I have to solve this sum, but I can't seem to finish it.
$$\sum_{n=-\infty}^\infty\left(\frac1{n^2+a^2}\right)\left(\frac1{n^2+b^2}\right)$$
Here's my ideas:
I've been suggested to use the Poisson summation formula as follows: $\sum_{n=-\infty}^\infty f(n)=\sum_{m=-\infty}^\infty\hat{f}(2\pi m),$ where $\hat{f}$ is the Fourier transform of $f$. But I am not entirely sure how to proceed afterwards.
Best Answer
We first show that, for $a \in \mathbb{R}$,
$$ S(a) := \sum_{n=-\infty}^{\infty} \frac{1}{n^2 + a^2} = \frac{\pi \coth(\pi a)}{a}. $$
Since both sides of the equality are even functions of $a$, it suffices to prove this for $a \geq 0$ and we do so.
Starting off, we realize the sum $S(a)$ in the form $\sum_{n=-\infty}^{\infty} f(n)$. Obviously, we can choose $f(x) = \frac{1}{x^2+a^2}$. Then, in order to invoke Poisson summation formula, we need to compute
\begin{align*} \hat{f}(2\pi m) &= \int_{-\infty}^{\infty} \frac{e^{-2\pi i m x}}{x^2+a^2} \, \mathrm{d}x = \int_{-\infty}^{\infty} \frac{\cos(2\pi m x)}{x^2+a^2} \, \mathrm{d}x, \end{align*}
where the last step holds becaue the imaginary term $\frac{i\sin(2\pi m x)}{x^2+a^2}$ is an odd integrable function and hence its integral over $\mathbb{R}$ vanishes by symmetry. Then, applying the substitution $x = at$ and invoking the formula $\int_{-\infty}^{\infty}\frac{\cos(at)}{1+t^2}\,\mathrm{d}t=\pi e^{-|a|}$, we get
\begin{align*} \hat{f}(2\pi m) &= \frac{1}{a} \int_{-\infty}^{\infty} \frac{\cos(2\pi m a t)}{t^2 + 1} \, \mathrm{d}t = \frac{\pi}{a} e^{-2\pi |m|a}. \end{align*}
Plugging this to Poisson summation formula, we get
\begin{align*} S(a) &= \sum_{m=-\infty}^{\infty} \hat{f}(2\pi m) = \sum_{m=-\infty}^{\infty} \frac{\pi}{a}e^{-2\pi|m|a} = \frac{\pi}{a} \left( 1 + 2 \sum_{m=1}^{\infty} e^{-2\pi ma} \right) \\ &= \frac{\pi}{a} \left( 1 + \frac{2e^{-2\pi a}}{1 - e^{-2\pi a}} \right) = \frac{\pi \coth (\pi a)}{a}, \end{align*}
proving the claim. Then, finally, the original sum can be written as
\begin{align*} \sum_{n=-\infty}^{\infty} \frac{1}{(n^2 + a^2)(n^2 + b^2)} &= \sum_{n=-\infty}^{\infty} \frac{1}{b^2 - a^2} \left[ \frac{1}{n^2 + a^2} - \frac{1}{n^2 + b^2} \right] \\ &= \frac{1}{b^2 - a^2}[S(a) - S(b)] \\ &= \frac{1}{b^2 - a^2}\left( \frac{\pi \coth(\pi a)}{a} - \frac{\pi \coth(\pi b)}{b}\right). \end{align*}