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i have posted this question on MO, and they referred me to post here . one starts with the formal definition of zeta :

$\displaystyle \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 : $\displaystyle 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 !?!?

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    thanks for the correction , still i am not able to deduce the integral representation of $\frac{in\zeta(s)}{s} $ in terms of $f(x)$2012-02-14

1 Answers 1

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Note, per my comment above, you left out a $\frac{1}{n}$ in the formula for $\frac{\log\zeta(s)}{s}$. It should have been:

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

Re-arrange the sum as: $\frac{\log{\zeta(s)}}{s}=\sum_{n=1}^\infty \frac{1}{n}\sum_p \int_{p^n}^\infty x^{-s-1}dx$ Now, in general, for any function $g$: $\sum_p \int_{p^n}^\infty g(x) dx = \int_0^\infty \pi(x^{1/n})g(x)dx$

I'll leave that step to to you. So we get:

$\frac{\log{\zeta(s)}}{s}=\sum_{n=1}^\infty \frac{1}{n}\int_0^{\infty} \pi(x^{1/n})x^{-s-1}dx = \int_0^\infty f(x) x^{-s-1}dx$