TITLE: A Function Meromorphic with Poles at the Primes
QUESTION [9 upvotes]: Is there a "naturally occurring" function $f$ which is meromorphic in the complex plane such that the poles of $f$ on the real axis are precisely at the primes? I say "naturally occurring" since we can always cook up a function with the right poles of the right order, but I would hope that it comes from number theory.
Alternatively, is there a naturally arising meromorphic function whose poles appear at all primes and powers of primes?

REPLY [5 votes]: Let $P(s) = \sum_{p} p^{-s}$ be the Prime zeta function.
Claim:  the series expansion $$f(z) = 2 \sum_{n=1}^\infty P(2n)z^{2n-1}$$ defines a meromorphic function whose only poles are simple poles of residue $1$ at the primes and their negatives.
Let $F(z) = \prod_ p (1-z^2/p^2)^{-1}$. The product converges uniformly on compact subsets of $\mathbf C - P$, where $P$ is the set of all primes and their negatives. Therefore it is holomorphic there. It has simple poles at the points of $P$. (Note the special value $F(1) = \zeta(2) = \pi^2/6$).
Using the series expansion for $\log(1-x)$, we have the series expansion for $\log F$, $$\log F(z) = \sum_{n=1}^\infty \frac{P(2n)}{n}z^{2n}.$$
Taking the derivative we get $f(z)$. Since the poles of $F$ are simple and $F$ is never $0$ on $\mathbf C - P$ (on account of the convergence of the product), $f = F'/F$ has simple poles with residue $1$ at the points of $P$, and no other poles.