1
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ This is explained in every text, the difficulty is to find a sequence $T_l$ such that $\lim_{l \to \infty} \int_{2+iT_l}^{-1+iT_l} \frac{\zeta'}{\zeta}(s) \frac{x^s}{s^2}ds = 0$, you know it exists by the density of zeros $O(T \ln T)$ and $\frac{\zeta'}{\zeta}(s) = \sum_{|\rho-s| < 1} \frac{1}{s-\rho} + O(\ln t)$, once you have it $\sum_{n \le x} \Lambda(n) (1-\frac{\ln n}{\ln x})=\int_{2-i\infty}^{2+i\infty}\frac{\zeta'}{\zeta}(s) \frac{x^s}{s^2}ds =\lim_{l \to \infty} \int_{\partial ([2-T_l,2] +i [-T_l,T_l])}\frac{\zeta'}{\zeta}(s) \frac{x^s}{s^2}ds = \lim_{l \to \infty} \sum Res(...)$. $\endgroup$ – reuns May 11 at 14:46
  • $\begingroup$ In which text I can find it explained in detail? $\endgroup$ – asd May 11 at 14:49
  • $\begingroup$ Also $\sum_{n \le x} \Lambda(n) (1-\frac{\ln n}{\ln x}) x^{-2}$ is integrable and continuous and so is its Mellin transform (on vertical lines) so the Mellin inversion converges locally uniformly for every $x>0$. What makes a difference is that for $x \le 1$ you'll apply the residue theorem to $\int_{2-i\infty}^{2+i\infty}\frac{\zeta'}{\zeta}(s) \frac{x^s}{s^2}ds =-\lim_{l \to \infty} \int_{\partial ([2,2+\color{red}{+} T_l] +i [-T_l,T_l])}\frac{\zeta'}{\zeta}(s) \frac{x^s}{s^2}ds = 0$, obtaining $\sum_\rho \frac{x^\rho}{\rho^2}1_{x>1}$ in the explicit formula $\endgroup$ – reuns May 11 at 14:56
  • $\begingroup$ can I use the arguments of Chapter 12 - Multiplicative umber Theory by Montgomery and Vaughan? Overthere they work with the integral of $-\frac{\zeta^{\prime}}{\zeta}(s)\frac{x^s}{s}$ $\endgroup$ – asd May 11 at 14:59

Your Answer

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

Browse other questions tagged or ask your own question.