Let \(d\theta\) be the point measure that assigns weight \(\log p\) to prime \(p\). Then \[\int_0^x\frac{d\theta (t)}{\log t} = \sum_{p\lt x}\frac{\log p}{\log p} = \pi(x).\] So for \(x\lt y\), \begin{equation} \label{pi} \pi(y)-\pi(x) =\int_x^y\frac{d\theta(t)}{\log t} = \frac{\theta(y)}{\log y} - \frac{\theta(x)}{\log x} +\int_x^y\theta(t)\left(-\frac{1}{\log t}\right)'dt.\end{equation}
Suppose that \(\theta(x)\sim x\) for large \(x\). Let \(\epsilon \gt 0\) be given, and take \(x\) large enough so that \[(1-\epsilon) x \lt \theta(x) \lt (1+\epsilon) x.\] Then from (\ref{pi}) we find \begin{eqnarray*} \pi(y)-\pi(x) &\leq & 2\epsilon \frac{x}{\log x}+ (1+\epsilon)\frac{y}{\log y} -(1+\epsilon)\frac{x}{\log x} + (1+\epsilon)\int_x^y t\left(-\frac{1}{\log t}\right)' dt\\[2ex] &=&2\epsilon \frac{x}{\log x}+(1+\epsilon) \left.\frac{t}{\log t}\right|_x^y - (1+\epsilon)\left.\frac{t}{\log t}\right|_x^y+(1+\epsilon) \int_x^y\frac{dt}{\log t} =2\epsilon \frac{x}{\log x}+(1+\epsilon)(\mbox{Li}(y)-\mbox{Li}(x)). \end{eqnarray*} So we find \[\frac{\pi(y)}{\mbox{Li}(y)}\leq \frac{\pi(x)}{\mbox{Li}(y)}+\frac{2\epsilon x}{\mbox{Li}(y) \log x}-(1+\epsilon)\frac{\mbox{Li}(x)}{\mbox{Li}(y)} + 1 + \epsilon.\] For a fixed \(x\), when \(y\) is large enough, we get \[\frac{\pi(y)}{\mbox{Li}(y)}\leq 1+2\epsilon.\] In a similar way, \[\frac{\pi(y)}{\mbox{Li}(y)}\geq 1-2\epsilon .\] So we conclude that \[\lim_{x\rightarrow \infty} \frac{\pi(x)}{\mbox{Li}(x)}=1.\]

By definition, \[\int_0^\infty \frac{d\psi(x)}{x^s}=\sum_{n=2}^\infty \frac{1}{n^s}\Lambda(n).\] The zeta function \[\zeta(s) =\sum_{n=1}^\infty \frac{1}{n^s}, \hspace{2ex}\mbox{Re}(s)\gt 1 \] converges locally uniformly for \(\mbox{Re}(s)\gt 1\), and using Euler's product formula we obtain by logarithmic differentiation \[-\frac{\zeta'(s)}{\zeta(s)} = \sum_p \frac{\log p}{p^s-1} =\sum_p \log p \sum_{n=1}^\infty p^{-sn}=\sum_{n=2}\frac{1}{n^s}\Lambda(n)=\int_0^\infty \frac{1}{x^s} d\psi , \hspace{2ex}\mbox{Re}(s)\gt 1 .\]

Suppose that \(\psi(x)\sim x\) as \(x\rightarrow \infty\). Note that \[\psi(x)=\sum_{n=1}^\infty \theta\left(x^{1/n}\right),\] and the sum is finite, because \(\theta\left(x^{1/n}\right)\) is zero as soon as \( 2^n\gt x \). Since \(\theta(x)=0\) for \(n\gt \log x /\log 2\), and \(\theta(x)\geq \theta\left(x^{1/n}\right)\) for \(n\geq 2\), we find \[\psi(x)\leq \theta(x) +\frac{\log x}{\log 2} \theta\left(x^{1/2}\right).\] So \begin{equation} \label{theta} \psi(x)-\frac{ \theta\left(x^{1/2}\right) \log x}{\log 2} \lt \theta(x) \lt \psi(x). \end{equation} Note that since \(\psi(x)\sim x\), if \(\epsilon \gt 0\), \[\frac{\theta(x)}{x^{1+\epsilon}} \leq \frac{\psi(x)}{x^{1+\epsilon}} \rightarrow 0 \mbox{ as } x\rightarrow \infty.\] So for small \(\epsilon \gt 0\), \[\frac{ \theta\left(x^{1/2}\right)\log x}{x}= \frac{\theta\left(x^{1/2}\right)}{\left(x^{1/2}\right)^{1+\epsilon}} \frac{\log x}{\left(x^{1/2}\right)^{1-\epsilon}}\rightarrow 0 \mbox{ as }x\rightarrow \infty .\] Dividing (\ref{theta}) by \(x\), we conclude that \[\lim_{x\rightarrow \infty}\frac{\theta(x)}{x} =1.\]

The starting point is the extension of the factorial (\ref{prodform}). The integral representation \begin{equation} \label{gamma} \Pi(s)=\int_0^\infty e^{-s}x^sdx. \end{equation} valid for \(\mbox{Re}(s)\gt -1\), was also known to Euler. Using (\ref{gamma}) with \(s\) replaced by \(s-1\), making the substitution \(x\mapsto nx\), and summing over \(n\geq 1\), we obtain the integral representation \begin{equation} \label{IntRep} \int_0^\infty \frac{x^{s-1}}{e^x-1}dx = \Pi(s-1)\sum_{n=1}^\infty \frac{1}{n^s}, \end{equation} valid for \(s\gt 1\).
In view of the integral representation (\ref{IntRep}), Riemann considers the contour integral \[ \int_{C} \frac{(-z)^s}{e^z-1}\frac{dz}{z},\] where the integral is along the Hankel path \(C=C(\delta,\varepsilon)\) that starts at distance \(\varepsilon\) above the positive real axis at infinity, circles counterclockwise around the origin with radius \(\delta\gt \varepsilon\), and returns to infinity traveling at distance \(\varepsilon\) below the positive real axis (see figure below).

\(\img{Fig1.svg}{}{35em}{}\)

The Hankel path \(C(\delta,\varepsilon)\)

It is not hard to see that the integral does not depend on the values of \(\varepsilon\) and \(\delta\) (as long as they are small enough so that no singularities are enclosed). This can be proved by truncating the path at \(x=R\) for some large \(R\), then joining it via vertical segments to a similar path \(C(\delta_1,\varepsilon_1)\), where \(\delta_1\lt \delta\), \(\varepsilon_1\lt \varepsilon\), in order to form a closed contour without singularities inside, so that by Cauchy's theorem the integral will be zero, then let \(R\rightarrow \infty\) and notice that the integral over the vertical segments will go to zero. This will prove that the integral over \(C(\delta,\varepsilon)\) is the same as the integral over \(C(\delta_1,\varepsilon_1)\).
On \(L_+\) we have \(-z=-x-i\varepsilon\) and \(\log(-x-i\varepsilon)\rightarrow \log|z| -i\pi\) as \(\varepsilon \rightarrow 0\), while on \(L_-\), \(-z=-x+i\varepsilon\), and \(\log(-x+i\varepsilon)\rightarrow \log|z| +i\pi\) as \(\varepsilon \rightarrow 0\). So the integral over \(L_+\) approaches \(\displaystyle e^{-i\pi s} \int_0^\infty \frac{x^s}{e^x-1}\frac{dx}{x}\) and the integral over \(L_-\) approaches \(\displaystyle e^{i\pi s} \int_0^\infty \frac{x^s}{e^x-1}\frac{dx}{x}\) as \(\varepsilon, \delta \rightarrow 0\). Also, if we write \(z=\delta e^{i\theta}\) on \(C_0\), the integrand is bounded by \(\displaystyle \frac{\delta^s}{|e^{\delta e^{i\theta}}-1|}=\frac{\delta^{s-1}}{|e^{i\theta} +O(\delta)|}\) and so the integral over \(C_0\) will approach zero as long as \(\mbox{Re}(s)\gt 1\). So we conclude that, for \(\mbox{Re}(s) \gt 1\), \[\int_C \frac{(-z)^s}{e^z-1}\frac{dz}{z} =2i\sin(\pi s)\int_0^\infty \frac{x^s}{e^x-1}\frac{dx}{x}=2i\sin(\pi s) \zeta(s) \Pi(s-1).\] But the left side of this equation is an entire function of \(s\). Using standard identities for the factorial function, this leads to the definition of the Riemann zeta function as in the statement of the theorem: \[ \zeta(s)=\frac{\Pi(-s)}{2\pi i}\int_{C} \frac{(-z)^s}{e^z-1}\frac{dz}{z}. \]
The integral on the right side defines an entire function. So the only singularities can come from the simple poles of \(\Pi(-s)\) at the positive integers. But for \(s=2,3,4,\ldots\) the function coincides with \(\sum_{n\geq 1} 1/n^s\), so the simple poles of \(\Pi(-s)\) must be cancelled by zeros of the integral, while at \(s=1\) we know that \(\sum_{n\geq 1} 1/n^s\) diverges, so there must be a simple pole (with residue 1) coming from \(\Pi(-s)\). So formula (\ref{zeta}) defines a function for all complex values of \(s\) except \(s=1\), where it has a simple pole, and it coincides with the previous definition for \(\mbox{Re}(s)\gt 1\).

The Maclaurin expansion \[\frac{x}{e^x-1}=\sum_{n=0}^\infty \frac{B_n}{n!}x^n\] introduces the Bernoulli numbers \(B_n\) in the evaluation of \(\zeta(s)\). For \(s=-n\), \(n=0,1,2,3,\ldots\), the integrals over \(L_+\) and \(L_-\) will cancel each other, and the integral over \(L_0\) as \(\delta, \varepsilon \rightarrow 0\) can easily be evaluated to obtain \[ \zeta(-n)=(-1)^n\frac{B_{n+1}}{n+1}. \] In particular, since \(B_n=0\) for \(n\) odd and greater than 1, we find \(\zeta(-2n)=0\) for all positive integers \(n\).

1. Formula (\ref{IntRep}) in the proof of Theorem 1 that led to the analytic continuation of \(\zeta(s)\) from \(s\gt 1\) to the whole complex plane (except \(s=1\)) was obtained by substituting \(nx\) for \(x\) in the integral representation (\ref{gamma}) for \(\Pi(s-1)\), and then summing over \(n\). If we instead make the substitution \(x\mapsto n^2\pi x\) in the integral representation for \(\Pi(s/2-1)\) and sum, we find \begin{equation} \label{intpsi} \Pi\left(\frac{s}{2}-1\right)\pi^{-s/2} \zeta(s) = \int_0^\infty \phi(x)x^{s/2}\frac{dx}{x}, \end{equation} where \[\phi(x)= \sum_{n=1}^\infty e^{-\pi n^2x}.\] The function \(\phi(x)\) is related to the Jocobi theta function \( \displaystyle G(u)=\vartheta(0;iu^2)=\sum_{n\in {\mathbb Z}} e^{-\pi n^2 u^2}\) by \[2\phi(u^2) +1 =G(u)\] and so the functional equation for the theta function Jacobi theta function \(\vartheta(z;\tau)\) is defined for \(z\in {\mathbb C}\) and \(\mbox{Im} \tau \gt 0\) to be \[\vartheta(z;\tau)=\sum_{n\in {\mathbb Z}} e^{\pi i n^2\tau +2\pi i nz}.\] Consider the special case \[G(u)=\vartheta(0;iu^2)=\sum_{n\in {\mathbb Z}} e^{-\pi n^2 u^2}, u\gt 0\] We prove that \(G\) satisfies the functional equation \begin{equation} \label{functional}G(u)=\frac{1}{u}G\left(\frac{1}{u}\right).\end{equation} Let \(u\gt 0\) be fixed, and consider the function \[f(x)=e^{-\pi x^2/u^2}.\] Let \[\hat{f}(s)=\int_{-\infty}^\infty f(x) e^{2\pi i xs}dx\] be its Fourier transform. Then by Parseval's theorem Let \(f: {\mathbb R} \rightarrow {\mathbb C}\) be a function that goes to zero sufficiently fast at \(\pm \infty\). The Fourier transform of \(f\) is \[\hat{f}(u) =\int_{-\infty}^\infty f(x) e^{2\pi i x u} dx.\] We can "periodify" \(f\) by defining \begin{equation}\label{periodify} F(x) =\sum_{n\in {\mathbb Z}} f(x+n).\end{equation} Then by the theory of Fourier series [see Rudin, Real and Complex Analysis], \begin{equation} \label{Fourier} F(x) = \sum_{n\in {\mathbb Z}} a_n e^{2\pi i nx},\end{equation} where \[a_n= \int_0^1F(x) e^{-2\pi i nx}dx.\] We then find \[a_n=\int_0^1 \sum_{m\in{\mathbb Z}} f(x+m)e^{-2\pi i nx}dx =\sum_{m\in{\mathbb Z}}\int_0^1 f(x+m) e^{-2\pi i nx} dx \\ =\sum_{m\in{\mathbb Z}} \int_m^{m+1} f(t) e^{-2\pi i n(t-m)}dt = \sum_{m\in{\mathbb Z}} \int_m^{m+1} f(t) e^{-2\pi i nt}dt = \int_{-\infty}^\infty f(t) e^{-2\pi i nt} dt = \hat{f}(-n).\] Setting \(x=0\) in (\ref{periodify}) and (\ref{Fourier}), we find \[ \sum_{n\in{\mathbb Z}} f(n) =F(0) = \sum_{n\in{\mathbb Z}} a_n=\sum_{n\in{\mathbb Z}}\hat{f}(-n)=\sum_{n\in{\mathbb Z}}\hat{f}(n).\] This proves Parseval's Theorem: \[\sum_{n\in{\mathbb Z}}f(n) = \sum_{n\in{\mathbb Z}} \hat{f}(n).\]
   \begin{equation} \label{Parseval}\sum_{n\in {\mathbb Z}} f(n) =\sum_{n\in{\mathbb Z}} \hat{f}(n).\end{equation} We find \[\sum_{n\in {\mathbb Z}} f(n) =\sum_{n\in {\mathbb Z}}e^{-\pi n^2/u^2}=G\left(\frac{1}{u}\right).\] Using the integral formula \[\int_{-\infty}^\infty e^{-\pi x^2}dx=1\] and Cauchy's theorem to change the path of integration from \((-\infty +iu, \infty+iu)\) to \((-\infty,\infty)\), we also find \[\hat{f}(n) =\int_{-\infty}^\infty e^{-\pi x^2/u^2}e^{2\pi i xn}dx= u\int_{-\infty}^\infty e^{-\pi t^2}e^{2\pi i u n t}dt= ue^{-\pi u^2 n^2}\int_{-\infty}^{\infty} e^{-\pi(t-iu)^2} dt\\ =ue^{-\pi u^2n^2} \int_{-\infty +iu}^{\infty +iu}e^{-\pi z^2} dz=ue^{-\pi u^2n^2}\int_{-\infty}^\infty e^{-\pi x^2}dx=ue^{-\pi u^2n^2}.\] So we conclude that \[\sum_{n\in {\mathbb Z}} \hat{f}(n) =u G(u),\] and so by Parseval's theorem (\ref{Parseval}) the functional equation (\ref{functional}) follows.
   
   \(uG(u)=G(1/u)\) for \(G(u)\) gives \begin{equation} \label{functpsi} \phi(x) =\frac{1}{\sqrt{x}}\phi\left(\frac{1}{x}\right) +\frac{1}{\sqrt{x}}-\frac{1}{2}. \end{equation} Differentiation of this identity gives \begin{equation} \label{psi} \phi(1)+4\phi'(1)+\frac{1}{2}=0. \end{equation} By splitting the integral (\ref{intpsi}) from \(0\) to \(1\) and \(1\) to \(\infty\), and then using the functional equation (\ref{functpsi}), some routine calculations give \begin{equation} \label{entint} \Pi\left(\frac{s}{2}-1\right) \pi^{-s/2} \zeta(s) = \int_1^\infty \phi(x) \left(x^{s/2}+x^{(1-s)/2}\right) \frac{dx}{x} -\left(\frac{1}{s}+\frac{1}{1-s}\right). \end{equation} The integral on the right side of (\ref{entint}) is an entire function of \(s\). So multiplying the equation by \(s(s-1)/2\) we get an entire function, denoted by \(\xi(s)\): \begin{equation} \label{xi} \xi(s)=(s-1)\Pi\left(\frac{s}{2}\right)\pi^{-s/2}\zeta(s) =\frac{1}{2} -\frac{s(1-s)}{2}\int_1^\infty \phi(x)\left(x^{s/2}+x^{(1-s)/2}\right)\frac{dx}{x}. \end{equation}
2. The right side of (\ref{xi}) is invariant under \(s\rightarrow 1-s\), and so we obtain the functional equation \[\xi(s)=\xi(1-s).\]
3. We show that \(\zeta(s)\) (and hence \(\xi(s)\)) has no zeros for \(\mbox{Re}(s)\gt 1\). Let \(s\in {\mathbb C}\) with \(\sigma=\mbox{Re}(s)\gt 1\), and let \((p_n:n\geq 1)\) be the prime numbers. Let \(a_n=(p_n^s-1)^{-1}\), so that \(1+a_n=(1-p_n^{-s})^{-1}\). Since \[\sum_{n=1}^\infty |a_n|=\sum_{n=1}^\infty \frac{1}{|p_n^s-1|}\leq \sum_{n=2}^\infty \frac{1}{n^\sigma-1}\lt \infty ,\] we conclude from basic results on infinite products
   Suppose \(z_k, k\geq 1\) are complex numbers, and let \(\displaystyle p_n=z_1z_2\cdots z_n=\prod_{k=1}^nz_k\).

   1. If \( \displaystyle \lim_{n\rightarrow \infty}p_n\) exists and is non-zero, we denote it by \(\displaystyle \prod_{k=1}^\infty z_k\), we call it the infinite product of the sequence \(z_k\), and say that the infinite product converges. In this case, clearly we must have \(z_k\neq 0\) for all \(k\), and if we let \( \displaystyle p= \prod_{k=1}^\infty z_k\), we have \[z_n=\frac{p_n}{p_{n-1}}\rightarrow \frac{p}{p} = 1 \mbox{ as } n\rightarrow \infty.\] So a necessary condition for the infinite product \( \displaystyle \prod_{k=1}^\infty z_k \) to converge is that \(z_k\rightarrow 1\) as \(k\rightarrow \infty\).
   2. If \(z_k\neq 0\) for all \(k\) but \(\displaystyle \lim_{n\rightarrow \infty} p_n=0\), we say that the infinite product diverges to zero. An example is the case \(z_k=a\) for all \(k\), where \(|a|\lt 1\).
   3. If \(z_k=0\) for some \(k\), but there is some \(m\) such that \(z_k\neq 0\) for \(k\geq m\), and the infinite product \(z_m z_{m+1} z_{m+2}\cdots\) converges, we say that the infinite product converges to zero. Note that in this terminology (used for example by Whittaker & Watson in A course of modern analysis) an infinite product that converges to zero does not converge. Other sources will say that the infinite product diverges as soon as the \(\displaystyle \lim_{n\rightarrow \infty}p_n=0\).
   
   
   Since \[\log\prod_{k=1}^n(1+a_k) =\sum_{k=1}^n \log(1+a_k),\] we see from the definitions that \[\prod_{n=1}^\infty (1+a_n)\mbox{ converges } \Longleftrightarrow \sum_{n=1}^\infty \log(1+a_n) \mbox{ converges }.\] Note that if \(\displaystyle \sum_{n=1}^\infty \log(1+a_n)\) converges, then \[\prod_{n=1}^\infty (1+a_n) =\exp\left(\sum_{n=1}^\infty \log(1+a_n)\right) \neq 0.\]
   
   
   Let \(a_n\) be a sequence of complex numbers, with \(a_n\neq -1\) for all \(n\). So \(\log(1+a_n)\) is defined for all \(n\). Suppose \(a_n\rightarrow 0\) as \(n\rightarrow \infty\). Find \(m\) such that \(|a_k|\leq 1/2\) for \(k\geq m\). Then we find, for \(k\geq m\), \[\log(1+a_k)=a_k\left(1+\sum_{n=1}^\infty \frac{(-1)^n}{n+1}a_k^n\right).\] Since \[ \left|\sum_{n=1}^\infty \frac{(-1)^n}{n+1}a_k^n\right|\leq \sum_{n=1}^\infty \frac{1}{n+1}|a_k|^n\leq \frac{1}{2} \sum_{n=1}^\infty \frac{1}{2^n}=\frac{1}{2},\] we find \[\frac{1}{2}|a_k|\leq |\log(1+a_k)|\leq\frac{3}{2} |a_k|.\] So by the comparison test we see that \(\displaystyle \sum_{n=1}^\infty a_n\) converges absolutely if and only if \(\displaystyle \sum_{n=1}^\infty \log(1+a_n)\) converges absolutely.
   
   
   We say that the infinite product \(\displaystyle \prod_{n=1}^\infty (1+a_n)\) converges absolutely if the infinite series \(\displaystyle \sum_{n=1}^\infty \log(1+a_n)\) converges absolutely. If \(I_+=\{i:a_i\geq 0\}\) and \(I_-=\{i:a_i\lt 0\}\), absolute convergence of \(\displaystyle \prod_{n=1}^\infty (1+a_n)\) means that both \(\displaystyle \prod_{i\in I_+}(1+a_i) \) and \( \displaystyle \prod_{i\in I_-}(1+a_i)\) converge. By the previous item, \(\displaystyle \prod_{n=1}^\infty (1+a_n)\) converges absolutely if and only if \(\displaystyle \sum_{n=1}^\infty a_n\) converges absolutely.
   that \[\sum_{n=1}^\infty \log(1+a_n) = \log \prod_{n=1}^\infty \left(1-\frac{1}{p_n^s}\right)^{-1} =L \in {\mathbb C},\] and so \[\zeta(s)=\prod_{n=1}^\infty \left(1-\frac{1}{p_n^s}\right)^{-1}=e^L\neq 0.\] Using the functional equation \(\xi(s)=\xi(1-s)\) it follows that \(\xi(s)\) has no zeros outside \(0\leq \mbox{Re}(s) \leq 1\).
   
   
4. Using the estimates \[\phi(x)=\sum_{n=1}^\infty e^{-\pi n^2x}\leq \sum_{n=1}^\infty e^{-\pi nx}=\frac{1}{e^{\pi x}-1}\] and \[|\phi'(x)|=\pi \sum_{n=1}^\infty n^2 e^{-\pi n^2x}\leq \pi \sum_{n=1}^\infty n^2 e^{-\pi nx}= \pi e^{-\pi x}\frac{1+e^{-\pi x}}{(1-e^{-\pi x})^3},\] we see that for every \(s\), \[ |x^s \phi(x)|\rightarrow 0 \mbox{ and } |x^s\phi'(x)|\rightarrow 0 \mbox{ as } x\rightarrow \infty . \] Integrating (\ref{xi}) by parts twice and using (\ref{psi}) we then find \[\xi(s) = 2 \int_1^\infty \left(x^{s/2-1/2} +x^{-s/2}\right) \left(x^{3/2} \phi'(x)\right)'dx\\ =4\int_1^\infty x^{-1/4}\cosh\left(\frac{1}{2}\left(s-\frac{1}{2}\right)\log x \right) \left(x^{3/2} \phi'(x)\right)'dx.\] Using the Maclaurin expansion for \(\cosh\), this gives \[ \xi(s)=\sum_{n=0}^\infty c_n \left(s-\frac{1}{2}\right)^{2n}, \mbox{ or } \xi\left(\frac{1}{2}+it\right) =\sum_{n=0}^\infty (-1)^n c_n t^{2n}, \] where \[c_n=\frac{4}{(2n)!}\int_1^\infty x^{-1/4}\left(\frac{\log x}{2}\right)^{2n} \left(x^{3/2}\phi'(x)\right)' dx.\] We find \[\frac{d}{dx}\left(x^{3/2}\phi'(x)\right)= \pi x^{1/2} \sum_{n=1}^\infty n^2\left(\pi xn^2-\frac{3}{2}\right) e^{-\pi n^2 x}\] and so all coefficients \(c_n\) are positive.

It is reasonable to expect that in order to prove that a product \[\prod_{\rho} \left(1-\frac{s}{\rho}\right)\] converges we need to have an estimate of how many zeros \(\rho\) there are. The first step in the proof will be to put a bound on the growth of \(\xi(s)\), as described in the next result.

Lemma

For all large enough \(R\) and \(|s-1/2|\leq R\), we have \(|\xi(s)|\leq R^R.\)

This will be used to provide the estimate on the number of zeros. Proof of Lemma . According to Theorem 2 d., the coefficients \(c_n\) for the power series of \(\xi\) as an even function of \(s-1/2\) are positive real numbers. Hence if \(|s-1/2|\leq R\), we find \[|\xi(s)|=\left|\sum_{n=0}^\infty c_n \left(s-\frac{1}{2}\right)^{2n}\right|\leq \sum_{n=0}^\infty c_n \left|s-\frac{1}{2}\right|^{2n}\leq \sum_{n=0}^\infty c_n R^{2n}=\xi(R+1/2). \] For a fixed \(R\gt 0\), let \(N\in {\mathbb N}\) be such that \[\frac{1}{4} +\frac{R}{2} \leq N \lt \frac{1}{4}+\frac{R}{2}+1.\] Recall that \[\xi(s) =(s-1) \Pi\left(\frac{s}{2}\right) \pi^{-s/2} \zeta(s).\] Since \(\displaystyle \xi(t) = \sum_{n=0}^\infty c_n (t-1/2)^{2n}\) is increasing for \(t\) real and \(t\geq 1/2\), we have \[\xi\left(\frac{1}{2}+R\right) \leq \xi(2N)=(2N-1) \Pi(N) \pi^{-N}\zeta(2N)\leq (2N) N! \zeta(2) \\ \leq 2\zeta(2) N^{N+1}\leq 2\zeta(2) \left(\frac{1}{4}+\frac{R}{2} +1\right)^{1/4+R/2+2} \leq R^R\] for \(R\) large enough. \(\blacksquare\)

Lemma

For \(R\gt 0\), let \(n(R)\) be the number of zeros of \(\xi(s)\) such that \(|s-1/2|\leq R\). Then \(n(R)\leq 3R\log R\) for all large enough \(R\).

Proof of Lemma . If \(R\gt 0\) and \(f\) is a function defined and analytic for \(|z|\leq R\) and with \(f(0)\neq 0\), let \[Z(f;R)=\{z\in {\mathbb C}: f(z)=0 \mbox{ and } |z|\lt R\}\] \[Z_0(f;R)=\{z\in{\mathbb C}: f(z)=0 \mbox{ and } |z|\leq R\}.\] Then \(Z_0(f;R)\subset Z(f;2R)\). If \(z\in Z(f;2R)\), then \(2R/|z|\gt 1\), so \(\log(|2R/z|)\gt 0\), and we find \[\sum_{z\in Z(f;2R)}\log\left|\frac{2R}{z}\right| \geq \sum_{z\in Z_0(f;R)}\log\left|\frac{2R}{z}\right|\geq \log(2) |Z_0(f;R)|,\] because \(2R/|z|\geq 2\) for \(z\in Z_0(f;R)\). If \(f\) has no zeros on the circle \(|z|=2R\), Jensen's formula Suppose that \(f\) is analytic on an open set containing \(\overline{B}(0,R)\), and \(f(z)\neq 0\) in \(B(0,R)\). Integrating the Maclaurin series for \(f\) we find \[f(0)=\frac{1}{2\pi}\int_0^{2\pi}f\left(Re^{i\theta}\right)d\theta.\] Note that for any \(w\neq 0\) we have \(\mbox{Re}(\log w)=\log|w|\). So applying the above formula to \(\log(f(z))\) and taking the real part, we find \[\log|f(0)|=\frac{1}{2\pi}\int_0^{2\pi}\log\left|f(R e^{i\theta})\right|d\theta.\] Now suppose \(f\) has a simple zero at \(z_0\in B(0,R)\), and \(z_0\neq 0\). If we can find some \(\phi(z)\) with a simple pole at \(z_0\), without zeros for \(|z|\leq R\), and such that \(|\phi(z)|=1\) on \(|z|=R\), then the function \(f(z)\phi(z)\) has no zeros in \(B(0,R)\), and since \(|f(z)\phi(z)|=|f(z)|\) for \(|z|=R\), applying the above formula to \(f(z)\phi(z)\) we have \[\log(|f(0)\phi(0)|) =\frac{1}{2\pi} \int_0^{2\pi} \log(|f(Re^{i\theta})| d\theta.\] Such a factor is \[\phi(z) =\frac{R^2-\overline{z_0}z}{R(z-z_0)}, \] as we can see by noting that \(R^2-\overline{z_0}z\neq 0\) in \(B(0,R)\), and for \(|z|=R\), \[\left|\frac{R^2-\overline{z_0}z}{R(z-z_0)}\right|=\left|\frac{(R^2-\overline{z_0}z)\overline{z}}{R(z-z_0)\overline{z}}\right|= \left|\frac{R^2(\overline{z}-\overline{z_0})}{R^2(z-z_0)}\right|=1.\] Since \(\phi(0)=R/|z_0|\), we find \[\log\left(|f(0)|\frac{R}{|z_0|}\right)=\frac{1}{2\pi}\int_0^{2\pi} \log\left|f(R e^{i\theta})\right| d\theta.\] In the same way, if \(f(0)\neq 0\) and \(f\) has zeros \(z_1,\ldots, z_k\) in \(B(0,R)\) (repeated according to multiplicity), consider \(\displaystyle f(z)\frac{R^2-\overline{z_1}z}{R(z-z_1)}\cdots \frac{R^2-\overline{z_k}z}{R(z-z_k)}\). We then find Jensen's formula: \[ \log\left(|f(0)|\frac{R}{|z_1|}\cdots \frac{R}{|z_k|}\right)=\frac{1}{2\pi}\int_0^{2\pi} \log\left|f(R e^{i\theta})\right| d\theta.\] applies and we find \begin{equation} \label{Jen} \log|f(0)|+\log(2)|Z_0(f;R)|\leq \log|f(0)|+\sum_{z\in Z(f;2R)} \log\left|\frac{2R}{z}\right|\\ \hspace{25ex} =\displaystyle \log\left(|f(0)|\prod_{z\in Z(f;2R)}\frac{2R}{|z|}\right)=\frac{1}{2\pi}\int_0^{2\pi} \log\left|f(2R e^{i\theta})\right|d\theta. \end{equation} Now let \(f(z)=\xi(z+1/2)\). Then \(n(R)=|Z_0(f;R)|\), and by Lemma 1, we find \[\left|f\left(2Re^{i\theta}\right)\right| =\left|\xi\left(\frac{1}{2}+2Re^{i\theta}\right)\right|\leq \left(2R\right)^{2R}.\] If \(\xi\) has no zeros on the circle \(|s-1/2|= 2R\), we find from (\ref{Jen}) \[\log\left|\xi\left(\frac{1}{2}\right)\right|+\log(2) n(R) \leq 2R\log(2R) =2R\log R +2R\log(2),\] or \[n(R) \leq \frac{2}{\log 2} R\log R +2R -\frac{\log|\xi(1/2)|}{\log 2}\] Let \(R_0\) be such that the right side of the above inequality is \(\leq 3 R \log R\) for \(R\geq R_0\). So \(n(R)\leq 3R\log R\) if \(R\geq R_0\) and there are no zeros on the circle \(|s-1/2|=2R\). If \(R\geq R_0\) and there are some zeros on the circle \(|s-1/2|=2R\), we can find some \(\epsilon_R \gt 0\) such that \(f\) has no zeros on \(|s-1/2|=2(R+\epsilon)\) for all \(0\lt \epsilon \leq \epsilon_R\), and then \[n(R)\leq n(R+\epsilon)\leq 3(R+\epsilon)\log(R+\epsilon) \] for all \(0\lt \epsilon \leq \epsilon_R\). Letting \(\epsilon\rightarrow 0\), we find \(n(R)\leq 3R\log R\). So the inequality holds for all \(R\geq R_0\). \(\blacksquare\)

The above estimate of the number of zeros of \(\xi(s)\) allows us to prove the next result.

Lemma

Let \(\rho_k: k=1,2,3,\ldots\) be the zeros of \(\xi(s)\) ordered so that \(|\rho_k -1/2|\leq |\rho_{k+1} -1/2|\). Let \(\varepsilon \gt 0\) be given. Then the series

\[\sum_{k=1}^\infty \frac{1}{|\rho_k-1/2|^{1+\varepsilon}}\]

converges, and so in particular the infinite product

\[P(s)=\prod_{k=1}^\infty \left(1-\frac{(s-1/2)^2}{(\rho-1/2)^2}\right)\]

converges.

Proof of Lemma . Define \(R_n\) implicitly by \(4R_n \log R_n=n\). Then if \(n(R)\) is as defined in the previous lemma, we find \(n(R_n)\leq 3R_n\log R_n =3n/4\lt n\). This means that the \(n\)-th zero is outside the circle of radius \(R_n\) centered at \(1/2\), so \(|\rho_n-1/2|\gt R_n\), and \[\frac{1}{|\rho_n-1/2|^{1+\epsilon}}\lt \frac{1}{R_n^{1+\epsilon}}=\frac{(4\log R_n)^{1+\epsilon}}{n^{1+\epsilon}}= \frac{(4\log R_n)^{1+\epsilon}}{n^{1+\epsilon/2}}\frac{1}{n^{1+\epsilon/2}}.\] Since \(\log R_n \lt \log 4+\log R_n +\log \log R_n = \log(4R_n \log R_n)=\log n\), we find \[\frac{(4\log R_n)^{1+\epsilon}}{n^{1+\epsilon/2}} \lt \frac{(4\log n)^{1+\epsilon}}{n^{1+\epsilon/2}}\lt 1\] for large \(n\), and so \[\frac{1}{|\rho_n-1/2|^{1+\epsilon}}\lt \frac{1}{n^{1+\epsilon/2}}\] holds for all large enough \(n\), and the series \(\displaystyle \sum_{n=1}^\infty \frac{1}{|\rho_n-1/2|^{1+\epsilon}}\) converges.\(\blacksquare\)

The infinite product \(P(s)\) has exactly the same zeros as \(\xi(s)\). We will show that in fact it is the same as \(\xi(s)\), up to a multiplicative constant. Our main tool is the next result from complex analysis, that can be thought of as an analogue of Liouville's theorem when an entire function is also even.

Lemma

Suppose that \(f(s)\) is entire and even and for each \(\varepsilon \gt 0\), \(\mbox{Re} (f(s))\leq \varepsilon |s|^2\) for all large enough \(|s|\). Then \(f(s)\) is a constant function.

Proof of Lemma . The proof of Lemma depends on the following result, that provides a bound on \(|f(s)|\) on a disk \(|s|\leq r\) if we are given a bound for \(\mbox{Re}f(s)\) on a larger disk \(|s|\leq R\).

Lemma A
Suppose \(f(s)\) is analytic on \(|s|\leq R\), \(f(0)=0\), and \(r\lt R\). If \(\mbox{Re}f(s)\leq M \) for \(|s|\leq R\), then \[ |f(s)|\leq \frac{2Mr}{R-r} \mbox{ for } |s|\leq r. \] Proof of Lemma A. Write \(f(s)=u(s)+iv(s)\). Since \(u(s)\) is harmonic, its maximum on \(|s|\leq R\) is achieved at the boundary. Let \[\phi(s)=\frac{f(s)}{s(2M-f(s))}\] Note that \[u-2M \leq u \leq M \leq 2M-u,\] and so \[|u(s)|\leq 2M-u(s).\] Then we find \[|2M-f(s)|^2=(2M-u(s))^2+v(s)^2\geq u(s)^2+v(s)^2=|f(s)|^2.\] Hence \[|\phi(s)|\leq \frac{1}{|s|}=\frac{1}{R} \mbox{ on } |s|\leq R.\] Then \[|f(s)|=\left|\frac{2Ms\phi(s)}{1+s\phi(s)}\right|\leq \frac{2Mr/R}{1-r/R}=\frac{2Mr}{R-r} \mbox{ on } |s|\leq r. \blacksquare\]
To prove Lemma , suppose that \(f(s)\) is entire and even, and for every \(\epsilon \gt 0\), \(\mbox{Re}f(s)\leq \epsilon |s|^2\) for large enough |s|. Assume first \(f(0)=0\). Write \[f(s)=\sum_{n=2}a_ns^n.\] Then given \(r\gt 0\) and using Lemma A with \(R=2r\), we find \[|a_n|=\left|\frac{1}{2\pi i}\int_{|s|=r}\frac{f(re^{i\theta}}{s^{n+1}} ds\right| \leq \frac{1}{2\pi} \int_0^{2\pi} \frac{|f(re^{i\theta})|}{r^n}d\theta \leq \frac{1}{2\pi} \frac{\epsilon 4r^2}{r^n}=\frac{2\epsilon}{\pi r^{n-2}}.\] Since \(n\geq 2\) and \(\epsilon \) is arbitrary, we must have \(a_n=0\).
If \(f(0)\neq 0\), we can consider \(g(s)=f(s)-f(0)\), that satisfies the same growth condition, and conclude that \(f(s)=f(0)\). \(\blacksquare\)

The next result provides the asymptotic estimate needed to make use of Lemma . Its proof depends on the estimate given by Lemma .

Lemma

Let \(\varepsilon \gt 0 \) be given. Then for all large enough \(|s-1/2|\),

\[ \mbox{Re} \log \frac{\xi(s)}{P(s)} \leq \left|s-\frac{1}{2}\right|^{1+\varepsilon}. \]

Proof of Lemma . To prove Lemma we will make use of three Lemmas.

Lemma B If \(|z|\leq 1/2\), then \(-\mbox{Re}\log (1-z) \leq 2|z|\).
Proof of Lemma B. \[-\log(1-z)=\int_0^z\frac{dw}{1-w}.\] So \[-\mbox{Re}\log(1-z)=\mbox{Re}\int_0^z\frac{dw}{1-w}\leq \left|\int_0^z\frac{dw}{1-w}\right|\leq |z|\max\left\{\frac{1}{|1-w|}: |w|\leq \frac{1}{2}\right\} \leq 2|z| \hspace{2ex}\blacksquare\]
Lemma C Suppose \(a_n\geq 0\), and \(\displaystyle \sum_{n=1}^\infty a_n\) converges. For \(t\gt 0\), let \(A(t)=\{n\in {\mathbb N}: a_n\leq t\}\). Then \[\lim_{t\rightarrow 0^+} \sum_{n\in A(t)} a_n=0.\]
Proof of Lemma C. Let \(\epsilon \gt 0\). Find \(m\) such that \(\displaystyle \sum_{n=m}^\infty a_n\lt \epsilon\). If \(a_n=0\) for \(1\leq n \leq m\), then \(\displaystyle \sum_{n\in A(t)}a_n \leq \sum_{n=m}^\infty a_n\lt \epsilon\). Otherwise, let \(\delta = \min\{a_n: a_n\gt 0, 1\leq n \leq m\}\). Suppose \(t\lt \delta\). If \(n\leq m\), then either \(a_n=0\), or \(a_n\geq \delta \gt t\), and then \(n\not \in A(t)\), so \[\sum_{n\in A(t)}a_n \leq \sum_{n=m}^\infty a_n \lt \epsilon \hspace{2ex} \blacksquare\]
Lemma D Suppose \(z_n\) is such that for every \(\epsilon \gt 0\), \(\displaystyle \sum_{n=1}^\infty \frac{1}{|z_n|^{1+\epsilon}}\) converges. For \(R\gt 0\) and \(s\in {\mathbb C}\), let \[v_R(s)=\mbox{Re}\log \frac{1}{\displaystyle \prod_{|z_n|\geq 2R}\left(1-\frac{s^2}{z_n^2}\right)}.\] Then given \(\epsilon \gt 0\), if \(|s|=R\) and \(R\) is large enough, \(v_R(s) \leq R^{1+\epsilon}\).

Proof of Lemma D. Let \(\epsilon\gt 0\), and suppose \(|s|=R\). Then \begin{eqnarray*} v_R(s)&=&-\mbox{Re} \sum_{|z_n|\geq 2R}\log\left(1-\frac{s^2}{z_n^2}\right)\\ &\leq &2 \sum_{|z_n|\geq 2R}\frac{R^2}{|z_n|^2} & \mbox{ from Lemma B}\\ &=& 2\sum_{|z_n|\geq 2R} \left(\frac{R}{|z_n|}\right)^{1-\epsilon}\left(\frac{R}{|z_n|}\right)^{1+\epsilon}\\ &\leq & 2\left(\frac{1}{2}\right)^{1-\epsilon} \sum_{|z_n|\geq 2R} \left(\frac{R}{|z_n|}\right)^{1+\epsilon}\\ &=&2^\epsilon R^{1+\epsilon} \sum_{|z_n|\geq 2R} \left(\frac{1}{|z_n|}\right)^{1+\epsilon}. \end{eqnarray*} From Lemma C, the sum will go to zero as \(R\rightarrow \infty\), so \(v_R(s) \leq R^{1+\epsilon}\) when \(|s|=R\) and \(R\) is large enough.\( \blacksquare\)



To prove Lemma , consider the function \begin{equation} \label{F} F(s)=\frac{\xi(s)}{P(s)}. \end{equation} For each zero \(\rho\) of \(\xi(s)\), the factor \(1-(s-1/2)/(\rho-1/2)=2(s-\rho)/(1-2\rho)\) in the product \(P(s)\) will cancel the zero of \(\xi(s)\) at \(\rho\), and so \(F(s) \) is entire and without zeros. Fix \(R\gt 0\), and write \[A_R(s)=\prod_{|\rho-1/2|\leq 2R}\left(1-\left(\frac{s-1/2}{\rho-1/2}\right)^2\right),\] \[B_R(s)=\prod_{|\rho-1/2|\gt 2R}\left(1-\left(\frac{s-1/2}{\rho-1/2}\right)^2\right),\] so that \(P(s)=A_R(s)B_R(s)\). Then on the disk \(|s-1/2|\leq R\), \(\xi(s)/A_R(s)\) is analytic and has no zeros, because the zeros of \(\xi(s)\) are cancelled by those of \(A_R(s)\) , and \(1/B_R(s)\) is analytic, because all the zeros of \(B_R\) are in \(|s-1/2|\gt 2R\). So we can consider \[ u_R(s)=\mbox{Re} \log \frac{\xi(s)}{A_R(s)}=\log \frac{|\xi(s)|}{\displaystyle \prod_{|\rho-1/2|\leq 2R}\left|1-\frac{(s-1/2)^2}{(\rho-1/2)^2}\right|}. \] \[ v_R(s)=\mbox{Re} \log \frac{1}{B_R(s)}=\log \frac{1}{\displaystyle \prod_{|\rho-1/2|\gt 2R}\left|1-\frac{(s-1/2)^2}{(\rho-1/2)^2}\right|}, \] both defined and harmonic on \(|s-1/2|\leq R\), and \[u_R(s)+v_R(s)=\mbox{Re} \log F(s).\] Now consider \(u_R(s)\) on \(|s-1/2|\leq 4R\). Then \(u_R\) is harmonic there except at the points \(s=\rho\) with \(2R\lt |\rho-1/2|\leq 4R\). But at those points \(u_R(s)\rightarrow -\infty\). If we draw closed disks of small radius \(\epsilon\) around each point of the finite set of zeros of \(\xi(s)\) in the disk \(|s-1/2|\leq 4R\), then \(u_R\) is harmonic on the same disk with those small disks removed, and so it will achieve its maximum on the boundary. By taking the radius of the disks small enough, since \(u\rightarrow -\infty\) on the boundary of each small disk, we conclude that the maximum of \(u\) is achieved on the circle \(|s-1/2|= 4R\). But if \(|s-1/2|=4R\), then for \(|\rho-1/2|\leq 2R\), we have \[\left|1-\frac{s-1/2}{\rho-1/2}\right|\geq \left|\frac{s-1/2}{\rho-1/2}\right|-1 =\frac{4R}{|\rho-1/2|}-1\geq \frac{4R}{2R}-1=1.\] So each factor in the denominator has absolute value at least \(1\), and

\begin{equation}\label{uR} \max \{u_R(s):|s-1/2|=R\}\leq \max\{u_R(s) : |s-1/2|\leq 4R\} = \max\{u_R(s): |s-1/2|=4R \\ \hspace{10ex} \leq \max\{\log|\xi(s)|: |s-1/2|=4R\} \leq \log(4R)^{4R}=4R\log(4R)\leq R^{1+\epsilon} \end{equation} for large \(R\).

We now prove the same inequality for \(v_R(s)\). By taking \(z_n=\rho-1/2\) in Lemma D, we find that for all \(\epsilon \gt 0\),

\begin{equation} \label{vR} v_R(s)=\mbox{Re} \log \frac{1}{B_R(s)}=\log \frac{1}{\displaystyle \prod_{|\rho-1/2|\gt 2R}\left|1-\frac{(s-1/2)^2}{(\rho-1/2)^2}\right|}\leq R^{1+\epsilon} \end{equation} for \(|s-1/2|=R\) and all large \(R\). Let now \(\epsilon \gt 0\) be given, and let \(F(s) =\xi(s)/P(s)\) as given in (\ref{F}). From (\ref{uR}) and (\ref{vR}), we find \(\mbox{Re}\log F(s) =u_R(s)+v_R(s) \leq 2|s-1/2|^{1+\epsilon/2}\leq |s-1/2|^{1+\epsilon}\) for large enough \(|s-1/2|\). So we conclude that
\[ \mbox{Re} \log F(s) \leq \left|s-\frac{1}{2}\right|^{1+\epsilon} \] for all large enough \(|s-1/2|\). \(\blacksquare\)

We can now derive the product expansion for \(\xi(s)\) by applying Lemma to the function \(F(s)=\xi(s)/P(s)\). This completes the proof of Theorem 3.

Taking the logarithmic derivative of (\ref{xiprod0}), and using the product formula (\ref{prodform}) for \(\Pi(s)\), we find \begin{equation} \label{logder} -\frac{\zeta'(s)}{\zeta(s)}=\sum_{n=1}^\infty \left[\frac{1}{2}\log\left(1+\frac{1}{n}\right)-\frac{1}{2n+s}\right]-\frac{1}{2} \log \pi+\frac{1}{s-1}+\sum_\rho\frac{1}{\rho -s}. \end{equation} Subtracting from (\ref{logder}) the same equation evaluated at \(s=-b\), we find (\ref{logderb}).

To justify the pointwise differentiation of infinite series, it is enough to show that the two infinite series in (\ref{logder}) converge locally uniformly. We find \[\frac{1}{2}\log\left(1+\frac{1}{n}\right) -\frac{1}{2n+s} =\frac{1}{2n^2}\sum_{k=0}^\infty \frac{(-1)^k}{n^k}\left(\frac{s^{k+1}}{2^{k+1}}-\frac{1}{k+2}\right)\] and so \[\left|\frac{1}{2}\log\left(1+\frac{1}{n}\right) -\frac{1}{2n+s} \right| \leq \frac{c}{n^2}\] for some constant \(c\), all \(s\) in a compact set, and all large enough \(n\). This proves that the pointwise differentiation of the first series is justified.
For the second series, we need to pair terms with \(\rho\) and \((1-\rho)\). Then we find \[\frac{1}{s-\rho}+\frac{1}{s-(1-\rho)} = \frac{2(s-1/2)}{(s-1/2)^2-(\rho-1/2)^2}\] and so \[\left|\frac{1}{s-\rho}+\frac{1}{s-(1-\rho)}\right|\leq \frac{K}{|\rho-1/2|^2}\] for a constant \(K\), \(s\) in a compact set, and \(|\rho -1/2|\) large enough. Then the series converges by Lemma in the proof of Theorem 3.

We will often need to consider the integral \[F_h(x,\beta)=\frac{1}{2\pi i}\int_{a-ih}^{a+ih}\frac{x^{s-\beta}}{s-\beta}ds\] where \(a\gt 0\), \(x\gt 0\), \(h\gt 0\) and \(\beta\) is a complex number with \(a\gt \mbox{Re}(\beta)\). The next lemma describes the rate of convergence of \(F_h(x,0)\) as \(h\rightarrow \infty\), and it will be used to justify taking limits inside integrals or infinite series. These estimates are derived by evaluating the integral along suitable large rectangles and making use of the residue theorem.

Lemma

\[ F_h(1,0) = \frac{1}{\pi} \tan^{-1}\left(\frac{h}{a}\right) \] \begin{equation} \label{xlt1} \mbox{If \(0\lt x \lt 1\), then }\hspace{5ex} |F_h(x,0)| \leq \frac{x^a}{\pi h |\log x|}. \end{equation} \begin{equation} \label{xgt1} \mbox{If \(x \gt 1\), then } \hspace{5ex} | F_h(x,0)-1| \leq \frac{x^a}{\pi h \log x} . \end{equation} In particular, \begin{equation} \label{main} \lim_{h\rightarrow \infty}\frac{1}{2\pi i}\int_{a-ih}^{a+ih}x^s\frac{ds}{s} = \begin{cases} 0 & \mbox{if } 0\lt x \lt 1 \\ \frac{1}{2} & \mbox{if } x=1\\ 1 & \mbox{if } x\gt 1 \end{cases}, \hspace{3ex} a\gt 0. \end{equation}

Proof of Lemma . \[F_h(1,0)= \frac{1}{2\pi i}\int_{a-ih}^{a+ih} \frac{ds}{s} = \frac{1}{2\pi}\int_{-h}^h \frac{a-it}{a^2+t^2}dt = \frac{a}{2\pi}\int_{-h}^h \frac{dt}{a^2+t^2}=\frac{1}{\pi} \tan^{-1}\left(\frac{h}{a}\right) .\] If \(0\lt x \lt 1\), consider the integral over the rectangle shown: \[\img{Ch3fig1.png}{}{}{16em}.\] The integral over the closed path is zero, because the integrand is analytic inside. The integral over the top or bottom sides of the rectangle are easily seen to be bounded by \[\frac{1}{2\pi}\frac{x^K-x^a}{h|\log x|},\] and the integral on the right side is bounded by \[\frac{1}{2\pi} \frac{x^K}{K}(2h).\] Letting \(K\rightarrow \infty\), we find \[ |F_h(x,0)|=\left|\frac{1}{2\pi i} \int_{a-ih}^{a+ih} \frac{x^s}{s} ds\right| \leq \frac{x^a}{\pi h |\log x|}.\] If \(x\gt 1\), consider the integral over the rectangle: \[\img{Ch3fig2.png}{}{}{16em}\] The integrand has a simple pole at \(s=0\), and the integral over the closed path is \(1\). The integral over the top or bottom is bounded by \[\frac{1}{2\pi}\frac{x^a-x^{-K}}{h\log x}\] and the integral over the left side is bounded by \[\frac{1}{2\pi}\frac{2h}{K x^K}.\] Letting \(K\rightarrow \infty\), we find the bound \[ \left|F_h(x,0)-1\right|=\left|\frac{1}{2\pi i}\int_{a-ih}^{a+ih} \frac{x^s}{s}ds -1 \right| \leq \frac{x^a}{\pi h \log x}. \blacksquare \]
The next two lemmas are simple consequences of the above estimates. They will be needed to justify taking some limits inside integrals.

Lemma

\[ \mbox{If \(x \gt 1\) and \(a\gt \mbox{Re}(\beta)\), then } \hspace{2ex} \lim_{h\rightarrow \infty} F_h(x,\beta)=1. \]

Proof of Lemma . Consider the integral \[\frac{1}{2\pi i}\int_{a-i\infty}^{a+i\infty} x^s\frac{ds}{s-\beta},\] where \(x\gt 1 \), \(\beta=\sigma+i\tau\) and \(a\gt \sigma\). Let \(t=s-\beta\). Then \[\int_{a-ih}^{a+ih}x^s\frac{ds}{s-\beta} = x^\beta \int_{a-\sigma-i(h+\tau)}^{a-\sigma+i(h-\tau)}x^t\frac{dt}{t} =x^\beta \int_{a-\sigma-i(h+\tau)}^{a-\sigma+i(h+\tau)}x^t\frac{dt}{t} + x^\beta \int_{a-\sigma+i(h+\tau)}^{a-\sigma+i(h-\tau)}x^t\frac{dt}{t}.\] The limit of the first term as \(h\rightarrow \infty\) is \(2\pi i x^\beta\), by (\(*\)). For the second term, we find \[\left|x^\beta \int_{a-\sigma+i(h+\tau)}^{a-\sigma+i(h-\tau)}x^t\frac{dt}{t}\right|\leq x^{a-\sigma}\max\left\{\frac{1}{\sqrt{(a-\sigma)^2+y^2}}: h-\tau\leq y \leq h+\tau \right\}2\tau \leq \frac{2\tau x^{a-\sigma}}{h-\tau} \rightarrow 0 \mbox{ as } h\rightarrow \infty.\] So we conclude that \[\frac{1}{2\pi i}\int_{a-i\infty}^{a+i\infty} x^s\frac{ds}{s-\beta} = x^\beta, \hspace{2ex} x\gt 1, \hspace{2ex} a\gt \mbox{Re}(\beta). \blacksquare\]

Lemma

\[ \mbox{If \(x \gt 1, d\gt c \geq 0\), then } \hspace{5ex} \left|\frac{1}{2\pi i} \int_{a+ic}^{a+id}\frac{x^s}{s}ds\right| \leq K \frac{x^a}{(a+c)\log x}, \mbox{ where } K=\sqrt{2}\left(\frac{1}{\pi}+\frac{1}{4}\right). \]

Proof of Lemma . Integrating by parts we find \begin{equation} \label{parts} \int_{a+ic}^{a+id} \frac{x^s}{s}ds =\frac{x^{a+id}}{(a+id)\log x} - \frac{x^{a+ic}}{(a+ic)\log x} + \frac{ix^a}{\log x} \int_c^d\frac{x^{it}}{(a+it)^2}dt. \end{equation} Using the inequality \[\sqrt{a^2+c^2}\geq \frac{a+c}{\sqrt{2}},\] we find \[|a+id|\geq |a+ic|\geq \frac{a+c}{\sqrt{2}}\] and so the first two terms on the right side of (\ref{parts}) are bounded by \[\frac{x^a\sqrt{2}}{(a+c)\log x}.\] The integral on the right side of (\ref{parts}) is bounded by \[\left|\int_c^d\frac{x^{it}}{(a+it)^2}dt\right| \leq \int_0^\infty \frac{du}{a^2+(c+u)^2}\leq \int_0^\infty \frac{du}{a^2+c^2+u^2}=\frac{1}{\sqrt{a^2+c^2}}\frac{\pi}{2} \leq \frac{\sqrt{2}}{a+c} \frac{\pi}{2}.\] So we conclude that \[ \left|\frac{1}{2\pi i} \int_{a+ic}^{a+id}\frac{x^s}{s}ds\right| \leq K \frac{x^a}{(a+c)\log x},\] where \[K=\sqrt{2}\left(\frac{1}{\pi}+\frac{1}{4}\right). \blacksquare\]

The next lemma is obtained by carrying out the first substitution of \(-\zeta'(s)/\zeta(s)\) in \(I(b)\), as mentioned in item , using equation (\ref{int}).

Lemma

\begin{equation} \label{I0} I(0)=\psi(x)=\sum_{n\lt x} \Lambda(n) \end{equation} \begin{equation} \label{I1} I(1)=\frac{1}{x} \sum_{n\lt x} n\Lambda(n) \end{equation}

Proof of Lemma . Let \(x\gt 2\) be fixed, and \(m=\lceil x \rceil\), \(a\gt 1\). Then for each \(h\gt 0\), \begin{array}{l} \displaystyle \frac{1}{2\pi i}\int_{a-ih}^{a+ih}\left(-\frac{\zeta'(s)}{\zeta(s)}\right)x^s\frac{ds}{s+b}=\frac{1}{2\pi i}\int_{a-ih}^{a+ih}\sum_{n=2}^\infty \Lambda(n) \frac{x^s}{n^s}\frac{ds}{s+b}\\ \displaystyle = \frac{1}{2\pi i}\int_{a-ih}^{a+ih}\sum_{n=2}^{m-1}\Lambda(n)\left(\frac{x}{n}\right)^s\frac{ds}{s+b}+\Lambda(m)\frac{1}{2\pi i}\int_{a-ih}^{a+ih}\left(\frac{x}{m}\right)^s\frac{ds}{s+b} +\frac{1}{2\pi i}\int_{a-ih}^{a+ih}\sum_{n=m+1}^\infty \Lambda(n) \left(\frac{x}{n}\right)^s\frac{ds}{s+b}. \end{array} The first term involves a finite sum and its limit as \(h\rightarrow \infty\) is \[\sum_{n=2}^{m-1}\Lambda(n) \frac{1}{2\pi i}\int_{a-i\infty}^{a+i\infty}\left(\frac{x}{n}\right)^s\frac{ds}{s+b} = \sum_{n=2}^{m-1}\Lambda(n) \frac{n^b}{2\pi i x^b}\int_{a+b-i\infty}^{a+b+i\infty}\left(\frac{x}{n}\right)^t\frac{dt}{t}=\frac{1}{x^b}\sum_{n=2}^{m-1}n^b\Lambda(n) = \frac{1}{x^b}\sum_{n\lt x}n^b\Lambda(n),\] where we have used (\ref{main}). In a similar way, the limit of the second term is \[\Lambda(m) \frac{1}{2}[x=m].\] So if \(b=0\), the sum of the first two terms gives \[ \frac{1}{2}\sum_{n\lt x}\Lambda(n) + \frac{1}{2}\sum_{n\lt x}\Lambda(n) +\frac{1}{2}\Lambda(n)[n=x]=\frac{1}{2}\left(\sum_{n\lt x}\Lambda(n)+\sum_{n\leq x}\Lambda(n)\right) =\psi(x), \] while if \(b=1\) we get \[\frac{1}{2x}\left(\sum_{n\lt x}n\Lambda(n)+\sum_{n\leq x}n\Lambda(n)\right) \] For the third term, since the series converges uniformly on any halfplane \(\mbox{Re}(s)\geq a \gt 1\), we can integrate termwise on finite paths. Using \(\Lambda(n)\leq \log(n)\) and the estimate (\ref{xlt1}) we find \[\left|\int_{a-ih}^{a+ih} \sum_{n=m+1}^\infty \Lambda(n) \left(\frac{x}{n}\right)^s\frac{ds}{s+b}\right|\leq \sum_{n=m+1}^\infty\Lambda(n) \frac{n^b}{x^b}\left|\int_{a+b-ih}^{a+b+ih}\left(\frac{x}{n}\right)^t\frac{dt}{t}\right|\leq \sum_{n=m+1}^\infty \Lambda (n) \frac{x^a}{n^a \pi h \log(1+1/x)}\leq \frac{1}{h}\sum_{n=m+1}^\infty \frac{c}{n^a}\] for some constant \(c\). So the limit of the third term is zero as \(h\rightarrow \infty\). This proves that \[ I(0)=\frac{1}{2\pi i}\int_{a-i\infty}^{a+i\infty} \left(-\frac{\zeta'(s)}{\zeta(s)}\right)x^s\frac{ds}{s}=\psi(x) = \sum_{n\lt x}\Lambda(n)\] and \[ I(1)=\frac{1}{2\pi i}\int_{a-i\infty}^{a+i\infty} \left(-\frac{\zeta'(s)}{\zeta(s)}\right)x^s\frac{ds}{s+1}= \frac{1}{x}\left(\sum_{n\lt x}n \Lambda(n)\right). \blacksquare \]

The analytic formulas for \(\psi(x)\) and its antiderivative will now be derived by substituting formula (\ref{logderb}) into \(I(b)\). There are four terms in (\ref{logderb}). The calculations, and the switching of infinite sums and integral, will be relatively straightforward for all terms except the one involving the sum of the roots \(\rho\). That term will produce the limit \[\lim_{h\rightarrow \infty} \sum_\rho \frac{x^\rho}{\rho+b} F_h(x,\rho),\] because the infinite series (that is locally uniformly convergent) can be integrated termwise on finite paths. So we know that the limit exists, since all other terms in the formula obtained by substituting (\ref{logderb}) into \(I(b)\) exist. But the difficult part (as mentioned earlier) will be in proving that the same limit is equal to \[ \lim_{h\rightarrow \infty}\sum_{|{\rm Im}(\rho)\leq h}\frac{x^\rho}{\rho+b}.\] One ingredient of the proof is the estimate of the vertical density of the roots of \(\xi(s)\) given in the next lemma. Note that this formula can be considered a refinement of the formula for \(n(R)\) given in Lemma (in the proof of Theorem 3), because it is easy to derive that formula if we have the vertical density estimate. But the proof of the vertical density given in the next Lemma depends on the product formula for \(\xi(s)\), and the formula for \(n(R)\) was needed in order to prove the product formula. We will also need Stirling's formula for \(\Pi(s)\), and in fact it will be necessary to use not just the dominant term, but also the fact that the relative error is of order \(1/|s|\) for large \(|s|\).

Lemma

Let \(D(T)\) be the number of roots of \(\xi(s)\) with imaginary part between \(T\) and \(T+1\). Then \begin{equation} \label{DT} D(T)\leq 2 \log T \hspace{5ex}\mbox{ for all large }T. \end{equation}

Proof of Lemma . We first prove a lemma that will be needed in order to put an upper bound on the number of roots \(\rho\) in a horizontal strip of height 1.

Lemma E Let \(\rho=\sigma +i\tau\) be a root of \(\zeta(s)\) with \(0\leq \sigma \leq 1\), and let \(T\geq 0\). Then

1. \( \displaystyle \mbox{Im} \int_{2+iT}^{2+i(T+1)}\frac{ds}{s-\rho} \geq 0\)
   
   
2. If \(T\leq \tau \leq T+1\), then \( \displaystyle \mbox{Im} \int_{2+iT}^{2+i(T+1)}\frac{ds}{s-\rho} \geq \tan^{-1}\left(\frac{1}{2}\right)\)

Proof of Lemma E. From a geometric point view, part a. follows from the fact that the left side is the angle \(\theta_2-\theta_1\) in the following picture, \[\img{Ch3p1.png}{}{}{12em}\] and part b. from the fact that in case \(T\leq \tau \leq T+1\), the same angle is at least the angle \(\varphi\) in the following picture. \[\img{Ch3p2.png}{}{}{12em}\]

1. Let \(s=\rho+re^{i\theta}\). Then (using Cauchy's theorem to replace the segment \([2+iT,2+i(T+1)]\) with an arc centered at \(\rho\)), we find \[\mbox{Im} \int_{2+iT}^{2+i(T+1)}\frac{ds}{s-\rho} = \theta_2 -\theta_1,\] where \[\tan \theta_1 =\frac{T-\tau}{2-\sigma} \lt \frac{T+1-\tau}{2-\sigma} = \tan \theta_2.\] So \(\theta_1 \lt \theta_2\).
   
   
2. Let \(s=2+i(u+T)\), \(\rho=\sigma +i(b+T)\). Then \(0\leq \sigma \leq 1\), \(0\leq b \leq 1\), and \begin{eqnarray*} \mbox{Im} \int_{2+iT}^{2+i(T+1)}\frac{ds}{s-\rho}& =&\mbox{Im} \int_0^1\frac{idu}{2-\sigma+i(u-b)}\\[2ex] &=&\mbox{Im } i \int_0^1\frac{2-\sigma -i(u-b)idu}{(2-\sigma)^2+(u-b)^2}\\[2ex] &=&\int_0^1 \frac{2-\sigma}{(2-\sigma)^2+(u-b)^2} \\[2ex] &=& \tan^{-1}\frac{b}{2-\sigma} -\tan^{-1}\frac{b-1}{2-\sigma}\\[2ex] &=&\tan^{-1} \frac{2-\sigma}{(2-\sigma)^2+b(b-1)} \end{eqnarray*} Now the lemma follows from the fact that the inequality \(\sigma (2-\sigma) +b(1-b)\geq 0\) is equivalent to \(\displaystyle \frac{2-\sigma}{(2-\sigma)^2+b(b-1)}\geq \frac{1}{2} \hspace{1ex} \blacksquare \)

Since the Hadamard product \[\xi(s)=\xi(0) \prod_\rho \left(1-\frac{s}{\rho}\right)\] was shown to converge locally uniformly, we can take the logarithmic derivative and then integrate termwise over finite paths. So using Lemma E, we find \begin{equation} \label{Im} \mbox{Im} \int_{2+iT}^{2+i(T+1)} \frac{\xi'(s)}{\xi(s)} ds = \mbox{Im} \sum_\rho \int_{2+iT}^{2+i(T+1)}\frac{ds}{s-\rho} \geq \sum_{T\leq \textrm{Im}(\rho) \leq T+1}\int_{2+iT}^{2+i(T+1)}\frac{ds}{s-\rho} \geq D(T) \tan^{-1} \left(\frac{1}{2}\right). \end{equation} So we can get an upper bound for \(D(T)\) from an upper bound for \(\displaystyle \mbox{Im} \int_{2+iT}^{2+i(T+1)} \frac{\xi'(s)}{\xi(s)} ds\). This is done in the next lemma.
We denote by \(O(s^0)\) any function \(f(s)\) that remains bounded as \(|s|\rightarrow \infty\).

Lemma F For \(T\geq 0\), we have \[\int_{2+iT}^{2+i(T+1)} \frac{\xi'(s)}{\xi(s)} ds = \frac{i}{2}\log T +O(T^0).\] Proof of Lemma F. We will use the Stirling's approximation \[\Pi(s) = s^se^{-s} \sqrt{2\pi s} \left(1+\frac{O(s^0)}{s}\right).\] From the definition of \(\xi(s)\), we find \[ \xi(s) = \prod_{k=0}^5f_k(s) \] where \begin{eqnarray*} f_0(s)&=&\left(\frac{s}{2}\right)^{s/2}\\[1ex] f_1(s)&=& (e\pi)^{-s/2}\\[1ex] f_2(s)&=& \sqrt{\pi s}\\[1ex] f_3(s)&=& (s-1)\\[1ex] f_4(s)&=& \zeta(s)\\[1ex] f_5(s)&=& \left(1+\frac{O(s^0)}{s}\right) \end{eqnarray*} For any non-zero differentiable function \(f(s)\), let \(\displaystyle L(f)(s)=\frac{f'(s)}{f(s)}\). Then \(L(fg)=L(f)+L(g)\). So we find \[L(\xi)(s))=\sum_{k=0}^5L(f_k)(s),\] and we need to show that \[\sum_{k=0}^5\int_{2+iT}^{2+i(T+1)} L(f_k)(s)ds = \frac{i}{2}\log T +O(T^0).\] We will show that all terms in the sum with \(k\gt 0\) are bounded as \(T\rightarrow \infty\), while the term with \(k=0\) is \((i/2) \log T +O(T^0)\). For \(k=1\), we have \[L\left((e\pi)^{-s/2}\right)=-\frac{1}{2}\log(e\pi)\] and so \[\int_{2+iT}^{2+i(T+1)} L(f_1)(s)ds=-\frac{i}{2}\log(e\pi)\] For \(k=2\), \[L(\sqrt{\pi s}) = \frac{1}{2s},\] and \[\int_{2+iT}^{2+i(T+1)} L(f_2)(s)ds=\frac{1}{2}\log\left(1+\frac{i}{2+iT}\right) = O(T^0).\] For \(k=3\), \[L(s-1)=\frac{1}{s-1},\] and \[\int_{2+iT}^{2+i(T+1)} L(f_3)(s)ds=\log\left(1+\frac{i}{1+iT}\right) = O(T^0).\] For \(k=4\), \[L(\zeta(s))=\frac{d}{ds}\log \zeta(s)\] and for \(\mbox{Re}(s)=2\), \[|\log\zeta(2+it)|=\left|\sum_p\sum_{n=1}^\infty\frac{1}{n}\frac{1}{p^{2n+nit}}\right| \leq \sum_p\sum_{n=1}^\infty \frac{1}{n}\frac{1}{p^{2n}}=\log \zeta(2),\] so \[\int_{2+iT}^{2+i(T+1)} L(f_4)(s)ds = O(T^0).\] For \(k=5\), \[\int_{2+iT}^{2+i(T+1)} L(f_5)(s)ds =\left.\log\left(1+\frac{O(s^0)}{s}\right)\right|_{2+iT}^{2+i(T+1)}=O(T^0).\] For \(k=0\), \[\int L\left(\frac{s}{2}\right)^{s/2}=\frac{1}{2}s \log\left(\frac{s}{2}\right)=\frac{1}{2}s\log s-\frac{1}{2}s \log 2,\] and \[\left.\frac{s \log 2}{2}\right|_{2+iT}^{2+i(T+1)}=O(T^0).\] So it remains to show that \[\left.\frac{1}{2}s\log s\right|_{2+iT}^{2+i(T+1)} = \frac{1}{2} i\log T + O(T^0).\] This is easily seen from the computation \begin{eqnarray*} (2+i(T+1))\log(2+i(T+1))-(2+iT)\log(2+iT)&=& (2+iT)\log(2+i(T+1))+i \log(2+i(T+1))-(2+iT)\log(2+iT)\\[1ex] &=&(2+iT)\log\left(1+\frac{i}{2+iT}\right)+i\log(2+i(T+1)) \\[1ex] &=& O(T^0) + i\log(2+i(T+1))\\[1ex] &=& i\log T +O(T^0) \hspace{1ex} \blacksquare \end{eqnarray*} Combining Lemma F and (\ref{Im}), we find that for large \(T\), \[D(T)\leq \frac{\log T}{2 \tan^{-1}(1/2)} +c\leq 1.1 \log T +c\] for some constant \(c\), and so \[ D(T)\leq 2 \log T \mbox{ for all large } T \hspace{1ex} \blacksquare \]

The last step in the proof of Theorem 4 is the derivation of the term that results from the sum involving the roots of \(\xi(s)\) when substituting (\ref{logderb}) into \(I(b)\). The main idea in von Mangoldt's proof is to split the difference \[ \sum_\rho \frac{x^\rho}{\rho+b} F_h(x,\rho)- \sum_{|{\rm Im}(\rho)|\leq h}\frac{x^\rho}{\rho+b}\] as \begin{equation} \label{split} \left(\sum_\rho \frac{x^\rho}{\rho+b} F_h(x,\rho)-\sum_{|{\rm Im}(\rho)|\leq h} \frac{x^\rho}{\rho+b} F_h(x,\rho)\right) + \left(\sum_{|{\rm Im}(\rho)|\leq h} \frac{x^\rho}{\rho+b} F_h(x,\rho)- \sum_{|{\rm Im}(\rho)|\leq h}\frac{x^\rho}{\rho+b}\right) \end{equation} and then use the estimate for \(D(T)\) to prove that the differences in both parentheses go to zero as \(h\rightarrow \infty\). The term that is added and subtracted is what Edwards calls a "diagonal" term.

Lemma

\[ \lim_{h\rightarrow \infty} \sum_\rho \frac{x^\rho}{\rho+b} F_h(x,\rho)= \lim_{h\rightarrow \infty}\sum_{|{\rm Im}(\rho)|\leq h}\frac{x^\rho}{\rho+b}\]

Proof of Lemma . Note that \(|F_h(\rho)|=|F_h(\bar{\rho})|\). Write \(\rho=\beta+i\gamma\). We will consider the first two and the last two terms of (\ref{split}) separately.
For the first two terms, we find \begin{eqnarray*} && \left|\sum_\rho \frac{x^\rho}{\rho+b} F_h(\rho)-\sum_{{\rm |Im(\rho)|} \leq h} \frac{x^\rho}{\rho+b} F_h(\rho)\right|=\left|\sum_{{\rm |Im(\rho)|} \gt h} \frac{x^\rho}{\rho+b} F_h(\rho)\right|\\ &\leq &\sum_{\gamma \gt h} \frac{x^\beta}{\gamma}\left|F_h(\beta+i\gamma)\right| +\sum_{\gamma \lt -h} \frac{x^\beta}{|\gamma|}\left|F_h(\beta+i\gamma)\right|\\ &=&\sum_{\gamma \gt h} \frac{x^\beta}{\gamma}\left|F_h(\beta+i\gamma)\right|+ \sum_{\gamma \gt h} \frac{x^\beta}{\gamma}\left|F_h(\beta-i\gamma)\right|\\ &=&2\sum_{\gamma \gt h} \frac{x^\beta}{\gamma}\left|F_h(\beta-i\gamma)\right| \end{eqnarray*} Since \(0\leq \mbox{Re}(\rho) \leq 1\), and \(a\gt 1\), we have \(a-\beta\geq a-1 \gt 0\). Using Lemma , we have, for some constant \(K\), \[|F_h(\beta-i\gamma)|=\left|\frac{1}{2\pi}\int_{a-ih}^{a+ih}\frac{x^{s-\beta+i\gamma}}{s-\beta+i\gamma}ds\right| =\left|\frac{1}{2\pi}\int_{a-\beta+i(\gamma-h)}^{a-\beta+i(\gamma+h)}\frac{x^t}{t}dt\right| \leq \frac{K x^{a-\beta}}{(a-\beta+\gamma -h)\log x}\leq \frac{K x^{a-\beta}}{(a-1+\gamma-h)\log x}\] and so \[\sum_{\gamma\gt h}\frac{x^\beta}{\gamma} |F_h(\beta-i\gamma)|\leq \frac{Kx^a}{\log x}\sum_{\gamma\gt h}\frac{1}{\gamma (a-1+\gamma-h)}.\] Let \(T\gt 0\) be large enough so that (according to Lemma ) \(D(h)\leq 2\log h\) for \(h\geq T\). By increasing \(T\) if necessary, we can also assume that \(\log(T+j)\leq (T+j)^{1/2}\) for all \(j\geq 0\). Then if \(h\geq T\), we have \begin{eqnarray*} &&\sum_{\gamma\gt h}\frac{1}{\gamma (a-1+\gamma-h)}=\sum_{j=0}^\infty\sum_{h+j\lt \gamma \leq h+j+1} \frac{1}{\gamma (a-1+\gamma-h)}\leq \sum_{j=0}^\infty \frac{2\log(h+j)}{(h+j)(a-1+j)}\\ &&\leq 2\sum_{j=0}^\infty\frac{(h+j)^{1/2}}{(h+j)(j+a-1)}\leq 2\sum_{j=0}^\infty\frac{1}{(h+j)^{1/4} (h+j)^{1/4}(j+a-1)} \leq \frac{2}{h^{1/4}}\sum_{j=0}^\infty\frac{1}{(h+j)^{1/4}(j+a-1)}. \end{eqnarray*} The last infinite series converges, and so we conclude that \[\lim_{h\rightarrow \infty} \left|\sum_\rho \frac{x^\rho}{\rho+b} F_h(\rho)-\sum_{{\rm |Im(\rho)|} \leq h} \frac{x^\rho}{\rho+b} F_h(\rho)\right|=0.\] Before considering the last two terms of (\ref{split}), we prove a lemma. We will use the fact that the harmonic numbers \(\displaystyle H_n=\sum_{k=1}^n 1/k\) are of order \(\log n\) as \(n\rightarrow \infty\), so that we can write \(H_n=\log n (1+\epsilon_n)\) where \(\epsilon_n\rightarrow 0\) as \(n\rightarrow \infty\).

Lemma G Denote by \(\gamma\) the imaginary part of a root of \(\xi(s)\). Then \begin{equation} \label{pta} \lim_{h\rightarrow \infty} \frac{1}{h}\sum_{0\lt \gamma \leq h}\frac{1}{\gamma} =0 \end{equation} and if \(c\gt 0\), \begin{equation} \label{ptb} \lim_{h\rightarrow \infty} \frac{1}{h}\sum_{0\lt \gamma \leq h}\frac{1}{c+h-\gamma} =0 \end{equation} Proof of Lemma G.

Let \(N\) be an integer large enough so that (according to Lemma ) the number of roots in \([N+j,N+j+1]\) is at most \(2\log(N+j)\), for all \(j\geq 0\). We then find \[\frac{1}{h}\sum_{0\lt \gamma \leq h} \frac{1}{\gamma} = \frac{1}{h}\sum_{0\lt \gamma \leq N} \frac{1}{\gamma} +\frac{1}{h}\sum_{N\lt \gamma \leq h} \frac{1}{\gamma}\] and the limit of the first (finite) sum is \(0\) as \(h\rightarrow \infty\). For the second sum, we find \begin{eqnarray*} &&\frac{1}{h}\sum_{N\lt \gamma \leq h} \frac{1}{\gamma}= \frac{1}{h}\sum_{j=0}^{\lfloor h-N\rfloor} \sum_{N+j\lt \gamma \leq N+j+1} \frac{1}{\gamma} \leq \frac{1}{h}\sum_{j=0}^{\lfloor h-N\rfloor}\frac{2\log(N+j)}{N+j}\\ &&\leq \frac{2\log h}{h} \sum_{j=0}^{\lfloor h \rfloor} \frac{1}{N+j}\leq \frac{2\log h}{h} H_{N+\lfloor h \rfloor} =\frac{2\log h}{h}\log(N+ h)(1+\epsilon_h). \end{eqnarray*} Since \((\log h)^2/h \rightarrow 0\) as \(h\rightarrow \infty\), this proves (\ref{pta}).

To prove (\ref{ptb}), it is enough to consider (as for (\ref{pta})) the sum for \(N\lt \gamma \leq h\). We find \[\sum_{N\lt \gamma \leq h}\frac{1}{c+h-\gamma}=\sum_{j=0}^{\lfloor h-N\rfloor-1} \sum_{N+j\lt \gamma \leq N+j+1}\frac{1}{c+h-\gamma}+\sum_{N+\lfloor h-N\rfloor -1\lt \gamma \leq h}\frac{1}{c+h-\gamma}.\] The second sum is at most \((2\log h)/c\). The first sum is at most \[\sum_{j=0}^{\lfloor h-N\rfloor-1} \frac{2\log h}{c+h-N-j-1}\leq 2 \log h\left(\frac{1}{c} +H_{\lfloor h-N\rfloor}\right) \leq 2\log h\left( \frac{1}{c} +\log h (1+\epsilon_h)\right).\] This proves (\ref{ptb}) \(\blacksquare\)

We now consider the last two terms of (\ref{split}). We have \begin{eqnarray*} F_h(\rho)-1 &=&\frac{1}{2\pi i}\int_{a-ih}^{a+ih}\frac{x^{s-\rho}}{s-\rho}ds -1 = \frac{1}{2\pi i} \int_{a-\beta -i \gamma -ih}^{a-\beta -i\gamma +ih}\frac{x^t}{t}dt -1\\[2ex] &=&\frac{1}{2\pi i} \int_{a-\beta -i (\gamma +h)}^{a-\beta +i(\gamma +h)}\frac{x^t}{t}dt -1- \frac{1}{2\pi i} \int_{a-\beta +i(h-\gamma)}^{a-\beta +i(h+\gamma)}\frac{x^t}{t}dt. \end{eqnarray*} Using the estimate (\ref{xgt1}) and Lemma , we find \begin{eqnarray*} &&\frac{x^\beta}{\gamma}|F_h(\rho)-1|\leq \frac{x^a}{\pi\log x}\frac{1}{\gamma(\gamma +h)} + \frac{Kx^a}{\log x}\frac{1}{\gamma(c+h-\gamma)}\\[2ex] &&=\frac{x^a}{\pi h\log x}\left(\frac{1}{\gamma}-\frac{1}{\gamma +h}\right)+\frac{Kx^a}{(c+h)\log x}\left(\frac{1}{\gamma}+ \frac{1}{c+h-\gamma}\right)\\[2ex] &&\leq \left(\frac{x^a}{\pi \log x}+\frac{Kx^a}{\log x}\right) \frac{1}{h\gamma} +\frac{Kx^a}{\log x}\frac{1}{h(c+h-\gamma)}. \end{eqnarray*} So using (\ref{pta}) and (\ref{ptb}) we find \[\lim_{h\rightarrow \infty} \sum_{|{\rm Im}(\gamma)|\leq h} \frac{x^\rho}{\rho+b}(F_h(\rho)-1) =0.\] This completes the proof of Lemma \(\blacksquare\).



We are now ready to substitute (\ref{logderb}) into (\ref{Ib}). Since the series in (\ref{logderb}) converge locally uniformly, we can integrate them termwise over finite paths and we find \begin{eqnarray*} &&\frac{1}{2\pi i}\lim_{h\rightarrow \infty}\int_{a-ih}^{a+ih} \left(-\frac{\zeta'(s)}{\zeta(s)}\right)x^s\frac{ds}{s+b}\\ &=& \frac{1}{2\pi i}\lim_{h\rightarrow \infty}\int_{a-ih}^{a+ih} \left(\frac{s+b}{(s-1)(1+b)} -\frac{\zeta'(-b)}{\zeta(-b)}+\sum_{n=1}^\infty \frac{s+b}{(2n-b)(2n+s)} +\sum_\rho \frac{s+b}{(\rho+b)(\rho-s)} \right)x^s\frac{ds}{s+b}\\ &=&\frac{1}{2\pi i(1+b)}\int_{a-i\infty}^{a+i\infty}\frac{x^s}{s-1}ds+\frac{1}{2\pi i}\int_{a-i\infty}^{a+i\infty}\left(- \frac{\zeta'(-b)}{\zeta(-b)}\right)x^s \frac{ds}{s+b}\\ &+&\lim_{h\rightarrow \infty}\sum_{n=1}^\infty \frac{1}{(2n-b)}\frac{1}{2\pi i}\int_{a-ih}^{a+ih}\frac{x^s}{s+2n}ds +\lim_{h\rightarrow \infty}\sum_\rho\frac{1}{(\rho+b)}\frac{1}{2\pi i}\int_{a-ih}^{a+ih} \frac{x^s}{\rho-s} ds\\ &=& \frac{x}{1+b}- x^{-b}\frac{\zeta'(-b)}{\zeta(-b)}+\lim_{h\rightarrow \infty}\sum_{n=1}^\infty \frac{1}{(2n-b)}\frac{1}{2\pi i}\int_{a-ih}^{a+ih}\frac{x^s}{s+2n}ds+ \lim_{h\rightarrow \infty}\sum_\rho\frac{1}{(\rho+b)}\frac{1}{2\pi i}\int_{a-ih}^{a+ih} \frac{x^s}{\rho-s} ds \end{eqnarray*} where we have used Lemma to evaluate the first two terms. We now need to show that we can switch the limit with the two infinite series.
First consider the series in \(n\). We will show that the it converges uniformly in \(h\). We find \[\int_{a-ih}^{a+ih}\frac{x^{s+2n}}{s+2n}ds=\int_0^h\left(\frac{x^{a+2n+iy}}{a+2n+iy}+ \frac{x^{a+2n-iy}}{a+2n-iy}\right)dy=2\mbox{Re}\int_0^h\frac{x^{a+2n+iy}}{a+2n+iy}dy =2\mbox{Re}\int_{a+2n}^{a+2n+ih}\frac{x^t}{t}dt\] and so using Lemma , \[\left|\int_{a-ih}^{a+ih}\frac{x^{s+2n}}{s+2n}ds\right|\leq 2 \left|\int_{a+2n}^{a+2n+ih} \frac{x^t}{t}dt\right|\leq \frac{K x^{a+2n}}{(a+2n)\log x}\] for a constant \(K\). It follows that the \(n\)-th term of the series in \(n\) is bounded by \(c/n^2\) for a constant \(c\) independent of \(h\). So the series converges uniformly in \(h\), and it follows from Lemma that \[\lim_{h\rightarrow \infty}\sum_{n=1}^\infty \frac{1}{2n-b}\frac{1}{2\pi i}\int_{a-ih}^{a+ih}\frac{x^s}{s+2n}ds=\sum_{n=1}^\infty \frac{1}{2n-b}\frac{1}{2\pi i}\int_{a-i\infty}^{a+i\infty}\frac{x^s}{s+2n}ds=\sum_{n=1}^\infty \frac{x^{-2n}}{2n-b}.\] So we have now shown that \begin{equation} \label{exist} I(b)=\frac{x}{1+b}- x^{-b}\frac{\zeta'(-b)}{\zeta(-b)}+\sum_{n=1}^\infty \frac{x^{-2n}}{2n-b}- \lim_{h\rightarrow \infty}\sum_\rho\frac{x^\rho}{\rho+b}F_h(x,\rho) \end{equation} and using Lemma we obtain the formula for \(I(b)\) \[I(b)=\frac{x}{1+b}- x^{-b}\frac{\zeta'(-b)}{\zeta(-b)}+\sum_{n=1}^\infty \frac{x^{-2n}}{2n-b}- \lim_{h\rightarrow \infty}\sum_{|{\rm Im}(\rho)|\leq h}\frac{x^\rho}{\rho+b}. \] Setting \(b=0\) we find, according to (\ref{I0}), van Mangoldt's formula for \(\psi(x)\) \[ I(0)=\psi(x)=x-\frac{\zeta'(0)}{\zeta(0)}+\sum_{n=1}^\infty \frac{x^{-2n}}{2n} -\lim_{h\rightarrow \infty}\sum_{|{\rm Im}(\rho)\leq h}\frac{x^\rho}{\rho} =x-\frac{\zeta'(0)}{\zeta(0)}+\frac{1}{2}\log\left(\frac{x^2}{x^2-1}\right) -\lim_{h\rightarrow \infty}\sum_{|{\rm Im}(\rho)|\leq h}\frac{x^\rho}{\rho} \] while setting \(b=1\) we find, according to (\ref{I1}), \[ I(1)=\frac{1}{2x}\left(\sum_{n\lt x}n \Lambda(n)+\sum_{n\leq x} n \Lambda(n)\right)= \frac{x}{2}- \frac{\zeta'(-1)}{x\zeta(-1)} + \sum_{n=1}^\infty \frac{x^{-2n}}{2n-1}- \lim_{h\rightarrow \infty}\sum_{|{\rm Im}(\rho)|\leq h}\frac{x^\rho}{\rho+1} \] and the formula for the antiderivative of \(\psi(x)\) follows by computing \(xI(0)-xI(1)\). This completes the proof of Theorem 4.

Let \(\displaystyle F(s) = \sum_p \frac{1}{p^s}\), for \(\mbox{Re}(s)\gt 1\). We claim that \(F(\sigma) +\log (\sigma -1)\) is bounded as \(\sigma \rightarrow 1^+\).
Using the representation \[\Pi(s) =\lim_{n\rightarrow \infty}\frac{n! (n+1)^s}{(s+1)\cdots (s+n)}\] we find \[\lim_{s\rightarrow 1} (s-1) \Pi(-s)=-1.\] Since \[\frac{1}{2\pi i}\int_C\frac{-z}{e^z-1}\frac{dz}{z} =-1\] (where \(C\) is the Hankel path from Chapter 1), we find from the integral representation \[\zeta(s)=\frac{\Pi(-s)}{2\pi i}\int_C\frac{(-z)^s}{e^z-1}\frac{dx}{z} \] that \[\lim_{s\rightarrow 1} (s-1)\zeta(s) =1.\] As a consequence of Euler's product formula, Let \(p_1,p_2,p_3,\ldots\) be the prime numbers, and \(P_m=\{p_i: 1\leq i \leq m\}\). Then \[\log\sum_{k\not \in P_n}\frac{1}{k^s}=\sum_{i=1}^n\log\left(1-\frac{1}{p_i^s}\right)^{-1}, s\gt 1.\] The sum on the right is \[\sum_{i=1}^n\sum_{m=1}^\infty \frac{1}{m}\frac{1}{p_i^{ms}}=\sum_{i=1}^n\frac{1}{p_i^s}+\sum_{i=1}^n\sum_{m=2}^\infty \frac{1}{m}\frac{1}{p_i^{ms}}\] and the second sum on the right is bounded by \[\sum_{r=2}^{p_n}\sum_{m=2}^\infty \frac{1}{m}\frac{1}{r^{ms}}\leq \frac{1}{2}\sum_{r=2}^{p_n}\sum_{m=2}^\infty \left(\frac{1}{r^s}\right)^m=\frac{1}{2}\sum_{r=2}^{p_n}\frac{1}{r^{2s}}\frac{1}{1-r^{-s}} =\frac{1}{2}\sum_{r=2}^\infty\frac{1}{r^s(r^s-1)}\leq \frac{1}{2}.\] So letting \(n\rightarrow \infty\) we see that \[\log\zeta(s)=\sum_{i=1}^\infty\frac{1}{p_i^s}+B(s) \mbox{ for } s\gt 1,\] where \(|B(s)|\leq 1/2\) for \(s\geq 1\). Hence the sum \(\displaystyle \sum_{i=1}^\infty \frac{1}{p_i^s}\) diverges as \(s\rightarrow 1^+\) like \(\displaystyle \log\zeta(s)\).

we find \[\log \zeta(s) =F(s) + B(s), |B(s)|\leq \frac{1}{2} \mbox{ for Re}(s)\gt 1.\] So \[\log((\sigma -1)\zeta(\sigma))=\log(\sigma -1) +F(\sigma) +B(\sigma) \rightarrow 0 \mbox{ as } \sigma\rightarrow 1^+\] and the claim follows.

Suppose now \(t\neq 0\) is such that \(\zeta(1+it)=0\). Let \(\delta\gt 0\) be such that \(1-\cos\delta \gt 0\) and \(\cos(2\delta)\gt 0\). Consider the set \[A=\{p \mbox{ prime}: |(2n+1)\pi -t \log p|\lt \delta \mbox{ for some } n\in {\mathbb Z}\}\] Let \[F_1(\sigma) = \sum_{p\in A}\frac{1}{p^\sigma}, \hspace{2ex} F_2(\sigma) = \sum_{p\not \in A}\frac{1}{p^\sigma}, \hspace{2ex} \sigma \gt 1.\] Since \(\zeta(s)/(s-1-it)\) is analytic, using \(s=\sigma +it\) we find that \(\zeta(\sigma +it)/(\sigma -1)\) is bounded as \(\sigma \rightarrow 1^+\), and so \(\mbox{Re}\log \zeta(\sigma +it) -\log(\sigma -1) \) is bounded. This implies \[\mbox{Re} F(\sigma +it) -\log(\sigma -1)=\sum_p\frac{\cos(t \log p)}{p^\sigma} -\log(\sigma -1) \mbox{ is bounded as } \sigma\rightarrow 1^+.\] If \(p \not \in A\), then \(\cos(t \log p)\geq 1-\cos(\delta)\), and so \[\sum_p\frac{\cos(t \log p)}{p^\sigma}\geq -\sum_{p\in A}\frac{1}{p^\sigma}-\cos(\delta) \sum_{p\not \in A}\frac{1}{p^\sigma} = -F_1(\sigma) -\cos(\delta) F_2(\sigma).\] Hence \begin{equation} \label{1} -F_1(\sigma) -\cos(\delta) F_2(\sigma)-\log(\sigma -1) \leq \sum_p\frac{\cos(t \log p)}{p^\sigma} -\log(\sigma -1) \mbox{ is bounded as }\sigma\rightarrow 1^+.\end{equation} According to the claim, \begin{equation} \label{2} F(\sigma)+\log(\sigma-1) =F_1(\sigma) + F_2(\sigma) +\log(\sigma -1) \mbox{ is bounded as } \sigma\rightarrow 1^+ . \end{equation} Adding (\ref{1}) and (\ref{2}), we find \[(1-\cos(\delta))F_2(\sigma) \mbox{ is bounded as } \sigma\rightarrow1^+.\] So we conclude that \(F_2(\sigma)\) is bounded as \(\sigma \rightarrow 1^+\). But \(\zeta(s)\) is analytic at \(s=1+2it\), and so \(\mbox{Re}F(\sigma +2it)\) is bounded as \(\sigma \rightarrow 1^+\). If \(p\in A\), then \(|2(2n+1)\pi -2t\log p|\lt 2\delta\) for some \(n\), and so \(\cos(2t\log p)\geq \cos(2\delta)\). So \[\mbox{Re} F(\sigma +2it) \geq \cos(2\delta) F_1(\sigma) -F_2(\sigma)\] implies that \(F_1(\sigma)\) is also bounded as \(\sigma\rightarrow 1^+\). But this contradicts the fact that \(F(\sigma) = F_1(\sigma) + F_2(\sigma)\) is unbounded as \(\sigma \rightarrow 1^+\blacksquare \).

From (\ref{antider}), we find \[\frac{1}{x^2}\left(\int_0^x\psi(t)dt-\frac{x^2}{2}\right)= -\frac{\zeta'(0)}{x \zeta(0)} +\frac{\zeta'(-1)}{x^2\zeta(-1)} -\sum_{n=1}^\infty \frac{x^{-2n-1}}{2n(2n-1)} -\lim_{h\rightarrow \infty} \sum_{|{\rm Im}(\rho)\leq h}\frac{x^{\rho-1}}{\rho(\rho+1)} \] Since this last series in \(\rho\) now converges absolutely and uniformly in \(x\), we can evaluate the limit as \(x\rightarrow \infty\) termwise and since \(\mbox{Re}(\rho)-1 \lt 0\), we find \[\lim_{x\rightarrow \infty} \frac{1}{x^2}\left(\int_0^x\psi(t)dt-\frac{x^2}{2}\right)=0.\blacksquare\]

Write \(\displaystyle F(x)=\int_0^x f(t) dt\). Let \(\epsilon \gt 0\) be given. Then for large enough \(x\), \[(1-\epsilon)\frac{x^2}{2} \leq F(x) \leq (1+\epsilon) \frac{x^2}{2}.\] If \(x\) is large enough and \(y=\beta x\), where \(\beta \gt 1\), we have \[(1-\epsilon)\frac{y^2}{2} \leq f(y)(y-x)+(1+\epsilon)\frac{x^2}{2}=(y-x)f(y)+(1-\epsilon) \frac{x^2}{2} + \epsilon x^2,\] and then we find \[\frac{1}{2}(1-\epsilon)(y^2-x^2)\leq (y-x)f(y) +\epsilon x^2\] and using \(y=\beta x\), \[\frac{f(y)}{y}\geq \frac{1}{2}(1-\epsilon)\frac{\beta +1}{\beta}-\frac{\epsilon}{\beta(\beta-1)}.\] Hence \[\liminf_{x\rightarrow \infty} \frac{f(x)}{x} \geq \frac{1}{2}(1-\epsilon)\frac{\beta +1}{\beta}-\frac{\epsilon}{\beta(\beta-1)}.\] Letting \(\epsilon \) go to zero, we find \[\liminf_{x\rightarrow \infty} \frac{f(x)}{x} \geq\frac{\beta+1}{2 \beta},\] and letting \(\beta \) approach 1, \[\liminf_{x\rightarrow \infty} \frac{f(x)}{x}\geq 1.\] In a similar way, we find \[\frac{f(x)}{x} \leq \frac{1}{2} (1-\epsilon)(\beta+1) +\epsilon \frac{\beta^2}{\beta-1}\] and so \[\limsup_{x\rightarrow \infty} \frac{f(x)}{x} \leq \frac{1}{2} (1-\epsilon)(\beta+1) +\epsilon \frac{\beta^2}{\beta-1}.\] Letting \(\epsilon \) go to zero, we find \[\limsup_{x\rightarrow \infty}\frac{f(x)}{x} \leq \frac{\beta+1}{2},\] and then letting \(\beta \) approach 1, \[\limsup_{x\rightarrow \infty}\frac{f(x)}{x} \leq 1\leq \liminf_{x\rightarrow \infty} \frac{f(x)}{x}.\] So we conclude that \[\lim_{x\rightarrow \infty}\frac{f(x)}{x} =1.\blacksquare \]