How does it follow $s\int_1^{\infty}\frac{\psi(x)}{x^{s+1}}dx$? I have two relations:
1)$-\frac{\zeta'(s)}{\zeta(s)}=\sum_{1}^{\infty}\frac{\Lambda(n)}{n^s}$.
2)$\psi(x)=\sum_{n\leq x}\Lambda(n)$.
From these two how does it follow that $-\frac{\zeta'(s)}{\zeta(s)}=s\int_1^{\infty}\frac{\psi(x)}{x^{s+1}}dx$, for $s>1$.
Can anyone explain how does it follow?
 A: Let's start with the definition of the von Mangoldt function ($p$ will always mean a prime) :
$$\ \Lambda(n):=\begin{cases}
\log\, p & \text{if}\ n=p^k\ \text{and}\ k>0\\
0 & \text{else}
\end{cases}$$
We may use the Euler product in $\ \displaystyle\log \zeta(s)=-\sum_{p\ \text{prime}}\log(1-p^{-s})=\sum_p\sum_{k=1}^\infty \frac{p^{-ks}}k\ $ to obtain your first relation (after derivation) :
$$\tag{1}-\frac{\zeta'(s)}{\zeta(s)}=\sum_p\sum_{k=1}^\infty \frac{\log\,p}{p^{ks}}=\sum_{n=1}^\infty \frac{\Lambda(n)}{n^s}\quad\text{for}\ \ \Re(s)>1$$
After that we want to use the definition of the second Chebyshev function :
$$\tag{2} \psi(x)=\sum_{n\leq x}\Lambda(n)$$
But, using Abel's sum formula with $a(n):=\Lambda(n)$ and $\phi(n):=n^{-s}$, we get :
$$\sum_{n=1}^\infty \frac{\Lambda(n)}{n^s}=s\int_1^\infty \frac {\sum_{n\leq x}\Lambda(n)}{x^{s+1}}\;dx$$
that is, using $(1)$ and $(2)$, the wished formula :
$$-\frac{\zeta'(s)}{\zeta(s)}=s\int_1^{\infty}\frac{\psi(x)}{x^{s+1}}dx$$
Inverting this Mellin transform to express $\psi(x)$ would produce Perron's formula as shown in this derivation of the 'explicit formula' with better handling of the discontinuities.   
