^1−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:

Γs 2

Γ 1 − s 2

= 2^1−sπ^−1/2cosπs 2

Γ(s). (1.31)

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

Γ(s) =√

2πe^−ss^s−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 the approximation 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]).

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1.3 The functional equation

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

θ(s) :=

∞

X

n=−∞

e^−πn²^s (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

F_xh e^−αx²i

(ξ) =r π

αe^−(πξ)²^/α (1.35)

for complex numbers α with Re(α) > 0, where F_x denotes the Fourier transform with respect to the variable x.

Setting α = πs, we get

Fx

he^−πx²^si

(ξ) = 1

√se^−πξ²^/s. (1.36)

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

θ(s) =

∞

X

n=−∞

e^−πn²^s=

∞

X

k=−∞

√1

se^−πk²^/s= 1

√sθ 1 s

. (1.37)

We now define the helper function ω by

ω(s) :=

∞

X

n=1

e^−πn²^s=θ(s) − 1

2 . (1.38)

It follows that

ω 1 s

= θ(¹_s) − 1

2 =

√sθ(s) − 1

2 =

√s

2 (2ω(s) + 1) −1 2 = −1

2+1 2

√s +√

sω(s). (1.39)

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

h

e^−πn²^xi (s) =

ˆ ∞ 0

x^s−1e^−πn²^xdx =t := πn²x = (πn²)^−s ˆ ∞

0

t^s−1e^−tdt = (πn²)^−sΓ(s) (1.40) for Re(s) > 0, whence

M_xh

e^−πn²^xi s 2

= ˆ ∞

0

x^s/2−1e^−πn²^xdx = π^−s/2Γs 2

n^−s. (1.41)

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

π^−s/2Γs 2

ζ(s) =

∞

X

n=1

π^−s/2Γs 2

n^−s=

∞

X

n=1

ˆ ∞ 0

x^s/2−1e^−πn²^xdx = ˆ ∞

0

x^s/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 ∞:

M_x[ω(x)]s 2

= ˆ 1

0

x^s/2−1ω(x) dx + ˆ ∞

1

x^s/2−1ω(x) dx

= ˆ ∞

1

x^−s/2−1ω 1 x

dx +

ˆ ∞ 1

x^s/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/2Γs 2

ζ(s) = M_x[ω(x)]s 2

= − 1

s (1 − s)+ ˆ ∞

1

x^s/2−1+ x^−(s+1)/2

ω(x) dx, (1.45)

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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 any bounded 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 may use 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/2Γs

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)^−scosπs

2

Γ(s)ζ(s), (1.48)

which is known as the Riemann functional equation. Substituting 1 − s for s yields ζ(s) = 2^sπ^s−1sinπs

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 the region Re(s) > 0, where the integral representation (1.5) of ζ is valid:

ζ(0) = lim

s→02^sπ^s−1sinπs 2

Γ(1 − s)ζ(1 − s)

= lim

s→0

(2π)^s π

∞

X

n=0

(−1)ⁿ (2n + 1)!

πs 2

2n+1!

Γ(1 − s)

1 − s

(1 − s) − 1− (1 − s) ˆ ∞

1

{x}x^{−(1−s)−1}dx

= lim

s→0

(2π)^s π

πs 2

X^∞

n=0

(−1)ⁿ (2n + 1)!

πs 2

2n!

Γ(1 − s) −1 + s

s − (1 − s) ˆ ∞

1

{x}x^s−2dx

= lim

s→0

(2π)^s

2 1 +

∞

X

n=1

(−1)ⁿ (2n + 1)!

πs 2

²ⁿ

!

Γ(1 − s)

−1 + s − s(1 − s) ˆ ∞

1

{x}x^s−2dx

= 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)Γs 2

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 = −2n for 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 about the real axis. This, together with the Riemann functional equation, shows that the zeros are also symmetric about the line Re(s) = ¹₂. 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 critical strip. The Riemann hypothesis asserts that all zeros in the critical strip lie on the line Re(s) = ¹₂, which we name the critical line. Figure 1.1 illustrates our recent conclusions.

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

The black spots on the critical line Re(s) = ¹₂ represent the non-trivial zeros.

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

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

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

ζ ¹₂+ it

for −50 ≤ t ≤ 50.

Figure 1.1: Visualization of some complex-valued functions using the domain coloring technique. The magnitude of the output is represented by the brightness, where black represents zero and white represents infinity, while the argument 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 ¹⁹₂₇ of the zeros are simple. Thus far, all zeros that have been located are simple.

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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 numbers α for which this inequality holds for some number C the function order of f . We first show that a finite-ordered function f with no zeros must be of the form e^g, where g is a polynomial. Moreover, we also show that the degree of g 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_f⁰ is itself entire. Furthermore, it is known that any antiderivative of an entire function is entire, from which it follows that the single-valued function g(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)| by the 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)| =

z 2Rψ(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

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

Re(g(z)) = ln |f (z)| ≤ ln C + |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= 2^2+αR^α. (2.7)

It follows that |h(z)| ≤ 2M (R) < 2^3+αR^αfor |z| ≤ R. Hence, there exists a constant k such that

|g(z)| = |g(0) + h(z)| < |g(0)| + 2^3+αR^α< 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 the term |g(0)|.

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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) =

∞

X

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. It follows that g is a polynomial of degree at most α, which is the order of f . Since α was defined as the infimum of all numbers 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 open disk 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 {z_k}ⁿ_k=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 (Re^iθ)| dθ − ln |f (0)| =

n

X

k=1

ln

R

|zk|

= ln Rⁿ

|z1| . . . |zn|, (2.11) from where we see that the modulus of f (Re^iθ) depends on the moduli of the zeros in the disk. This enables us to prove 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

ln Rⁿ

|z1| . . . |zn| = ln|z2|

|z1|+ 2 ln|z3|

|z2|+ · · · + n ln R

|zn| = ˆ R

0

n(r)

r dr, (2.12)

so that Jensen’s formula becomes 1 2π

ˆ 2π 0

ln |f (Re^iθ)| dθ − ln |f (0)| = ˆ R

0

n(r)

r dr. (2.13)

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

ln |f (Re^iθ)| ≤ ln C + R^α< R^α+ (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) r dr ≤

ˆ 2R 0

n(r)

r dr < (2R)^α+− ln |f (0)| < 2(2R)^α+ (2.15) by Jensen’s formula. But this means that n(R) = O(R^α+), which is what we wanted to show. It also follows that the sum

∞

X

n=1

|zn|^−β= ˆ ∞

0

dn(r)

r^β = n(r) r^β

^∞

0

+ β ˆ ∞

0

n(r) r^β+1 dr = β

ˆ ∞

|z1|

n(r)

r^β+1 dr β ˆ ∞

|z1|

r^α+−β−1dr < ∞ (2.16)

converges for β > α + . We note that the sum converges for any β > α if we choose < 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 avoid unnecessary details, we shall concern ourselves only with functions of order α = 1.

We previously observed that the sum P |zn|^−β converges for any β > α. By taking β = 2 and recalling the Maclaurin series for the exponential function,

e^z=

∞

X

n=0

zⁿ

n! = 1 + z + O(z²), (2.17)

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we find that the product

P (z) :=

∞

Y

n=1

1 − z

zn

e^z/zⁿ (2.18)

converges absolutely for all z. Now, if we take the logarithm of the product and recall that ln(1 − z) = −

∞

X

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

z_n

+ z

z_n

z z_n

2

|z_n|⁻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. In consequence, the convergence of the product (2.18) is uniform in any bounded region. Thus, P is an entire function with the same zeros as f , counting multiplicity, if we assume that f (0) 6= 0. Note that we make this assumption only for the sake of simplicity; otherwise, we would multiply the product by the factor z^m, where m is the multiplicity of the 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 > 0. Since we know that f is of order 1, it would suffice to prove that the estimate

1 P (z)

= O(exp(|z|¹⁺)) (2.22)

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

|F (z)| =

f (z) P (z)

= O(exp(|z|¹⁺)) (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⁺)) (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⁺)), (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 write P (z) = P1(z)P2(z)P3(z), where

P1(z) :=

∞

Y

k=1+n(2R)

1 − z

zk

e^z/z^k, (2.26)

P2(z) :=

n(2R)

Y

k=1

e^z/z^k= exp





n(2R)

X

k=1

z z_k



 (2.27)

and

P₃(z) :=

n(2R)

Y

k=1

1 − z

zk

. (2.28)

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

|P_i(z)|, because | ln |P_i(z)|| = ln

1 P_i(z)

. By our previous estimate (2.20), we know that the k^th term of ln |P₁(z)| is O

z z_k

2!

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

ln |P₁(z)| R²

∞

X

k=1+n(2R)

|zk|⁻² = R² ˆ ∞

2R

dn(r) r² R²

ˆ ∞ 2R

r⁻²dr = −R²(2R)⁻¹

− 1 R¹⁺. (2.29)

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We also note that the k^th term of ln |P2(z)| is O

z z_k

, from where we conclude that

ln |P2(z)| R

n(2R)

X

k=1

|zk|⁻¹= R ˆ 2R

|z1|

dn(r)

r R

ˆ 2R

|z1|

r⁻¹dr = R

((2R)− |z1|) R¹⁺. (2.30) We now consider the last subproduct P3. This time, due to the zeros of P3, we must look for a bound that does not necessarily hold everywhere in the annulus R < |z| < 2R. Since P |zk|⁻² converges, we know that the union of intervals S(|zk| − |zk|⁻², |zk| + |zk|⁻²) is of finite length. It follows that for all sufficiently large R, there exists a number R < r < 2R such that ||zk| − r| > |zk|⁻²for all zk. With this in mind, we find that every factor of P3satisfies

1 − z zk

=

zk− z zk

≥||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)

Y

k=1

1 − z zk

−1

< (2R)³^n(2R)

(2R)³^R¹⁺

= exp(3 ln(2R)R¹⁺) exp(R¹⁺²), (2.32)

because the number of factors in the product is O(R¹⁺) 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 ≡ e^g, where g is a polynomial of degree at most 1. We finally deduce that

f (z) = F (z)P (z) = e^A+Bz

∞

Y

n=1

1 − z

z_n

e^z/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 P |zn|⁻¹⁻ converges for any > 0 by Jensen’s formula.

However, the sumP |zn|⁻¹ may or may not converge. Since |1 − z| ≤ 1 + |z| ≤ e^|z| for any complex number z, we find that |(1 − z)e^z| ≤ e^2|z|. Hence, if the latter sum converges, then

|f (z)| = e^A+Bz

∞

Y

n=1

1 − z

z_n

e^z/zⁿ

≤ e^|A+Bz|

∞

Y

n=1

e^2|z/zⁿ^|= e^|A+Bz|exp 2|z|

∞

X

n=1

|z_n|⁻¹

!

< e^C|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)Γs 2

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 > 0, there exists a real number C such that

|ξ(s)| ≤ C exp(|s|¹⁺) (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) ln π e^{2 ln s}− e^{ln 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)

ln Γs 2

|s| ln |s|, (2.39)

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from where it follows that

Γs 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− s ˆ ∞

1

{x}x^−s−1dx, (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 that

ζ(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| for any > 0 completes the proof that ξ has order at most 1.

We now show that ξ is entire, which is a prerequisite for our latest conclusion. From the representation we obtained in (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 functional equation that there exists a complex number t with Re(t) ≥ 1 such that

ζ(s) = ζ(1 − t) = 2(2π)^−tcos 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 this region is a simple pole at t = 1 from the factor ζ(t), but at this point we have

cos πt 2

= cosπ 2

= 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 factor s(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

s→1lim(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 that ξ(0) = ξ(1) = ¹₂ 6= 0. It now follows that

ξ(s) = e^A+BsY

ρ

1 − s

ρ

e^s/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-fraction decomposition 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 +

∞

X

n=2

1

n^s, (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’s formula (1.33). It follows that ξ does not satisfy the inequality |ξ(s)| < e^C|s| found in (2.34) for any constant C.

We conclude thatP |ρ|⁻¹⁻converges for any > 0, but that P |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

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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 ξ⁰(s)

ξ(s) = B +X

ρ

1 s − ρ+1

ρ

. (2.49)

By the definition in (1.51), we have ζ(s) = 2

s(s − 1)π^s/2Γ⁻¹s 2

ξ(s) = 1

(s − 1)π^s/2Γ⁻¹s 2 + 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

1

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

ζ⁰(s)

ζ(s) = B − 1 s − 1+1

2ln π −1 2

Γ⁰ ^s₂+ 1 Γ ^s₂+ 1 +X

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 = 1 and the non-trivial zeros. Taking the logarithmic derivative of the Weierstrass product (1.17) for Γ, we get

−1 2

Γ⁰ ^s₂+ 1 Γ ^s₂+ 1 = 1

2γ +

∞

X

n=1

1

s + 2n− 1 2n

, (2.52)

which exhibits the trivial zeros. Here, γ := lim

n→∞

n

X

k=1

1 k − ln n

!

= 1 − lim

n→∞ ln n −

n

X

k=2

1 k

!

= 1 − lim

n→∞

n−1

X

k=1

(ln(k + 1) − ln k) −

n−1

X

k=1

1 k + 1

!

= 1 − lim

n→∞

n−1

X

k=1

ln(k + 1) − ln k − 1 k + 1

= 1 − lim

n→∞

n−1

X

k=1

ln(k + 1) − ln k + k k + 1− 1

= 1 − lim

n→∞

n−1

X

k=1

ln x +k x

^k+1

k

= 1 − lim

n→∞

n−1

X

k=1

ˆ k+1 k

x − k x² dx

= 1 − lim

n→∞

ˆ n 1

x − bxc x² dx

= 1 − ˆ ∞

1

x − bxc x² dx

= 1 − ˆ ∞

1

{x}x⁻²dx (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) = e^A, but we previously noted that ξ(0) =¹₂, so A = ln¹₂. For B, we take the logarithmic derivative (2.49) and observe that

ξ⁰(0)

ξ(0) = B = −ξ⁰(1)

ξ(1), (2.54)

where the latter equality follows from the functional equation (1.52). By the relation (2.50), we get ξ⁰(s)

ξ(s) = ζ⁰(s) ζ(s) + 1

s − 1−1

2ln π +1 2

Γ⁰ ^s₂+ 1

Γ ^s₂+ 1 . (2.55)

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

−1 2

Γ⁰ ³₂ Γ ³₂ =1

2γ − 1 + ln 2, (2.56)

whence

B = 1

2γ − 1 +1

2ln(4π) − lim

s→1

ζ⁰(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−1dx. (2.59)

Now, computation gives

s→1lim

ζ⁰(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 P |ρ|⁻¹ diverges, we note that P 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 +X

ρ

1

1 − s − ρ+1 ρ

= −B −X

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 = −X

ρ

1

ρ = −2X

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 pair of zeros is s ≈ ¹₂± 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 product formula, derived in (1.3) to be

ζ(s) =Y

p∈P

1

1 − p^−s, (2.65)

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

ln ζ(s) = −X

p∈P

ln(1 − p^−s) =X

p∈P

∞

X

n=1

p^−sn

n =X

p∈P

∞

X

n=1

p^−nσ

n exp(−itn ln p), (2.66)

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

Reln ζ(s) =X

p∈P

∞

X

n=1

p^−nσ

n cos(t ln(pⁿ)). (2.67)

Now, we consider the identity

3 + 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)

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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 contradiction to the inequality (2.70), because 4 > 3. We conclude that ζ(s) 6= 0 on σ = 1. By the functional equation (1.52), the same 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 difficulties with meromorphic continuation to the left of σ = 1; this is so because the former function has only simple poles at s = 1 for σ ≥ 1 and at the non-trivial zeros of ζ for 1 > σ > 0, while the latter function has logarithmic singularities at s = 1 and at the non-trivial zeros of ζ.

We define the von Mangoldt function by

Λ(n) :=

(ln p if n = p^mfor 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 product formula (1.3), we get

ln ζ(s) = −X

p∈P

ln 1 − p^−s . (2.74)

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

−ζ⁰(s) ζ(s) =X

p∈P

(ln p)p^−s 1 − p^−s =X

p∈P

(ln p)

1

1 − p^−s − 1

=X

p∈P

(ln p)

∞

X

n=1

p^−sn=X

p∈P

X

m∈Z>0

ln p p^ms =

∞

X

n=1

Λ(n)

n^s . (2.75) This formula will prove very useful to our future work. It is also instrumental to observe that

X

d|n

Λ(d) = ln n (2.76)

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

∞

X

n=1

1 n^s

X

d|n

Λ(d) =

∞

X

n=1

1 n^s

X

km=n

Λ(m) =

∞

X

k=1

∞

X

m=1

Λ(m) k^sm^s =

∞

X

n=1

1 n^s

! _∞ X

n=1

Λ(n) n^s

!

= ζ(s)

−ζ⁰(s) ζ(s)

=

∞

X

n=1

ln n n^s .

(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 two Dirichlet series in (2.77).

By the formula (2.75), we observe that

−ζ⁰(s) ζ(s) =

∞

X

n=1

Λ(n) n^s =

∞

X

n=1

Λ(n)

n^σ exp(−it ln n). (2.78)

Taking the real part, we get

− Reζ⁰(s) ζ(s) =

∞

X

n=1

Λ(n)

n^s cos(t ln n). (2.79)

Once again, we apply the identity (2.68). By the same token, we have 3

−ζ⁰(σ) ζ(σ)

+ 4

−Reζ⁰(σ + it) ζ(σ + it)

+

−Reζ⁰(σ + 2it) ζ(σ + 2it)

≥ 0 (2.80)

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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 a real number C such that

−ζ⁰(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 behaviour to be influenced by the possible zeros at a height near to t or 2t just to the left of σ = 1. This observation is made more precise by the partial-fraction decomposition of the logarithmic derivative of ζ, which we derived in (2.51) to be

−ζ⁰(s) ζ(s) = 1

s − 1− B −1

2ln π +1 2

Γ⁰(¹₂s + 1) Γ(¹₂s + 1) −X

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

Γ⁰(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

− Reζ⁰(s)

ζ(s) < O(ln |t|) −X

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 the right-hand side of the inequality (2.84). Putting s = σ + 2it in (2.84), we obtain

− Reζ⁰(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 _s−ρ¹ in the sum which corresponds to this zero, and leave out the rest. Putting s = σ + it, we get

− Reζ⁰(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 statement outside this region. Furthermore, it can be shown (cf. [Kadiri, 2005, p. 2]) that the inequality holds for C = _5.69693¹ . Figure 2.1 illustrates our conclusions.

An introduction to the Riemann hypothesis