Prove that the logarithmic derivative of the Riemann zeta function $$-\frac{\zeta^{\prime}(s)}{\zeta(s)}=\sum_{n=1}^{\infty}\frac{\Lambda(n)}{n^s}=s\int_{1}^{\infty}\psi(x)x^{-s-1}\, d\mathrm{x},$$ where $\Lambda(n)$ is the Mangoldt function and $\psi(x)=\sum_{n\leq x} \Lambda(n)$ (Chebychev function), $Re(s)>1$.
I found this result in page 172 of the book Number Theory an Introduction via the Distribution of Primes by Benjamin Fire but I am not able to prove it.
Best Answer
@reuns has summarised it already. I'll get you started. Since $-\ln\zeta(s)=\sum_{p\in\Bbb P}\ln(1-p^{-s})$,$$-\frac{\zeta^\prime(s)}{\zeta(s)}=\sum_{p\in\Bbb P}\frac{p^{-s}\ln p}{1-p^{-s}}=\sum_{p\in\Bbb P,\,k\ge1}p^{-ks}\ln p=\sum_{n\ge1}\Lambda(n)n^{-s}=s\int_n^\infty\sum_{n\ge1}\Lambda(n)x^{-s-1}dx.$$You can rewrite this in terms of $\psi$ with the Iverson bracket, which I'll leave you to try.