Analytic continuation of the prime zeta series
Matthew Martinez 
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:
References Behind pay wall:
$\endgroup$ 32 Answers
$\begingroup$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 $\Re(s)>0$.
$\endgroup$ 2 $\begingroup$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 $$L'(s) = -\int_2^\infty x^{-s}dx = \frac{2^{1-s}}{s-1}$$
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