prove that, if $a \notin \mathbb{Z}$, $a>0$ so

Question

$\sum_{n=-\infty}^{\infty} \frac{(-1)^n}{n^4-a^4}=\frac{1}{2a^4}-\frac{\pi}{4a^3}(\frac{1}{\sin{\pi a}}+\frac{1}{\sinh{\pi a}})$

Answer 1

Use the standard sums" $$\sum \frac{(-1)^n}{n^2+x^2}=\frac{\pi x~ csch \pi x-1}{2x^2},~ \sum \frac{(-1)^n}{n^2-x^2}=\frac{1-\pi x~ csc \pi x}{2x^2}~~~~(1)$$

These are also known as exansion of $csch x$ and $csc x$ respectively.

Next $$S=\sum_{n=-\infty}^{\infty} \frac{(-1)^n}{n^4-a^4}= -\frac{1}{a^4}+2 \sum_{n=1}^{\infty}\frac{(-1)^n}{n^4-a^4}=-\frac{1}{a^4}+ \frac{2}{2a^2}\sum_{n=1}^{\infty} \left[\frac{(-1(^n}{n^2-a^2}-\frac{(-1)^n}{n^2+a^2}\right]~~~~(2)$$ So using (1) in (2), we get $$S=-\frac{\pi}{4a^3}\left[\frac{1}{\sin \pi a}+ \frac{1}{\sinh \pi a}\right].$$ I do not get the additive term $\frac{1}{2a^4}$ and my result also matches with Mathematica.