For real $s>0$, let $$S(s):=\sum_{n=-\infty}^\infty e^{-n^2/(2s^2)} =\vartheta _3\left(0,e^{-1/(2 s^2)}\right),$$ where $\vartheta$ is the elliptic theta function.

Plotting suggests that the identity \begin{equation} S(s)=s\sqrt{2\pi} \end{equation} is true at least for $s\ge3/2$. Is it indeed?

This conjecture, with a plot, appeared as a part of this answer, but seems to warrant separate posting. Mathematica cannot prove or disprove this identity.

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    $\begingroup$ if it holds for $s \ge 3/2$, then it must hold for $s > 0$ (holomorphic extension). $\endgroup$ Jul 15, 2021 at 5:43
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    $\begingroup$ No, the equality you wrote doesn’t hold exactly, although it probably looks like it numerically because the error is exponentially small as $s\to \infty$, e.g. by Euler-Maclaurin or complex analysis (see math.stackexchange.com/questions/719401/…). $\endgroup$
    – Zachary
    Jul 15, 2021 at 5:47
  • $\begingroup$ In Mathematica, With[{s=2},N[EllipticTheta[3,0,E^(-1/(2s^2))]/(s Sqrt[2Pi]),50]] gives the answer 1.0000000000000000000000000000000001024500455847086 $\endgroup$ Jul 15, 2021 at 6:22
  • $\begingroup$ To add to my comment, you can probably see the non-equality and understand the smallness of the error even more explicitly by using Poisson summation. $\endgroup$
    – Zachary
    Jul 15, 2021 at 6:39
  • $\begingroup$ Thank you everyone for your comments. $\endgroup$ Jul 15, 2021 at 11:15

1 Answer 1


To give an answer, adding to my comments, your formula doesn’t hold true, although the error is exponentially small as $s\to \infty$, as can be seen by Poisson summing, which transforms your sum to $$\sqrt{2\pi}s\left(1+\mathcal{O}(e^{-2\pi^2 s^2})\right).$$

  • $\begingroup$ I don't get it. How did you answer the question? You just said the identity is false, but you didn't explain why... $\endgroup$ Jul 15, 2021 at 14:49
  • $\begingroup$ By Poisson summation, you can show that $S(s) -\sqrt{2\pi}s=\sqrt{2\pi}s\sum_{n\neq 0} e^{-2\pi^2 n^2 s^2} \neq 0$, which also explains why equality seems to hold numerically, because the right hand side is exponentially small (up to an overall factor of $s$ in front). $\endgroup$
    – Zachary
    Jul 15, 2021 at 15:00

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