---
abstract: |
    Let $\Lambda\left(n\right)$ be the von Mangoldt function and $r_{G}\left(n\right):=\sum_{m_{1}+m_{2}=n}\Lambda\left(m_{1}\right)\Lambda\left(m_{2}\right)$ be the counting function for the numbers that can be written as sum of two primes (that we will call “Goldbach numbers”, for brevity) and let $\widetilde{S}\left(z\right):=\sum_{n\geq1}\Lambda\left(n\right)e^{-nz}$, with $z\in\mathbb{C}$, $\mathrm{Re}\left(z\right)>0$. In this paper we will prove the identity $$\begin{aligned}
    \widetilde{S}\left(z\right)= & 2e^{-2z}+\frac{e^{-2z}}{z}-\sum_{\rho}z^{-\rho}\Gamma\left(\rho\right)\\
    + & \sum_{\rho}\left(z^{-\rho}\gamma\left(\rho,2z\right)-\frac{2^{\rho}e^{-z}}{\rho}\right)-\frac{\zeta^{\prime}}{\zeta}\left(0\right)e^{-2z}\\
    + & \frac{e^{-z}}{2}\left(-i\pi\textrm{sgn}\left(\textrm{Im}\left(-z\right)\right)+\textrm{Ei}\left(-z\right)\right)\\
    + & \frac{e^{z}}{2}\left(-i\pi\textrm{sgn}\left(\textrm{Im}\left(-z\right)\right)-\log\left(3\right)e^{-3z}+\textrm{Ei}\left(-3z\right)\right)\\
    + & i\pi\textrm{sgn}\left(\textrm{Im}\left(-z\right)\right)+\log\left(2\right)e^{-2z}-\textrm{Ei}\left(-2z\right)\end{aligned}$$ where $\gamma\left(\rho,2z\right)$ is the lower incomplete Gamma function, $\rho=\beta+i\gamma$ runs over the non-trivial zeros of the Riemann Zeta function and $$\mathrm{Ei}\left(z\right)=-\int_{-z}^{\infty}\frac{e^{-t}}{t}dt,\ \left|\arg\left(z\right)\right|<\pi$$ is the Exponential integral function. In addition we will prove that $$\begin{aligned}
    \sum_{n\leq N}r_{G}\left(n\right)\left(N-n\right)= & \frac{N^{3}}{6}-2\sum_{\rho}\frac{\left(N-2\right)^{\rho+2}}{\rho\left(\rho+1\right)\left(\rho+2\right)}\\
    + & \sum_{\rho_{1}}\sum_{\rho_{2}}\frac{\Gamma\left(\rho_{1}\right)\Gamma\left(\rho_{2}\right)}{\Gamma\left(\rho_{1}+\rho_{2}+2\right)}N^{\rho_{1}+\rho_{2}+1}-2\sum_{\rho_{1}}\sum_{\rho_{2}}\frac{N^{\rho_{1}+\rho_{2}+1}}{\rho_{1}\rho_{2}}B_{2/N}\left(\rho_{1}+1,\rho_{2}+1\right)\\
    + & F\left(N\right)\end{aligned}$$ where $N>4$ is a natural number, $B_{x}\left(a,b\right)$ is the incomplete Beta function and $F\left(N\right)$ is a sum of (explicitly calculate) elementary functions, dilogarithms and sums over non-trivial zeros of the Riemann Zeta function involving the incomplete Beta function. Furthemore we will prove that $$F\left(N\right)=O\left(N^{2}\right)$$ as $N\rightarrow\infty.$
author:
- Marco Cantarini
title: Some identities involving the Cesàro average of Goldbach numbers
---

2010 Mathematics Subject Classification:[ Primary 11P32; Secondary 44A10]{}

Introduction
============

In this paper we prove some identities inherent the study to Cesàro average of Goldbach numbers. The main result in this direction is the paper of Languasco e Zaccagnini [@Langzac]. The authors proved the formula $$\begin{aligned}
\sum_{n\leq N}r_{G}\left(n\right)\frac{\left(N-n\right)^{k}}{\Gamma\left(k+1\right)}= & \frac{N^{2+k}}{\Gamma\left(k+3\right)}-2\sum_{\rho}\frac{\Gamma\left(\rho\right)}{\Gamma\left(\rho+k+2\right)}N^{\rho+k+1}\nonumber \\
+ & \sum_{\rho_{1}}\sum_{\rho_{2}}\frac{\Gamma\left(\rho_{1}\right)\Gamma\left(\rho_{2}\right)}{\Gamma\left(\rho_{1}+\rho_{2}+k+1\right)}N^{\rho_{1}+\rho_{2}+k}+O_{k}\left(N^{k+1}\right)\label{langzac formula}\end{aligned}$$ for $k>1$, where $N$ is a large natural number and $\rho$, with or without subscripts, runs over the non-trivial zeros of the Riemann Zeta function $\zeta\left(s\right)$. The parameter $k$ plays a central role in the study of these problems and we would like to keep it as small as possible since, for $k=0$, the Cesàro weight disappears. The method of Languasco and Zaccagnini allows to study many types of additive problems; see [@Canta1], [@Canta2] and [@LangZac2]. The crucial points of their technique are the relation $$\sum_{n\leq N}r_{G}\left(n\right)\frac{\left(N-n\right)^{k}}{\Gamma\left(k+1\right)}=\frac{1}{2\pi i}\int_{\left(1/N\right)}e^{Nz}z^{-k-1}\widetilde{S}\left(z\right)^{2}dz\label{identit=0000E0 fondamentale}$$ which holds for $k>0,$ where $$\widetilde{S}\left(z\right):=\sum_{n\geq1}\Lambda\left(n\right)e^{-nz},\,z=a+iy,\,\mathrm{with}\,a>0,\,y\in\mathbb{R}.$$ is the power series of the Von Mangoldt function and $\int_{\left(1/N\right)}$ means $\int_{1/N-i\infty}^{1/N+i\infty}$, and the estimation $$\widetilde{S}\left(z\right)=\frac{1}{z}-\sum_{\rho}z^{-\rho}\Gamma\left(\rho\right)+E\left(a,y\right)\label{eq:asint s tilda zac}$$ where $$E\left(a,y\right)\ll\begin{cases}
1+\left|z\right|^{1/2}, & \left|y\right|\leq a\\
1+\left|z\right|^{1/2}\left(1+\log^{2}\left(\left|y\right|/a\right)\right), & \left|y\right|>a.
\end{cases}\label{errore s tilda}$$ The proof of is present in [@Langzac]. We recall that the study of $\widetilde{S}\left(z\right)$ is classical, see for example [@HardLit]. One of the Recently Goldston and Yang [@GolYan] proved, assuming RH, that $$\sum_{n\leq N}r_{G}\left(n\right)\left(1-\frac{n}{N}\right)=\frac{N^{2}}{6}-2\sum_{\rho}\frac{\Gamma\left(\rho\right)}{\Gamma\left(\rho+3\right)}N^{\rho+1}+O\left(N\right)\label{eq:GoldeYan}$$ which correspond to what Languasco and Zaccagnini’s formula implies if one take $k=1$. Languasco and Zaccagnini [@LangZacNew] also proved (assuming RH) that $$\begin{aligned}
\sum_{n=N-H}^{N+H}r_{G}\left(n\right)\left(1-\frac{\left|n-N\right|}{H}\right)= & HN-\frac{2}{H}\sum_{\rho}\frac{\left(N+H\right)^{\rho+2}-2N^{\rho+2}+\left(N-H\right)^{\rho+2}}{\rho\left(\rho+1\right)\left(\rho+2\right)}\\
+ & O\left(N\log^{2}\left(\frac{2N}{H}\right)+H\log^{2}\left(N\right)\log\left(2H\right)\right)\end{aligned}$$ where $N\geq2,\,1\leq H\leq N$. $k.$ In a very recent paper Brüdern, Kaczorowski and Perelli [@Perelli] were able to find an explicit formula which holds for all $k>0.$ Similar averages of arithmetical functions are common in literature, see, e.g., Chandrasekharan - Narasimhan [@Chana] and Berndt [@Ber] who built on earlier classical work. In In this paper we will prove an explicit formula for $\widetilde{S}\left(z\right)$ which will allow us to find a smaller error respect to . Kunik and Lucht [@Kunik] wrote an explicit formula for $\widetilde{S}\left(z\right)$ but in their closed form there is a term regarding another summation of Von Mangoldt function, i.e. the series $\sum_{n\geq1}\Lambda\left(n\right)\left(1-\exp\left(-z/n\right)\right)/n$. We will able to rewrite this series in terms of the lower incomplete Gamma function and sum over non-trivial zeros of the Riemann Zeta function. We first prove the following

\[lem:5\]Let $z=a+iy,\,a>0,\,y\in\mathbb{R}.$ Let us consider the function $$\widetilde{S}\left(z\right)=\sum_{m\geq1}\Lambda\left(m\right)e^{-mz}.$$ Then $$\begin{aligned}
\widetilde{S}\left(z\right)= & 2e^{-2z}+\frac{e^{-2z}}{z}-\sum_{\rho}z^{-\rho}\Gamma\left(\rho\right)\\
+ & \sum_{\rho}\left(z^{-\rho}\gamma\left(\rho,2z\right)-\frac{2^{\rho}e^{-z}}{\rho}\right)-\frac{\zeta^{\prime}}{\zeta}\left(0\right)e^{-2z}\\
+ & \frac{e^{-z}}{2}\left(-i\pi\mathrm{sgn}\left(\mathrm{Im}\left(-z\right)\right)+\mathrm{Ei}\left(-z\right)\right)\\
+ & \frac{e^{z}}{2}\left(-i\pi\mathrm{sgn}\left(\mathrm{Im}\left(-z\right)\right)-\log\left(3\right)e^{-3z}+\mathrm{Ei}\left(-3z\right)\right)\\
+ & i\pi\mathrm{sgn}\left(\mathrm{Im}\left(-z\right)\right)+\log\left(2\right)e^{-2z}-\mathrm{Ei}\left(-2z\right)\end{aligned}$$ where $\gamma\left(\rho,2z\right)$ is the lower incomplete Gamma function, $\rho=\beta+i\gamma$ runs over the non-trivial zeros of the Riemann Zeta function and $$\mathrm{Ei}\left(z\right)=-\int_{-z}^{\infty}\frac{e^{-t}}{t}dt,\ \left|\arg\left(z\right)\right|<\pi$$ is the Exponential integral function. Furthermore if we put $$\widetilde{S}\left(z\right)=\frac{e^{-2z}}{z}-\sum_{\rho}\Gamma\left(\rho\right)z^{-\rho}+\sum_{\rho}\left(z^{-\rho}\gamma\left(\rho,2z\right)-\frac{2^{\rho}e^{-2z}}{\rho}\right)+E\left(a,y\right)$$ we get the bound $$E\left(a,y\right)\ll\begin{cases}
1+\left|z\right|^{1/2}, & \left|y\right|\leq a\\
1+\left|z\right|^{1/2}\left(1+\log^{2}\left(\left|y\right|/a\right)\right), & a<\left|y\right|\leq1\\
1, & \left|y\right|>1.
\end{cases}$$

Note that we improved the error term as $\left|y\right|\rightarrow\infty.$ This is interesting since, following the Languasco and Zaccagnini approach, the size of the error term $E\left(a,y\right)$ is linked to the size of $k$ (see Section $3$ of [@Langzac] for more details). We also obtain the extra term $\sum_{\rho}\left(z^{-\rho}\gamma\left(\rho,2z\right)-\frac{2^{\rho}e^{-2z}}{\rho}\right)$; following, again, the technique in [@Langzac], we see that it would be nice to have a sharp uniform estimation of this series. Unfortunately, even if this series, as we will see, converges absolutely, it is not simple to get a good uniform estimation. A deeper analysis of this fact will be done in future research. Also note that the presence of $e^{-2z}/z$ as “main term” instead of $1/z$ does not significantly alter the asymptotic formula . For example if $z\in\mathbb{R}^{+}$ is clear that $$\widetilde{S}\left(z\right)\sim e^{-2z}/z\sim1/z$$ as $z\rightarrow0^{+},$ which is one of the form of the Prime Number Theorem. In addition we have the following

\[thm:k=00003D1\]Let $N>4$ be a natural number. Then $$\begin{aligned}
\sum_{n\leq N}r_{G}\left(n\right)\left(N-n\right)= & \frac{N^{3}}{6}-2\sum_{\rho}\frac{\left(N-2\right)^{\rho+2}}{\rho\left(\rho+1\right)\left(\rho+2\right)}\\
+ & \sum_{\rho_{1}}\sum_{\rho_{2}}\frac{\Gamma\left(\rho_{1}\right)\Gamma\left(\rho_{2}\right)}{\Gamma\left(\rho_{1}+\rho_{2}+2\right)}N^{\rho_{1}+\rho_{2}+1}-2\sum_{\rho_{1}}\sum_{\rho_{2}}\frac{N^{\rho_{1}+\rho_{2}+1}}{\rho_{1}\rho_{2}}B_{2/N}\left(\rho_{1}+1,\rho_{2}+1\right)\\
+ & F\left(N\right)\end{aligned}$$ where $F\left(N\right)$ is a sum of (explicitly calculate) elementary functions, dilogarithms and sum over non-trivial zeros of the Riemann Zeta function involving the incomplete Beta function. Furthermore we will prove that $$F\left(N\right)=O\left(N^{2}\right)$$ as $N\rightarrow\infty.$

This explicit formula extend, in some sense, the main result in [@Langzac] (since their formula holds for $k>1$) and in [@GolYan] (since the authors assume the Riemann Hypothesis). Furthermore it provides different way to write the explicit formula of the Cesàro average of Goldbach representations in the case $k=1$ respect to the main formula in [@Perelli]. The study of the double series $\sum_{\rho_{1}}\sum_{\rho_{2}}\frac{N^{\rho_{1}+\rho_{2}+1}}{\rho_{1}\rho_{2}}B_{2/N}\left(\rho_{1}+1,\rho_{2}+1\right)$ is quite delicate since it is not obvious if converges absolutely or not but it is clear that it plays a central role in formulae like . A deeper analysis of this type of double series will be the subject of future research.

I thank my mentor Alessandro Zaccagnini, Jacopo “Jack” D’Aurizio and Matthias Kunik for several conversation on this topic.

Notations
=========

To avoid ambiguity but keeping the standard notations, we specify some symbols:

$\gamma,$ with or without subscripts, will be the imaginary part of the non-trivial zeros of the Riemann Zeta function,

$\gamma\left(a,b\right)$ will be the lower incomplete Gamma function,

$\mathcal{L}\left(f\left(t\right)\right)\left(z\right)$ will be the Laplace transform of $f\left(t\right).$

Proof of Theorem 1
==================

We first recall that the upper incomplete Gamma function and the lower incomplete Gamma function are defined as $$\Gamma\left(a,z\right)=\int_{z}^{\infty}t^{a-1}e^{-t}dt\label{up gamma}$$ $$\gamma\left(a,z\right)=\int_{0}^{z}t^{a-1}e^{-t}dt,\,\textrm{Re}\left(a\right)>0\label{low gamma}$$ (see [@Olver], formula $8.2.1$ and $8.2.2$ at page $174$). Furthermore we have $$\Gamma\left(a,z\right)+\gamma\left(a,z\right)=\Gamma\left(a\right)\label{eq:G1}$$ $$\Gamma\left(a+1,z\right)=a\Gamma\left(a,z\right)+e^{-z}z^{a}\label{eq:G2}$$ (see [@Olver], formula $8.2.3$ at page $174$ and formula $8.8.2$ at page $178$).

Let us consider the finite sum $\sum_{m=2}^{M}\Lambda\left(m\right)e^{-mz},\,M>2.$ By Abel summation we have $$\sum_{m=2}^{M}\Lambda\left(m\right)e^{-mz}=\psi\left(M\right)e^{-Mz}+z\int_{2}^{M}\psi\left(t\right)e^{-tz}dt$$ where $\psi\left(M\right)$ is the Chebyshev psi function. From PNT we have $$\left|\psi\left(M\right)e^{-Mz}\right|\leq\psi\left(M\right)e^{-Ma}\ll Me^{-Ma}\rightarrow0$$ as $M\rightarrow\infty$ so $$\widetilde{S}\left(z\right)=\sum_{m\geq1}\Lambda\left(m\right)e^{-mz}=z\int_{2}^{\infty}\psi\left(t\right)e^{-tz}dt\label{eq:Laplace psi}$$ so $z^{-1}\widetilde{S}\left(z\right)$ is the Laplace transform of $\psi\left(t\right).$ From the explicit formula for $\psi\left(t\right)$ (see for example [@daven], page $104$) we obtain $$\begin{aligned}
z\int_{2}^{\infty}\psi\left(t\right)e^{-tz}dt= & z\int_{2}^{\infty}te^{-tz}dt-z\int_{2}^{\infty}\sum_{\rho}\frac{t^{\rho}}{\rho}e^{-tz}dt\nonumber \\
- & \frac{\zeta^{\prime}}{\zeta}\left(0\right)z\int_{2}^{\infty}e^{-tz}dt-\frac{z}{2}\int_{2}^{\infty}\log\left(1-t^{-2}\right)e^{-tz}dt.\label{eq:9}\end{aligned}$$ Now we integrate term by term and we assume for now that we can switch the series over the non-trivial zeros and the integral (we will show that it is possible at the end of this proof). We have immediately that $$z\int_{2}^{\infty}te^{-tz}dt=2e^{-2z}+\frac{e^{-2z}}{z}$$ $$-\frac{\zeta^{\prime}}{\zeta}\left(0\right)z\int_{2}^{\infty}e^{-tz}dt=-\frac{\zeta^{\prime}}{\zeta}\left(0\right)e^{-2z}.$$ Now, from and , we have $$\begin{aligned}
-z\int_{2}^{\infty}\sum_{\rho}\frac{t^{\rho}}{\rho}e^{-tz}dt= & -\sum_{\rho}\frac{1}{\rho}z\int_{2}^{\infty}t^{\rho}e^{-tz}dt\nonumber \\
= & -\sum_{\rho}\frac{1}{\rho}z^{-\rho}\int_{2z}^{\infty}t^{\rho}e^{-t}dt\nonumber \\
= & -\sum_{\rho}\frac{1}{\rho}z^{-\rho}\Gamma\left(\rho+1,2z\right)\nonumber \\
= & -\sum_{\rho}\left(z^{-\rho}\Gamma\left(\rho,2z\right)+\frac{2^{\rho}e^{-2z}}{\rho}\right)\nonumber \\
= & -\sum_{\rho}\left(z^{-\rho}\Gamma\left(\rho\right)-z^{-\rho}\gamma\left(\rho,2z\right)+\frac{2^{\rho}e^{-2z}}{\rho}\right)\nonumber \\
= & -\sum_{\rho}z^{-\rho}\Gamma\left(\rho\right)+\sum_{\rho}\left(z^{-\rho}\gamma\left(\rho,2z\right)-\frac{2^{\rho}e^{-2z}}{\rho}\right).\label{eq:identit=0000E0}\end{aligned}$$

Note that the series in converges absolutely; for $-\sum_{\rho}z^{-\rho}\Gamma\left(\rho\right)$ see [@Kunik], formula $(2.2)$ and for $\sum_{\rho}\left(z^{-\rho}\gamma\left(\rho,2z\right)-\frac{2^{\rho}e^{-2z}}{\rho}\right)$, using the integral representation and integrating by pars, we observe that $$\sum_{\rho}\left(z^{-\rho}\gamma\left(\rho,2z\right)-\frac{2^{\rho}e^{-2z}}{\rho}\right)=z\sum_{\rho}\frac{1}{\rho}\int_{0}^{2}t^{\rho}e^{-tz}dt\ll\left(\left|z\right|+\left|z\right|^{2}\right)\sum_{\rho}\frac{1}{\left|\rho\right|^{2}}.$$ Now we consider the integral involving $\log\left(1-t^{-2}\right).$ Integrating by parts, we observe that $$-\frac{z}{2}\int\log\left(t\right)e^{-zt}dt=\frac{1}{2}\left(\log\left(t\right)e^{-zt}-\textrm{Ei}\left(-zt\right)\right)+K\label{eq:primitivaEi}$$ where $\textrm{Ei}\left(x\right)$ is the Exponential integral function and $K$ is a constant. So $$\begin{aligned}
-\frac{z}{2}\int_{2}^{\infty}\log\left(1-t^{-2}\right)e^{-tz}dt= & -\frac{z}{2}\int_{2}^{\infty}\log\left(t-1\right)e^{-tz}dt-\frac{z}{2}\int_{2}^{\infty}\log\left(t+1\right)e^{-tz}dt\\
+ & z\int_{2}^{\infty}\log\left(t\right)e^{-tz}dt\\
= & \mathcal{J}_{1}+\mathcal{J}_{2}+\mathcal{J}_{3},\end{aligned}$$ say. Let us study $\mathcal{J}_{1}$. By we have that $$\begin{aligned}
\mathcal{J}_{1}= & -\frac{e^{-z}}{2}z\int_{1}^{\infty}\log\left(t\right)e^{-tz}dt\\
= & \frac{e^{-z}}{2}\left(\log\left(t\right)e^{-tz}-\textrm{Ei}\left(-tz\right)\right)_{1}^{\infty}.\end{aligned}$$ From the asymptotic $$\textrm{Ei}\left(w\right)\sim i\pi\textrm{sgn}\left(\textrm{Im}\left(w\right)\right)-\frac{e^{w}}{w}\label{eq:eiasimpt}$$ as $\left|w\right|\rightarrow\infty$ (see for example [@Peg], equation $(26)$ at page $192$) we get $$\lim_{t\rightarrow\infty}\left(\log\left(t\right)e^{-tz}-\textrm{Ei}\left(-tz\right)\right)=-i\pi\textrm{sgn}\left(\textrm{Im}\left(-z\right)\right).\label{eq:limitEi}$$ So $$\mathcal{J}_{1}=\frac{e^{-z}}{2}\left(-i\pi\textrm{sgn}\left(\textrm{Im}\left(-z\right)\right)+\textrm{Ei}\left(-z\right)\right).$$ Let us consider $\mathcal{J}_{2}$. Using again we get $$\begin{aligned}
\mathcal{J}_{2}= & -\frac{e^{z}}{2}z\int_{3}^{\infty}\log\left(t\right)e^{-tz}dt\\
= & \frac{e^{z}}{2}\left(\log\left(t\right)e^{-tz}-\textrm{Ei}\left(-tz\right)\right)_{3}^{\infty}\\
= & \frac{e^{z}}{2}\left(-i\pi\textrm{sgn}\left(\textrm{Im}\left(-z\right)\right)-\log\left(3\right)e^{-3z}+\textrm{Ei}\left(-3z\right)\right).\end{aligned}$$ We have now to analyze $\mathcal{J}_{3}$. From we have $$\begin{aligned}
\mathcal{J}_{3}= & -\left(\log\left(t\right)e^{-tz}-\textrm{Ei}\left(-tz\right)\right)_{2}^{\infty}\\
= & i\pi\textrm{sgn}\left(\textrm{Im}\left(-z\right)\right)+\log\left(2\right)e^{-2z}-\textrm{Ei}\left(-2z\right).\end{aligned}$$ Finally $$\begin{aligned}
-\frac{z}{2}\int_{2}^{\infty}\log\left(1-t^{-2}\right)e^{-tz}dt= & \frac{e^{-z}}{2}\left(-i\pi\textrm{sgn}\left(\textrm{Im}\left(-z\right)\right)+\textrm{Ei}\left(-z\right)\right)\\
+ & \frac{e^{z}}{2}\left(-i\pi\textrm{sgn}\left(\textrm{Im}\left(-z\right)\right)-\log\left(3\right)e^{-3z}+\textrm{Ei}\left(-3z\right)\right)\\
+ & i\pi\textrm{sgn}\left(\textrm{Im}\left(-z\right)\right)+\log\left(2\right)e^{-2z}-\textrm{Ei}\left(-2z\right).\end{aligned}$$ So $$\begin{aligned}
\widetilde{S}\left(z\right)= & 2e^{-2z}+\frac{e^{-2z}}{z}-\sum_{\rho}z^{-\rho}\Gamma\left(\rho\right)\\
+ & \sum_{\rho}\left(z^{-\rho}\gamma\left(\rho,2z\right)-\frac{2^{\rho}e^{-2z}}{\rho}\right)-\frac{\zeta^{\prime}}{\zeta}\left(0\right)e^{-2z}\\
+ & \frac{e^{-z}}{2}\left(-i\pi\textrm{sgn}\left(\textrm{Im}\left(-z\right)\right)+\textrm{Ei}\left(-z\right)\right)\\
+ & \frac{e^{z}}{2}\left(-i\pi\textrm{sgn}\left(\textrm{Im}\left(-z\right)\right)-\log\left(3\right)e^{-3z}+\textrm{Ei}\left(-3z\right)\right)\\
+ & i\pi\textrm{sgn}\left(\textrm{Im}\left(-z\right)\right)+\log\left(2\right)e^{-2z}-\textrm{Ei}\left(-2z\right).\end{aligned}$$ We have almost proved the first part of the theorem. Now we have to show that we may exchange the series with the integral in . Let us define $$h_{T}\left(t\right)=e^{-tz}\sum_{\rho:\,\left|\gamma\right|\leq T}\frac{t^{\rho}}{\rho}\label{eq:convdominata}$$ where $T>2.$ From the truncated explicit formula (see [@daven] at page $109$) we have $$\psi\left(t\right)=t-\sum_{\rho:\,\left|\gamma\right|\leq T}\frac{t^{\rho}}{\rho}+O\left(\frac{t\log^{2}\left(tT\right)}{T}+\log\left(t\right)\min\left(1,\frac{t}{T\left\langle t\right\rangle }\right)\right),\,t>2,\,T>2\label{psi troncata}$$ where $\left\langle t\right\rangle $ is the distance from $t$ to the nearest prime power, and so $$\begin{aligned}
\left|h_{T}\left(t\right)\right|\ll & e^{-ta}\left(\left|\psi\left(t\right)-t\right|+\frac{t\log^{2}\left(tT\right)}{T}+\log\left(t\right)\right)\\
\ll & e^{-ta}\left(t+\frac{t\log^{2}\left(t\right)+2t\log\left(t\right)\log\left(T\right)+t\log^{2}\left(T\right)+T\log\left(t\right)}{T}\right)\end{aligned}$$ since $\left|\psi\left(t\right)-t\right|\ll t.$ But $\log^{2}\left(T\right)/T\ll1$ as $T\rightarrow\infty$, then $$\left|h_{T}\left(t\right)\right|\ll e^{-ta}t\log^{2}\left(t\right)$$ so we have to prove that $$\int_{2}^{\infty}t\log^{2}\left(t\right)e^{-ta}dt$$ converges. Trivially we have $$\int_{2}^{\infty}t\log^{2}\left(t\right)e^{-ta}dt\ll_{a}\int_{0}^{\infty}t^{2}e^{-t}dt=\Gamma\left(3\right)$$ so by the dominated convergence theorem we may exchange the integral with the series. It remains to prove the estimation of $E\left(a,y\right)$. From Lemma 1 of [@Langzac] (The bound has been corrected in [@Lang]) we observe that $$\sum_{\rho}\left(z^{-\rho}\gamma\left(\rho,2z\right)-\frac{2^{\rho}e^{-2z}}{\rho}\right)\ll\begin{cases}
1+\left|z\right|^{1/2}, & \left|y\right|\leq a\\
1+\left|z\right|^{1/2}\left(1+\log^{2}\left(\left|y\right|/a\right)\right), & a<\left|y\right|\leq1
\end{cases}$$ and $$E\left(a,y\right)\ll\begin{cases}
1+\left|z\right|^{1/2}, & \left|y\right|\leq a\\
1+\left|z\right|^{1/2}\left(1+\log^{2}\left(\left|y\right|/a\right)\right), & a<\left|y\right|\leq1
\end{cases}$$ so the estimation follows. If $\left|y\right|>1$ from we see that $$\left|\textrm{Ei}\left(-z\right)-i\pi\textrm{sgn}\left(\textrm{Im}\left(-z\right)\right)\right|\ll\left|\frac{e^{-z}}{z}\right|\ll1$$ as $y\rightarrow\pm\infty$ and this concludes the proof.

Proof of Theorem 2
==================

Let $z=1/N+iy$, $N\in\mathbb{N}$, $N>4$ and $y\in\mathbb{R}.$ From we have that $$\mathcal{L}\left(\psi\left(t\right)\right)\left(z\right)=z^{-1}\widetilde{S}\left(z\right)$$ so by the convolution theorem and the inversion theorem of the Laplace transform (see [@Folland] theorems $8.1$ and $8.4$), we get $$\begin{aligned}
\frac{1}{2\pi i}\int_{\left(1/N\right)}e^{uz}z^{-2}\widetilde{S}\left(z\right)^{2}dz= & \begin{cases}
\int_{2}^{u-2}\psi\left(t\right)\psi\left(u-t\right)dt, & u>4\\
0, & \mathrm{otherwise}
\end{cases}\end{aligned}$$ then from the explicit formula we have to evaluate $$\int_{2}^{u-2}\left(t-\sum_{\rho}\frac{t^{\rho}}{\rho}-\frac{\zeta^{\prime}}{\zeta}\left(0\right)-\frac{1}{2}\log\left(1-\frac{1}{t^{2}}\right)\right)\left(u-t-\sum_{\rho}\frac{\left(u-t\right)^{\rho}}{\rho}-\frac{\zeta^{\prime}}{\zeta}\left(0\right)-\frac{1}{2}\log\left(1-\frac{1}{\left(u-t\right)^{2}}\right)\right)dt$$ $$=\sum_{m=1}^{10}\int_{2}^{u-2}s_{m}\left(t,u\right)dt,$$ say, where the functions $s_{m}\left(t,u\right)$ will be defined in the next sections.

Integral of $s_{1}\left(t,u\right)$
-----------------------------------

We take $s_{1}\left(t,u\right):=t\left(u-t\right).$ We get immediately $$\int_{2}^{u-2}s_{1}\left(t,u\right)dt=\frac{u^{3}-24u+32}{6}.$$

Integral of $s_{2}\left(t,u\right)$
-----------------------------------

We take $s_{2}\left(t,u\right):=-2\left(u-t\right)\sum_{\rho}\frac{t^{\rho}}{\rho}.$ We have already seen in that $$\sum_{\rho:\,\left|\gamma\right|\leq T}\frac{t^{\rho}}{\rho}\ll t\log^{2}\left(t\right)\label{bound esplicito somma zeri troncata}$$ then by the dominated convergence theorem we can write $$\begin{aligned}
\int_{2}^{u-2}s_{2}\left(t,u\right)dt= & -2\sum_{\rho}\frac{1}{\rho}\int_{2}^{u-2}\left(u-t\right)t^{\rho}dt\\
= & -4\sum_{\rho}\frac{\left(u-2\right)^{\rho+1}}{\rho\left(\rho+1\right)}+\left(u-2\right)\sum_{\rho}\frac{2^{\rho+2}}{\rho\left(\rho+1\right)}\\
- & 2\sum_{\rho}\frac{1}{\rho\left(\rho+1\right)}\int_{2}^{u-2}t^{\rho+1}dt\\
= & -4\sum_{\rho}\frac{\left(u-2\right)^{\rho+1}}{\rho\left(\rho+1\right)}+\left(u-2\right)\sum_{\rho}\frac{2^{\rho+2}}{\rho\left(\rho+1\right)}\\
- & 2\sum_{\rho}\frac{\left(u-2\right)^{\rho+2}}{\rho\left(\rho+1\right)\left(\rho+2\right)}+\sum_{\rho}\frac{2^{\rho+3}}{\rho\left(\rho+1\right)\left(\rho+2\right)}.\end{aligned}$$ Note that the all the series in this Section converges absolutely.

Integral of $s_{3}\left(t,u\right)$
-----------------------------------

We consider $s_{3}\left(t,u\right):=-2t\frac{\zeta^{\prime}}{\zeta}\left(0\right).$ We immediately obtain $$\int_{2}^{u-2}s_{3}\left(t,u\right)dt=\frac{\zeta^{\prime}}{\zeta}\left(0\right)\left(4u-u^{2}\right).$$

Integral of $s_{4}\left(t,u\right)$
-----------------------------------

We take $s_{4}\left(t,u\right):=-\left(u-t\right)\log\left(1-\frac{1}{t^{2}}\right).$ We get $$\begin{aligned}
\int_{2}^{u-2}s_{4}\left(t,u\right)dt= & -\int_{2}^{u-2}\left(u-t\right)\log\left(1-\frac{1}{t^{2}}\right)dt\\
= & -\int_{2}^{u-2}\left(u-t\right)\log\left(t-1\right)dt-\int_{2}^{u-2}\left(u-t\right)\log\left(t+1\right)dt\\
+ & 2\int_{2}^{u-2}\left(u-t\right)\log\left(t\right)dt\end{aligned}$$ then, after simple but boring calculations, we obtain $$\begin{aligned}
\int_{2}^{u-2}s_{4}\left(t,u\right)dt= & \frac{1}{4}\left(\left(u-4\right)\left(3u-2\right)-2\left(u-3\right)\left(u+1\right)\log\left(u-3\right)\right.\\
- & 8+3u^{2}-6\log\left(3\right)+2u\left(\log\left(729\right)-5\right)-2\log\left(u-1\right)\left(u^{2}+2u-3\right)\\
- & \left.6u^{2}-8u\left(\log\left(4\right)-3\right)+2\log\left(256\right)+4\log\left(u-2\right)\left(u^{2}-4\right)\right).\end{aligned}$$ Also it is quite simple to observe that $$\int_{2}^{u-2}s_{4}\left(t,u\right)dt\ll u^{2}.$$

Integral of $s_{5}\left(t,u\right)$
-----------------------------------

We now take $s_{5}\left(t,u\right):=\sum_{\rho_{1}}\frac{\left(u-t\right)^{\rho_{1}}}{\rho_{1}}\sum_{\rho_{2}}\frac{t^{\rho_{2}}}{\rho_{2}}.$ Then we have to calculate $$\int_{2}^{u-2}s_{5}\left(t,u\right)dt=\int_{2}^{u-2}\sum_{\rho_{1}}\frac{\left(u-t\right)^{\rho_{1}}}{\rho_{1}}\sum_{\rho_{2}}\frac{t^{\rho_{2}}}{\rho_{2}}dt.$$ We have to prove that it is possible to exchange the integral and the series over the non-trivial zeros of the Riemann Zeta function. Let us define $$S_{T,u}^{1}\left(t\right)=\sum_{\rho_{1}:\,\left|\gamma_{1}\right|\leq T}\frac{\left(u-t\right)^{\rho_{1}}}{\rho_{1}}\sum_{\rho_{2}}\frac{t^{\rho_{2}}}{\rho_{2}},\,T>2.$$ From and the trivial bound $$\left|\sum_{\rho_{2}}\frac{t^{\rho_{2}}}{\rho_{2}}\right|=\left|t-\psi\left(t\right)-\frac{\zeta^{\prime}}{\zeta}\left(0\right)-\frac{1}{2}\log\left(1-\frac{1}{t^{2}}\right)\right|\ll t\label{trivial}$$ for $t>2,$ we have $$\left|S_{T,u}^{1}\left(t\right)\right|\ll t\left(u-t\right)\log^{2}\left(u-t\right)$$ then by the dominated convergence theorem $$\int_{2}^{u-2}\sum_{\rho_{1}}\frac{\left(u-t\right)^{\rho_{1}}}{\rho_{1}}\sum_{\rho_{2}}\frac{t^{\rho_{2}}}{\rho_{2}}dt=\sum_{\rho_{1}}\frac{1}{\rho_{1}}\int_{2}^{u-2}\left(u-t\right)^{\rho_{1}}\sum_{\rho_{2}}\frac{t^{\rho_{2}}}{\rho_{2}}dt.$$ Now if we take $$S_{T,u,\rho_{1}}^{2}\left(t\right)=\left(u-t\right)^{\rho_{1}}\sum_{\rho_{2}:\,\left|\gamma_{2}\right|\leq T}\frac{t^{\rho_{2}}}{\rho_{2}},\,T>2$$ from we have $$\left|S_{T,u,\rho_{1}}^{2}\left(t\right)\right|\ll\left(u-t\right)t\log^{2}\left(t\right)$$ then again by the dominated convergence theorem we can conclude that $$\sum_{\rho_{1}}\frac{1}{\rho_{1}}\int_{2}^{u-2}\left(u-t\right)^{\rho_{1}}\sum_{\rho_{2}}\frac{t^{\rho_{2}}}{\rho_{2}}dt=\sum_{\rho_{1}}\sum_{\rho_{2}}\frac{1}{\rho_{1}\rho_{2}}\int_{2}^{u-2}\left(u-t\right)^{\rho_{1}}t^{\rho_{2}}dt.$$

Then $$\begin{aligned}
\int_{2}^{u-2}\sum_{\rho_{1}}\frac{\left(u-t\right)^{\rho_{1}}}{\rho_{1}}\sum_{\rho_{2}}\frac{t^{\rho_{2}}}{\rho_{2}}dt= & \sum_{\rho_{1}}\sum_{\rho_{2}}\frac{u^{\rho_{1}+\rho_{2}+1}}{\rho_{1}\rho_{2}}\frac{\Gamma\left(\rho_{1}+1\right)\Gamma\left(\rho_{2}+1\right)}{\Gamma\left(\rho_{1}+\rho_{2}+2\right)}\\
- & \sum_{\rho_{1}}\sum_{\rho_{2}}\frac{u^{\rho_{1}+\rho_{2}+1}}{\rho_{1}\rho_{2}}B_{2/u}\left(\rho_{1}+1,\rho_{2}+1\right)\\
- & \sum_{\rho_{1}}\sum_{\rho_{2}}\frac{u^{\rho_{1}+\rho_{2}+1}}{\rho_{1}\rho_{2}}B_{2/u}\left(\rho_{2}+1,\rho_{1}+1\right)\\
= & \sum_{\rho_{1}}\sum_{\rho_{2}}u^{\rho_{1}+\rho_{2}+1}\frac{\Gamma\left(\rho_{1}\right)\Gamma\left(\rho_{2}\right)}{\Gamma\left(\rho_{1}+\rho_{2}+2\right)}-2\sum_{\rho_{1}}\sum_{\rho_{2}}\frac{u^{\rho_{1}+\rho_{2}+1}}{\rho_{1}\rho_{2}}B_{2/u}\left(\rho_{1}+1,\rho_{2}+1\right)\end{aligned}$$ where $B_{x}\left(a,b\right)$ is the incomplete Beta function (for more details see for example [@Olver], ch. $8.17$). We recall that $$\sum_{\rho_{1}}\sum_{\rho_{2}}u^{\rho_{1}+\rho_{2}+1}\frac{\Gamma\left(\rho_{1}\right)\Gamma\left(\rho_{2}\right)}{\Gamma\left(\rho_{1}+\rho_{2}+2\right)}$$ converges absolutely (see [@Langzac]), so $\sum_{\rho_{1}}\sum_{\rho_{2}}\frac{u^{\rho_{1}+\rho_{2}+1}}{\rho_{1}\rho_{2}}B_{2/u}\left(\rho_{1}+1,\rho_{2}+1\right)$ converges (at least conditionally), since $$\begin{aligned}
\left|2\sum_{\rho_{1}}\sum_{\rho_{2}}\frac{u^{\rho_{1}+\rho_{2}+1}}{\rho_{1}\rho_{2}}B_{2/u}\left(\rho_{1}+1,\rho_{2}+1\right)\right|\ll & \sum_{\rho_{1}}\sum_{\rho_{2}}\left|u^{\rho_{1}+\rho_{2}+1}\frac{\Gamma\left(\rho_{1}\right)\Gamma\left(\rho_{2}\right)}{\Gamma\left(\rho_{1}+\rho_{2}+2\right)}\right|\\
+ & \left|\int_{2}^{u-2}\sum_{\rho_{1}}\frac{\left(u-t\right)^{\rho_{1}}}{\rho_{1}}\sum_{\rho_{2}}\frac{t^{\rho_{2}}}{\rho_{2}}dt\right|\end{aligned}$$ and trivially $$\left|\int_{2}^{u-2}\sum_{\rho_{1}}\frac{\left(u-t\right)^{\rho_{1}}}{\rho_{1}}\sum_{\rho_{2}}\frac{t^{\rho_{2}}}{\rho_{2}}dt\right|\ll\int_{2}^{u-2}\left(u-t\right)tdt<\infty$$ from .

Integral of $s_{6}\left(t,u\right)$
-----------------------------------

We take $s_{6}\left(t,u\right):=2\frac{\zeta^{\prime}}{\zeta}\left(0\right)\sum_{\rho}\frac{t^{\rho}}{\rho}.$ From we know that it is possible to switch the integral and the series, so $$\begin{aligned}
\int_{2}^{u-2}s_{6}\left(t,u\right)dt= & 2\frac{\zeta^{\prime}}{\zeta}\left(0\right)\sum_{\rho}\frac{1}{\rho}\int_{2}^{u-2}t^{\rho}dt\\
= & 2\frac{\zeta^{\prime}}{\zeta}\left(0\right)\sum_{\rho}\frac{\left(u-2\right)^{\rho+1}}{\rho\left(\rho+1\right)}-\frac{\zeta^{\prime}}{\zeta}\left(0\right)\sum_{\rho}\frac{2^{\rho+2}}{\rho\left(\rho+1\right)}.\end{aligned}$$ Again it is quite simple to observe that $$\int_{2}^{u-2}s_{6}\left(t,u\right)dt\ll u^{2}$$ due to the absolute convergence of $\sum_{\rho}\frac{1}{\rho\left(\rho+1\right)}.$

Integral of $s_{7}\left(t,u\right)$
-----------------------------------

We consider $s_{7}\left(t,u\right):=\sum_{\rho}\frac{\left(u-t\right)^{\rho}}{\rho}\log\left(1-\frac{1}{t^{2}}\right)$. Using the dominated convergence theorem we can prove that we can exchange the integral and the series, so $$\begin{aligned}
\int_{2}^{u-2}s_{7}\left(t,u\right)dt= & \sum_{\rho}\frac{1}{\rho}\int_{2}^{u-2}\left(u-t\right)^{\rho}\log\left(1-\frac{1}{t^{2}}\right)dt\\
= & \sum_{\rho}\frac{1}{\rho}\int_{2}^{u-2}\left(u-t\right)^{\rho}\log\left(t-1\right)dt+\sum_{\rho}\frac{1}{\rho}\int_{2}^{u-2}\left(u-t\right)^{\rho}\log\left(t+1\right)dt\\
- & 2\sum_{\rho}\frac{1}{\rho}\int_{2}^{u-2}\left(u-t\right)^{\rho}\log\left(t\right)dt.\end{aligned}$$ Let us consider the first integral. Integrating by parts we get $$\begin{aligned}
H_{1}\left(u\right):= & \sum_{\rho}\frac{1}{\rho}\int_{2}^{u-2}\left(u-t\right)^{\rho}\log\left(t-1\right)dt\\
+ & \sum_{\rho}\frac{2^{\rho+1}\log\left(u-3\right)}{\rho\left(\rho+1\right)}-\sum_{\rho}\frac{1}{\rho\left(\rho+1\right)}\int_{2}^{u-2}\left(u-t\right)^{\rho+1}\left(t-1\right)^{-1}dt\\
= & \sum_{\rho}\frac{2^{\rho+1}\log\left(u-3\right)}{\rho\left(\rho+1\right)}-\sum_{\rho}\frac{\left(u-1\right)^{\rho+1}}{\rho\left(\rho+1\right)}\int_{2/\left(u-1\right)}^{1-1/\left(u-1\right)}s^{\rho+1}\left(1-s\right)^{-1}ds\\
= & \sum_{\rho}\frac{2^{\rho+1}\log\left(u-3\right)}{\rho\left(\rho+1\right)}-\sum_{\rho}\frac{\left(u-1\right)^{\rho+1}}{\rho\left(\rho+1\right)}\left[B_{1-1/\left(u-1\right)}\left(\rho+2,0\right)-B_{2/\left(u-1\right)}\left(\rho+2,0\right)\right]\end{aligned}$$ In a similar way we have $$\begin{aligned}
H_{2}\left(u\right):= & \sum_{\rho}\frac{1}{\rho}\int_{2}^{u-2}\left(u-t\right)^{\rho}\log\left(t+1\right)dt\\
= & \sum_{\rho}\frac{2^{\rho+1}\log\left(u-1\right)}{\rho\left(\rho+1\right)}-\sum_{\rho}\frac{\left(u-2\right)^{\rho+1}\log\left(3\right)}{\rho\left(\rho+1\right)}\\
- & \sum_{\rho}\frac{1}{\rho\left(\rho+1\right)}\int_{2}^{u-2}\left(u-t\right)^{\rho+1}\left(t+1\right)^{-1}dt\\
= & \sum_{\rho}\frac{2^{\rho}\log\left(u-1\right)}{\rho\left(\rho+1\right)}-\sum_{\rho}\frac{\left(u-2\right)^{\rho}\log\left(3\right)}{\rho\left(\rho+1\right)}\\
- & \sum_{\rho}\frac{\left(u+1\right)^{\rho+1}}{\rho\left(\rho+1\right)}\left[B_{1-3/\left(u+1\right)}\left(\rho+2,0\right)-B_{2/\left(u+1\right)}\left(\rho+2,0\right)\right]\end{aligned}$$ and $$\begin{aligned}
H_{3}\left(u\right):= & -2\sum_{\rho}\frac{1}{\rho}\int_{2}^{u-2}\left(u-t\right)^{\rho}\log\left(t\right)dt\\
= & -\sum_{\rho}\frac{2^{\rho+2}\log\left(u-2\right)}{\rho\left(\rho+1\right)}+2\sum_{\rho}\frac{\left(u-2\right)^{\rho+1}\log\left(2\right)}{\rho\left(\rho+1\right)}\\
+ & 2\sum_{\rho}\frac{1}{\rho\left(\rho+1\right)}\int_{2}^{u-2}\frac{\left(u-t\right)^{\rho+1}}{t}dt\\
= & -\sum_{\rho}\frac{2^{\rho+2}\log\left(u-2\right)}{\rho\left(\rho+1\right)}+2\sum_{\rho}\frac{\left(u-2\right)^{\rho+1}\log\left(2\right)}{\rho\left(\rho+1\right)}\\
+ & 2\sum_{\rho}\frac{u^{\rho+1}}{\rho\left(\rho+1\right)}\left[B_{1-2/u}\left(\rho+2,0\right)-B_{2/u}\left(\rho+2,0\right)\right]\end{aligned}$$ Finally we can write $$\int_{2}^{u-2}s_{7}\left(t,u\right)dt=H_{1}\left(u\right)+H_{2}\left(u\right)+H_{3}\left(u\right).$$ From the well known estimation $$\sum_{\rho}\frac{x^{\rho}}{\rho}\ll x\exp\left(-C\sqrt{\log\left(x\right)}\right),\,x>1,\,C>0$$ (see [@daven], chapter $18$) we can easily conclude that $$\int_{2}^{u-2}s_{7}\left(t,u\right)dt\ll u^{2}.$$ Again we can observe that all the series over the non-trivial zeros of $\zeta\left(s\right)$ in this Section converges absolutely; to see this fact in the case of the series involving the incomplete Beta function (which is, probably, the less evident absolute convergence) note that $$2\sum_{\rho}\frac{u^{\rho+1}}{\rho\left(\rho+1\right)}\left[B_{1-2/u}\left(\rho+2,0\right)-B_{2/u}\left(\rho+2,0\right)\right]=2\sum_{\rho}\frac{1}{\rho\left(\rho+1\right)}\int_{2}^{u-2}\frac{\left(u-t\right)^{\rho+1}}{t}dt$$ $$\ll\sum_{\rho}\frac{1}{\left|\rho\right|\left|\rho+1\right|}\int_{2}^{u-2}\frac{\left(u-t\right)^{2}}{t}dt<\infty.$$

Integral of $s_{8}\left(t,u\right)$
-----------------------------------

We take $s_{8}\left(t,u\right):=\frac{\zeta^{\prime}}{\zeta}\left(0\right)^{2}.$ We immediately get $$\int_{2}^{u-2}s_{8}\left(t,u\right)dt=\frac{\zeta^{\prime}}{\zeta}\left(0\right)^{2}\left(u-4\right).$$

Integral of $s_{9}\left(t,u\right)$
-----------------------------------

We consider $s_{9}\left(t,u\right):=\frac{\zeta^{\prime}}{\zeta}\left(0\right)\log\left(1-\frac{1}{t^{2}}\right).$ We obtain $$\begin{aligned}
\int_{2}^{u-2}s_{9}\left(t,u\right)dt= & \frac{\zeta^{\prime}}{\zeta}\left(0\right)\left(\left(u-3\right)\log\left(u-3\right)\right.\\
+ & \log\left(\frac{1}{27\left(u-1\right)}\right)+u\log\left(u-1\right)\\
+ & \left.2\log\left(4\right)-2\left(u-2\right)\log\left(u-2\right)\right)\end{aligned}$$ and again we trivially get the bound $$\int_{2}^{u-2}s_{9}\left(t,u\right)dt\ll u^{2}.$$

Integral of $s_{10}\left(t,u\right)$
------------------------------------

We take $s_{10}\left(t,u\right):=\frac{1}{4}\log\left(1-\frac{1}{t^{2}}\right)\log\left(1-\frac{1}{\left(u-t\right)^{2}}\right).$ We have that $$\begin{aligned}
\int_{2}^{u-2}s_{10}\left(t,u\right)dt=\frac{1}{4} & \int_{2}^{u-2}\log\left(1-\frac{1}{t^{2}}\right)\log\left(1-\frac{1}{\left(u-t\right)^{2}}\right)dt\nonumber \\
= & \frac{1}{4}\int_{2}^{u-2}\log\left(\frac{t^{2}-1}{t^{2}}\right)\log\left(\frac{\left(u-t\right)^{2}-1}{\left(u-t\right)^{2}}\right)dt\nonumber \\
= & \frac{1}{4}\int_{2}^{u-2}\left(\log\left(t-1\right)+\log\left(t+1\right)-2\log\left(t\right)\right)\nonumber \\
\times & \left(\log\left(u-t-1\right)+\log\left(u-t+1\right)-2\log\left(u-t\right)\right)dt\label{integrali prodotti log}\end{aligned}$$ so we have to evaluate all the combinations of . We will calculate explicitly only the first, since the others are similar. We consider $$\begin{aligned}
\int_{2}^{u-2}\log\left(t-1\right)\log\left(u-t-1\right)dt= & \int_{1}^{u-3}\log\left(v\right)\log\left(u-v-2\right)dv\\
= & \left(u-2\right)\int_{1/\left(u-2\right)}^{1-1/\left(u-2\right)}\log\left(\left(u-2\right)s\right)\log\left(\left(u-2\right)\left(1-s\right)\right)ds\\
= & \left(u-2\right)\log^{2}\left(u-2\right)\left(1-\frac{2}{u-2}\right)\\
+ & \left(u-2\right)\log\left(u-2\right)\int_{1/\left(u-2\right)}^{1-1/\left(u-2\right)}\log\left(1-s\right)ds\\
+ & \left(u-2\right)\log\left(u-2\right)\int_{1/\left(u-2\right)}^{1-1/\left(u-2\right)}\log\left(s\right)ds\\
+ & \left(u-2\right)\int_{1/\left(u-2\right)}^{1-1/\left(u-2\right)}\log\left(s\right)\log\left(1-s\right)ds.\end{aligned}$$ Maybe it is useful to recall that, integrating by parts, we get $$\begin{aligned}
\left(u-2\right)\int_{1/\left(u-2\right)}^{1-1/\left(u-2\right)}\log\left(s\right)\log\left(1-s\right)ds= & \left(u-4\right)\log\left(1-\frac{1}{u-2}\right)\log\left(\frac{1}{u-2}\right)\\
- & \left(u-2\right)\int_{1/\left(u-2\right)}^{1-1/\left(u-2\right)}s\left(\frac{\log\left(1-s\right)}{s}-\frac{\log\left(s\right)}{1-s}\right)ds\end{aligned}$$ and $$\begin{aligned}
-\left(u-2\right)\int_{1/\left(u-2\right)}^{1-1/\left(u-2\right)}s\left(\frac{\log\left(1-s\right)}{s}-\frac{\log\left(s\right)}{1-s}\right)ds= & -\left(u-2\right)\int_{1/\left(u-2\right)}^{1-1/\left(u-2\right)}\log\left(1-s\right)ds\\
- & \left(u-2\right)\int_{1/\left(u-2\right)}^{1-1/\left(u-2\right)}\log\left(s\right)ds\\
+ & \left(u-2\right)\int_{1/\left(u-2\right)}^{1-1/\left(u-2\right)}\frac{\log\left(s\right)}{1-s}ds\\
= & -2\left(u-3\right)\log\left(u-3\right)+2\left(u-4\right)\left(1+\log\left(u-2\right)\right)\\
+ & \left(u-2\right)\left(\mathrm{Li}_{2}\left(\frac{1}{u-2}\right)-\mathrm{Li}_{2}\left(1-\frac{1}{u-2}\right)\right)\end{aligned}$$ where $\mathrm{Li}_{2}\left(x\right)$ is the Dilogartihm function. After a boring calculation we obtain $$\begin{aligned}
V_{1}\left(u\right):=\frac{1}{4}\int_{2}^{u-2}\log\left(t-1\right)\log\left(u-t-1\right)dt= & \frac{1}{4}\left(2\left(u-4\right)+\log\left(u-3\right)\left(6-2u+\left(u-2\right)\log\left(u-2\right)\right)\right.\\
+ & \left.\left(u-2\right)\mathrm{Li}_{2}\left(\frac{1}{u-2}\right)-\left(u-2\right)\mathrm{Li}_{2}\left(1-\frac{1}{u-2}\right)\right).\end{aligned}$$ For the explicit calculations of the other integrals see the appendix. Furthermore it is clear that $$\int_{2}^{u-2}s_{10}\left(t,u\right)dt\ll u^{2}$$ holds.

To finish the proof we have only to take $u=N$, recalling that $$\sum_{n\leq N}r_{G}\left(n\right)\left(N-n\right)=\frac{1}{2\pi i}\int_{\left(1/N\right)}e^{Nz}z^{-2}\widetilde{S}\left(z\right)^{2}dz.$$

Appendix
========

In this section we calculate explicitly all the integrals from the Section $4.10$. Using the same ideas that we used for $V_{1}\left(u\right)$ (and the Mathematica’s help) we get $$\begin{aligned}
V_{2}\left(u\right):= & \frac{1}{4}\int_{2}^{u-2}\log\left(t-1\right)\log\left(u-t+1\right)dt\\
= & \frac{1}{4}\left(-8+2u+\log\left(27\right)+\log\left(u-1\right)+\log\left(u-3\right)\left(u\log\left(u\right)+3-3\log\left(3\right)\right)\right.\\
- & \left.u\log\left(u^{2}-4u+3\right)+u\mathrm{Li}_{2}\left(\frac{1}{u}\right)-u\mathrm{Li}_{2}\left(1-\frac{3}{u}\right)\right),\end{aligned}$$ $$\begin{aligned}
V_{3}\left(u\right):= & \frac{1}{4}\int_{2}^{u-2}\log\left(t+1\right)\log\left(u-t-1\right)dt\\
= & \frac{1}{4}\left(-8+2u-3\left(-1+\log\left(3\right)\right)\log\left(u-3\right)+\log\left(27(u-1)\right)+u\log\left(3\right)\log\left(1-\frac{3}{u}\right)\right.\\
+ & \left.u\log\left(u-1\right)\log\left(u\right)-u\log\left(u^{2}-4u+3\right)+u\mathrm{Li}_{2}\left(\frac{3}{u}\right)-u\mathrm{Li}_{2}\left(1-\frac{1}{u}\right)\right),\end{aligned}$$ $$\begin{aligned}
V_{4}\left(u\right):= & \frac{1}{4}\int_{2}^{u-2}\log\left(t+1\right)\log\left(u-t+1\right)dt\\
= & \frac{1}{4}\left(-8+2u+\log\left(729\right)+2\log\left(u-1\right)-2u\log\left(u-1\right)-4\log\left(3\right)\log\left(u-1\right)\right.\\
+ & u\log\left(3\right)\log\left(u-1\right)-u\log\left(3\right)\log\left(u+2\right)-\log\left(9\right)\log\left(u+2\right)+2\log\left(u-1\right)\log\left(u+2\right)\\
- & \left.u\log\left(u-1\right)\log\left(u+2\right)+\left(2+u\right)\mathrm{Li}_{2}\left(\frac{3}{u+2}\right)-\left(2+u\right)\mathrm{Li}_{2}\left(1-\frac{3}{u+2}\right)\right),\end{aligned}$$ $$\begin{aligned}
V_{5}\left(u\right):= & -\frac{1}{2}\int_{2}^{u-2}\log\left(t-1\right)\log\left(u-t\right)dt\\
= & \frac{1}{2}\left(8-2u-\log\left(4\right)-2\log\left(u-2\right)-\log\left(u-3\right)\left(3+u\log\left(u-1\right)-\log\left(4\left(u-1\right)\right)\right)\right.\\
+ & \left.u\log\left(u^{2}-5u+6\right)-\left(u-1\right)\mathrm{Li}_{2}\left(\frac{1}{u-1}\right)+\left(u-1\right)\mathrm{Li}_{2}\left(1-\frac{2}{u-1}\right)\right)\end{aligned}$$ $$\begin{aligned}
V_{6}\left(u\right):= & -\frac{1}{2}\int_{2}^{u-2}\log\left(t+1\right)\log\left(u-t\right)dt\\
= & \frac{1}{2}\left(8-2u-2\log\left(u-2\right)+u\log\left(u-2\right)-u\log\left(3\right)\log\left(u-2\right)\right.\\
+ & \log\left(9\right)\log\left(u-2\right)+\log\left(\frac{1}{108\left(u-1\right)}\right)+u\log\left(u-1\right)+\log\left(4\right)\log\left(u-1\right)\\
+ & \log\left(3\right)\log\left(u+1\right)+u\log\left(3\right)\log\left(u+1\right)-\log\left(u-1\right)\log\left(u+1\right)\\
- & \left.u\log\left(u-1\right)\log\left(u+1\right)-\left(u+1\right)\mathrm{Li}_{2}\left(\frac{3}{u+1}\right)+\left(u+1\right)\mathrm{Li}_{2}\left(1-\frac{2}{u+1}\right)\right),\end{aligned}$$ $$\begin{aligned}
V_{7}\left(u\right):= & -\frac{1}{2}\int_{2}^{u-2}\log\left(t\right)\log\left(u-t-1\right)dt\\
= & \frac{1}{2}\left(8-2u-\log\left(4\right)-3\log\left(u-3\right)+u\log\left(u-3\right)-u\log\left(2\right)\log\left(u-3\right)\right.\\
+ & \log\left(8\right)\log\left(u-3\right)-2\log\left(u-2\right)+u\log\left(u-2\right)-\log\left(2\right)\log\left(u-1\right)\\
+ & u\log\left(2\right)\log\left(u-1\right)+\log\left(u-2\right)\log\left(u-1\right)-u\log\left(u-2\right)\log\left(u-1\right)\\
- & \left.\left(u-1\right)\mathrm{Li}_{2}\left(\frac{2}{u-1}\right)+\left(u-1\right)\mathrm{Li}_{2}\left(1-\frac{1}{u-1}\right)\right),\end{aligned}$$ $$\begin{aligned}
V_{8}\left(u\right):= & -\frac{1}{2}\int_{2}^{u-2}\log\left(t\right)\log\left(u-t+1\right)dt\\
= & \frac{1}{2}\left(8-2u-2\log\left(u-2\right)+u\log\left(u-2\right)]+\log\left(27\right)\log\left(u-2\right)+\log\left(\frac{1}{108\left(u-1\right)}\right)\right.\\
+ & u\log\left(u-1\right)-\log\left(2\right)\log\left(u-1\right)-u\log\left(2\right)\log\left(u-1\right)+\log\left(4\right)\log\left(u-1\right)\\
+ & \log\left(2\right)\log\left(u+1\right)+u\log\left(2\right)\log\left(u+1\right)-\log\left(u-2\right)\log\left(u+1\right)\\
- & \left.u\log\left(u-2\right)\log\left(u+1\right)-\left(u+1\right)\mathrm{Li}_{2}\left(\frac{2}{u+1}\right)+\left(u+1\right)\mathrm{Li}_{2}\left(1-\frac{3}{u+1}\right)\right)\end{aligned}$$ and finally $$\begin{aligned}
V_{9}\left(u\right):=\int_{2}^{u-2}\log\left(t\right)\log\left(u-t\right)dt= & -8+2u+\log\left(16\right)-u\log\left(2\right)\log\left(u\right)\\
+ & \log\left(u-2\right)\left(-2\left(-2+u+\log\left(4\right)\right)+u\log\left(2u\right)\right)\\
+ & u\left(\mathrm{Li}_{2}\left(\frac{2}{u}\right)-\mathrm{Li}_{2}\left(1-\frac{2}{u}\right)\right).\end{aligned}$$ So we can write $$\int_{2}^{u-2}s_{10}\left(t,u\right)dt=\sum_{m=1}^{9}V_{m}\left(u\right).$$

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