Integral representation of prime zeta function

Question

The prime zeta function is defined for $\Re(s)>1$ as $$P(s)=\sum_{p\in\text{primes}}\frac{1}{p^s}$$ Is there any known integral representation of $P(2)$?

$$P(2)=\sum_{p\in\text{primes}}\frac{1}{p^2}$$ By Fubini-Tonelli theorem, $$P(2)=\sum_{p\in\text{primes}}\int_0^1\int_0^1 x^{p-1} y^{p-1} \ dx dy $$ and $$P(2)=\int_0^1\int_0^1\sum_{p\in\text{primes}} x^{p-1} y^{p-1} \ dx dy $$ Thank you!

Answer 1

This answer does not aim to provide an integral representation, but it points out a connection with the Riemann $\zeta$-function and indicates that an analytic continuation of $P(s)$ via any method (including integral representation) to the left half plane is impossible.

From the definition, we have for $\Re(s)>1$ that

$$ \log\zeta(s)=\sum_{k\ge1}\log(1-p^{-s})^{-1}=\sum_{k\ge1}{1\over k}\sum_p{1\over p^{ks}}=\sum_{k\ge1}{P(ks)\over k}. $$

By Möbius inversion, we have

$$ P(s)=\sum_{n\ge1}{\mu(n)\over n}\log\zeta(ns). $$

Due to the presence nontrivial zeros of $\zeta$ in the critical strip, the singularities of $P(s)$ are dense in the imaginary axis (for a proof, see Chapter IX of Titchmarsh's The theory of the Riemann zeta-function), so the domain of $P(s)$ cannot be extended beyond the right half plane.

Answer 2

You could always do summation by parts: For $\mathfrak{Re}(s) > 1$,

$$P(s) = s \int_1^{+\infty} \frac{\pi(t)}{t^{s+1}} \, \mathrm{d}t$$ where $\pi(t) = \sum_{p \leq t} 1$ is the number of prime numbers less than $t$.