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.

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

1 Answer 1


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.

For Eisenstein summation, see also

Remmert, Classical topics in complex function theory, Springer 1998. He uses it to define trigonometric functions.

  • $\begingroup$ Thanks. I see.. I'm a little unsure how the definition above of "Eisenstein summation" is different to normal summation? I think I'm missing something. $\endgroup$
    – zooby
    May 20 at 19:43
  • 3
    $\begingroup$ Eisenstein summation extends symmetrically (m=-M..M). For ordinary summation, it must not matter how you sum. $\endgroup$
    – Marcus
    May 20 at 19:45
  • 1
    $\begingroup$ The problem here is not just with the definition of the sum but with proving that it is doubly periodic. $\endgroup$ May 20 at 19:53
  • $\begingroup$ The inner sum (in Eisenstein summation) is by the way equal to $\pi^2/\sin(π(x+n))^2$ or equivalently $(2πi)^2 \frac{\exp(2\pi i(x+n))}{(\exp(2\pi i(x+n))−1)^2}$ $\endgroup$
    – Marcus
    May 21 at 11:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.