I want to prove that for any $x\geq 2$ we have $$ \begin{split} -\frac{\zeta^{\prime}}{\zeta}(s)&=\sum_{n\leq x}\frac{\Lambda(n)}{n^s}\frac{\log(x/n)}{\log x}+\frac{1}{\log x}\left(\frac{\zeta^{\prime}}{\zeta}(s)\right)^{\prime}+\frac{1}{\log x}\sum_{\rho}\frac{x^{\rho-s}}{(\rho-s)^2}\\ &\qquad\qquad\qquad-\frac{x^{1-s}}{(1-s)^2\log x}+\frac{1}{\log x}\sum_{k=1}^{\infty}\frac{x^{-2k-s}}{(2k+s)^2}. \end{split} $$ The idea of the proof is to consider to express $\frac{\zeta^{\prime}}{\zeta}$ as a Dirichlet series and make use of the identity (for $c>0$) $$ \frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}x^s\frac{ds}{s^2}= \begin{cases} \log x &\text{if }x\geq 1,\\ 0 &\text{if } 0\leq x <1. \end{cases} $$ Indeed in this way, setting $c=\max\{1,2-\sigma\}$ and interchanging the order of summation and integration (which is justified b absolute convergence), we get $$ \begin{split} \frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}-\frac{\zeta^{\prime}}{\zeta}(s+w)\frac{x^w}{w^2}\,dw&=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\left[\sum_{n=1}^{\infty}\frac{\Lambda(n)}{n^{s+w}}\right]\frac{x^w}{w^2}\,dw\\ &=\frac{1}{2\pi i}\sum_{n=1}^{\infty}\frac{\Lambda(n)}{n^s}\int_{c-i\infty}^{c+i\infty}\left(\frac{x}{n}\right)^w\frac{dw}{w^2}\\ &=\sum_{n\leq x}\frac{\Lambda(n)}{n^s}\log(x/n). \end{split} $$ Now I would to estimate the integral in another way: moving the line of integration to the left ($c\to \infty$) and using Cauchy residue theorem. The residue I get is $$ -\frac{\zeta^{\prime}}{\zeta}(s)\log x-\left(\frac{\zeta^{\prime}}{\zeta}(s)\right)^{\prime}-\sum_{\rho}\frac{x^{\rho-s}}{(\rho-s)^2}+\frac{x^{1-s}}{(1-s)^2}-\sum_{k=1}^{\infty}\frac{x^{-2k-s}}{(2k+s)^2}. $$ Therefore I would get my claim if I was able to show that the other integrals go to zero. How can I show it? Let $$ f(w)=-\frac{\zeta^{\prime}}{\zeta}(s+w)\frac{x^w}{w^2} $$ I have to fix $K>0$ and show that $$ \int_{c+iK}^{-K+iK}f(w)\,dw,\qquad \int_{-K+iK}^{-K-iK}f(w)\,dw,\qquad \int_{-K-iK}^{c-iK}f(w)\,dw, $$ all tend to zero as $K\to \infty$. Do you have any hint on how to proceed? Which bound should I use? Thanks for your help!

## 1 Answer

As suggested by @Peter we can prove the claim as follows. We will use of the following results from the book "Multiplicative Number Theory, I" by Montgomery-Vaughan.

Lemma 12.2For each real number $T\geq 2$ there is a $T_1$ with $T\leq T_1\leq T+1$ such that $$ \frac{\zeta^{\prime}}{\zeta}(\sigma+iT_1)\ll(\log T)^2 $$ uniformly for $-1\leq \sigma\leq 2$.

and

Lemma 12.4Let $\mathcal{A}$ denote the set of points of $s\in\mathbb{C}$ such that $\sigma\leq -1$ and $|s+2k|\geq 1/4$ for every positive integer $k$. Then $$ \frac{\zeta^{\prime}}{\zeta}(s)\ll\log(|s|+1) $$ uniformly for $s\in\mathcal{A}$.

Using these two results we proceed as follows (basically following the proof of **Theorem 12.5** in the same book): let $K$ be a positive integer and consider the contour of integration consisting of the line segments connecting
$$
c-iT_1,\quad -K-iT_1,\quad -K+iT_1,\quad c+iT_1
$$
Moreover we split the horizontal segments
$$
[-K\pm iT_1,\,c\pm iT_1]
$$
in
$$
[-K\pm iT_1,\,-1-\Re(s)\pm iT_1]\cup [-1-\Re(s)\pm iT_1,\,c\pm iT_1]
$$
Since $|\sigma+iT_1|\geq T$, by Lemma 12.2 if follows that
$$
\int_{-1-\Re(s)\pm iT_1}^{c\pm iT_1}-\frac{\zeta^{\prime}}{\zeta}(s+w)\frac{x^w}{w^2}\,dw\ll\frac{(\log T)^2}{T^2}\int_{-1-\Re(s)}^{c}x^{\sigma}d\sigma\ll\frac{x(\log T)^2}{T^2 \log x}
$$
which goes to 0 as $T\to \infty$. Similarly, by Lemma 12.4 we have
$$
\int_{-K\pm iT_1}^{-1-\Re(s)\pm iT_1}-\frac{\zeta^{\prime}}{\zeta}(s+w)\frac{x^w}{w^2}\,dw\ll \frac{\log T}{T^2}\int_{-K}^{-1-\Re(s)}x^{\sigma}d\sigma\ll\frac{\log T}{T^2}\int_{-\infty}^{-1}x^{\sigma}d\sigma\\ \ll\frac{\log T}{xT^2 \log x}
$$
which again tends to 0 as $T\to \infty$. It remains to bound the vertical segment. Using again Lemma 12.4 and subsadditivity of the logarithm, we get
$$
\int_{-K-iT_1}^{-K+iT_1}-\frac{\zeta^{\prime}}{\zeta}(s+w)\frac{x^w}{w^2}\,dw\ll\frac{\log (K+T)}{K^2}x^{-K}\int_{-T_1}^{T_1}1\,dt\ll\frac{T\log KT}{K^2x^K}
$$
which tends to 0 as $K\to \infty$, proving our claim.

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