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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?

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en.wikipedia.org/wiki/… –  Erick Wong May 19 '13 at 18:31

1 Answer 1

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.

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+1 very nice, Raymond... –  draks ... May 19 '13 at 22:41
    
Thanks @draks even if you should know much about this now ! :-) This is only a small (edited) part of this expanded answer after all... –  Raymond Manzoni May 19 '13 at 22:48

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