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.

• How would you evaluate it without any restrictions? Feb 8, 2013 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 $\frac{1}{8}$. Feb 8, 2013 at 13:10
• This sum converges extremely quickly. The first two terms give $0.12499999998527185286$. Feb 8, 2013 at 23:05
• This is equivalent to showing that $\theta_3^{(0,1)}(0,e^{-\pi})=\dfrac{\theta_3(0,e^{-\pi})}{4~\pi~e^{-\pi}}$ Jan 23, 2015 at 20:32

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?) May 19, 2013 at 12:02
• Yes $2(s-1)>1$ gives $s>3/2$ (half plane of convergence of the Dirichlet series). May 19, 2013 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. May 19, 2013 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. May 19, 2013 at 22:34
• Am I alone or not? I don't see a full conection between $\pi^{-s}\Gamma(s)\zeta(2s)(s-\frac14)$ with your arument about vanishment of integral of $\pi^{-s}\Gamma(s) \left(\pi \zeta(2(s-1)) - \frac{1}{4} \zeta(2s) \right)$ at $s=\frac14 + it$. Can you give me a hint why integral of the latter vanishes? Aug 4, 2016 at 19:05