# How to work out this elliptic function?

Let $$f(x) = \sum\limits_{(n,m)\in\mathbb{Z}^2} \frac{1}{(x+ n + i m )^2}$$

If feel it should be $$1/E(x)$$ where $$E$$ is some elliptic function, like $$sn^2$$. But Wolfram Alpha is giving me some strange expression in terms of q-digamma functions.

But I would rather like to find it in terms of theta functions or elliptic functions.

• The series you wrote is not absolutely convergent, therefore it does not define any function without additional comments how to sum it. May 20 at 19:19
• Pretty sure it converges when $x$ is not a Gaussian integer. Just sum it up from smallest $(n,m)$ to largest. May 20 at 19:25

This is a divergent series. But if one applies summation in the sense of Eisenstein, $$\lim_{N\to\infty}\sum_{n=-N}^N\left(\lim_{M\to\infty}\sum_{m=-M}^M\right)$$ then the sum is doubly periodic. Since the poles are at the lattice and residues are equal to $$1$$, it is equal $$\wp(z)+C$$. Looking at the Laurent expansion at $$0$$ we obtain $$C=0$$. So your sum is the Weierstrass function (if it is understood in the sense of Eisenstein).

Remark. The inner sum in parentheses is absolutely convergent.

Ref. A. Weil, Elliptic functions according to Eisenstein and Kronecker, Springer, 1976.

• The inner sum (in Eisenstein summation) is by the way equal to $\pi^2/\sin(π(x+n))^2$ or equivalently $(2πi)^2 \frac{\exp(2\pi i(x+n))}{(\exp(2\pi i(x+n))−1)^2}$ May 21 at 11:44