Analytic continuation of the prime zeta series

Question

The prime zeta series is denoted by $ \sum_p \frac{1}{p^s} $, where $ p $ is a prime number. It is absolutely convergent in the half plane right of the abscissa at $ \sigma_a = 1 $. I have seen several resources asserting it is possible to extend it analytically to $ \sigma_c = 0 $ but not beyond.

However, I haven’t been able to find how exactly the convergence plane can be extended. Does anyone know how to extend the half plane of convergence for prime zeta beyond $ \sigma_a = 1 $?

References: https://en.m.wikipedia.org/wiki/Prime_zeta_function http://mathworld.wolfram.com/PrimeZetaFunction.html

References Behind pay wall: https://link.springer.com/article/10.1007%2FBF01933420 https://page-one.live.cf.public.springer.com/pdf/preview/10.1007/BF03014596

Answer 1

Let me add this to Klangen's answer :

Assuming the RH then $$P(s) = s \int_2^\infty \pi(x) x^{-s-1}dx = s \int_2^\infty (\pi(x)-Li(x)) x^{-s-1}dx+ L(s) \\ = s \int_2^\infty (\pi(x)-Li(x)) x^{-s-1}dx+L(2) + \log(s-1)+ \int_2^s \frac{2^{1-u}-1}{u-1}du$$ where the latter integrals converge and are analytic for $\Re(s) > 1/2$ (the proof is similar to the PNT).

$$ Li(x) = \int_2^x \frac{dt}{\log t}$$ $$L(s) = s\int_2^\infty Li(x) x^{-s-1}dx = \int_2^\infty \frac{x^{-s}}{\log x}dx $$ $$ = L(2)+\int_2^s L'(u)du = L(2) + \log(s-1)+ \int_2^s \frac{2^{1-u}-1}{u-1}du$$

since $$L'(s) = -\int_2^\infty x^{-s}dx = \frac{2^{1-s}}{s-1}$$

Answer 2

Let $P(s)$ denote the prime zeta function and $\zeta(s)$ the Riemann Zeta function. Using the Euler product of Riemann zeta function, i.e., $\zeta(s)=(1-p^{-s})^{-1}$, we have that

$$ {\displaystyle \log \zeta (s)=\sum _{n>0}{\frac {P(ns)}{n}}}. $$

By Möbius inversion, we get

$$ {\displaystyle P(s)=\sum _{n>0}\mu (n){\frac {\log \zeta (ns)}{n}}}. $$

As shown in the comment section, this continues $P(s)$ analytically to $\Re(s)>0$.