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
    $\begingroup$ 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. $\endgroup$ – Peter Humphries May 11 '19 at 15:04
  • $\begingroup$ Basically I have to follow the proof of Theorem 12.5 in Montgomery-Vaughan to get the claim, right? $\endgroup$ – asd May 11 '19 at 15:56
  • 1
    $\begingroup$ pretty much - basicallu the bottom half of p.401 $\endgroup$ – Peter Humphries May 11 '19 at 15:57
  • $\begingroup$ great! many thanks! $\endgroup$ – asd May 11 '19 at 15:59
  • 1
    $\begingroup$ I suggest that either @Peter or asd post an answer based on these comments. $\endgroup$ – Gerry Myerson May 11 '19 at 23:27

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 12.2 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.