I am a PhD student in electrical engineering. I need to find a closed form formula for the following series: $$\sum_{k=1}^{\infty}\frac{1}{2}A_k^2e^{-k^2\sigma_m^2}(e^{k^2\sigma_m^2}-1)$$where $A_k= \frac{4\sin(\frac{\pi}{2}k)}{\pi k}$ and $\sigma_m^2$ is a constant. This is a very important result if I can find it.
Thanks a lot.
First of all, the $\sin$ term is a red herring, since it is equal to $1$ for odd $k,$ and $0$ for even $k.$ Second, expanding the second term, you will get two sums. The first is
$$\sum_{k=1}^\infty \frac{8}{\pi k}^2 \sin^2(k\pi/2),$$ which is a multiple of the sum of inverses of odd squares, and is easy to evaluate.
The second sum is $$(8/\pi^2)\sum_{\mbox{odd $k$}} e^{-k^2/\sigma^2}/k^2.$$ Make the exponent $(-x^2 k^2/\sigma^2),$ and differentiate with respect to $x.$ You will get a sum of the form
$$a x\sum e^{-x^2 b k^2},$$ so a multiple of a linear function and a theta function, so you sum can be expressed via the integrals of such. I leave the rest up to you (since I have no idea what it is you are trying to get).