Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Zeta function over primes

  1. May 15, 2009 #1
    where could i get some info about the function

    [tex] \sum_{p} p^{-s}=P(s) [/tex]

    * the functional equation relating P(s) and P(1-s)

    * the relation with Riemann zeta
  2. jcsd
  3. May 15, 2009 #2
    You can express it as a summation over Riemann zeta's multiplied by a Möbius function. We have:

    [tex]\zeta(s) = \sum_{r_{1},r_{2}\ldots}\prod_{j}p_{j}^{-sr_{j}}[/tex]

    where [itex]p_{j}[/itex] is the jth prime and the [itex]r_{j}[/itex] in the summation range from zero to infinity. Summing over the [itex]r_{j}[/itex] gives:

    [tex]\zeta(s)= \prod_{p}\frac{1}{1-p^{-s}}[/tex]

    Take the log of both sides:

    [tex]\log\left[\zeta(s)\right]= -\sum_{p}\log\left(1-p^{-s}\right)[/tex]

    Expand the logarithm and sum over the primes p:


    You can then invert this relation to find the [itex]P(s)[/itex] using Möbius inversion.
  4. May 15, 2009 #3
    So, you find:

    [tex]P(s) = \log\left[\zeta(s)\right] - \sum_{p}\frac{\log\left[\zeta(ps)\right]}{p} + \sum_{p_{1}<p_{2}}\frac{\log\left[\zeta(p_{1}p_{2}s)\right]}{p_{1}p_{2}}- \sum_{p_{1}<p_{2}<p_{3}}\frac{\log\left[\zeta(p_{1}p_{2}p_{3}s)\right]}{p_{1}p_{2}p_{3}}+\cdots[/tex]
  5. May 16, 2009 #4
    I think [tex]P(s)[/tex] as defined above by Count Iblis, can be written as

    [tex]P(s)= \log \zeta(s)+\sum_{m=0}^{\infty}\sum_{n=1}^{\infty}(-1)^{m+1}\frac{\log\zeta\Big(s\prod_{k=0}^{m}p_{n+k}\Big)}{\prod_{k=0}^{m} p_{n+k}}.

    I assume that's for [tex]\textnormal{Re}(s)>1[/tex].
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook