zeta function over primes

**zetafunction** · May15-09, 10:37 AM

where could i get some info about the function

$LaTeX Code: \\sum_{p} p^{-s}=P(s)$

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

* the relation with Riemann zeta

**Count Iblis** · May15-09, 11:30 AM

You can express it as a summation over Riemann zeta's multiplied by a Möbius function. We have:

$LaTeX Code: \\zeta(s) = \\sum_{r_{1},r_{2}\\ldots}\\prod_{j}p_{j}^{-sr_{j}}$

where $LaTeX Code: p_{j}$ is the jth prime and the $LaTeX Code: r_{j}$ in the summation range from zero to infinity. Summing over the $LaTeX Code: r_{j}$ gives:

$LaTeX Code: \\zeta(s)= \\prod_{p}\\frac{1}{1-p^{-s}}$

Take the log of both sides:

$LaTeX Code: \\log\\left[\\zeta(s)\\right]= -\\sum_{p}\\log\\left(1-p^{-s}\\right)$

Expand the logarithm and sum over the primes p:

$LaTeX Code: \\log\\left[\\zeta(s)\\right]=\\sum_{k=1}^{\\infty}\\frac{P(ks)}{k}$

You can then invert this relation to find the

using Möbius inversion.

**Count Iblis** · May15-09, 12:30 PM

So, you find:

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

**squidsoft** · May16-09, 04:37 PM

I think

as defined above by Count Iblis, can be written as

$LaTeX Code: P(s)= \\log \\zeta(s)+\\sum_{m=0}^{\\infty}\\sum_{n=1}^{\\infty}(-1)^{m+1}\\frac{\\log\\zeta\\Big(s\\prod_{k=0}^{m}p_{n+k }\\Big)}{\\prod_{k=0}^{m} p_{n+k}}.<BR>$

I assume that's for $LaTeX Code: \\textnormal{Re}(s)>1$ .