- Prove that
\[ \frac {1}{\pi} \sum_ {n = - \infty} ^ {\infty} \frac {a}{a ^ {2} + n ^ {2}} = \sum_ {n = - \infty} ^ {\infty} e ^ {- 2 \pi a | n |} \]
whenever \(a > 0\) . Hence show that the sum equals coth \(\pi a\)
Solution
From Exercise 3, for \[ f(x)=\frac1\pi\frac{a}{a^2+x^2},\qquad a>0, \] we have \[ \widehat f(\xi)=e^{-2\pi a|\xi|}. \] Both \(f\) and \(\widehat f\) are of moderate decrease, so the Poisson summation formula applies: \[ \sum_{n\in\mathbb Z}f(n)=\sum_{n\in\mathbb Z}\widehat f(n). \] Therefore \[ \frac1\pi\sum_{n=-\infty}^{\infty}\frac{a}{a^2+n^2} =\sum_{n=-\infty}^{\infty}e^{-2\pi a|n|}. \] The latter sum is geometric: \[ \sum_{n\in\mathbb Z}e^{-2\pi a|n|} =1+2\sum_{n=1}^{\infty}e^{-2\pi an} =1+\frac{2e^{-2\pi a}}{1-e^{-2\pi a}} =\frac{e^{\pi a}+e^{-\pi a}}{e^{\pi a}-e^{-\pi a}} =\coth(\pi a). \] Hence both displayed sums equal \(\coth(\pi a)\).