Previous set of notes: Notes 3. Next set of notes: 246C Notes 1.

One of the great classical triumphs of complex analysis was in providing the first complete proof (by Hadamard and de la Vallée Poussin in 1896) of arguably the most important theorem in analytic number theory, the prime number theorem:

Theorem 1 (Prime number theorem) Let {\pi(x)} denote the number of primes less than a given real number {x}. Then

\displaystyle \lim_{x \rightarrow \infty} \frac{\pi(x)}{x/\ln x} = 1

(or in asymptotic notation, {\pi(x) = (1+o(1)) \frac{x}{\ln x}} as {x \rightarrow \infty}).

(Actually, it turns out to be slightly more natural to replace the approximation {\frac{x}{\ln x}} in the prime number theorem by the logarithmic integral {\int_2^x \frac{dt}{\ln t}}, which happens to be a more precise approximation, but we will not stress this point here.)

The complex-analytic proof of this theorem hinges on the study of a key meromorphic function related to the prime numbers, the Riemann zeta function {\zeta}. Initially, it is only defined on the half-plane {\{ s \in {\bf C}: \mathrm{Re} s > 1 \}}:

Definition 2 (Riemann zeta function, preliminary definition) Let {s \in {\bf C}} be such that {\mathrm{Re} s > 1}. Then we define

\displaystyle \zeta(s) := \sum_{n=1}^\infty \frac{1}{n^s}. \ \ \ \ \ (1)

Note that the series is locally uniformly convergent in the half-plane {\{ s \in {\bf C}: \mathrm{Re} s > 1 \}}, so in particular {\zeta} is holomorphic on this region. In previous notes we have already evaluated some special values of this function:

\displaystyle \zeta(2) = \frac{\pi^2}{6}; \quad \zeta(4) = \frac{\pi^4}{90}; \quad \zeta(6) = \frac{\pi^6}{945}. \ \ \ \ \ (2)

However, it turns out that the zeroes (and pole) of this function are of far greater importance to analytic number theory, particularly with regards to the study of the prime numbers.

The Riemann zeta function has several remarkable properties, some of which we summarise here:

Theorem 3 (Basic properties of the Riemann zeta function)
  • (i) (Euler product formula) For any {s \in {\bf C}} with {\mathrm{Re} s > 1}, we have

    \displaystyle \zeta(s) = \prod_p (1 - \frac{1}{p^s})^{-1} \ \ \ \ \ (3)

     where the product is absolutely convergent (and locally uniform in {s}) and is over the prime numbers {p = 2, 3, 5, \dots}.
  • (ii) (Trivial zero-free region) {\zeta(s)} has no zeroes in the region {\{s: \mathrm{Re}(s) > 1 \}}.
  • (iii) (Meromorphic continuation) {\zeta} has a unique meromorphic continuation to the complex plane (which by abuse of notation we also call {\zeta}), with a simple pole at {s=1} and no other poles. Furthermore, the Riemann xi function

    \displaystyle \xi(s) := \frac{1}{2} s(s-1) \pi^{-s/2} \Gamma(s/2) \zeta(s) \ \ \ \ \ (4)

     is an entire function of order {1} (after removing all singularities). The function {(s-1) \zeta(s)} is an entire function of order one after removing the singularity at {s=1}.
  • (iv) (Functional equation) After applying the meromorphic continuation from (iii), we have

    \displaystyle \zeta(s) = 2^s \pi^{s-1} \sin(\frac{\pi s}{2}) \Gamma(1-s) \zeta(1-s) \ \ \ \ \ (5)

     for all {s \in {\bf C}} (excluding poles). Equivalently, we have

    \displaystyle \xi(s) = \xi(1-s) \ \ \ \ \ (6)

     for all {s \in {\bf C}}. (The equivalence between the (5) and (6) is a routine consequence of the Euler reflection formula and the Legendre duplication formula, see Exercises 26 and 31 of Notes 1.)

Proof: We just prove (i) and (ii) for now, leaving (iii) and (iv) for later sections.

The claim (i) is an encoding of the fundamental theorem of arithmetic, which asserts that every natural number {n} is uniquely representable as a product {n = \prod_p p^{a_p}} over primes, where the {a_p} are natural numbers, all but finitely many of which are zero. Writing this representation as {\frac{1}{n^s} = \prod_p \frac{1}{p^{a_p s}}}, we see that

\displaystyle \sum_{n \in S_{x,m}} \frac{1}{n^s} = \prod_{p \leq x} \sum_{a=0}^m \frac{1}{p^{as}}

whenever {x \geq 1}, {m \geq 0}, and {S_{x,m}} consists of all the natural numbers of the form {n = \prod_{p \leq x} p^{a_p}} for some {a_p \leq m}. Sending {m} and {x} to infinity, we conclude from monotone convergence and the geometric series formula that

\displaystyle \sum_{n=1}^\infty \frac{1}{n^s} = \prod_{p} \sum_{a=0}^\infty \frac{1}{p^{s}} =\prod_p (1 - \frac{1}{p^s})^{-1}

whenever {s>1} is real, and then from dominated convergence we see that the same formula holds for complex {s} with {\mathrm{Re} s > 1} as well. Local uniform convergence then follows from the product form of the Weierstrass {M}-test (Exercise 19 of Notes 1).

The claim (ii) is immediate from (i) since the Euler product {\prod_p (1-\frac{1}{p^s})^{-1}} is absolutely convergent and all terms are non-zero. \Box

We remark that by sending {s} to {1} in Theorem 3(i) we conclude that

\displaystyle \sum_{n=1}^\infty \frac{1}{n} = \prod_p (1-\frac{1}{p})^{-1}

and from the divergence of the harmonic series we then conclude Euler’s theorem {\sum_p \frac{1}{p} = \infty}. This can be viewed as a weak version of the prime number theorem, and already illustrates the potential applicability of the Riemann zeta function to control the distribution of the prime numbers.

The meromorphic continuation (iii) of the zeta function is initially surprising, but can be interpreted either as a manifestation of the extremely regular spacing of the natural numbers {n} occurring in the sum (1), or as a consequence of various integral representations of {\zeta} (or slight modifications thereof). We will focus in this set of notes on a particular representation of {\zeta} as essentially the Mellin transform of the theta function {\theta} that briefly appeared in previous notes, and the functional equation (iv) can then be viewed as a consequence of the modularity of that theta function. This in turn was established using the Poisson summation formula, so one can view the functional equation as ultimately being a manifestation of Poisson summation. (For a direct proof of the functional equation via Poisson summation, see these notes.)

Henceforth we work with the meromorphic continuation of {\zeta}. The functional equation (iv), when combined with special values of {\zeta} such as (2), gives some additional values of {\zeta} outside of its initial domain {\{s: \mathrm{Re} s > 1\}}, most famously

\displaystyle \zeta(-1) = -\frac{1}{12}.

If one formally compares this formula with (1), one arrives at the infamous identity

\displaystyle 1 + 2 + 3 + \dots = -\frac{1}{12}

although this identity has to be interpreted in a suitable non-classical sense in order for it to be rigorous (see this previous blog post for further discussion).

From Theorem 3 and the non-vanishing nature of {\Gamma}, we see that {\zeta} has simple zeroes (known as trivial zeroes) at the negative even integers {-2, -4, \dots}, and all other zeroes (the non-trivial zeroes) inside the critical strip {\{ s \in {\bf C}: 0 \leq \mathrm{Re} s \leq 1 \}}. (The non-trivial zeroes are conjectured to all be simple, but this is hopelessly far from being proven at present.) As we shall see shortly, these latter zeroes turn out to be closely related to the distribution of the primes. The functional equation tells us that if {\rho} is a non-trivial zero then so is {1-\rho}; also, we have the identity

\displaystyle \zeta(s) = \overline{\zeta(\overline{s})} \ \ \ \ \ (7)

for all {s>1} by (1), hence for all {s} (except the pole at {s=1}) by meromorphic continuation. Thus if {\rho} is a non-trivial zero then so is {\overline{\rho}}. We conclude that the set of non-trivial zeroes is symmetric by reflection by both the real axis and the critical line {\{ s \in {\bf C}: \mathrm{Re} s = \frac{1}{2} \}}. We have the following infamous conjecture:

Conjecture 4 (Riemann hypothesis) All the non-trivial zeroes of {\zeta} lie on the critical line {\{ s \in {\bf C}: \mathrm{Re} s = \frac{1}{2} \}}.

This conjecture would have many implications in analytic number theory, particularly with regard to the distribution of the primes. Of course, it is far from proven at present, but the partial results we have towards this conjecture are still sufficient to establish results such as the prime number theorem.

Return now to the original region where {\mathrm{Re} s > 1}. To take more advantage of the Euler product formula (3), we take complex logarithms to conclude that

\displaystyle -\log \zeta(s) = \sum_p \log(1 - \frac{1}{p^s})

for suitable branches of the complex logarithm, and then on taking derivatives (using for instance the generalised Cauchy integral formula and Fubini’s theorem to justify the interchange of summation and derivative) we see that

\displaystyle -\frac{\zeta'(s)}{\zeta(s)} = \sum_p \frac{\ln p/p^s}{1 - \frac{1}{p^s}}.

From the geometric series formula we have

\displaystyle \frac{\ln p/p^s}{1 - \frac{1}{p^s}} = \sum_{j=1}^\infty \frac{\ln p}{p^{js}}

and so (by another application of Fubini’s theorem) we have the identity

\displaystyle -\frac{\zeta'(s)}{\zeta(s)} = \sum_{n=1}^\infty \frac{\Lambda(n)}{n^s}, \ \ \ \ \ (8)

 for {\mathrm{Re} s > 1}, where the von Mangoldt function {\Lambda(n)} is defined to equal {\Lambda(n) = \ln p} whenever {n = p^j} is a power {p^j} of a prime {p} for some {j=1,2,\dots}, and {\Lambda(n)=0} otherwise. The contribution of the higher prime powers {p^2, p^3, \dots} is negligible in practice, and as a first approximation one can think of the von Mangoldt function as the indicator function of the primes, weighted by the logarithm function.

The series {\sum_{n=1}^\infty \frac{1}{n^s}} and {\sum_{n=1}^\infty \frac{\Lambda(n)}{n^s}} that show up in the above formulae are examples of Dirichlet series, which are a convenient device to transform various sequences of arithmetic interest into holomorphic or meromorphic functions. Here are some more examples:

Exercise 5 (Standard Dirichlet series) Let {s} be a complex number with {\mathrm{Re} s > 1}.
  • (i) Show that {-\zeta'(s) = \sum_{n=1}^\infty \frac{\ln n}{n^s}}.
  • (ii) Show that {\zeta^2(s) = \sum_{n=1}^\infty \frac{\tau(n)}{n^s}}, where {\tau(n) := \sum_{d|n} 1} is the divisor function of {n} (the number of divisors of {n}).
  • (iii) Show that {\frac{1}{\zeta(s)} = \sum_{n=1}^\infty \frac{\mu(n)}{n^s}}, where {\mu(n)} is the Möbius function, defined to equal {(-1)^k} when {n} is the product of {k} distinct primes for some {k \geq 0}, and {0} otherwise.
  • (iv) Show that {\frac{\zeta(2s)}{\zeta(s)} = \sum_{n=1}^\infty \frac{\lambda(n)}{n^s}}, where {\lambda(n)} is the Liouville function, defined to equal {(-1)^k} when {n} is the product of {k} (not necessarily distinct) primes for some {k \geq 0}.
  • (v) Show that {\log \zeta(s) = \sum_{n=1}^\infty \frac{\Lambda(n)/\ln n}{n^s}}, where {\log \zeta} is the holomorphic branch of the logarithm that is real for {s>1}, and with the convention that {\Lambda(n)/\ln n} vanishes for {n=1}.
  • (vi) Use the fundamental theorem of arithmetic to show that the von Mangoldt function is the unique function {\Lambda: {\bf N} \rightarrow {\bf R}} such that

    \displaystyle \ln n = \sum_{d|n} \Lambda(d)

    for every positive integer {n}. Use this and (i) to provide an alternate proof of the identity (8). Thus we see that (8) is really just another encoding of the fundamental theorem of arithmetic.

Given the appearance of the von Mangoldt function {\Lambda}, it is natural to reformulate the prime number theorem in terms of this function:

Theorem 6 (Prime number theorem, von Mangoldt form) One has

\displaystyle \lim_{x \rightarrow \infty} \frac{1}{x} \sum_{n \leq x} \Lambda(n) = 1

(or in asymptotic notation, {\sum_{n\leq x} \Lambda(n) = x + o(x)} as {x \rightarrow \infty}).

Let us see how Theorem 6 implies Theorem 1. Firstly, for any {x \geq 2}, we can write

\displaystyle \sum_{n \leq x} \Lambda(n) = \sum_{p \leq x} \ln p + \sum_{j=2}^\infty \sum_{p \leq x^{1/j}} \ln p.

The sum {\sum_{p \leq x^{1/j}} \ln p} is non-zero for only {O(\ln x)} values of {j}, and is of size {O( x^{1/2} \ln x )}, thus

\displaystyle \sum_{n \leq x} \Lambda(n) = \sum_{p \leq x} \ln p + O( x^{1/2} \ln^2 x ).

Since {x^{1/2} \ln^2 x = o(x)}, we conclude from Theorem 6 that

\displaystyle \sum_{p \leq x} \ln p = x + o(x)

as {x \rightarrow \infty}. Next, observe from the fundamental theorem of calculus that

\displaystyle \frac{1}{\ln p} - \frac{1}{\ln x} = \int_p^x \frac{1}{\ln^2 y} \frac{dy}{y}.

Multiplying by {\log p} and summing over all primes {p \leq x}, we conclude that

\displaystyle \pi(x) - \frac{\sum_{p \leq x} \ln p}{\ln x} = \int_2^x \sum_{p \leq y} \ln p \frac{1}{\ln^2 y} \frac{dy}{y}.

From Theorem 6 we certainly have {\sum_{p \leq y} \ln p = O(y)}, thus

\displaystyle \pi(x) - \frac{x + o(x)}{\ln x} = O( \int_2^x \frac{dy}{\ln^2 y} ).

By splitting the integral into the ranges {2 \leq y \leq \sqrt{x}} and {\sqrt{x} < y \leq x} we see that the right-hand side is {o(x/\ln x)}, and Theorem 1 follows.

Exercise 7 Show that Theorem 1 conversely implies Theorem 6.

The alternate form (8) of the Euler product identity connects the primes (represented here via proxy by the von Mangoldt function) with the logarithmic derivative of the zeta function, and can be used as a starting point for describing further relationships between {\zeta} and the primes. Most famously, we shall see later in these notes that it leads to the remarkably precise Riemann-von Mangoldt explicit formula:

Theorem 8 (Riemann-von Mangoldt explicit formula) For any non-integer {x > 1}, we have

\displaystyle \sum_{n \leq x} \Lambda(n) = x - \lim_{T \rightarrow \infty} \sum_{\rho: |\hbox{Im}(\rho)| \leq T} \frac{x^\rho}{\rho} - \ln(2\pi) - \frac{1}{2} \ln( 1 - x^{-2} )

where {\rho} ranges over the non-trivial zeroes of {\zeta} with imaginary part in {[-T,T]}. Furthermore, the convergence of the limit is locally uniform in {x}.

Actually, it turns out that this formula is in some sense too precise; in applications it is often more convenient to work with smoothed variants of this formula in which the sum on the left-hand side is smoothed out, but the contribution of zeroes with large imaginary part is damped; see Exercise 22. Nevertheless, this formula clearly illustrates how the non-trivial zeroes {\rho} of the zeta function influence the primes. Indeed, if one formally differentiates the above formula in {x}, one is led to the (quite nonrigorous) approximation

\displaystyle \Lambda(n) \approx 1 - \sum_\rho n^{\rho-1} \ \ \ \ \ (9)

or (writing {\rho = \sigma+i\gamma})

\displaystyle \Lambda(n) \approx 1 - \sum_{\sigma+i\gamma} \frac{n^{i\gamma}}{n^{1-\sigma}}.

Thus we see that each zero {\rho = \sigma + i\gamma} induces an oscillation in the von Mangoldt function, with {\gamma} controlling the frequency of the oscillation and {\sigma} the rate to which the oscillation dies out as {n \rightarrow \infty}. This relationship is sometimes known informally as “the music of the primes”.

Comparing Theorem 8 with Theorem 6, it is natural to suspect that the key step in the proof of the latter is to establish the following slight but important extension of Theorem 3(ii), which can be viewed as a very small step towards the Riemann hypothesis:

Theorem 9 (Slight enlargement of zero-free region) There are no zeroes of {\zeta} on the line {\{ 1+it: t \in {\bf R} \}}.

It is not quite immediate to see how Theorem 6 follows from Theorem 8 and Theorem 9, but we will demonstrate it below the fold.

Although Theorem 9 only seems like a slight improvement of Theorem 3(ii), proving it is surprisingly non-trivial. The basic idea is the following: if there was a zero at {1+it}, then there would also be a different zero at {1-it} (note {t} cannot vanish due to the pole at {s=1}), and then the approximation (9) becomes

\displaystyle \Lambda(n) \approx 1 - n^{it} - n^{-it} + \dots = 1 - 2 \cos(t \ln n) + \dots.

But the expression {1 - 2 \cos(t \ln n)} can be negative for large regions of the variable {n}, whereas {\Lambda(n)} is always non-negative. This conflict eventually leads to a contradiction, but it is not immediately obvious how to make this argument rigorous. We will present here the classical approach to doing so using a trigonometric identity of Mertens.

In fact, Theorem 9 is basically equivalent to the prime number theorem:

Exercise 10 For the purposes of this exercise, assume Theorem 6, but do not assume Theorem 9. For any non-zero real {t}, show that

\displaystyle -\frac{\zeta'(\sigma+it)}{\zeta(\sigma+it)} = o( \frac{1}{\sigma-1})

as {\sigma \rightarrow 1^+}, where {o( \frac{1}{\sigma-1})} denotes a quantity that goes to zero as {\sigma \rightarrow 1^+} after being multiplied by {\sigma-1}. Use this to derive Theorem 9.

This equivalence can help explain why the prime number theorem is remarkably non-trivial to prove, and why the Riemann zeta function has to be either explicitly or implicitly involved in the proof.

This post is only intended as the briefest of introduction to complex-analytic methods in analytic number theory; also, we have not chosen the shortest route to the prime number theorem, electing instead to travel in directions that particularly showcase the complex-analytic results introduced in this course. For some further discussion see this previous set of lecture notes, particularly Notes 2 and Supplement 3 (with much of the material in this post drawn from the latter).

— 1. Meromorphic continuation and functional equation —

We now focus on understanding the meromorphic continuation of {\zeta}, as well as the functional equation that that continuation satisfies. The arguments here date back to Riemann’s original paper on the zeta function. The general strategy is to relate the zeta function {\zeta(s)} for {\mathrm{Re}(s) > 1} to some sort of integral involving the parameter {s}, which is manipulated in such a way that the integral makes sense for values of {s} outside of the halfplane {\{ s: \mathrm{Re}(s) > 1 \}}, and can thus be used to define the zeta function meromorphically in such a region. Often the Gamma function {\Gamma} is involved in the relationship between the zeta function and integral. There are many such ways to connect {\zeta} to an integral; we present some of the more classical ones here.

One way to motivate the meromorphic continuation {\zeta} is to look at the continuous analogue

\displaystyle \frac{1}{s-1} = \int_1^\infty \frac{1}{t^s}\ dt, \quad \mathrm{Re} s > 1

of (1). This clearly extends meromorphically to the whole complex plane. So one now just has to understand the analytic continuation properties of the residual

\displaystyle \frac{1}{s-1} - \zeta(s) = \int_1^\infty \frac{1}{t^s}\ dt - \sum_{n=1}^\infty \frac{1}{n^s}, \quad \mathrm{Re} s > 1.

For instance, using the Riemann sum type quadrature

\displaystyle \int_n^{n+1} \frac{1}{t^s}\ dt = \frac{1}{n^s} + \int_n^{n+1} \frac{1}{t^s} - \frac{1}{n^s}\ dt

one can write this residual as

\displaystyle \sum_{n=1}^\infty \int_n^{n+1} \frac{1}{t^s} - \frac{1}{n^s}\ dt;

since {\frac{1}{t^s} - \frac{1}{n^s} = O_s( \frac{1}{n^{\mathrm{Re} s+1}} )}, it is a routine application of the Fubini and Morera theorems to establish analytic continuation of the residual to the half-plane {\mathrm{Re} s > 0}, thus giving a meromorphic extension

\displaystyle \zeta(s) = \frac{1}{s-1} - \sum_{n=1}^\infty \int_n^{n+1} \frac{1}{t^s} - \frac{1}{n^s}\ dt \ \ \ \ \ (10)

 of {\zeta} to the region {\{ s: \mathrm{Re}(s) > 0 \}}. Among other things, this shows that (the meromorphic continuation of) {\zeta} has a simple pole at {s=1} with residue {1}.

Exercise 11 Using the trapezoid rule, show that for any {s} in the region {\{ s: \hbox{Re}(s) > -1 \}} with {s \neq 1}, there exists a unique complex number {\zeta(s)} for which one has the asymptotic

\displaystyle \sum_{n=1}^N \frac{1}{n^s} = \zeta(s) + \frac{N^{1-s}}{1-s} + \frac{1}{2} N^{-s} + O( \frac{|s| |s+1|}{\sigma+1} N^{-s-1} )

for any natural number {N}, where {s=\sigma+it}. Use this to extend the Riemann zeta function meromorphically to the region {\{ s: \hbox{Re}(s) > -1 \}}. Conclude in particular that {\zeta(0)=-\frac{1}{2}} and {\xi(0)=\xi(1)= \frac{1}{2}}.

Exercise 12 Obtain the refinement

\displaystyle \sum_{y \leq n \leq x} f(n) = \int_y^x f(t)\ dt + \frac{1}{2} f(x) + \frac{1}{2} f(y)

\displaystyle + \frac{1}{12} (f'(x) - f'(y)) + O( \int_x^y |f'''(t)|\ dt )

to the trapezoid rule when {y < x} are integers and {f: [y,x] \rightarrow {\bf C}} is continuously three times differentiable. Then show that for any {s} in the region {\{ s: \hbox{Re}(s) > -2 \}} with {s \neq 1}, there exists a unique complex number {\zeta(s)} for which one has the asymptotic

\displaystyle \sum_{n=1}^N \frac{1}{n^s} = \zeta(s) + \frac{N^{1-s}}{1-s} + \frac{1}{2} N^{-s} - \frac{s}{12} N^{-s-1}

\displaystyle + O( \frac{|s| |s+1| |s+2|}{\sigma+2} N^{-s-2} )

for any natural number {N}, where {s=\sigma+it}. Use this to extend the Riemann zeta function meromorphically to the region {\{ s: \hbox{Re}(s) > -2 \}}. Conclude in particular that {\zeta(-1)=-\frac{1}{12}}.

One can keep going in this fashion using the Euler-Maclaurin formula (see this previous blog post) to extend the range of meromorphic continuation to the rest of the complex plane. However, we will now proceed in a different fashion, using the theta function

\displaystyle \theta(ix) = \sum_{n \in {\bf Z}} e^{-\pi n^2 x} \ \ \ \ \ (11)

that made an appearance in previous notes, and try to transform this function into the zeta function. We will only need this function for imaginary values {ix} of the argument in the upper half-plane (so {x>0}); from Exercise 7 of Notes 2 we have the modularity relation

\displaystyle \theta(i/x) = x^{1/2} \theta(ix). \ \ \ \ \ (12)

 In particular, since {\theta(ix)} decays exponentially to {1} as {x \rightarrow \infty}, {\theta(ix)} blows up like {x^{-1/2}} as {x \rightarrow 0}.

We will attempt to apply the Mellin transform (Exercise 11 from Notes 2) to this function; formally, we have

\displaystyle {\mathcal M}[\theta(i\cdot)](s) := \int_0^\infty x^s \theta(ix) \frac{dx}{x}.

There is however a problem: as {x} goes to infinity, {\theta(ix)} converges to one, and the integral here is unlikely to be convergent. So we will compute the Mellin transform of {\theta(ix)-1}:

\displaystyle {\mathcal M}(\theta(i\cdot)-1)(s) := \int_0^\infty x^s (\theta(ix)-1) \frac{dx}{x}. \ \ \ \ \ (13)

 The function {\theta-1} decays exponentially as {x \rightarrow \infty}, and blows up like {O(x^{-1/2})} as {x \rightarrow 0}, so this integral will be absolutely integrable when {\mathrm{Re} s > 1/2}. Since

\displaystyle \theta(ix)-1 = 2 \sum_{n=1}^\infty e^{-\pi n^2 x}

we can write

\displaystyle {\mathcal M}(\theta(i\cdot)-1)(s) = 2 \int_0^\infty \sum_{n=1}^\infty x^s e^{-\pi n^2 x} \frac{dx}{x}.

By the Fubini–Tonelli theorem, the integrand here is absolutely integrable, and hence

\displaystyle {\mathcal M}(\theta(i\cdot)-1)(s) = 2 \sum_{n=1}^\infty \int_0^\infty x^s e^{-\pi n^2 x} \frac{dx}{x}.

From the Bernoulli definition of the Gamma function (Exercise 29(ii) of Notes 1) and a change of variables we have

\displaystyle \int_0^\infty x^s e^{-\pi n^2 x} \frac{dx}{x} = \frac{\Gamma(s)}{(\pi n^2)^s}

and hence by (1) we obtain the identity

\displaystyle {\mathcal M}(\theta(i\cdot)-1)(s) = \frac{2 \Gamma(s) \zeta(2s)}{\pi^s}

whenever {\mathrm{Re}(s) > 1/2}. Replacing {s} by {s/2}, we can rearrange this as a formula for the {\xi} function (4), namely

\displaystyle \xi(s) = \frac{s(s-1)}{4} {\mathcal M}(\theta(i\cdot)-1)(s/2)

whenever {\mathrm{Re}(s) > 1}.

Now we exploit the modular identity (12) to improve the convergence of this formula. The convergence of {\theta(ix)-1} is much better near {x=\infty} than near {x=0}, so we use (13) to split

\displaystyle \xi(s) = \frac{s(s-1)}{4} ( \int_0^1 x^{s/2} (\theta(ix)-1) \frac{dx}{x} + \int_1^\infty x^{s/2} (\theta(ix)-1) \frac{dx}{x} )

and then transform the first integral using the change of variables {x \mapsto 1/x} to obtain

\displaystyle \xi(s) = \frac{s(s-1)}{4} ( \int_1^\infty x^{-s/2} (\theta(i/x)-1) \frac{dx}{x} + \int_1^\infty x^{s/2} (\theta(ix)-1) \frac{dx}{x} ).

Using (12) we can write this as

\displaystyle \xi(s) = \frac{s(s-1)}{4} \int_1^\infty (x^{(1-s)/2} + x^{s/2}) (\theta(ix)-1) + (x^{(1-s)/2}-x^{- s/2})\frac{dx}{x}.

Direct computation shows that

\displaystyle \int_1^\infty (x^{(1-s)/2}-x^{-s/2})\frac{dx}{x} = \frac{2}{s-1} - \frac{2}{s} = \frac{2}{s(s-1)}

and thus

\displaystyle \xi(s) = \frac{1}{2} + \frac{s(s-1)}{4} \int_1^\infty (x^{(1-s)/2} + x^{s/2}) (\theta(ix)-1) \frac{dx}{x}

whenever {\mathrm{Re}(s) > 1}. However, the integrand here is holomorphic in {x} and exponentially decaying in {x}, so from the Fubini and Morera theorems we easily see that the right-hand side is an entire function of {s}; also from inspection we see that it is symmetric with respect to the symmetry {s \mapsto 1-s}. Thus we can define {\xi} as an entire function, and hence {\zeta} as a meromorphic function, and one verifies the functional equation (6).

It remains to establish that {\xi} is of order {1}. From (11) we have {\theta(ix)-1 = O( e^{-\pi x} )} so from the triangle inequality

\displaystyle \xi(s) \lesssim 1 + (1+|s|)^2 \int_1^\infty x^{\frac{1+|s|}{2}} e^{-\pi x} \frac{dx}{x}

\displaystyle = 1 + (1+|s|)^2 \pi^{-\frac{1+|s|}{2}} \Gamma( \frac{1+|s|}{2} ).

From the Stirling approximation (Exercise 30(v) from Notes 1) we conclude that

\displaystyle \xi(s) \lesssim \exp( O( |s| \log |s| ) )

for {|s| \geq 2} (say), and hence {\xi} is of order at most {1} as required. (One can show that {\xi} has order exactly one by inspecting what happens to {\xi(s)} as {s \rightarrow +\infty}, using that {\zeta(s) \rightarrow 1} in this regime.) This completes the proof of Theorem 3.

Exercise 13 (Alternate derivation of meromorphic continuation and functional equation)
  • (i) Establish the identity

    \displaystyle \Gamma(s) \zeta(s) = \int_0^\infty \frac{t^s}{e^t-1} \frac{dt}{t}

    whenever {\mathrm{Re}(s) > 1}.
  • (ii) Establish the identity

    \displaystyle \zeta(s) = \frac{1}{\Gamma(s) (1 - e^{2\pi i(s-1)})} \lim_{R \rightarrow \infty} \int_{C_{R,\varepsilon}} \frac{z^s}{e^z-1} \frac{dz}{z} \ \ \ \ \ (14)

     whenever {\mathrm{Re}(s) > 1}, {s} is not an integer, {\varepsilon>0}, {z^s := \exp( s \mathrm{Log}_{[0,2\pi)} z)} where {\mathrm{Log}_{[0,2\pi)}} is the branch of the logarithm with real part in {[0,2\pi)}, and {C_{R,\varepsilon}} is the contour consisting of the line segment {\gamma_{R-i\varepsilon \rightarrow -i\varepsilon}}, the semicircle {\{ \varepsilon e^{i(3\pi/2 - \theta)}: 0 \leq \theta \leq \pi \}}, and the line segment {\gamma_{i\varepsilon \rightarrow R+i\varepsilon}}.
  • (iii) Use (ii) to meromorphically continue {\zeta} to the entire complex plane {{\bf C}}.
  • (iv) By shifting the contour {C_{R,\varepsilon}} to the contour {\tilde C_{R,N} := \gamma_{R - 2\pi i (N+1/2) \rightarrow -1 - 2\pi i(N+1/2) \rightarrow -1 + 2\pi i (N+1/2) \rightarrow R + 2\pi i (N+1/2)}} for a large natural number {N} and applying the residue theorem, show that

    \displaystyle \lim_{R \rightarrow \infty} \int_{C_{R,\varepsilon}} \frac{z^s}{e^z-1} \frac{dz}{z} = 2\pi i \sum_{n \in {\bf Z} \backslash \{0\}} (2\pi i n)^{s-1}

    again using the branch {\mathrm{Log}_{[0,2\pi)}} of the logarithm to define {(2\pi i n)^{s-1}}.
  • (v) Establish the functional equation (5).

Exercise 14 Use the formula {\zeta(-1)=\frac{-1}{12}} from Exercise 12, together with the functional equation, to give yet another proof of the identity {\zeta(2) = \frac{\pi^2}{6}}.

Exercise 15 (Relation between zeta function and Bernoulli numbers)
  • (i) For any complex number {z} with {\hbox{Re}(z)>0}, use the Poisson summation formula (Proposition 3(v) from Notes 2) to establish the identity

    \displaystyle \sum_{n \in {\bf Z}} e^{-|n|z} = \sum_{m \in {\bf Z}} \frac{2z}{z^2+(2\pi m)^2}.

  • (ii) For {z} as above and sufficiently small, show that

    \displaystyle 2\sum_k (-1)^{k+1} \zeta(2k) (z/2\pi)^k = \frac{z}{1-e^{-z}} - 1 - \frac{z}{2}.

    Conclude that

    \displaystyle \zeta(2k) = \frac{(-1)^{k+1} (2\pi)^k}{2 (2k!)} B_{2k}

    for any natural number {k}, where the Bernoulli numbers {B_2, B_4, B_6, \dots} are defined through the Taylor expansion

    \displaystyle \frac{z}{1-e^{-z}} = 1 + \frac{z}{2} + \sum_k \frac{B_{2k}}{(2k)!} z^{2k}.

    Thus for instance {B_2 = 1/6}, {B_4 = -1/30}, and so forth.
  • (iii) Show that

    \displaystyle \zeta(-n) = -\frac{B_{n+1}}{n+1} \ \ \ \ \ (15)

     for any odd natural number {n}. (This identity can also be deduced from the Euler-Maclaurin formula, which generalises the approach in Exercise 12; see this previous post.)
  • (iv) Use (14) and the residue theorem (now working inside the contour {\tilde C_{R,N}}, rather than outside) to give an alternate proof of (15).

Exercise 16 (Convexity bounds)
  • (i) Establish the bounds {\zeta(\sigma+it) \lesssim_\sigma 1} for any {\sigma > 1} and {t \in {\bf R}} with {|t|>1}.
  • (ii) Establish the bounds {\zeta(\sigma+it) \lesssim_\sigma |t|^{1/2-\sigma}} for any {\sigma < 0} and {t \in {\bf R}} with {|t|>1}. (Hint: use the functional equation.)
  • (iii) Establish the bounds {\zeta(\sigma+it) \lesssim_{\sigma,\varepsilon} |t|^{\frac{1-\sigma}{2}+\varepsilon}} for any {0 \leq \sigma \leq 1, \varepsilon > 0} and {t \in {\bf R}} with {|t|>1}. (Hint: use the Phragmén-Lindelöf principle, Exercise 19 from Notes 2, after dealing somehow with the pole at {s=1}.)
It is possible to improve the bounds (iii) in the region {0 < \sigma < 1}; such improvements are known as subconvexity estimates. For instance, it is currently known that {\zeta(\frac{1}{2}+it) \lesssim_\mu (1+|t|)^\mu} for any {\mu > 13/84} and {t \in {\bf R}}, a result of Bourgain; the Lindelöf hypothesis asserts that this bound in fact holds for all {\mu>0}, although this remains unproven (it is however a consequence of the Riemann hypothesis).

Exercise 17 (Riemann-von Mangoldt formula) Show that for any {T \geq 2}, the number of zeroes of {\zeta} in the rectangle {\{ \sigma+it: 0 \leq \sigma \leq 1; 0 \leq t \leq T \}} is equal to {\frac{T}{2\pi} \log \frac{T}{2\pi} - \frac{T}{2\pi} + O( \log T )}. (Hint: apply the argument principle to {\xi} evaluated at a rectangle {\gamma_{2-iT' \rightarrow 2+iT' \rightarrow -1+iT' \rightarrow -1-iT' \rightarrow 2-iT'}} for some {T' = T + O(1)} that is chosen so that the horizontal edges of the rectangle do not come too close to any of the zeroes (cf. the selection of the radii {R_k} in the proof of the Hadamard factorisation theorem in Notes 1), and use the functional equation and Stirling’s formula to control the asymptotics for the horizontal edges.)

We remark that the error term {O(\log T)}, due to von Mangoldt in 1905, has not been significantly improved despite over a century of effort. Even assuming the Riemann hypothesis, the error has only been reduced very slightly to {O(\log T/\log\log T)} (a result of Littlewood from 1924).

Remark 18 Thanks to the functional equation and Rouche’s theorem, it is possible to numerically verify the Riemann hypothesis in any finite portion {\{ \sigma+it: 0 \leq \sigma \leq 1; |t| \leq T \}} of the critical strip, so long as the zeroes in that strip are all simple. Indeed, if there was a zero {\sigma+it} off of the critical line {\sigma \neq 1/2}, then an application of the argument principle (and Rouche’s theorem) in some small contour around {\sigma+it} but avoiding the critical line would be capable of numerically determining that there was a zero off of the line. Similarly, for each simple zero {\frac{1}{2}+it} on the critical line, applying the argument principle for some small contour around that zero and symmetric around the critical line would numerically verify that there was exactly one zero within that contour, which by the functional equation would then have to lie exactly on that line. (In practice, more efficient methods are used to numerically verify the Riemann hypothesis over large finite portions of the strip, but we will not detail them here.)

— 2. The explicit formula —

We now prove Riemann-von Mangoldt explicit formula. Since {\xi} is a non-trivial entire function of order {1}, with zeroes at the non-trivial zeroes of {\zeta} (the trivial zeroes having been cancelled out by the Gamma function), we see from the Hadamard factorisation theorem (in the form of Exercise 35 from Notes 1) that

\displaystyle \frac{\xi'(s)}{\xi(s)} = B + \sum_\rho \frac{1}{s-\rho} + \frac{1}{\rho}

away from the zeroes of {\xi}, where {\rho} ranges over the non-trivial zeroes of {\zeta} (note from Exercise 11 that there is no zero at the origin), and {B} is some constant. From (4) we can calculate

\displaystyle \frac{\xi'(s)}{\xi(s)} = \frac{\zeta'(s)}{\zeta(s)} + \frac{1}{s} + \frac{1}{s-1} - \frac{1}{2} \ln \pi - \frac{1}{2} \frac{\Gamma'(s/2)}{\Gamma(s/2)}

while from Exercise 27 of Notes 1 we have

\displaystyle \frac{1}{2} \frac{\Gamma'(s/2)}{\Gamma(s/2)} = -\frac{1}{2} \gamma + \sum_{k=0,-2,-4,\dots} \frac{1}{2-k} - \frac{1}{s-k}

and thus (after some rearranging)

\displaystyle -\frac{\zeta'}{\zeta}(s) = B' + \frac{1}{s-1} - \sum_\rho (\frac{1}{s-\rho} + \frac{1}{\rho}) - \sum_{n=1}^\infty (\frac{1}{s+2n} - \frac{1}{2n}) \ \ \ \ \ (16)

 where

\displaystyle B' = -B - \frac{1}{2} \log \pi - \frac{1}{2} \gamma.

One can compute the values of {B,B'} explicitly:

Exercise 19 By inspecting both sides of (16) as {s \rightarrow 0}, show that {B' = - \log 2\pi + 1}, and hence {B = \log 2 + \frac{1}{2} \log \pi - 1 - \frac{1}{2} \gamma}.

Jensen’s formula tells us that the number of non-trivial zeroes of {\zeta} in a disk {D(0,R)} is at most {O_\varepsilon(R^{1+\varepsilon})} for any {\varepsilon>0} and {R>1}. One can obtain a local version:

Exercise 20 (Local bound on zeroes)
  • (i) Establish the upper bound {|\zeta(\sigma+it)| \lesssim |t|^{O(1)}} whenever {\frac{1}{4} \leq \sigma \lesssim 1} and {t \in {\bf R}} with {|t| \geq 1}. (Hint: use (10). More precise bounds are available with more effort, but will not be needed here.)
  • (ii) Establish the bounds {|\zeta(2+it)| \sim 1} uniformly in {t}. (Hint: use the Euler product.)
  • (iii) Show that for any {T \geq 2}, the number of non-trivial zeroes with imaginary part in {[T,T+1]} is {O(\log T)}. (Hint: use Jensen’s formula and the functional equation.)
  • (iv) For {T \geq 2}, {|\mathrm{Re} s| \leq 10}, and {|\mathrm{Im} s| \in [T,T+1]}, with {s} not a zero of {\zeta}, show that

    \displaystyle \frac{\zeta'(s)}{\zeta(s)} = \sum_{\rho: |\rho-s| \leq 1} \frac{1}{s-\rho} + O(\log T).

    (Hint: use Theorem 9 of Notes 1.)

Meanwhile, from Perron’s formula (Exercise 12 of Notes 2) and (8) we see that for any non-integer {x>1}, we have

\displaystyle \sum_{n \leq x} \Lambda(n) = \frac{1}{2\pi i} \lim_{T \rightarrow \infty} \int_{\gamma_{2-iT \rightarrow 2+iT}} (B' + \frac{1}{s-1} - \sum_\rho (\frac{1}{s-\rho} + \frac{1}{\rho})

\displaystyle - \sum_{n=1}^\infty (\frac{1}{s+2n} - \frac{1}{2n})) \frac{x^s}{s}\ ds.

We can compute individual terms here and then conclude the Riemann-von Mangoldt explicit formula:

Exercise 21 (Riemann-von Mangoldt explicit formula) Let {T \geq 2} and {x > 1}. Establish the following bounds:
  • (i) {\frac{1}{2\pi i} \int_{\gamma_{2-iT \rightarrow 2+iT}} B' \frac{x^s}{s}\ ds = B' + O_x( 1/T )}.
  • (ii) {\frac{1}{2\pi i} \int_{\gamma_{2-iT \rightarrow 2+iT}} \frac{1}{s-1} \frac{x^s}{s}\ ds = x - 1 + O_x( 1/T )}.
  • (iii) For any positive integer {n}, we have

    \displaystyle \frac{1}{2\pi i} \int_{\gamma_{2-iT \rightarrow 2+iT}} (\frac{1}{s+2n} - \frac{1}{2n}) \frac{x^s}{s}\ ds

    \displaystyle = \frac{x^{-2n}}{2n} + O_x( \frac{1}{n(n+T)} ).

  • (iv) For any non-trivial zero {\rho}, we have

    \displaystyle \frac{1}{2\pi i} \int_{\gamma_{2-iT \rightarrow 2+iT}} (\frac{1}{s-\rho} - \frac{1}{\rho}) \frac{x^s}{s}\ ds

    \displaystyle = 1_{|\mathrm{Im}(\rho)| \leq T} \frac{x^{\rho}}{\rho} + O_x( \frac{1}{(1+||\mathrm{Im}(\rho)| - T|) (|\mathrm{Im}(\rho)|+T)} ).

  • (v) We have {\lim_{T \rightarrow \infty} \sum_{n=1}^\infty \frac{1}{n(n+T)} = 0}.
  • (vi) We have {\lim_{T \rightarrow \infty} \sum_\rho \frac{1}{(1+||\mathrm{Im}(\rho)| - T|) (|\mathrm{Im}(\rho)|+T)} = 0}.
(Hint: for (i)-(iii), shift the contour {\gamma_{2-iT \rightarrow 2+iT}} to {\gamma_{2-iT \rightarrow -R-iT \rightarrow -R+iT \rightarrow 2+iT}} for an {R} that gets sent to infinity, and using the residue theorem. The same argument works for (iv) except when {\rho} is really close to {2+iT}, in which case a detour to the contour may be called for. For (vi), use Exercise 20 and partition the zeroes depending on what unit interval {\mathrm{Im}(\rho)} falls into.)
  • (viii) Using the above estimates, conclude Theorem 8.

The explicit formula in Theorem 8 is completely exact, but turns out to be a little bit inconvenient for applications because it involves all the zeroes {\rho}, and the series involving them converges very slowly (indeed the convergence is not even absolute). In practice it is preferable to work with a smoothed version of the formula. Here is one such smoothing:

Exercise 22 (Smoothed explicit formula)
  • (i) Let {g: {\bf R} \rightarrow {\bf C}} be a smooth function compactly supported on {(0,+\infty)}. Show that {\hat g} is entire and obeys the bound

    \displaystyle \hat g(\frac{i}{2\pi} (\sigma+it)) \lesssim_g e^{-c \sigma} (1+|t|)^{-2}

    (say) for some {c>0}, all {\sigma \leq 10}, and all {t \in {\bf R}}.
  • (ii) With {g} as in (i), establish the identity

    \displaystyle \sum_n \Lambda(n) g(\ln n) = \hat g(\frac{i}{2\pi}) - \sum_\rho \hat g(\frac{i}{2\pi} \rho)

    \displaystyle - \sum_n \hat g(-2n \frac{i}{2\pi})

    with the summations being absolutely convergent by applying the Fourier inversion formula to {g}, shifting the contour to frequencies {\frac{i}{2\pi} (\sigma+it)} for some {\sigma>1}, applying (8), and then shifting the contour again (using Exercise 20 and (i) to justify the contour shifting).
  • (iii) Show that

    \displaystyle \sum_n \Lambda(n) \eta(n) = \int_{\bf R} (1 - \frac{1}{y^3-y}) \eta(y)\ dy

    \displaystyle - \sum_\rho \int_{\bf R} \eta(y) y^{\rho-1}\ dy

    whenever {\eta: {\bf R} \rightarrow {\bf C}} is a smooth function, compactly supported in {(1,+\infty)}, with the summation being absolutely convergent.
  • (iv) Explain why (iii) is formally consistent with Theorem 8 when applied to the non-smooth function {\eta = 1_{[1,x]}}.

— 3. Extending the zero free region, and the prime number theorem —

We now show how Theorem 9 implies Theorem 6. Let {2 \leq T \leq x} be parameters to be chosen later. We will apply Exercise 22 to a function {\eta = \eta_{T,x}} which equals one on {[2,x]}, is supported on {[1.5, x + x/T]}, and obeys the derivative estimates

\displaystyle \eta^{(j)}(y) \lesssim_j 1

for all {y \in [1.5,2]} and {j \geq 0}, and

\displaystyle \eta^{(j)}(y) \lesssim_j (T/x)^j

for all {y \in [x,x+T]} and {j \geq 0}. Such a function can be constructed by gluing together various rescaled versions of (antiderivatives of) standard bump functions. For such a function, we have

\displaystyle \sum_{n \leq x} \Lambda(n) \leq \sum_n \Lambda(n) \eta(n).

On the other hand, we have

\displaystyle \int_{\bf R} \eta(y)\ dy = x + O(x/T)

and

\displaystyle \int_{\bf R} \frac{1}{y^3-y} \eta(y)\ dy = O(1)

and hence

\displaystyle \sum_{n \leq x} \Lambda(n) \leq x + O(x/T) + \sum_\rho O( |\int_{\bf R} \eta(y) y^{\rho-1}\ dy| ). \ \ \ \ \ (17)

 We split into the two cases {|\rho| \leq T_*} and {|\rho| > T_*}, where {T_* \geq T} is a parameter to be chosen later. For {|\rho| \leq T_*}, there are only {O_T(1)} zeros, and all of them have real part strictly less than {1} by Theorem 9. Hence there exists {\varepsilon = \varepsilon_{T_*} > 0} such that {\mathrm{Re} \rho \leq 1-\varepsilon} for all such zeroes. For each such zero, we have from the triangle inequality

\displaystyle \int_{\bf R} \eta(y) y^{\rho-1}\ dy \lesssim_{T_*} x^{1-\varepsilon}

and so the total contribution of these zeroes to (17) is {O_{T_*}( x^{1-\varepsilon})}. For each zero {\rho} with {|\rho| \geq T_*}, we integrate parts twice to get some decay in {\rho}:

\displaystyle \int_{\bf R} \eta(y) y^{\rho-1}\ dy = \frac{-1}{\rho} \int_{\bf R} \eta'(y) y^\rho\ dy

\displaystyle = \frac{1}{\rho(\rho+1)} \int_{\bf R} \eta''(y) y^{\rho+1}\ dy,

and from the triangle inequality and the fact that {\mathrm{Re} \rho \leq 1} we conclude

\displaystyle \int_{\bf R} \eta(y) y^{\rho-1}\ dy \lesssim \frac{1}{|\rho|^2} \frac{T^2}{x^2} \frac{x}{T} x^2 = \frac{T}{|\rho|^2} x.

Since {\sum_\rho \frac{1}{|\rho|^2}} is convergent (this follows from Exercise 20 we conclude (for {T_*} large enough depending on {T}) that the total contribution here is {O(x/T)}. Thus, after choosing {T_*} suitably, we obtain the bound

\displaystyle \sum_{n \leq x} \Lambda(n) \leq x + O(x/T) + O_T(x^{1-\varepsilon})

and thus

\displaystyle \sum_{n \leq x} \Lambda(n) \leq x + O(x/T)

whenever {x} is sufficiently large depending on {T} (since {\varepsilon} depends only on {T_*}, which depends only on {T}). A similar argument (replacing {x,x+x/T} by {x-x/T, x} in the construction of {\eta}) gives the matching lower bound

\displaystyle \sum_{n \leq x} \Lambda(n) \geq x - O(x/T)

whenever {x} is sufficiently large depending on {T}. Sending {T \rightarrow \infty}, we obtain Theorem 6.

Exercise 23 Assuming the Riemann hypothesis, show that

\displaystyle \sum_{n \leq x} \Lambda(n) = x + O_\varepsilon(x^{1/2+\varepsilon})

for any {\varepsilon>0} and {x>1}, and that

\displaystyle \pi(x) = \int_2^x \frac{dt}{\log t} + O_\varepsilon(x^{1/2+\varepsilon})

for any {\varepsilon>0} and {x>1}. Conversely, show that either of these two estimates are equivalent to the Riemann hypothesis. (Hint: find a holomorphic continuation of {-\frac{\zeta'(s)}{\zeta(s)} - \frac{1}{s-1}} to the region {\mathrm{Re}(s) > 1/2} in a manner similar to how {\zeta(s)-\frac{1}{s-1}} was first holomorphically continued to the region {\mathrm{Re}(s) > 0}).

It remains to prove Theorem 9. The claim is clear for {t=0} thanks to the simple pole of {\zeta} at {s=1}, so we may assume {t \neq 0}. Suppose for contradiction that there was a zero of {\zeta} at {1+it}, thus

\displaystyle \zeta(\sigma+it) = O_t( \sigma-1)

for {\sigma>1} sufficiently close to {1}. Taking logarithms, we see in particular that

\displaystyle \ln |\zeta(\sigma+it)| \leq - \ln \frac{1}{\sigma-1} + O_t(1).

Using Lemma 5(v), we conclude that

\displaystyle \sum_{n=2}^\infty \frac{\Lambda(n)/\ln n}{n^\sigma} \cos(t \ln n) \leq - \ln \frac{1}{\sigma-1} + O_t(1).

Note that the summands here are oscillatory due to the cosine term. To manage the oscillation, we use the simple pole at {s=1} that gives

\displaystyle \zeta(\sigma) \sim \frac{1}{\sigma-1}

for {\sigma>1} sufficiently close to one, and on taking logarithms as before we get

\displaystyle \sum_{n=2}^\infty \frac{\Lambda(n)/\ln n}{n^\sigma} = \ln \frac{1}{\sigma-1} + O(1).

These two estimates come close to being contradictory, but not quite (because we could have {\cos(t\ln n)} close to {-1} for most numbers {n} that are weighted by {\frac{\Lambda(n)/\ln n}{n^\sigma}}. To get the contradiction, we use the analytic continuation of {\zeta} to {1+2it} to conclude that

\displaystyle \zeta(\sigma+2it) = O_t( 1)

and hence

\displaystyle \sum_{n=2}^\infty \frac{\Lambda(n)/\ln n}{n^\sigma} \cos(2 t \ln n) \leq O_t(1).

Now we take advantage of the Mertens inequality

\displaystyle 3 + 4 \cos(\theta) + \cos(2\theta) = 2 (1 + \cos(\theta))^2 \geq 0

(which is a quantitative variant of the observation that if {\cos(\theta)} is close to {-1} then {\cos(2\theta)} has to be close to {+1}) as well as the non-negative nature of {\Lambda(n)} to conclude that

\displaystyle \sum_{n=2}^\infty \frac{\Lambda(n)/\ln n}{n^\sigma} (3 + 4\cos(2 t \ln n) + \cos(2\theta) \geq 0

and hence

\displaystyle 0 \leq - \ln \frac{1}{\sigma-1} + O_t(1).

This leads to the desired contradiction by sending {\sigma \rightarrow 1^+}, and proves the prime number theorem.

Exercise 24 Establish the inequality

\displaystyle \zeta(\sigma)^3 |\zeta(\sigma+it)|^4 |\zeta(\sigma+2it)| \geq 1

for any {t \in {\bf R}} and {\sigma>1}.

Remark 25 There are a number of ways to improve Theorem 9 that move a little closer in the direction of the Riemann hypothesis. Firstly, there are a number of zero-free regions for the Riemann zeta function known that give lower bounds for {|\zeta(s)|} (and in particular preclude the existence of zeros) a small amount inside the critical strip, and can be used to improve the error term in the prime number theorem; for instance, the classical zero-free region shows that there are no zeroes in the region {\{ \sigma+it: \sigma \geq 1 - \frac{c}{\ln(1+|t|)}} for some sufficiently small absolute constant {c}, and lets one improve the {o(x)} error term in Theorem 6 to {O( x \exp( -c\sqrt{\ln x}))} (with a corresponding improvement in Theorem 1, provided that one replaces {x/\ln x} with the logarithmic integral {\int_2^x \frac{dt}{\ln t}}). A further improvement in the zero free region and in the prime number theorem error term was subsequently given by Vinogradov. We also mention a number of important zero density estimates which provide non-trivial upper bounds for the number of zeroes in other, somewhat larger regions of the critical strip; the bounds are not strong enough to completely exclude zeroes as is the case with zero-free regions, but can at least limit the collective influence of such zeroes. For more discussion of these topics, see the various lecture notes to this previous course.