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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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If I remember right you simply write $\sum_p$ as $\sum_n (\pi(n)-\pi(n-1))$. Note that the Bracket is $1$ when n prime and 0 otherwise. Now, the difference becomes a difference of two series, the second one being indexed one step lower, reindex it , combine and done. – N. S. Feb 14 '12 at 13:08
Can you define the weighted prime counting function? Is it: $f(x)=\sum_{p<x} \lfloor \log_p x\rfloor$? – Thomas Andrews Feb 14 '12 at 13:37
thanks for the reply, but i am afraid i am not following the logic . the sum over $\pi(n)-\pi(n-1)$ is quite clear . but the integral part is what puzzles me. could you show the work step by step please . – Mohammad Al Jamal Feb 14 '12 at 13:42
$ f(x) = \sum_{n=1}^{\infty} \frac{\pi(x^{1/n})}{n} $ it's also called the riemann prime counting function – Mohammad Al Jamal Feb 14 '12 at 13:44
It becomes a little more obvious if you realize your formula after the formula form $p^{-sn}$ is wrong - it is missing a factor of $\frac{1}{n}$. It should be: $$\frac{\ln\zeta (s)}{s} = \sum_{p}\sum_{n=1}^{\infty}\frac{1}{n}\int_{p^{n}}^{\infty}x^{-s-1}dx$$ – Thomas Andrews Feb 14 '12 at 14:00
up vote 6 down vote accepted

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:

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

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thanks ... it's clear now – Mohammad Al Jamal Feb 14 '12 at 14:17

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