# Compute $\sum_{k=1}^{\infty}e^{-\pi k^2}\left(\pi k^2-\frac{1}{4}\right)$

How may I evaluate the below series? $$\sum_{k=1}^{\infty}e^{-\pi k^2}\left(\pi k^2-\frac{1}{4}\right)$$ I'm supposed to come up with a solution by only using high school knowledge.

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How would you evaluate it without any restrictions? –  Ishan Banerjee Feb 8 '13 at 10:45
Without restriction, one can evaluate this by studying Fourier transform of $\sum_{n=-\infty}^{\infty} exp(-\frac{(x-2n\pi)^2}{4\pi\beta})$, establish the functional equation $\sum_{n=-\infty}^{\infty} \exp( -\pi k^2/\beta ) = \sqrt{\beta} \sum_{n=-\infty}^{\infty} \exp( -\pi \beta k^2 )$ and then look at the derivative of the functional equation at $\beta = 1$. The final answer is $\frac{1}{8}$. –  achille hui Feb 8 '13 at 13:10
This sum converges extremely quickly. The first two terms give $0.12499999998527185286$. –  George V. Williams Feb 8 '13 at 23:05

I don't know about high school math, but there is an answer using Mellin transforms. First compute the Mellin transform of the sum, then invert to get a closed form expression. Introduce $$f(x) = \sum_{k\ge 1} e^{- k^2 x} \left(\pi k^2 - \frac{1}{4} \right),$$ so that we are looking for $f(\pi).$
We have straightforwardly (using the definition of the Mellin transform) that the Mellin transform $f^*(s)$ of $f(x)$ is given by $$f^*(s) = \mathfrak{M}\left(f(x); s\right) = \Gamma(s) \sum_{k\ge 1} \left(\frac{\pi}{k^{2(s-1)}} - \frac{1}{4} \frac{1}{k^{2s}} \right) = \Gamma(s) \left(\pi \zeta(2(s-1)) - \frac{1}{4} \zeta(2s) \right).$$ Now the Mellin inversion integral (which we'll evaluate at $x=\pi$) is $$\frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} f^*(s) x^{-s} ds = \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} \Gamma(s) \left(\pi \zeta(2(s-1)) - \frac{1}{4} \zeta(2s) \right) x^{-s} ds.$$ Now the only singularity of the first zeta term is at $s=3/2$, with residue $$\operatorname{Res}\left(\Gamma(s) \pi \zeta(2(s-1)) x^{-s}; s=3/2\right) = 1/2\,{\frac {\Gamma \left( 3/2 \right) \pi }{{x}^{3/2}}}.$$ The only singularity of the second zeta term is at $s=1/2$, with residue $$\operatorname{Res}\left(\Gamma(s) \frac{1}{4} \zeta(2s) x^{-s}; s=1/2\right) = 1/8\,{\frac {\Gamma \left( 1/2 \right) }{\sqrt {x}}}.$$ It follows that $$f(x) = 1/2\,{\frac {\Gamma \left( 3/2 \right) \pi }{{x}^{3/2}}} - 1/8\,{\frac {\Gamma \left( 1/2 \right) }{\sqrt {x}}}.$$ Finally set $x=\pi$ to get $$\frac{1}{\sqrt{\pi}} \left(1/2\Gamma(3/2)-1/8\Gamma(1/2)\right) = \frac{1}{\sqrt{\pi}} \left(1/4\Gamma(1/2)-1/8\Gamma(1/2)\right) = \frac{1}{8} \frac{1}{\sqrt{\pi}} \Gamma(1/2) = \frac{1}{8}.$$ The reason why there is only one pole in every case is because the trivial zeros of the zeta function cancel the poles of the gamma function.
How did you apply the residue theorem to the last contour integral? (I suppose $c>\frac{3}{2}$, am I right?) –  Mizar May 19 '13 at 12:02
Yes $2(s-1)>1$ gives $s>3/2$ (half plane of convergence of the Dirichlet series). –  Marko Riedel May 19 '13 at 18:10
Ok, thank you, but how did you apply the residue theorem to the infinite contour? I'm wondering about that since clearly at $\infty$ we have an essential singularity and also a pole at the origin which you don't mention, so I don't understand how you have drawn your (right) conclusion. –  Mizar May 19 '13 at 22:18
This is discussed in the comments here. You only shift to $\Re(s) = 1/4.$ You might want to prove that the Mellin inversion integrand of the transform of the sum is odd on that line. –  Marko Riedel May 19 '13 at 22:34