Zeta function the the orime counting function

  • Thread starter mmzaj
  • Start date
  • #1
107
0
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:

Answers and Replies

  • #2
107
0
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 !!!
 

Related Threads on Zeta function the the orime counting function

  • Last Post
Replies
2
Views
2K
Replies
2
Views
2K
  • Last Post
Replies
8
Views
3K
Replies
1
Views
593
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
2
Views
4K
  • Last Post
Replies
7
Views
5K
Top