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I believe that it is possible show the following

Fact. For real $x>e$ then $$-\frac{\zeta'(\log x)}{x\zeta(\log x)}=\sum_{n=1}^\infty\frac{\Lambda(n)}{n^{\log x}},$$ where $\zeta(x)$ is the evaluation of the Riemann zeta function in a real, and $\Lambda(n)$ is the von Mangoldt function.

When I tried use this fact I deduce the following, I don't know if it is useful or these computations were in the literature

Question. Can you prove and justify for integers $k>1$ that $$\sum_{n=1}^\infty\Lambda(n)\log\frac{e}{n}\int_{e^2}^{e^{k+1}}\frac{\log\zeta(\log x)}{n^{\log x}}dx$$ equals to $$(k-1)+e^2\frac{\zeta'(2)}{\zeta(2)}\log\zeta(2)-e^{k+1}\frac{\zeta'(k+1)}{\zeta(k+1)}\log\zeta(k+1)?$$ Thanks in advance.

My computations were tedious but I believe that if there are no mistakes is the only way to solve it:

First I've deduced $$-1=\int_{e^2}^{e^3}\frac{\zeta(\log x)}{\zeta'(\log x)}\sum_{n=1}^{\infty}\frac{\Lambda(n)}{n^{\log x}}dx,$$ after I did the change $dv=\frac{\zeta(\log x)}{x\zeta'(\log x)}dx$ with $u=n^{-\log x}$, to get by** integration by parts** $$-1=-e^3\frac{\zeta'(3)}{\zeta(3)}\log\zeta(3)+e^2\frac{\zeta'(2)}{\zeta(2)}\log\zeta(2)-\sum_{n=1}^\infty\Lambda(n)\log\frac{e}{n}\int_{e^2}^{e^3}\frac{\log\zeta(\log x)}{n^{\log x}}dx,$$ and thus we can write similar equations that telescoping get the result.

If there are mistakes please tell us.

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    $\begingroup$ You don't need to do tedious calculations. Since $x>e$ you have $\log(x)>1$ so you can prove your equality (without the $x$ at the denominator) taking the logarithm of the Euler's product of $\zeta(\log(x))$ and then differentiating. $\endgroup$ – Marco Cantarini May 6 '16 at 16:14
  • $\begingroup$ Then feel free to add a more simple proof, i understand your aproach. Thanks @MarcoCantarini $\endgroup$ – user243301 May 6 '16 at 17:19
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Take $s=\log\left(x\right) $. Since $x>e $ we have $s>1 $. So we can consider the Euler's product of the Zeta function $$\zeta\left(s\right)=\prod_{p}\left(1-\frac{1}{p^{s}}\right)^{-1} $$ then taking the logarithm we have$$\log\left(\zeta\left(s\right)\right)=-\sum_{p}\log\left(1-\frac{1}{p^{s}}\right) $$ and differentiating $$\frac{\zeta'\left(s\right)}{\zeta\left(s\right)}=-\sum_{p}\frac{\log\left(p\right)}{p^{s}}\left(1-\frac{1}{p^{s}}\right)^{-1}=-\sum_{p}\frac{\log\left(p\right)}{p^{s}}\sum_{k\geq1}\frac{1}{p^{sk}} $$ $$=-\sum_{n\geq1}\frac{\Lambda\left(n\right)}{n^{s}} $$ then $$\frac{\zeta'\left(\log\left(x\right)\right)}{\zeta\left(\log\left(x\right)\right)}=-\sum_{n\geq1}\frac{\Lambda\left(n\right)}{n^{\log\left(x\right)}}. $$ Note that the series start from $1$ but $\Lambda\left(1\right)=0 $ so in some text you can find that the series starts from $2$.

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  • $\begingroup$ Very thanks much for your answers, when your do it, I can say that it were good computations, yours, and the statement is right when you say it. $\endgroup$ – user243301 May 6 '16 at 18:38
  • $\begingroup$ @user243301 You're welcome, I'm glad to help. $\endgroup$ – Marco Cantarini May 6 '16 at 18:53

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