Is $\sum_{n=0}^{\infty} \frac{n^2}{(b-n^2)(a-n^2)}$ expressible in terms of trigonometric functions I recently ran into the sum
$$S=\sum_{n=0}^{\infty} \frac{n^2}{(\alpha-n^2)(\beta-n^2)}.$$
Mathematica gives it in terms of the Digamma function as
$$S=\frac{-\alpha  \psi ^{(0)}(1-\alpha )+\alpha  \psi ^{(0)}(\alpha +1)+\beta  (\psi ^{(0)}(1-\beta )-\psi ^{(0)}(\beta +1))}{2 \left(\alpha ^2-\beta ^2\right)}.$$
However I am working on a physics paper where $S$ mysteriously gets written in terms of trigonometric functions. I don't see how this is possible... Does the Digamma function expression above simplify in such a way? Or is there any other way to compute $S$ in terms of trigonometric functions?

$$S=\sum_{n=0}^{\infty} \frac{n^2}{(b-n^2)(a-n^2)}=\frac 1 {a-b}\left(\sum_{n=0}^{\infty} \frac{a}{n^2-a}-\sum_{n=0}^{\infty} \frac{b}{n^2-b}\right)$$ leading to $$S=\frac{\pi  \sqrt{b} \cot \left(\pi  \sqrt{b}\right)-\pi  \sqrt{a} \cot \left(\pi 
   \sqrt{a}\right)}{2 (a-b)}$$
