- #1
mmzaj
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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 ??
[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 ??
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