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.
Thanks in advance for your hints, suggestions!
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asked Feb 8, 2013 at 9:01
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1$\begingroup$ How would you evaluate it without any restrictions? $\endgroup$– Ishan BanerjeeFeb 8, 2013 at 10:45
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4$\begingroup$ 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 $\frac{1}{8}$. $\endgroup$– achille huiFeb 8, 2013 at 13:10
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1$\begingroup$ This sum converges extremely quickly. The first two terms give $0.12499999998527185286$. $\endgroup$– George V. WilliamsFeb 8, 2013 at 23:05
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$\begingroup$ This is equivalent to showing that $\theta_3^{(0,1)}(0,e^{-\pi})=\dfrac{\theta_3(0,e^{-\pi})}{4~\pi~e^{-\pi}}$ $\endgroup$– LucianJan 23, 2015 at 20:32
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1 Answer
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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.
answered Feb 8, 2013 at 22:40
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$\begingroup$ How did you apply the residue theorem to the last contour integral? (I suppose $c>\frac{3}{2}$, am I right?) $\endgroup$– MizarMay 19, 2013 at 12:02
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1$\begingroup$ Yes $2(s-1)>1$ gives $s>3/2$ (half plane of convergence of the Dirichlet series). $\endgroup$ May 19, 2013 at 18:10
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$\begingroup$ 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. $\endgroup$– MizarMay 19, 2013 at 22:18
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1$\begingroup$ 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. $\endgroup$ May 19, 2013 at 22:34
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1$\begingroup$ 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? $\endgroup$ Aug 4, 2016 at 19:05