# 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?

• Since you apparently use Mathematica, did you try FullSimplify[%] ? – Claude Leibovici Jun 16 '17 at 9:47
• @ClaudeLeibovici embarrassingly that worked too... I had only tried Simplify[%] before (for runtime reasons I'm used to using that) and that didn't do the job. – Wolpertinger Jun 16 '17 at 9:53
• Don't forget FullSimplify[%]. It helps a lot (as far as I can remember). – Claude Leibovici Jun 16 '17 at 9:56

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

• when there is a perfect answer 2 minutes after I got the question into its final form... +1 and thanks! – Wolpertinger Jun 16 '17 at 9:44
• @Wolpertinger. You are very welcome ! No merit at all since I had this summation to work sixty years ago ! – Claude Leibovici Jun 16 '17 at 9:45

Use partial fraction to decompose the sum $S$ to two sums of the form $\sum_{n=0}^{\infty}\frac{1}{n^2-a}$, then use the fact that $$\sum_{n=0}^{\infty}\frac{1}{n^2-a} = \frac{-1-\sqrt{a}\pi \cot(\sqrt{a}\pi)}{2a}$$

• +1 and thanks for pointing out how to do this simpler sum. This also nicely complements Claude's answer! – Wolpertinger Jun 16 '17 at 9:45

We have $\psi(a+1)=\frac{1}{a}+\psi(a)$ and $\psi(a)-\psi(1-a) = -\pi\cot(\pi a)$ since

$$\Gamma(x+1) = x\,\Gamma(x),\qquad \Gamma(x)\,\Gamma(1-x)=\frac{\pi}{\sin(\pi x)}$$ and $\psi(x)=\frac{d}{dx}\log\Gamma(x)=\frac{\Gamma'(x)}{\Gamma(x)}$.
This immediately explains the trigonometric form shown by Claude Leibovici.

• +1 and thanks for the connection to the Digamma function. this nicely complements Claude's answer! – Wolpertinger Jun 16 '17 at 9:44