# Zeta function over primes

by zetafunction
Tags: function, primes, zeta
 P: 399 where could i get some info about the function $$\sum_{p} p^{-s}=P(s)$$ * the functional equation relating P(s) and P(1-s) * the relation with Riemann zeta
 P: 2,157 You can express it as a summation over Riemann zeta's multiplied by a Möbius function. We have: $$\zeta(s) = \sum_{r_{1},r_{2}\ldots}\prod_{j}p_{j}^{-sr_{j}}$$ where $p_{j}$ is the jth prime and the $r_{j}$ in the summation range from zero to infinity. Summing over the $r_{j}$ gives: $$\zeta(s)= \prod_{p}\frac{1}{1-p^{-s}}$$ Take the log of both sides: $$\log\left[\zeta(s)\right]= -\sum_{p}\log\left(1-p^{-s}\right)$$ Expand the logarithm and sum over the primes p: $$\log\left[\zeta(s)\right]=\sum_{k=1}^{\infty}\frac{P(ks)}{k}$$ You can then invert this relation to find the $P(s)$ using Möbius inversion.
 P: 2,157 So, you find: $$P(s) = \log\left[\zeta(s)\right] - \sum_{p}\frac{\log\left[\zeta(ps)\right]}{p} + \sum_{p_{1}  P: 54 Zeta function over primes I think [tex]P(s)$$ as defined above by Count Iblis, can be written as $$P(s)= \log \zeta(s)+\sum_{m=0}^{\infty}\sum_{n=1}^{\infty}(-1)^{m+1}\frac{\log\zeta\Big(s\prod_{k=0}^{m}p_{n+k}\Big)}{\prod_{k=0}^{ m} p_{n+k}}.$$ I assume that's for $$\textnormal{Re}(s)>1$$.