An introduction to the

Author: Alexander Bielik [email protected]

Supervisor: P¨arKurlberg

SA104X – Degree Project in Engineering Physics Royal 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 , drawing upon existing literature by Davenport, Ahlfors, et al.

We begin with the meromorphic continuation of the Riemann zeta function ζ and the Γ. We then derive a 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 theorem and conclude with an implication of the Riemann hypothesis. Contents

Introduction 2

1 The statement of the Riemann hypothesis3 1.1 The Riemann zeta function ζ ...... 3 1.2 The gamma function Γ...... 4 1.3 The functional equation...... 7 1.4 The critical strip...... 8

2 Zeros in the critical strip 10 2.1 Functions of finite order...... 10 2.2 The Hadamard product for functions of order 1...... 11 2.3 Proving that ξ has order at most 1...... 13 2.4 A zero-free region for ζ ...... 16 2.5 The number of zeros in a rectangle...... 19

3 The distribution of prime numbers 23 3.1 Perron’s formula for Dirichlet series...... 23 3.2 An approximation of the Chebyshev function ψ ...... 25 3.3 von Mangoldt’s explicit formula...... 27 3.4 Proving the explicit formula...... 28 3.5 The ...... 30 3.6 The smallest possible error term...... 32

Appendices 33

A Additional proofs 34 A.1 The Weierstrass form of Γ...... 34 A.2 Stirling’s formula for ln Γ...... 35 A.3 Jensen’s formula for holomorphic functions...... 37 A.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 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 . 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 mathemati- cians 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 distri- bution 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].

2 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 ∞ X 1 1 1 ζ(s) := = 1 + + + ··· . (1.1) ns 2s 3s n=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 i X 1 f (s) := . (1.2) i ns n=1 ∞ Since {fi}i=1 converges uniformly to ζ on any compact subset of Re(s) > 1, it follows that the function is holomorphic there.

In 1737, Euler deduced that

∞ n Y 1 Y X  1  = 1 − p−s ps p∈P p∈P n=0 Y  1 1  = 1 + + + ··· ps p2s p∈P X 1 X 1 = 1 + + + ··· ps psqs p∈P p,q∈P ∞ 1 1 X 1 = 1 + + + ··· = = ζ(s) (1.3) 2s 3s ns n=1 for any integer s > 1, where P denotes the set of prime numbers, though his argument can be extended to any 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 , we find that

Y 1 X 1 ln = ln 1 − p−s 1 − p−s p∈P p∈P

s X p = ln ps − 1 p∈P

X s s = ln |p | − ln |p − 1|

p∈P

|ps| |ps| X dx X dx X 1 = ≤ < s , (1.4) ˆ|ps−1| x ˆ|ps|−1 x |p | − 1 p∈P p∈P p∈P

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 convergent product is different from zero.

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

∞ X ζ(s) = n−s n=1 ∞ X = n n−s − (n + 1)−s n=1 ∞ X n+1 = s n x−s−1 dx ˆ n=1 n ∞ = s bxcx−s−1 dx ˆ1 ∞ ∞ = s x−s dx − s {x}x−s−1 dx ˆ1 ˆ1 s ∞ = − s {x}x−s−1 dx. (1.5) s − 1 ˆ1 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)ζ(s) = 1. (1.6) s→1

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 ∞ ∞ ∞  2  X 1 X 2 X (−1)n−1 1 − ζ(s) = − = =: η(s), (1.7) 2s ns (2n)s ns n=1 n=1 n=1 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. 2πi While the right-hand side has singularities at s = 1+k 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) := ts−1e−t dt. (1.8) ˆ0 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 ∞ ∞   Γ(s) ds = ts−1e−t dt ds = e−t ts−1 ds dt (1.9) ‰C ‰C ˆ0 ˆ0 ‰C 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:

∞ ∞ s −t  s −t∞ s−1 −t Γ(s + 1) = t e dt = −t e t=0 + s t e dt = sΓ(s). (1.10) ˆ0 ˆ0

We also observe that ∞ Γ(1) = e−t dt = 1, (1.11) ˆ0

4 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 1 Γ (s) := Γ(s + 1), (1.13) 1 s 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 1 Γ (s) := Γ(s + k) (1.14) k s(s + 1) ... (s + k − 1)

for Re(s) > −k. We similarly find that Γk is 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)Γ(s) s→−k (s + k)Γ(s + k) = lim s→−k s(s + 1) ... (s + k − 1) Γ(s + k + 1) = lim s→−k s(s + 1) ... (s + k − 1) Γ(1) = (−k)(−k + 1) ... (−1) (−1)k = . (1.15) k!

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:

∞ e−γs Y  s −1 Γ(s) = 1 + es/n. (1.16) s n n=1 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 Y  s  = seγs 1 + e−s/n (1.17) Γ(s) n n = 1 and Y  s  sin (πs) = πs 1 − es/n, (1.18) n n 6=0 we find that 1 sin (πs) = . (1.19) −sΓ(−s)Γ(s) π Using the functional equation Γ(1 + s) = sΓ(s), it follows that π Γ (1 − s)Γ(s) = −sΓ(−s)Γ(s) = , (1.20) sin (πs)

1 for non-integers. We call this Euler’s reflection formula. If we set s = 2 , we get 1 √ Γ = π. (1.21) 2 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 Γ0(s) X 1 = . (1.22) ds Γ(s) (s + n)2 n=0 1 Furthermore, we note that Γ(s)Γ(s + 2 ) and Γ(2s) have the same poles. Taking the second derivative of the logarithm 1 of Γ(s)Γ(s + 2 ), we get

0 1 ! ∞ ∞ d Γ0(s) d Γ s + X 1 X 1 + 2 = + ds Γ(s) ds Γ s + 1  (s + n)2 1 2 2 n=0 n=0 s + n + 2 " ∞ ∞ # X 1 X 1 = 4 + (2s + 2n)2 (2s + 2n + 1)2 n=0 n=0 ∞ X 1 = 4 (2s + m)2 m=0 d Γ0(2s) = 2 , (1.23) ds Γ(2s) where we used the chain rule in the last step. Integrating both sides twice and then taking the exponential, we obtain the relation  1 Γ(s)Γ s + = eas+bΓ(2s), (1.24) 2 1 where a and b are undetermined constants. Taking s = 2 and s = 1, we get 1 Γ Γ(1) = ea/2+bΓ(1) (1.25) 2 and  1 Γ(1)Γ 1 + = ea+bΓ(2), (1.26) 2 1 √ 1 1 1 1 √ where Γ(1) = Γ(2) = 1, Γ( 2 ) = π and Γ(1 + 2 ) = 2 Γ( 2 ) = 2 π. Taking the exponential, we are led to the system of equations ( 1 1 2 a + b = 2 ln π, 1 (1.27) a + b = 2 ln π − ln 2. 1 It follows that a = −2 ln 2 and b = 2 ln π + ln 2. Insertion into the original relation gives  1 √ Γ(s)Γ s + = 21−2s πΓ(2s), (1.28) 2 which is known as Legendre’s duplication formula. Note that the formula is not valid at the non-positive integers and half-integers.

1−s By substituting s with 2 in Euler’s reflection formula (1.20), we get s + 1 1 − s π Γ Γ = πs . (1.29) 2 2 cos 2 s By substituting s with 2 in Legendre’s duplication formula (1.28), we get  s  s + 1 √ Γ Γ = 21−s πΓ(s). (1.30) 2 2 If we take the quotient of the two recently derived formulae (1.29) and (1.30), we obtain a new functional equation:  s  1 − s πs Γ Γ = 21−sπ−1/2 cos Γ(s). (1.31) 2 2 2 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 the approximation as  1 1 ln Γ(s) = s − ln s − s + ln 2π + O(|s|−1), (1.33) 2 2 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

∞ X 2 θ(s) := e−πn s (1.34) n=−∞ on the right half-plane Re(s) > 0, where it is holomorphic. We shall find a functional equation for θ, and then 2 transform it into a functional equation for ζ. Recall that e−αx is a fixed point of the Fourier transform with r h 2 i π 2 F e−αx (ξ) = e−(πξ) /α (1.35) x α 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 h 2 i 1 2 F e−πx s (ξ) = √ e−πξ /s. (1.36) x s Thus, we obtain the following functional equation by the Poisson summation formula: ∞ ∞   X 2 X 1 2 1 1 θ(s) = e−πn s = √ e−πk /s = √ θ . (1.37) s s s n=−∞ k=−∞ We now define the helper function ω by

∞ X 2 θ(s) − 1 ω(s) := e−πn s = . (1.38) 2 n=1 It follows that √ √ 1 θ( 1 ) − 1 sθ(s) − 1 s 1 1 1√ √ ω = s = = (2ω(s) + 1) − = − + s + sω(s). (1.39) s 2 2 2 2 2 2

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

∞ ∞ h −πn2xi s−1 −πn2x  2 2 −s s−1 −t 2 −s Mx e (s) = x e dx = t := πn x = (πn ) t e dt = (πn ) Γ(s) (1.40) ˆ0 ˆ0 for Re(s) > 0, whence ∞ h −πn2xi  s  s/2−1 −πn2x −s/2  s  −s Mx e = x e dx = π Γ n . (1.41) 2 ˆ0 2 By the series representation (1.1) of ζ, it follows that

∞ ∞ ∞ ∞  s  X  s  X 2  s  π−s/2Γ ζ(s) = π−s/2Γ n−s = xs/2−1e−πn x dx = xs/2−1ω(x) dx = M [ω(x)] (1.42) 2 2 ˆ ˆ x 2 n=1 n=1 0 0 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, 1 one from 0 to 1, where we substitute x by x , and the other from 1 to ∞:

1 ∞  s  s/2−1 s/2−1 Mx [ω(x)] = x ω(x) dx + x ω(x) dx 2 ˆ0 ˆ1 ∞  1  ∞ = x−s/2−1ω dx + xs/2−1ω(x) dx. (1.43) ˆ1 x ˆ1 Next, using the functional equation for ω, we find that

∞  1  ∞  1 1√ √  x−s/2−1ω dx = x−s/2−1 − + x + xω(x) dx ˆ1 x ˆ1 2 2 1 1 ∞ = − + + x−(s+1)/2ω(x) dx, (1.44) s s − 1 ˆ1

and so ∞ −s/2  s   s  1  s/2−1 −(s+1)/2 π Γ ζ(s) = Mx [ω(x)] = − + x + x ω(x) dx, (1.45) 2 2 s (1 − s) ˆ1

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 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

 s  1 − s π−s/2Γ ζ(s) = π−(1−s)/2Γ ζ(1 − s). (1.47) 2 2

This, together with the functional equation for Γ in (1.31), shows that ζ satisfies πs ζ(1 − s) = 2(2π)−s cos Γ(s)ζ(s), (1.48) 2 which is known as the Riemann functional equation. Substituting 1 − s for s yields πs ζ(s) = 2sπs−1 sin Γ(1 − s)ζ(1 − s). (1.49) 2 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: πs ζ(0) = lim 2sπs−1 sin Γ(1 − s)ζ(1 − s) s→0 2 ∞ ! (2π)s X (−1)n πs2n+1  1 − s ∞  = lim Γ(1 − s) − (1 − s) {x}x−(1−s)−1 dx s→0 π (2n + 1)! 2 (1 − s) − 1 ˆ n=0 1 ∞ ! (2π)s πs X (−1)n πs2n −1 + s ∞  = lim Γ(1 − s) − (1 − s) {x}xs−2 dx s→0 π 2 (2n + 1)! 2 s ˆ n=0 1 ∞ ! (2π)s X (−1)n πs2n  ∞  = lim 1 + Γ(1 − s) −1 + s − s(1 − s) {x}xs−2 dx s→0 2 (2n + 1)! 2 ˆ n=1 1 1 = · 1 · Γ(1) · (−1) 2 1 = − . (1.50) 2 The Riemann functional equation also has a symmetric version. For Re(s) > 0, let 1  s  ξ(s) := π−s/2s(s − 1)Γ ζ(s) (1.51) 2 2 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 πs  for any positive integer n and note that sin 2 = 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 1 the line Re(s) = 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 critical 1 strip. The Riemann hypothesis asserts that all zeros in the critical strip lie on the line Re(s) = 2 , which we name the critical line. Figure 1.1 illustrates our recent conclusions.

8 (a) The Riemann zeta function ζ in the complex plane. The (b) The gamma function Γ in the complex plane. The white white spot at s = 1 represents the function’s pole. The black spots at the non-positive integers represent the poles of the spots at the negative even integers represent the trivial zeros. function. Note the lack of zeros. 1 The black spots on the critical line Re(s) = 2 represent the non-trivial zeros.

(c) The xi function ξ in the complex plane. Note the symmetry (d) The first few zeros of ζ on the critical line highlighted by 1  and the lack of poles. plotting ζ 2 + 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 , where represents zero.

We note that the Riemann hypothesis does not explicitly say anything about the multiplicity of the zeros. However, 19 it has been shown (cf. [Bui and Heath-Brown, 2013, pp. 1-10]) that the hypothesis implies that at least 27 of the zeros 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 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 eg, 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.

f 0 Since f is an entire function with no zeros, we know that its logarithmic derivative 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 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 Re(h(z)), (2.2) |z|≤2R

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

h(z) φ(z) := (2.3) 2M(R) − h(z)

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

2Rφ(z) ψ(z) := (2.4) z 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 z 1 |φ(z)| = ψ(z) ≤ (2.5) 2R 2 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 Re(h(z)) = sup Re(g(z)) − Re(g(0)) < 2(2R)α + 2(2R)α = 22+αRα. (2.7) |z|≤2R |z|≤2R

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 the term |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

∞ X g(n)(0) g(z) = zn. (2.9) n! n=0 Since |g(z)| < k|z|α on the circle |z| = R, Cauchy’s estimate gives

n!k|z|α |g(n)(0)| < (2.10) Rn 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.

n Suppose that {zk}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

n 1 2π X  R  Rn ln |f(Reiθ)| dθ − ln |f(0)| = ln = ln , (2.11) 2π ˆ |zk| |z1| ... |zn| 0 k=1 from where we see that the modulus of f(Reiθ) 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

Rn |z | |z | R R n(r) ln = ln 2 + 2 ln 3 + ··· + n ln = dr, (2.12) |z1| ... |zn| |z1| |z2| |zn| ˆ0 r so that Jensen’s formula becomes 1 2π R n(r) ln |f(Reiθ)| dθ − ln |f(0)| = dr. (2.13) 2π ˆ0 ˆ0 r Choose a number  > 0. By the definition of function order, we know that

ln |f(Reiθ)| ≤ ln C + Rα < Rα+ (2.14) for all sufficiently large R. From here it follows that

2R dr 2R n(r) 2R n(r) n(R) ln 2 = n(R) ≤ dr ≤ dr < (2R)α+ − ln |f(0)| < 2(2R)α+ (2.15) ˆR r ˆR r ˆ0 r by Jensen’s formula. But this means that n(R) = O(Rα+), which is what we wanted to show. It also follows that the sum

∞ ∞ X ∞ dn(r) n(r) ∞ n(r) ∞ n(r) ∞ |z |−β = = + β dr = β dr  β rα+−β−1 dr < ∞ (2.16) n ˆ rβ rβ ˆ rβ+1 ˆ rβ+1 ˆ n=1 0 0 0 |z1| |z1| 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.

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

∞ X zn ez = = 1 + z + O(z2), (2.17) n! n=0

11 we find that the product ∞ Y  z  P (z) := 1 − ez/zn (2.18) z n=1 n converges absolutely for all z. Now, if we take the logarithm of the product and recall that

∞ X zn ln(1 − z) = − = −z + O(z2) (2.19) n n=1 for |z| < 1, then it follows that each term in the obtained sum satisfies   2 z z z −2 ln 1 − +   |zn| (2.20) zn zn zn 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 zm, where m is the multiplicity of the zero at z = 0. Hence, the function f(z) F (z) := (2.21) P (z) 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 1+ = O(exp(|z| )) (2.22) P (z) is valid as |z| → ∞ in order to establish that the bound

f(z) 1+ |F (z)| = = O(exp(|z| )) (2.23) P (z) 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 1+ = O(exp(R )) (2.24) P (z) 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) 1+ |F (z)| = = O(exp(R )), (2.25) P (z) 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 ∞  z  Y z/zk P1(z) := 1 − e , (2.26) zk k=1+n(2R)

n(2R) n(2R)  z Y z/zk X P2(z) := e = exp   (2.27) zk k=1 k=1 and n(2R) Y  z  P3(z) := 1 − . (2.28) zk k=1 3 We now find upper bounds on the moduli of {ln |Pi(z)|} . These will yield both upper and lower bounds on i=1 1 th |Pi(z)|, because | ln |Pi(z)|| = ln . By our previous estimate (2.20), we know that the k term of ln |P1(z)| is Pi(z) 2! z O . Using the same tricks of integration as in (2.16), we get zk

∞ X ∞ dn(r) ∞ (2R)−1 ln |P (z)|  R2 |z |−2 = R2  R2 r−2 dr = −R2  R1+. (2.29) 1 k ˆ r2 ˆ  − 1 k=1+n(2R) 2R 2R

12   th z We also note that the k term of ln |P2(z)| is O , from where we conclude that zk

n(2R) X 2R dn(r) 2R R ln |P (z)|  R |z |−1 = R  R r−1 dr = ((2R) − |z |)  R1+. (2.30) 2 k ˆ r ˆ  1 k=1 |z1| |z1|

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

−2 z zk − z ||zk| − |z|| |zk| −3 −3 1 − = ≥ > = |zk| > (2R) (2.31) zk zk |zk| |zk| on the circle |z| = r, from which it follows that

n(2R) −1 1+ 1 Y z 3n(2R) 3R 1+ 1+2 = 1 − < (2R)  (2R) = exp(3 ln(2R)R )  exp(R ), (2.32) P3(z) zk k=1 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 ≡ eg, where g is a polynomial of degree at most 1. We finally deduce that

∞ Y  z  f(z) = F (z)P (z) = eA+Bz 1 − ez/zn , (2.33) z n=1 n where A and B are some constants. This factorization is known as the Hadamard product for functions of order 1.

P −1− We finish with an observation. We know that the sum |zn| converges for any  > 0 by Jensen’s formula. P −1 |z| However, the sum |zn| may or may not converge. Since |1 − z| ≤ 1 + |z| ≤ e for any complex number z, we find that |(1 − z)ez| ≤ e2|z|. Hence, if the latter sum converges, then

∞ ∞ ∞ !  z  A+Bz Y z/zn |A+Bz| Y 2|z/zn| |A+Bz| X −1 C|z| |f(z)| = e 1 − e ≤ e e = e exp 2|z| |zn| < e (2.34) z n=1 n n=1 n=1 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 1  s  ξ(s) = π−s/2s(s − 1)Γ ζ(s). (2.35) 2 2 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) 1 as |s| → ∞. By the symmetric version (1.52) of the functional equation, we know that ξ is an even function of s − 2 , 1 so we only need to consider Re(s) ≥ 2 .

We now check that each factor of ξ satisfies the bound. For the trivial factors, it is evident that 1 1 π−s/2s(s − 1) = e−(s/2) ln π e2 ln s − eln s  exp(|s| + ln |s|)  exp(|s|). (2.37) 2 2 For the Γ-factor, we can use Stirling’s formula, which we derived in (1.33) to be  1 1 ln Γ(s) = s − ln s − s + ln 2π + O(|s|−1) (2.38) 2 2

1 under the condition that |s| is large and | arg s| < π. Thanks to our assumption that Re(s) ≥ 2 , 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 given by the term s − 1  ln s. Hence, we have 2  s  ln Γ  |s| ln |s|, (2.39) 2

13 from where it follows that  s  Γ  exp(|s| ln |s|). (2.40) 2 It remains to estimate the ζ-factor. We do this by using the representation we obtained in (1.5) by partial summation,

s ∞ ζ(s) = − s {x}x−s−1 dx, (2.41) s − 1 ˆ1

1 which is valid for Re(s) > 0. In the region of interest, Re(s) ≥ 2 , 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

πt ζ(s) = ζ(1 − t) = 2(2π)−t cos Γ(t)ζ(t). (2.44) 2

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

πt π  cos = cos = 0, (2.45) 2 2 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(s − 1) cancels the simple poles of Γ 2 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

1 1 1 ξ(1) = π−1/2Γ lim(s − 1)ζ(s) = , (2.46) 2 2 s→1 2 which we can plug into the function by continuity at s = 1. By the functional equation (1.52), we deduce that 1 ξ(0) = ξ(1) = 2 6= 0. It now follows that

Y  s  ξ(s) = eA+Bs 1 − es/ρ, (2.47) ρ ρ 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

∞ X 1 ζ(s) = 1 + , (2.48) ns n=2 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)| < eC|s| found in (2.34) for any constant C. We conclude that P |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.

1 We also observe that the trivial zeros of the factor sζ(s) are cancelled by the poles of Γ( 2 s), and that the zero 1 −s/2 s  of the factor s − 1 is cancelled by the pole of ζ at s = 1. We have previously remarked that the factor 2 π Γ 2 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 1 many zeros in the critical strip. For curiosity, we note that ξ( 2 + it) is purely real.

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

ξ0(s) X  1 1 = B + + . (2.49) ξ(s) s − ρ ρ ρ By the definition in (1.51), we have 2  s  1  s  ζ(s) = πs/2Γ−1 ξ(s) = πs/2Γ−1 + 1 ξ(s), (2.50) s(s − 1) 2 (s − 1) 2

s  s  so we also get a product formula for ζ. Note that we have shifted from Γ 2 to Γ 2 + 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

0 s  ζ0(s) 1 1 1 Γ + 1 X  1 1 = B − + ln π − 2 + + . (2.51) ζ(s) s − 1 2 2 Γ s + 1 s − ρ ρ 2 ρ 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

0 s  ∞ 1 Γ + 1 1 X  1 1  − 2 = γ + − , (2.52) 2 Γ s + 1 2 s + 2n 2n 2 n=1 which exhibits the trivial zeros. Here,

n ! X 1 γ := lim − ln n n→∞ k k=1 n ! X 1 = 1 − lim ln n − n→∞ k k=2 n−1 n−1 ! X X 1 = 1 − lim (ln(k + 1) − ln k) − n→∞ k + 1 k=1 k=1 n−1 X  1  = 1 − lim ln(k + 1) − ln k − n→∞ k + 1 k=1 n−1 X  k  = 1 − lim ln(k + 1) − ln k + − 1 n→∞ k + 1 k=1 n−1 k+1 X  k  = 1 − lim ln x + n→∞ x k=1 k n−1 X k+1 x − k = 1 − lim dx n→∞ ˆ x2 k=1 k n x − bxc = 1 − lim 2 dx n→∞ ˆ1 x ∞ x − bxc = 1 − 2 dx ˆ1 x ∞ = 1 − {x}x−2 dx (2.53) ˆ1 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 we 1 1 previously noted that ξ(0) = 2 , so A = ln 2 . For B, we take the logarithmic derivative (2.49) and observe that ξ0(0) ξ0(1) = B = − , (2.54) ξ(0) ξ(1) where the latter equality follows from the functional equation (1.52). By the relation (2.50), we get

0 0 0 s  ξ (s) ζ (s) 1 1 1 Γ 2 + 1 = + − ln π + s  . (2.55) ξ(s) ζ(s) s − 1 2 2 Γ 2 + 1

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

0 3  1 Γ 2 1 − 3  = γ − 1 + ln 2, (2.56) 2 Γ 2 2 whence 1 1 ζ0(s) 1  B = γ − 1 + ln(4π) − lim + . (2.57) 2 2 s→1 ζ(s) s − 1 By the integral representation (1.5) of ζ, we have s ζ(s) = − sI(s), (2.58) s − 1 where ∞ I(s) := {x}x−s−1 dx. (2.59) ˆ1 Now, computation gives ζ0(s) 1  lim + = 1 − I(1) = γ (2.60) s→1 ζ(s) s − 1 by the definition of γ in (2.53). It follows that 1 1 B = − γ − 1 + ln(4π) ≈ −0.023, (2.61) 2 2 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 1 2α 2 + = ≤ , (2.62) ρ ρ α2 + β2 |ρ|2 whose sum we know converges. It follows from the logarithmic derivative (2.49) and the functional equation (1.52) that X  1 1 X  1 1 B + + = −B − + . (2.63) 1 − s − ρ ρ s − ρ ρ ρ ρ 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,

X 1 X α B = − = −2 . (2.64) ρ α2 + β2 ρ β>0

From this and the numerical value of B, we can deduce that |β| > 6 for all non-trivial zeros. Indeed, the smallest pair 1 of zeros is s ≈ 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 product formula, derived in (1.3) to be Y 1 ζ(s) = , (2.65) 1 − p−s p∈P and the Maclaurin series (2.19) for the natural logarithm, we have

∞ ∞ X X X p−sn X X p−nσ ln ζ(s) = − ln(1 − p−s) = = exp(−itn ln p), (2.66) n n p∈P p∈P n=1 p∈P n=1 for σ > 1, where we used that |p−s| < 1. It follows that

∞ X X p−nσ Re ln ζ(s) = cos(t ln(pn)). (2.67) n p∈P n=1 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)

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 1 ζ(σ) ∼ . (2.71) σ − 1 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 by ( ln p if n = pm for some p ∈ and m ∈ , Λ(n) := P Z>0 (2.73) 0 otherwise.

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 X ln ζ(s) = − ln 1 − p−s . (2.74) p∈P Now, by taking the derivative and identifying the infinite geometric series, we obtain

∞ ∞ ζ0(s) X (ln p)p−s X  1  X X X X ln p X Λ(n) − = = (ln p) − 1 = (ln p) p−sn = = . (2.75) ζ(s) 1 − p−s 1 − p−s pms ns p∈P p∈P p∈P n=1 p∈P m∈Z>0 n=1 This formula will prove very useful to our future work. It is also instrumental to observe that X Λ(d) = ln n (2.76) d|n for σ > 1. To derive this result, we multiply the formula (2.75) by ζ(s) to get

∞ ∞ ∞ ∞ ∞ ! ∞ ! ∞ X 1 X X 1 X X X Λ(m) X 1 X Λ(n)  ζ0(s) X ln n Λ(d) = Λ(m) = = = ζ(s) − = . ns ns ksms ns ns ζ(s) ns n=1 d|n n=1 km=n k=1 m=1 n=1 n=1 n=1 (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) X Λ(n) X Λ(n) − = = exp(−it ln n). (2.78) ζ(s) ns nσ n=1 n=1 Taking the real part, we get ∞ ζ0(s) X Λ(n) − Re = cos(t ln n). (2.79) ζ(s) ns n=1 Once again, we apply the identity (2.68). By the same token, we have

 ζ0(σ)  ζ0(σ + it)  ζ0(σ + 2it) 3 − + 4 −Re + −Re ≥ 0 (2.80) ζ(σ) ζ(σ + it) ζ(σ + 2it)

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 a real number C such that ζ0(σ) 1 − < + C, (2.81) ζ(σ) σ − 1 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) 1 1 1 Γ0( 1 s + 1) X  1 1 − = − B − ln π + 2 − + , (2.82) ζ(s) s − 1 2 2 Γ( 1 s + 1) s − ρ ρ 2 ρ 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 , which we find in appendix A.2 to be

Γ0(s) 1 = ln s − + O(|s|−2), (2.83) Γ(s) 2s we conclude that the gamma term is O(ln |t|). It follows that

ζ0(s) X  1 1 − Re < O(ln |t|) − Re + . (2.84) ζ(s) s − ρ ρ ρ

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

 1  σ − α Re = > 0 (2.85) s − ρ |s − ρ|2

and 1 α Re = > 0, (2.86) ρ |ρ|2 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

ζ0(σ + 2it) − Re < O(ln |t|). (2.87) ζ(σ + 2it)

By (2.64), we know that |β| > 6. Thus, if we wish, we can choose t to coincide with β without leaving the region 1 |t| ≥ 2. We may also pick out just the one term s−ρ in the sum which corresponds to this zero, and leave out the rest. Putting s = σ + it, we get ζ0(σ + it) 1 − Re < O(ln |t|) − . (2.88) ζ(σ + it) σ − α Substituting these upper bounds in the inequality (2.80) gives 4 3 < + O(ln |t|). (2.89) σ − α σ − 1

1 Write α = 1 − δ, where 0 < δ ≤ 4 is near zero, and take σ = 1 + 4δ. This yields 1 < O(ln |t|), (2.90) 20δ which is equivalent to 1 δ  . (2.91) ln |t| By the definition of δ, it follows that there exists a positive number C such that C α < 1 − . (2.92) ln |t|

Hence, we have proven that C σ ≥ 1 − (2.93) ln |t| defines a zero-free region for ζ for |t| ≥ 2. Since we know that |β| > 6, it is irrelevant to consider this statement 1 outside this region. Furthermore, it can be shown (cf. [Kadiri, 2005, p. 2]) that the inequality holds for C = 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 from C C 1 the inequality ln |t| < α < 1 − ln |t| with C = 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 information about 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 we would call n(T ) in the context of the product formula for ξ. For the sake of simplicity, we choose to work with the function ξ instead of ζ due to its convenient functional equation; as we ascertained earlier, both functions have the same in the open critical strip. To avoid complications, we also choose T so that there are no zeros on 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 without zeros or poles on C. Then, the argument principle states that

f 0(z) dz = 2πi(Nf − Pf ), (2.94) ‰C f(z)

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

f 0(z)  d  dz = ln f(z) dz = d(Ln |f(z)| + i arg f(z)) = i∆C arg f(z), (2.95) ‰C f(z) ‰C dz ‰C

where ∆C arg f(z) stands for the total variation of the argument of f as z travels around C. The last equality is true because the single-valued function Ln |f(z)| has zero net excursion on any closed curve. Note that we capitalized Ln to point out that we deal with the principal value of the logarithm; we will use this convention when we want to emphasize 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 depicted in 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 zero on the real line, so there is no contribution to ∆R arg ξ(s) from the base of the rectangle. Furthermore, because of the 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)

1 It follows that the variation of the argument of ξ for the path at the top-right corner (from 2 through 2+iT to 2 +iT ) 1 is the same as the path at the top-left corner (from 2 + iT through −1 + iT to −1). We denote the former path by Q and conclude that πN(T ) = ∆Q arg ξ(s). (2.99) Recalling the definition in (1.51), we see that we can write 1  s   s  ξ(s) = π−s/2s(s − 1)Γ ζ(s) = π−s/2(s − 1)Γ + 1 ζ(s), (2.100) 2 2 2 because s  s   s  Γ = Γ + 1 . (2.101) 2 2 2 For curiosity, we observe that ∞ 1 X 1 π ξ(2) = π−11Γ(2)ζ(2) = = (2.102) π n2 6 n=1 by Euler’s solution to the Basel problem. Seeing that the argument of a product is the sum of the arguments, we now decide to study the change in the argument of each factor of ξ as s moves along Q. For the first two factors, we have

 s   t  T ∆ arg π−s/2 = ∆ arg exp − ln π = ∆ − ln π = − ln π (2.103) Q Q 2 Q 2 2 and  1  π ∆ arg(s − 1) = arg − + iT − arg 1 = π + arctan(−2T ) = + O(T −1). (2.104) Q 2 2

20 Using Stirling’s formula (1.33) for the Γ-factor, we get  s   s  ∆ arg Γ + 1 = ∆ Im ln Γ + 1 Q 2 Q 2 5 T  = Im ln Γ + i − Im ln Γ(2) 4 2 3 T  5 T  5 T 1  = Im + i ln + i − − i + ln(2π) + O(T −1) 4 2 4 2 4 2 2   3 5 T T 5 T T −1 = arg + i + ln + i − + O(T ) 4 4 2 2 4 2 2 3π T T T = + ln − + O(T −1). (2.105) 8 2 2 2 Collecting the variations, we find that 1   s   N(T ) = ∆ arg π−s/2 + arg(s − 1) + arg Γ + 1 + arg ζ(s) π Q 2 T T T 7 1 = ln − + + O(T −1) + ∆ arg ζ(s). (2.106) 2π 2π 2π 8 π Q 7 The term 8 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 X 1 = O(ln T ), (2.107) 1 + (T − β)2 ρ where ρ = α + iβ runs through the non-trivial zeros in the critical strip. To prove this lemma, we first recall that

ζ0(s) X  1 1 − Re < O(ln |t|) − Re + (2.108) ζ(s) s − ρ ρ ρ for 1 ≤ σ ≤ 2 and |t| ≥ 2; we derived this in (2.84) from the partial-fraction decomposition of the logarithmic derivative of ζ. Putting s = 2 + iT in this inequality, we obtain

X  1 1 Re + < O(ln T ), (2.109) 2 + iT − ρ ρ ρ since the logarithmic derivative of ζ is uniformly bounded in this neighborhood, because by (2.75), we have

0 ∞ ∞ √ ∞   ζ (s) X Λ(n) X n X 1 1 − = < = = ζ σ − , (2.110) ζ(s) ns |ns| nσ−1/2 2 n=1 n=1 n=1

3 which is obviously bounded for σ > 2 . 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  1  2 − α 1 1 1 1 1 Re = ≥ = > , (2.111) 2 2 2 1 2 2 2 + iT − ρ (2 − α) + (T − β) 4 + (T − β) 4 1 + 4 (T − β) 4 1 + (T − β) 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 most O(ln T ) non-trivial zeros. Secondly, we find that the sum over the non-trivial zeros with |T − β| ≥ 1 is also

X 1 X 1 + (T − β)2 1 X 1 = ≤ 2 = O(ln T ). (2.112) (T − β)2 (T − β)2 1 + (T − β)2 1 + (T − β)2 ρ ρ ρ

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

ζ0(σ + iT ) ζ0(2 + iT ) X  1 1  − = − + O(1). (2.113) ζ(σ + iT ) ζ(2 + iT ) σ + iT − ρ 2 + iT − ρ ρ

By the estimate (2.110), we have ζ0(2 + iT ) = O(1). (2.114) ζ(2 + iT )

21 1 In the sum on the right-hand side of (2.113), we first consider the terms of the form 2+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(ln T ). As for the terms with |t − β| ≥ 1 of any form, we have

1 1 2 − σ 3 − = ≤ , (2.115) σ + iT − ρ 2 + iT − ρ |(σ + iT − ρ)(2 + iT − ρ)| |T − β|2 so by the second immediate consequence of (2.107), we find that the sum of these is also O(ln T ). It follows that

ζ0(s) X 1 = + O(ln T ) (2.116) ζ(s) s − ρ ρ

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 by 1 V and the horizontal path from 2 + iT to 2 + iT by L. On the former path, we observe that Reζ(2 + it) > 0, because

∞ X 1 ζ(2 + it) = 1 + , (2.117) n2+it n=2 where ∞ ∞ X 1 X 1 π2 ≤ = − 1 < 1. (2.118) n2+it n2 6 n=2 n=2 Therefore, the variation along the vertical line σ = 2 contributes O(1). Recalling the argument principle, we find that

1 +iT 2 ζ0(s) ∆Q arg ζ(s) = ∆V arg ζ(s) + Im(i∆L arg ζ(s)) = O(1) + Im ds. (2.119) ˆ2+iT ζ(s) On the other hand, we also have 1 +iT 2 1 Im ds = ∆L arg(s − ρ). (2.120) ˆ2+iT s − ρ 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(ln T ) by the first immediate consequence of (2.107), we conclude that ∆Q arg ζ(s) = O(ln T ). We finally get the asymptotic relation T T T N(T ) = ln − + O(ln T ). (2.121) 2π 2π 2π Incidentally, if we enumerate the non-trivial zeros with β > 0 in order of increasing ordinate, we get n β ∼ 2π (2.122) n ln n 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  0 if 0 < y < 1, 1 c+i∞ ys  δ(y) := ds = 1 if y = 1, (3.1) 2πi ˆ s 2 c−i∞ 1 if y > 1, where the not absolutely convergent integral is regarded as a principal value in the sense that

c+i∞ ys c+iT ys ds := lim ds (3.2) ˆc−i∞ s T →∞ ˆc−iT s and c > 0 is arbitrary. This identity, which is called Perron’s formula, allows one to pick up the terms with n ≤ x in any Dirichlet series. We note that the integral is independent of c (as long as c > 0), which is essential. We also observe that the integral can be interpreted as a unit step function that takes the average value at the point of discontinuity. 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 extending to the right. For y > 1, we regard the integral as one side of a rectangle extending to the left, where we pick up the pole 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 c+iT 1 1 T dt ds = 2πi ˆc−iT s 2π ˆ−T c + it 1 T  1 1  = + dt 2π ˆ0 c + it c − it 1 T 2c = 2 2 dt 2π ˆ0 c + t 1   t T = 2 arctan 2π c 0 1 T  1 = arctan −−−−→ . (3.3) π c T →∞ 2

23 We move on to the case 0 < y < 1. We want to shift the line of integration to the right, so the integrand becomes small. By the residue theorem, we can write ! c+iT ys d−iT d+iT d+iT ys ds = + − ds, (3.4) ˆc−iT s ˆc−iT ˆd−iT ˆc+iT s where d > c > 0, because the integrand has no singularities to the right of the vertical line σ = c. We now estimate the three integrals on the right-hand side. We find that the horizontal integrals satisfy

d±iT ys d±iT ys d yσ 1 ∞ 1  yσ ∞ yc σ ds ≤ |ds| ≤ dσ ≤ y dσ = = , (3.5) ˆc±iT s ˆc±iT s ˆc T T ˆc T ln y c T | ln y| while the vertical integral satisfies

d+iT ys d+iT ys T yd   t T T  ds ≤ |ds| = √ dt = yd arsinh = 2yd arsinh . (3.6) 2 2 ˆd−iT s ˆd−iT s ˆ−T d + t d −T d

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

c+iT ys 2yc

ds ≤ . (3.7) ˆc−iT s T | ln y| Now, letting T → ∞, we get 1 c+iT ys ds −−−−→ 0. (3.8) 2πi ˆc−iT s T →∞ We continue with the last case y > 1. This time, we want to shift the line of integration to the left, so the integrand becomes small. By the residue theorem, we can write ! c+iT ys c−iT −d+iT c+iT ys ds = − + + ds + 2πi, (3.9) ˆc−iT s ˆ−d−iT ˆ−d−iT ˆ−d+iT s where d > c > 0, because the integrand has a simple pole at s = 0 with residue

ys  ys Res , 0 = lim s = y0 = 1. (3.10) s s→0 s

By the same token, we find that the horizontal integrals satisfy

c±iT ys yc

ds ≤ , (3.11) ˆ−d±iT s T | ln y| while the vertical integral satisfies

−d+iT ys T  −d ds ≤ 2y arsinh . (3.12) ˆ−d−iT s d Since y−d → 0 as d → ∞ for y > 1, it follows that

1 c+iT ys ds −−−−→ 1, (3.13) 2πi ˆc−iT s T →∞ 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 a conditionally convergent integral. Since this is not always permissible, we consider the truncated integral

1 c+iT ys δ(y, T ) := ds, (3.14) 2πi ˆc−iT s which legitimizes the manipulation, but at the same time introduces an error term. We claim that  0 + O(yc min(1,T −1| ln y|−1)) if 0 < y < 1,  1 −1 δ(y, T ) = 2 + O(cT ) if y = 1, (3.15) 1 + O(yc min(1,T −1| ln y|−1)) if y > 1,

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

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

1 c+iT 1 1 T 2c  t  1 T/c du 1 1 ∞ du : δ(1,T ) = ds = 2 2 dt = u = = 2 = − 2 , (3.16) 2πi ˆc−iT s 2π ˆ0 c + t c π ˆ0 1 + u 2 π ˆT/c 1 + u where ∞ du ∞ du  1 ∞ c 2 < 2 = − = . (3.17) ˆT/c 1 + u ˆT/c u u T/c T Hence, 1  c  δ(1,T ) = + O . (3.18) 2 T We move on to the case 0 < y < 1. By (3.7), we have

1 c+iT ys 2yc

|δ(y, T )| = ds ≤ . (3.19) 2πi ˆc−iT s T | ln y| It follows that  yc  δ(y, T ) = O , (3.20) T | ln y| which proves the first inequality.√ For the second inequality, we replace the extended rectangle used in (3.7) by a circular 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

s c 1 y 1 y 1 c |δ(y, T )| ≤ max `(C) = πR = y (3.21) 2π s∈C s 2π R 2 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 arc to 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 the distribution of prime numbers. Let the Chebyshev function be defined by X X ψ(x) := Λ(n) = ln p, (3.23) n≤x pm≤x

which can also be interpreted as the logarithm of the least common multiple of all positive integers up to bxc. To mitigate the jump discontinuities at the prime powers, we modify the function as follows:   ( 1 m 1 X X ψ(x) − 2 Λ(x) if x = p for some prime p and positive integer m, ψ0(x) :=  Λ(n) + Λ(n) = (3.24) 2 ψ(x) otherwise. n≤x n

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

∞ X Λ(n) ζ0(s) = − , (3.25) ns ζ(s) n=1 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:   1 X X ψ (x) = Λ(n) + Λ(n) 0 2   n≤x n 1.

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

∞ X  x c  1  c R(x, T )  Λ(n) min 1, + Λ(x) . (3.27) n T | ln x | T n=1, n6=x n

We observe that every variable and function that appears in the sum is non-negative. To avoid technicalities, we only 1 c consider x > 1. We choose c = 1 + ln x > 1 to ensure that x = ex  x. We also recall that Λ(n) ≤ ln n. It follows that ∞ X Λ(n)x  1  ln x R(x, T )  min 1, + . (3.28) nc T | ln x | T n=1, n6=x n 3 4 We now subdivide all terms into three different groups. We first consider the terms with n ≤ 4 x or n ≥ 3 x. For these x 4 x 3 x 4 terms, we have n ≥ 3 or n ≤ 4 , whence | ln n | ≥ | ln 3 | > 0 has a positive lower bound. We also recollect that ζ0(σ) −1 ∼ (3.29) ζ(σ) σ − 1 near the simple pole at s = 1. It follows from (3.28) that the contribution of these terms is

∞ " 0 1 # x X Λ(n) x  ζ0(c) x ζ 1 + x  = − = − ln x  ln x. (3.30) T nc T ζ(c) T ζ 1 + 1  T n=1 ln x

We now consider the terms with 3 x < n < x. We note that this set vanishes if it does not contain a prime power. 4 Therefore, let us assume that there exists a greatest prime power q in this set. Since x−q < x/4 < 1 , we see from x x 4 the Maclaurin series (2.19) for the natural logarithm that

x q  x − q  x − q ln = − ln = − ln 1 − ≥ . (3.31) q x x x

Hence, by (3.28), this term contributes

x  x   x   Λ(q) min 1,  (ln x) min 1, . (3.32) qc T (x − q) T (x − q)

1 For the other prime power terms in this set, we can write n = q − m for some integer 0 < m < 4 x. By the same token, x q q − m  m m ln ≥ ln = − ln = − ln 1 − ≥ . (3.33) n q − m q q q

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

d 1 x−1e d 1 x−1e 4X x  q  4X q x ln2 x  Λ(q − m) min 1,  (ln x)  , (3.34) (q − m)c T m T m T m=1 m=1

26 since n X 1 Γ0(n) = + γ  ln n. (3.35) m Γ(n) m=1 4 The terms with x < n < 3 x 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 hxi. Summing up the estimates, we conclude that the total error is

x ln2 x  x  R(x, T )  + (ln x) min 1, . (3.36) T T hxi

It follows that 1 c+iT  ζ0(s) xs x ln2 x  x  ψ0(x) = − ds + O + (ln x) min 1, , (3.37) 2πi ˆc−iT ζ(s) s T T hxi which will be our next starting point.

3.3 von Mangoldt’s explicit formula

Let us take a closer look at the formula 1 c+iT  ζ0(s) xs ψ0(x) = − ds + R(x, T ), (3.38) 2πi ˆc−iT ζ(s) s where R is the error term (3.36) and x > 1. We want to compute the finite integral. To this end, we regard the integral as one side of a rectangle extending to the left. The reason we want to move the vertical line of integration away 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 its logarithmic derivative on the resulting contour and predict where its zeros and poles may be encountered. Indeed, the main obstacle in carrying out this integration is the difficulty to choose T so that the horizontal sides of the rectangle avoid 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 integral as the sum of the residues of the integrand at its poles. We recall that each singularity of the logarithmic derivative of a meromorphic function is a simple pole, where the residue is simply the multiplicity of the corresponding zero, or minus the order of the corresponding pole, of the original function. We also note that the exponential function has no poles.

1 We first observe that the simple pole of s at s = 0 contributes  ζ0(s) xs   ζ0(s) xs ζ0(0) Res − , 0 = lim s − = − = − ln 2π, (3.39) ζ(s) s s→0 ζ(s) s ζ(0) where the value of the constant follows from the partial-fraction decomposition (2.51) of the logarithmic derivative of ζ. We also see that each zero of ζ at s = ρ, whether trivial or not, contributes

 ζ0(s) xs   ζ0(s) xs xρ Res − , ρ = lim(s − ρ) − = − . (3.40) ζ(s) s s→ρ ζ(s) s ρ

Considering the ∞ 1 X x−2n ln(1 − x−2) = , (3.41) 2 −2n n=1 we notice that the contribution from the trivial zeros may be written in a more compact way. We finally note that the simple pole of ζ at s = 1 contributes

 ζ0(s) xs   ζ0(s) xs x1 Res − , 1 = lim(s − 1) − = = x, (3.42) ζ(s) s s→1 ζ(s) s 1 where we changed sign because it is a pole.

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

X xρ 1 ψ (x) = x − − ln(1 − x−2) − ln 2π, (3.43) 0 ρ 2 ρ

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

X xρ X xρ := lim . (3.44) ρ T →∞ ρ ρ |Im(ρ)|

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

1 c+iT  ζ0(s) xs ψ0(x) = − ds + R(x, T ). (3.45) 2πi ˆc−iT ζ(s) s Let us regard the vertical line of integration as one side of a rectangle extending to the left, as illustrated in Figure 3.2. In this way, it follows from the residue theorem that

U ! d 2 −1e 1 c−iT −U+iT c+iT  ζ0(s) xs X xρ X x−2n ψ (x) = − + + − ds+x− − −ln 2π+R(x, T ), (3.46) 0 2πi ˆ ˆ ˆ ζ(s) s ρ −2n −U−iT −U−iT −U+iT |β|

where U > 0 and we write ρ = α + β. The contributions to the sum of the residues of the integrand at its poles inside the 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 . We make 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 left vertical side of the rectangle passes halfway between two trivial zeros, which turns out to be good choice. The choice of T demands more care. By the first immediate consequence of (2.107), we know that the number of non-trivial zeros in the horizontal strip T − 1 < β < T + 1 is  ln T for any large T . It follows that there exists a zero-free substrip of 1 height  ln T . Thus, by varying T by at most 1, we can guarantee that 1 |T − β|  . (3.47) ln T for every non-trivial zero.

We recall that we deduced in (2.116) that

ζ0(s) X 1 = + O(ln T ) (3.48) ζ(s) s − ρ |T −β|<1 for s ∈ {σ + iT | − 1 ≤ σ ≤ 2}. By the bound (3.47), we know that each term in the sum is O(ln T ). But the number of terms in the sum is also O(ln T ), whence ζ0(s) = O(ln2 T ) (3.49) ζ(s)

28 for −1 ≤ σ ≤ 2. We also recollect from (2.110) that

0     ζ (s) 1 3 − < ζ σ − ≤ ζ , (3.50) ζ(s) 2 2 for σ ≥ 2, which is negligible in comparison to O(ln2 T ). Hence, the contribution from the horizontal integrals in this range is

c±iT  0  s c±iT s c σ 2 c 2  σ c 2 ζ (s) x 2 x 2 x ln T σ ln T x x ln T − ds  ln T ds  ln T dσ  x dσ =  , ˆ−1±iT ζ(s) s ˆ−1±iT s ˆ−1 T T ˆ−∞ T ln x −∞ T ln x (3.51) 1 since c = 1 + ln x .

0 It remains to estimate ζ (s) for −U ≤ σ ≤ −1. Fortunately, this is easy to do. Since we only consider σ ≤ −1, ζ(s) 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 πw  ζ(s) = ζ(1 − w) = 2(2π)−w cos Γ(w)ζ(w). (3.52) 2 Taking the logarithmic derivative, we find that

ζ0(s) ζ0(1 − w) π πw  Γ0(w) ζ0(w) = − = − ln 2π − tan + + , (3.53) ζ(s) ζ(1 − w) 2 2 Γ(w) ζ(w) where we used the product rule for differentiation. The first term is a constant that can be dropped. Since the π function tan z has singularities at z = 2 (1 + 2n), where n is any integer, we see that the second term is bounded 1 1 if, say, |w − (1 + 2n)| ≥ 2 , which is equivalent to |(1 − w) − (−2n)| ≥ 2 . Hence, the second term is bounded when 1 s = 1 − w lies outside every circle of radius 2 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 is bounded by (3.50). It follows that ζ0(s) = O(ln 2|s|), (3.54) ζ(s) whence we see that the contribution from the horizontal integrals in this range is

−1±iT  0  s −1±iT s −1 σ −1  σ −1 ζ (s) x x x ln T σ ln T x ln T − ds  ln 2T ds  ln T dσ  x dσ = = , ˆ−U±iT ζ(s) s ˆ−U±iT s ˆ−U T T ˆ−∞ T ln x −∞ T x ln x (3.55) which is negligible in comparison to the contribution from (3.51). We also see that the contribution from the vertical integral is −U+iT  0  s −U+iT s T −U ζ (s) x x x T ln U − ds  ln 2U ds  ln U dt  U , (3.56) ˆ−U−iT ζ(s) s ˆ−U−iT s ˆ−T U Ux which vanishes as U → ∞.

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

X xρ 1 ψ (x) = x − − ln(1 − x−2) − ln 2π + R(x, T ), (3.57) 0 ρ 2 |β|

where x ln2(xT )  x  R(x, T )  + (ln x) min 1, . (3.58) T T hxi Making T → ∞ at constant x > 1, we see that R(x, T ) → 0, which proves the explicit formula (3.43). Note that the convergence 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. By P xρ varying T by a bounded amount, we add or remove O(ln T ) terms to the sum ρ . Since each term in the sum is x x ln T O( T ), the total variation is O( T ), 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 integral by x dt Li(x) := . (3.60) ˆ2 ln t 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 Vall´ee Poussin [de la Vall´eePoussin, 1896, pp. 183-256]. Another weaker version of the theorem states that x π(x) ∼ , (3.62) ln x but it can be shown that ∞ x X n! Li(x) ∼ (3.63) ln x (ln x)n n=0 is a better approximation. This is also the version we will prove. We shall conclude this paper by relating this theorem to the Riemann hypothesis, which significantly improves the error term in (3.61).

We first show that ψ(x) ∼ x, (3.64) which is not unlike the prime number theorem in the sense that an arithmetic function is approximated by a simple expression. We recall from (3.57) that

X xρ 1 ψ (x) = x − − ln(1 − x−2) − ln 2π + R(x, T ), (3.65) 0 ρ 2 |β|

where x ln2(xT )  x  R(x, T )  + (ln x) min 1, . (3.66) T T hxi We note that this formula is suggestive of the statement we wish to prove, still we need to find an estimate for the sum of the non-trivial zeros. To this end, we use the fact that the real part of a non-trivial zero cannot be arbitrarily close to 1. In particular, we proved in (2.92) that there exists a positive number C such that C α < 1 − . (3.67) ln T for every non-trivial zero with |β| < T . It follows that   ρ α 1− C ln x |x | = x < x ln T = x exp −C . (3.68) ln T

P 1 We now consider the sum |ρ| . By (2.64), we know that |β| > 6. Hence,

X 1 X 1 X 1 X 1 = < = 2 . (3.69) |ρ| |ρ| |β| β |β|

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

T X 1 T dN(t) 2N(T ) T N(t) dt T ln t dt ln2 t 2 = 2 = + 2  ln T + = ln T +  ln2 T, (3.70) β ˆ t T ˆ t2 ˆ t 2 6<β

X xρ  ln x   x(ln2 T ) exp −C . (3.71) ρ ln T |β|

30 Without loss of generality, we may assume that x is an integer that is not a prime power. In this case, we have hxi ≥ 1, so that x ln2(xT ) R(x, T )  (3.72) T by the estimate (3.58). It follows that x ln2(xT )  ln x  ψ(x) − x  + x(ln2 T ) exp −C . (3.73) T ln T √ We choose T = e ln x, so that ln2 T = ln x. Insertion in (3.73) gives  √   √  ψ(x) − x  x(ln2 x) exp − ln x + x(ln x) exp −C ln x . (3.74) √  ln x Let C1 = min(1,C) − 2 for some  > 0. Since ln x  e , we find that   √  ψ(x) = x + O x exp −C1 ln x , (3.75) which is what we wanted to show.

We now prove the equivalent result for π. We note that integration by parts yields x dt  t x x  1  x 2 x dt x x t dt Li(x) = = − t d = − + 2 = + 2 + O(1). (3.76) ˆ2 ln t ln t 2 ˆ2 ln t ln x ln 2 ˆ2 ln t ln x ˆ2 t ln t On the other hand, if we let X Λ(n) π (x) := , (3.77) 1 ln n n≤x we find that x X  1  1   1 X X x dt ψ(x) x ψ(t) dt π1(x) = Λ(n) + − = Λ(n) + Λ(n) = + . (3.78) ln x ln t ln x ˆ t ln2 t ln x ˆ t ln2 t n≤x n n≤x n≤x n 2 by the definition of ψ. The last step can be justified by   X x dt X x dt X m+1 X dt x ψ(t) dt Λ(n) = Λ(n) = Λ(n) = . (3.79) ˆ 2 ˆ 2 ˆ 2  ˆ 2 n≤x n t ln t 2≤n

∞ blb xc blb xc  X 1 √ X 1 √ X √ √ π (x) = π( m x) = π( m x) = π(x) + O m x = π(x) + O( x). (3.81) 1 m m   m=1 m=1 m=2 √ √ √ √ √ 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 lb x = log2 x. Thus, we have ψ(x) x ψ(t) dt √ π(x) = + 2 + O( x). (3.82) ln x ˆ2 t ln t We now replace ψ in (3.82) by the approximation (3.75), obtaining √ 4 !  √  x x  √  π(x) − Li(x)  x exp −C1 ln x + + √ exp −C1 ln t dt ˆ2 ˆ 4 x

√4  √  x x  1 √   x exp −C1 ln x + 1 dt + √ exp − C1 ln x dt ˆ2 ˆ 4 x 2  √  √  1 √   x exp −C ln x + 4 x + x exp − C ln x (3.83) 1 2 1 √ √ √ 1 4 In the second step, we used the fact that ln t ≥ 2 ln x for x ≤ t ≤ x. Note that the constant C1 is the same as in (3.75). It follows that   1 √  π(x) = Li(x) + O x exp − C ln x , (3.84) 2 1 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 that due 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

1 It is time to involve the Riemann hypothesis to the matter. Suppose that there exists a number 2 ≤ 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 P 1 2 |ρ| over zeros with |β| < T is O(ln T ), we find that

x ln2(xT ) ψ(x) − x  + xθ ln2 T (3.85) T √ 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)

whence ψ(x) = x + O(xθ ln2 x). (3.87) By the relation (3.82) and the estimate (3.87), it follows that

ψ(x) x ψ(t) dt √ π(x) = + 2 + O( x) ln x ˆ2 t ln t x x x θ dt θ−1 √ = + O(x ln x) + 2 + O(t ) dt + O( x) ln x ˆ2 ln t ˆ2 x  t x √ = + O(xθ ln x) + Li(t) − + O(xθ) + O( x) ln x ln t 2 = Li(x) + O(xθ ln x), (3.88)

1 since θ ≥ 2 . Note that the error term is lowered by a factor of ln x. In particular, the assumption of the Riemann hypothesis implies that √ π(x) = Li(x) + O( x ln x), (3.89) which is a significant improvement over (3.84).

1 We now prove the converse. Suppose that there exists a number 2 ≤ 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

∞ ζ0(s) X Λ(n) ∞ − = = x−s dψ(x) (3.91) ζ(s) ns ˆ n=1 1 for σ > 1. Integrating by parts, we get

0 ∞ ∞ ζ (s)  −s ∞ −s −s−1 − = x ψ(x) 1 − ψ(x) d(x ) = s ψ(x)x dx, (3.92) ζ(s) ˆ1 ˆ1 where we used that ψ(1) = 0 and ψ(x) = O(x). Rearranging the integrand, we deduce that

ζ0(s) ∞ ∞ ∞ s − = s (ψ(x) − x)x−s−1 dx + s x−s dx = s (ψ(x) − x)x−s−1 dx + (3.93) ζ(s) ˆ1 ˆ1 ˆ1 s − 1 for σ > 1. But we assumed that ψ(x) − x  xθ, so the resulting integral extends to an holomorphic function for σ > θ. Thus, ζ cannot have zeros there, and hence we have α ≤ θ for all zeros. We also note that the same conclusion holds when we assume that π(x) = Li(x) + O(xθ). It follows that our best estimate (3.89) is valid if and only if the Riemann hypothesis holds true. Hence, the Riemann hypothesis is equivalent to the smallest possible error term in the prime number theorem (3.61).

32 Appendices

33 A. Additional proofs

A.1 The Weierstrass form of Γ

Let us consider the product ∞ 1 Y  s −1 g(s) := 1 + es/n. (A.1) s n n=1 We first observe that ∞ ∞ X   s  s  X  s 2 ln g(s) = − ln s + − ln 1 + +  − ln s + , (A.2) n n n n=1 n=1 so the product in (A.1) converges absolutely and uniformly in any compact subset of C \{0, −1, −2,...}. We further Γ note that g has simple poles at the non-positive integers and no zeros, just like Γ. It follows that g is an entire Γ f function with no zeros. Thus, we can write g ≡ e for some entire function f. We wish to show that f(s) = −γs, where γ is the Euler–Mascheroni constant, for then

∞ e−γs Y  s −1 Γ(s) = 1 + es/n, (A.3) s n n=1 which is the Weierstrass form of Γ.

We now observe that

N g(s + 1) s Y 1 + s = lim n e1/n g(s) N→∞ s + 1 1 + s+1 n=1 n N ! N ! s Y n + s X 1 = lim exp N→∞ s + 1 n + s + 1 n n=1 n=1 N ! N X 1 = lim s · · exp − ln N + N→∞ N + s + 1 n n=1 = 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) = seγ g(s). This looks similar to Γ(s + 1) = sΓ(s), which we recall from (1.10).

We now introduce the entire function Γ(s) q(s) := (A.5) e−γsg(s) and note that Γ(s + 1) sΓ(s) Γ(s) q(s + 1) = = = = q(s), (A.6) e−γ(s+1)g(s + 1) e−γ(s+1)seγ g(s) e−γsg(s) so q is periodic with period 1. We recollect that we want to show that f(s) = −γs, which is equivalent to q ≡ 1 by the definition of f. We also observe that

Γ(s) lim sΓ(s) 1 lim q(0) = lim = s→0 = = 1 (A.7) s→0 s→0 e−γsg(s) lim se−γsg(s) 1 s→0

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

∞ σ g(σ + it) |σ| Y 1 + n = g(σ) |σ + it| 1 + σ+it n=1 n ∞ Y |n + σ| = |n + σ + it| n=0 ∞ p ! X (n + σ)2 = exp ln p 2 2 n=0 (n + σ) + t ∞ ! 1 X  t2  = exp − ln 1 + . (A.8) 2 (n + σ)2 n=0 We notice that the summand decreases with increasing n + σ. It follows that

∞ X  t2  ∞  t2  ∞  1  ∞ dx ln 1 + ≤ ln 1 + dx = |t| ln 1 + dx = 2|t| = π|t|. (A.9) (n + σ)2 ˆ x2 ˆ x2 ˆ 1 + x2 n=0 0 0 0 by integration by parts. We conclude that |g(σ + it)| ≥ |g(σ)|e−π|t|/2. We also note that

∞ ∞ ∞ σ+it−1 −x σ+it−1 −x σ−1 −x |Γ(σ + it)| = x e dx ≤ x e dx = x e dx = Γ(σ) (A.10) ˆ0 ˆ0 ˆ0 by the integral definition of Γ in (1.8), which is valid for σ > 0. We find that

Γ(σ + it) Γ(σ) π|t|/2 |q(σ + it)| = ≤ = |q(σ)|e (A.11) e−γ(σ+it)g(σ + it) e−γσg(σ)e−π|t|/2 for σ ≥ 1.

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

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

Now, we consider the function q(s − 1 ) − q(0) p(s) := 4 ; (A.13) sin(2πs) − 1 1 it is entire, since the zeros of the denominator at s = 4 + 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) → 0 π|t|/2 1 2πt −2πt as |t| → ∞ at |σ| ≤ B, because |q(σ + it)| ≤ Ce and | sin(2πs)| ≥ 2 e − e . 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 be that 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 the definition of the Euler–Mascheroni constant γ in (2.53), we have

∞ e−γs Y  s −1 Γ(s) = 1 + es/n s n n=1 N !! N ! N 1 X 1 X 1 Y n = lim exp −s − ln N exp s N→∞ s n n n + s n=1 n=1 n=1 N 1 Y n = lim N s . (A.14) N→∞ s n + s n=1

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

35 s Γ0(z) of the poles of Γ, we can define an holomorphic branch of ln Γ there by 1 Γ(z) dz [Bak and Newman, 1996, p. 100], which is real on the positive real axis. Taking this logarithm, we get ´

N N ! N ! X X X ln Γ(s) = lim − ln s + s ln N + ln n − ln(n + s) = lim s ln N + ln N! − ln(n + s) . (A.15) N→∞ N→∞ n=1 n=1 n=0 We recall the well-known Stirling’s formula for the factorial, which can be formulated as follows: √ N! = 2πe−N N N+1/2 1 + O(N −1) , (A.16)

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

 1 1 ln N! = N + ln N − N + ln 2π + O(N −1), (A.17) 2 2

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

N !  1 1 X ln Γ(s) = lim s + N + ln N − N + ln 2π − ln(n + s) + O(N −1) . (A.18) N→∞ 2 2 n=0

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

1 1 ! n+ 2 n n+ 2 f(x) dx = + 1 · f(x) dx ˆ 1 ˆ 1 ˆ n− 2 n− 2 n  1 n n  1 = x − n + f(x) − x − n + f 0(x) dx 2 1 ˆn− 1 2 n− 2 2 n+ 1 n+ 1  1  2 2  1 + x − n − f(x) − x − n − f 0(x) dx 2 n ˆn 2 n n+ 1  1 2  1 = f(n) − x − n + f 0(x) dx − x − n − f 0(x) dx ˆ 1 2 ˆ 2 n− 2 n n 2 n+ 1 2 ! 1  1 2  1 = f(n) + x − n + f 00(x) dx + x − n − f 00(x) dx 2 ˆ 1 2 ˆ 2 n− 2 n n+ 1 2 1 2  1 = f(n) + x + f 00(x) dx (A.19) 2 ˆ 1 2 n− 2 where n is any integer and hxi is the distance from x to the nearest integer. Taking f(x) = ln(s + x) and 0 ≤ n ≤ N, it follows that N 1 1 N+ 2 N+ 2 1 2 X 1 hx + 2 i ln(s + n) = ln(s + x) dx − 2 dx, (A.20) ˆ 1 2 ˆ 1 (s + x) n=0 − 2 − 2 1 1 provided that |s| ≥ 1 (which we may assume in the limit |s| → ∞), so that s + x 6= 0 on the interval − 2 ≤ x ≤ N + 2 . Computing the first integral, we find that

1 N+ 2 1 N+ 2 ln(s + x) dx = [(s + x) ln(s + x) − x]− 1 ˆ 1 2 − 2  1  1  1  1 = s + N + ln s + N + − s − ln s − − N − 1 2 2 2 2  1   1  1  1  1 = s + N + ln N + ln 1 + s + − s − ln s − − N − 1. (A.21) 2 N 2 2 2

We also observe that 1 N+ 2 1 2 1 hx + 2 i −1 2 dx  |s| . (A.22) 2 ˆ 1 (s + x) − 2

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

  1  1  1  1  1 1  ln Γ(s) = lim − s + N + ln 1 + s + + s − ln s − + 1 + ln 2π + O(N −1 + |s|−1) N→∞ 2 N 2 2 2 2  1  1 1 1 = s − ln s − − s + + ln 2π + O(|s|−1) 2 2 2 2  1  1  1 1 = s − ln s − + O(|s|−2) − s + + ln 2π + O(|s|−1) 2 2s 2 2  1 1 = s − ln s − s + ln 2π + O(|s|−1) (A.23) 2 2 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 the derivative that we can differentiate the asymptotic expansion for ln Γ term by term [Erd´elyi,1956, p. 21], obtaining Γ0(s) 1 = ln s − + O(|s|−2). (A.24) Γ(s) 2s We also note that the logarithmic derivative of Γ is often called the digamma function ψ in the literature, but alas we cannot use this notation because it conflicts with the Chebyshev function defined in (3.23).

A.3 Jensen’s formula for holomorphic functions

m Suppose that f is holomorphic in a region which contains the closed disk |z| ≤ R, where R > 0. Let {zk}k=1 be the zeros 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

R 0 iθ iθ R   f (re )e d iθ iθ  iθ Re iθ dr = Re ln f(re ) dr = Re ln f(Re ) − ln f(0) = ln |f(Re )| − ln |f(0)|, (A.25) ˆ0 f(re ) ˆ0 dr where we used the fundamental theorem of calculus in the region of holomorphicity of f. On the other hand, integrating over the angle θ, we have

2π R R 2π R R 1 f 0(reiθ)eiθ 1 f 0(reiθ) 1 f 0(z) n(r) Re dr dθ = Re ireiθ dθ dr = Re dz dr = Re dr 2π ˆ ˆ f(reiθ) ˆ 2πir ˆ f(reiθ) ˆ 2πir ‰ f(z) ˆ r 0 0 0 0 0 |z|=r 0 (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

R n(r) 1 2π dr = ln |f(Reiθ)| dθ − ln |f(0)|. (A.27) ˆ0 r 2π ˆ0

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

m m m m R n(r) X |zj+1| n(r) X |zj+1| j X X dr = dr = dr = j(ln |z | − ln |z |) = − ln |z | + m ln R (A.28) ˆ r ˆ r ˆ r j+1 j j 0 j=0 |zj | j=1 |zj | j=1 j=1 We finally obtain m X  R  1 2π ln |f(0)| = − ln + ln |f(Reiθ)| dθ, (A.29) |z | 2π ˆ j=1 j 0 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 assumed to be holomorphic there. Applying (A.29) to

Rh g(z) := f(z) , (A.30) z

37 we find that m X  R  1 2π ln |c| + h ln R = − ln + ln |f(Reiθ)| dθ, (A.31) |z | 2π ˆ j=1 j 0 because g(0) = cRh and |g(Reiθ)| = |f(Reiθ)|.

A.4 The uniqueness theorem for Dirichlet series

Suppose that ∞ X f(n) F (s) := (A.32) ns n=1 and ∞ X g(n) G(s) := (A.33) ns n=1

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

Now, we define H(s) := F (s) − G(s) (A.34) to be the Dirichlet series for h(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 and H ≡ 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

∞ X h(n) h(m) H(s) = = + H (s), (A.36) ns ms 1 n=m where ∞ X h(n) H (s) := . (A.37) 1 ns n=m+1

Choosing s with σ ≥ σ0, we find that s h(m) = −m H1(s), (A.38) since H(s) = 0 in this half-plane. Now, we observe that

∞ ∞ ∞ ∞ X h(n) X |h(n)| X |h(n)| 1 X |h(n)| |H1(s)| = ≤ = ≤ . (A.39) ns nσ nσ0 nσ−σ0 (m + 1)σ−σ0 nσ0 n=m+1 n=m+1 n=m+1 n=m+1

It follows that

mσ ∞ |h(n)|  m σ ∞ |h(n)|  m σ σ X σ0 X |h(m)| ≤ m |H1(s)| ≤ = (m + 1) = C, (A.40) (m + 1)σ−σ0 nσ0 m + 1 nσ0 m + 1 n=m+1 n=m+1

where ∞ X |h(n)| C := (m + 1)σ0 (A.41) nσ0 n=m+1

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

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