I am trying to find a closed form for $S=\sum_{n=-\infty}^{n=+\infty}\frac{1}{n^{2k}+a^{2k}}$, $k \in \mathbb{N^{*}}$, $a>0$
I don't even bother to look for a closed form with an odd exponent, since as far as I know, there isn't even a closed form for the sum of the cubic reciprocals (other than Apery's constant... if it counts as a closed form)
From the residue theorem, I find $\sum_{n=-\infty}^{n=+\infty}\frac{1}{n^{2k}+a^{2k}}=-\sum Res(\frac{\pi cot(\pi z)}{z^{2k}+a^{2k}})$ for non-real poles.
$z^{2k}+a^{2k}=0 \iff z_m=ae^{\frac{2im\pi}{2k}}=ae^{\frac{im\pi}{k}}, m=1,...,k-1,k+1,...2k-1.$
To make things easier to handle, I'll use some notations :
$z_m=ae^{\frac{im\pi}{k}}=ae_m$
$c_m=cot(\pi z_m)$
and $e_m-e_n=\Delta_{mn}$
So we have $$S=-\pi \sum_{m=1, m\neq k}^{m=2k-1} \frac{c_m}{\prod_{l=1,l\neq m}^{l=k-1}(z_m-z_l)}=-\frac{\pi}{a^{k-2}} \sum_{m=1, m \neq k}^{m=2k-1} \frac{c_m}{\prod_{l=1,l\neq m}^{l=k-1}\Delta{ml}}$$
We can notice that $e_m-e_n=\Delta_{mn}=e_m(1-e_{n-m})=e_me_{\frac{n-m}{2}}(e_{-\frac{n-m}{2}}-e_{\frac{n-m}{2}})=-2ie_{\frac{n+m}{2}}sin(\frac{n-m}{2} \frac{\pi}{k})$
So
So we have $$S=-\frac{\pi}{a^{2k-2}} \sum_{m=1, m \neq k}^{m=2k-1} \frac{c_m}{\prod_{l=1,l\neq m}^{l=k-1}\Delta{ml}}=\frac{\pi}{2ia^{2k-2}} \sum_{m=1, m \neq k}^{m=2k-1} \frac{c_m}{\prod_{l=1,l\neq m}^{l=k-1}e_{\frac{l+m}{2}}sin(\frac{l-m}{2} \frac{\pi}{k})}$$
$$=\frac{\pi}{2ia^{2k-2}} e_{-\frac{1+2+...+k-1}{2}}\sum_{m=1, m \neq k}^{m=2k-1} \frac{c_m}{\prod_{l=1,l\neq m}^{l=k-1}sin(\frac{l-m}{2} \frac{\pi}{k})}$$
$$=\frac{\pi}{2ia^{2k-2}} e_{\frac{1}{4}k(1-k)}\sum_{m=1, m \neq k}^{m=2k-1} \frac{c_m}{\prod_{l=1,l\neq m}^{l=k-1}sin(\frac{l-m}{2} \frac{\pi}{k})}$$
I'm not sure where to go after that. I have tried to split the last sum in half using $cot(-x)=-cot(x)$ and $sin(x+\pi)=-sin(x)$ but it was pretty messy, and got me nowhere. I also couldn't find an easy way to simplify the cots together.
Any ideas or hints please ? Thanks