
#1
May1509, 09:37 AM

P: 399

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(1s) * the relation with Riemann zeta 



#2
May1509, 10:30 AM

P: 2,159

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}{1p^{s}}[/tex] Take the log of both sides: [tex]\log\left[\zeta(s)\right]= \sum_{p}\log\left(1p^{s}\right)[/tex] Expand the logarithm and sum over the primes p: [tex]\log\left[\zeta(s)\right]=\sum_{k=1}^{\infty}\frac{P(ks)}{k}[/tex] You can then invert this relation to find the [itex]P(s)[/itex] using Möbius inversion. 



#3
May1509, 11:30 AM

P: 2,159

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] 



#4
May1609, 03:37 PM

P: 54

zeta function over primes
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}}. [/tex] I assume that's for [tex]\textnormal{Re}(s)>1[/tex]. 


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