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Zeta function the the orime counting function

  1. Feb 13, 2012 #1
    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 :
    [tex] \zeta (s)=\prod_{p}\frac{1}{1-p^{-s}} [/tex]
    then :
    [tex] ln(\zeta (s))= -\sum_{p}ln(1-p^{-s})=\sum_{p}\sum_{n=1}^{\infty}\frac{p^{-sn}}{n}[/tex]

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

    [tex] \frac{\ln\zeta(s)}{s}=\int_{0}^{\infty}f(x)x^{-s-1}dx[/tex]

    where [itex] f(x) [/itex] is the weighted-prime counting function .
    how is this formula derived !?!?
    Last edited: Feb 13, 2012
  2. jcsd
  3. Feb 14, 2012 #2
    i have done a mistake
    [tex]\frac{ln\zeta (s)}{s} = \sum_{p}\sum_{n=1}^{\infty}\frac{1}{n}\int_{p^{n}}^{\infty}x^{-s-1}dx[/tex]

    after some digging , the integral follows from the fact , that for any function [itex]g(x)[/itex] :

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

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