# A Function Meromorphic with Poles at the Primes

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?

• What you mean is «meromorphic function with poles at the primes», not «meromorphic at the primes». – Mariano Suárez-Álvarez Feb 14 '13 at 17:14
• no there isnt... – user58512 Feb 14 '13 at 17:38
• Fixed the title – mck Feb 14 '13 at 17:53
• Can you construct a function with poles at the integers? – Thom Tyrrell Feb 14 '13 at 18:25
• Yes, the Weierstrass $\mathfrak{p}$ function for the integer lattice will have poles at exactly the integers. The poles will be order 2, but at least for now I won't put any restrictions on that. – mck Feb 14 '13 at 18:29

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.
• @mrf Up to a unit of the ring of holomorphic functions on $\mathbf C$, it is unique! :) – Bruno Joyal Sep 27 '13 at 6:27