# Zeta function the the orime counting function

1. Feb 13, 2012

### mmzaj

i have a question about the relation between the riemann zeta function and the prime counting function . one starts with the formal definition of zeta :
$$\zeta (s)=\prod_{p}\frac{1}{1-p^{-s}}$$
then :
$$ln(\zeta (s))= -\sum_{p}ln(1-p^{-s})=\sum_{p}\sum_{n=1}^{\infty}\frac{p^{-sn}}{n}$$

using the trick :
$$p^{-sn}=s\int_{p^{n}}^{\infty}x^{-s-1}dx$$
then :
$$\frac{ln\zeta (s)}{s} = \sum_{p}\sum_{n=1}^{\infty}\int_{p^{n}}^{\infty}x^{-s-1}dx$$
up until now, things make perfect sense , but the following line is mysterious to me :

$$\frac{\ln\zeta(s)}{s}=\int_{0}^{\infty}f(x)x^{-s-1}dx$$

where $f(x)$ is the weighted-prime counting function .
how is this formula derived !?!?

Last edited: Feb 13, 2012
2. Feb 14, 2012

### mmzaj

i have done a mistake
$$\frac{ln\zeta (s)}{s} = \sum_{p}\sum_{n=1}^{\infty}\frac{1}{n}\int_{p^{n}}^{\infty}x^{-s-1}dx$$

after some digging , the integral follows from the fact , that for any function $g(x)$ :

$$\sum_{p}\int_{p^{n}}^{\infty}g(x)dx= \int_{0}^{\infty} \pi(x^{1/n}) g(x)dx$$

$\pi(x)$ being the prime counting function .
but this line isn't so clear to me !!!