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.