48
$\begingroup$

Wikipedia informs me that

$$S = \vartheta(0;i)=\sum_{n=-\infty}^\infty e^{-\pi n^2} = \frac{\sqrt[4] \pi}{\Gamma\left(\frac 3 4\right)}$$

I tried considering $f(x,n) = e^{-x n^2}$ so that its Mellin transform becomes $\mathcal{M}_x(f)=n^{-2z} \Gamma(z)$ so inverting and summing

$$\frac{1}{2}(S-1)=\sum_{n=1}^\infty f(\pi,n)=\sum_{n=1}^\infty \frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}n^{-2z} \Gamma(z)\pi^{-z}\,dz = \frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\zeta(2z) \Gamma(z) \pi^{-z}\,dz$$

However, this last integral (whose integrand has poles at $z=0,\frac{1}{2}$ with respective residues of $-\frac 1 2$ and $\frac 1 2$) is hard to evaluate due to the behavior of the function as $\Re(z)\to \pm\infty$ which makes a classic infinite contour over the entire left/right plane impossible.

How does one go about evaluating this sum?

$\endgroup$
  • 1
    $\begingroup$ Have you looked at Poisson Summation ( en.wikipedia.org/wiki/Poisson_summation_formula )? $\endgroup$ – Steven Stadnicki Jul 3 '13 at 0:35
  • 2
    $\begingroup$ @StevenStadnicki I have, but it got me nowhere, because I found that if $f(n) = e^{-\pi n^2}$ then $\hat f (n) = f(n)$. Is there another similar approach that is more fruitful? $\endgroup$ – Argon Jul 3 '13 at 0:40
  • 2
    $\begingroup$ Following @CameronWilliams 's hint, I was able to find the proof, which happens to be in Page 103 of the following link: 207.150.202.110/FTHumanEvolutionCourse/FTFreeLearningKits/… $\endgroup$ – Lord Soth Jul 3 '13 at 2:41
  • 2
    $\begingroup$ ... Unfortunately this "proof" refers to several different results that have been derived previously, and it seems that it will not be easy to put a self-sufficient proof together, or such a proof would probably be long. $\endgroup$ – Lord Soth Jul 3 '13 at 2:42
  • 2
    $\begingroup$ @Argon This article (mathworld.wolfram.com/JacobiThetaFunctions.html) explains how to do it by relating $\vartheta_3(q)$ to elliptic integrals, and thus $\vartheta_3(e^{-\pi})$ to $K(\frac1{\sqrt{2}})=\frac{\Gamma^2(\frac14)}{4\sqrt\pi}.$ $\endgroup$ – Kirill Jul 3 '13 at 5:39
20
$\begingroup$

This one is a direct evaluation of elliptic integrals. Jacobi's theta function $\vartheta_{3}(q)$ is defined via the equation $$\vartheta_{3}(q) = \sum_{n = -\infty}^{\infty}q^{n^{2}}\tag{1}$$ Let $0 < k < 1$ and $k' = \sqrt{1 - k^{2}}$ and we define elliptic integrals $K, K'$ via $$K(k) = \int_{0}^{\pi/2}\frac{dx}{\sqrt{1 - k^{2}\sin^{2}x}}, K = K(k), K' = K(k')\tag{2}$$ Then it is almost a miracle that we can get $k$ in terms of $K, K'$ via the variable $q = e^{-\pi K'/K}$ using equations $$k = \frac{\vartheta_{2}^{2}(q)}{\vartheta_{3}^{2}(q)}\tag{3}$$ where $\vartheta_{2}(q)$ is another theta function of Jacobi defined by $$\vartheta_{2}(q) = \sum_{n = -\infty}^{\infty}q^{(n + (1/2))^{2}}\tag{4}$$ Also the function $\vartheta_{3}(q)$ is directly related to $K$ via $$\vartheta_{3}(q) = \sqrt{\frac{2K}{\pi}}\tag{5}$$ The proofs of $(3)$ and $(5)$ are given in the linked post on my blog.


The sum in the question is $\vartheta_{3}(e^{-\pi})$ so that we have $q = e^{-\pi}$. This implies that $K'/K = 1$ so that $k = k'$ and from $k^{2} + k'^{2} = 1$ we get $k^{2} = 1/2$. And then $$\vartheta_{3}(q) = \sqrt{\frac{2K}{\pi}} = \sqrt{\frac{2}{\pi}\cdot\frac{\Gamma^{2}(1/4)}{4\sqrt{\pi}}} = \frac{\Gamma(1/4)}{\pi^{3/4}\sqrt{2}}$$ Now using $\Gamma(1/4)\Gamma(3/4) = \pi/\sin(\pi/4) = \pi\sqrt{2}$ we get $$\sum_{n = -\infty}^{\infty}e^{-\pi n^{2}} = \vartheta_{3}(e^{-\pi}) = \frac{\sqrt[4]{\pi}}{\Gamma(3/4)}$$ The value of $K = K(1/\sqrt{2})$ in terms of $\Gamma(1/4)$ is evaluated in this answer.

$\endgroup$
  • 5
    $\begingroup$ Why the downvote? $\endgroup$ – Paramanand Singh Jun 8 '17 at 14:53
2
$\begingroup$

see: Ramanujan's Notebooks Volume 3, Chapter 17, Example(i). pp 103.

see also: Ramanujan's Notebook Volume 5 chapter 35. Values of Theta-Functions P. 325.

(seems like many of the previous comments mention what I have.)

$\endgroup$
2
$\begingroup$

I am not sure if it will ever help, but the following identity can be proved:

$$ S^2 = 1 + 4 \sum_{n=0}^{\infty} \frac{(-1)^n}{\mathrm{e}^{(2n+1)\pi} - 1}. $$

$\endgroup$
  • $\begingroup$ Can you please explain how can we prove your assertion? $\endgroup$ – Henry Jul 12 '16 at 8:05
  • $\begingroup$ Hello! I have asked my question separately on MSE. Can you please answer it there? Thanks in advance. $\endgroup$ – Henry Sep 11 '16 at 10:02
0
$\begingroup$

Maybe you can use this relationship:

If $$\vartheta(x)=\sum_{n\in \mathbb{Z}}e^{-\pi n^2 x},$$

then:

$$\pi^{-s/2}\Gamma(s/2)\zeta(s)=\int_{0}^{\infty}x^{s/2-1}\frac{\vartheta(x)-1}{2}dx.$$

$\endgroup$

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.