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I have the following series:

$$\sum_{n=-\infty}^{\infty}\frac{1}{(2n+1)^2\pi^2+a^2}=\frac{1}{2a}\tanh\left(\frac{a}{2}\right)$$

and on the text it is written that it can be proven by means of either Poisson sum formula or digamma function. However I didn't manage to do it. Can anybody help?

Thank you in advance!

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  • $\begingroup$ $\frac{1-e^{-b}}{2i\pi n+b}$ is the Fourier series coefficient of $e^{-bx} \in L^2([0,1])$ and $\frac{1}{2i\pi n+b}-\frac{1}{2i\pi n-b}$ is the Fourier series coefficient of $\frac{e^{-bx}}{1-e^{-b}}-\frac{e^{bx}}{1-e^{b}} \in L^2([0,1])$ whose periodization is absolutely continuous $\endgroup$ – reuns Apr 19 at 5:11
  • $\begingroup$ Can you be a little more explicative please? $\endgroup$ – Dpolito Apr 19 at 8:08
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To make it short, start with $$(2n+1)^2\pi^2+a^2=( (2 n+1)\pi -i a)\, ( (2 n+1)\pi +i a)$$ Using partial fraction decomposition $$\frac 1 {(2n+1)^2\pi^2+a^2}=\frac i {2a}\left(\frac 1{(2 n+1)\pi +i a } -\frac 1{(2 n+1)\pi -i a } \right)$$ make $$S_m=\sum_{n=-m}^{m}\frac{1}{(2n+1)^2\pi^2+a^2}$$ $$S_m=\frac{2 \pi a+\left(a^2+(2m+1)^2 \pi^2 \right) \left(-i \psi ^{(0)}\left(-\frac{i a}{2 \pi }+m+\frac{1}{2}\right)+i \psi ^{(0)}\left(\frac{i a}{2 \pi }+m+\frac{1}{2}\right)+\pi \tanh \left(\frac{a}{2}\right)\right)}{2 \pi a \left(a^2+(2 \pi m+\pi )^2\right)}$$

Using asymptotics $$S_m=\frac{\tanh \left(\frac{a}{2}\right)}{2 a}-\frac{1}{2 \pi ^2 m}+\frac{1}{4 \pi ^2 m^2}+O\left(\frac{1}{m^3}\right)$$

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