# Complex integral of logarithmic derivative of $\zeta$

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!

• Use (10.27) in Montgomery-Vaughan and Stirling's formula to bound the second integral, and Lemma 12.2 in Montgomery-Vaughan to bound the first and third integrals. – Peter Humphries May 11 '19 at 15:04
• Basically I have to follow the proof of Theorem 12.5 in Montgomery-Vaughan to get the claim, right? – asd May 11 '19 at 15:56
• pretty much - basicallu the bottom half of p.401 – Peter Humphries May 11 '19 at 15:57
• great! many thanks! – asd May 11 '19 at 15:59
• I suggest that either @Peter or asd post an answer based on these comments. – Gerry Myerson May 11 '19 at 23:27

Lemma 12.2 For 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 $$-1\leq \sigma\leq 2$$.
Lemma 12.4 Let $$\mathcal{A}$$ 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.