# Zeta function over primes

1. May 15, 2009

### zetafunction

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

2. May 15, 2009

### Count Iblis

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.

3. May 15, 2009

### Count Iblis

So, you find:

$$P(s) = \log\left[\zeta(s)\right] - \sum_{p}\frac{\log\left[\zeta(ps)\right]}{p} + \sum_{p_{1}<p_{2}}\frac{\log\left[\zeta(p_{1}p_{2}s)\right]}{p_{1}p_{2}}- \sum_{p_{1}<p_{2}<p_{3}}\frac{\log\left[\zeta(p_{1}p_{2}p_{3}s)\right]}{p_{1}p_{2}p_{3}}+\cdots$$

4. May 16, 2009

### squidsoft

I think $$P(s)$$ 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 $$\textnormal{Re}(s)>1$$.