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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?

Argon
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    Have you looked at Poisson Summation ( http://en.wikipedia.org/wiki/Poisson_summation_formula )? – Steven Stadnicki Jul 03 '13 at 00:35
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    @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? – Argon Jul 03 '13 at 00:40
  • What about a contour integral involving $\displaystyle \frac{\pi \text{ cos} (\pi z)}{\text{ sin} (\pi z)} e^{-\pi z^2}$ on some contour which contains the whole real axis? – bryanj Jul 03 '13 at 00:40
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    @bryanj I didn't try it, but my concern here is that $e^{-\pi z^2}$ explodes along the imaginary axis. – Argon Jul 03 '13 at 00:42
  • @Argon whoops, right - I had my constants wrong, mea culpa. I believe there's a particular Poisson-esque identity that can be used to get essentially a 'reflection formula' here but it's been a long time since I've looked at this particularly. – Steven Stadnicki Jul 03 '13 at 00:44
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    @Argon The Wiki article referenced a paper which, in turn, referenced one of Rammanujan's notebooks in which the identity is proved. The fifth notebook, chapter 35, page 325. If I had to guess, I would say that the identity you're interested in is highly nontrivial. – Cameron Williams Jul 03 '13 at 01:00
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    Following @CameronWilliams 's hint, I was able to find the proof, which happens to be in Page 103 of the following link: http://207.150.202.110/FTHumanEvolutionCourse/FTFreeLearningKits/01-MA-Mathematics,%20Economics%20and%20Preparation%20for%20University/008-MA19-OP01-History%20of%20Mathematics,%20Oeuvres%20and%20Reference/Ramanujan/Bruce%20C.%20Berndt%20-%20Ramanujan's%20Notebooks%20-%20Vol%203.pdf – Lord Soth Jul 03 '13 at 02:41
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    ... 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. – Lord Soth Jul 03 '13 at 02:42
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    @Argon This article (http://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 $K(\frac1{\sqrt{2}})=\frac{\Gamma^2(\frac14)}{4\sqrt\pi}.$ – Kirill Jul 03 '13 at 05:39
  • @bryanj: I tried the $\pi\cot(\pi z)e^{-\pi z^2}$ method by expanding $\pi\cot(\pi z)$ into powers of $e^{-2\pi iz}$ only to come back to $\sum e^{-\pi n^2}$ again. – robjohn Sep 25 '13 at 23:37

4 Answers4

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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.

Paramanand Singh
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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.)

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

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

Sangchul Lee
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Maybe you can use this relationship:

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

then:

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

MrYouMath
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