Line integral of prime zeta function

Question

Need help solving the integral $$\int_{c-i\infty}^{c+i\infty} \frac{P(s)\cdot x^s}{s^2}ds$$, where $P(s)$ is the prime zeta function. If the denominator was simply $s$, then this integral would nicely evaluate to $\pi(x)$ (multiplying the last term of the sum when x is an integer), but with the square term I don't know how to handle this. Another thing that can be done is somehow find a Dirichlet series for $\frac{P(s)}{s}$ of the form $\sum_{n=1}^\infty \frac{a_n}{n^s}$, for some sequence $a_n$ (and no, letting $a_n$ equal $\frac{1}{s}$ whenever $n$ is a prime doesn't count, the sequence has to be in terms of n only), but that doesn't seem to be possible so yeah. Regardless, any help with any of the two approaches will be useful and appreciated.

Answer 1

For $\Re(s) > 1$ $$P(s)=\sum_p p^{-s}=\sum_p s\int_p^\infty x^{-s-1}dx=s\int_1^\infty \pi(x)x^{-s-1}dx=s\int_1^\infty x^{-1}\pi(x)x^{-s}dx$$ $$=s\int_1^\infty (\int_1^x y^{-1}\pi(y)dy) s x^{-s-1}dx$$ thus for $c > 1$ and $x> 0$ $$\frac1{2i\pi}\int_{c-i\infty}^{c+i\infty} \frac{P(s)\cdot x^s}{s^2}ds= \int_1^x y^{-1}\pi(y)dy = \sum_{p\le x} \int_p^x y^{-1}dy= \sum_{p\le x} (\log x-\log p)$$