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