Question

Prove that:-


`\rm \sum_(n = 1)^ \infty \frac{1}{ {n}^(2) + {x}^(2) } = \frac{\pi x - 1 }{2 {x}^(2) } + \frac{\pi}{x( {e}^(2 \pi x) - 1)} \\`​

Answer 1

Observe that

`\displaystyle \frac1{e^(2\pi x) - 1} = (e^(-\pi x))/(e^(\pi x) - e^(-\pi x)) = (\cosh(\pi x) - \sinh(\pi x))/(2\sinh(\pi x)) = \frac{\coth(\pi x) - 1}2`

We have the following series expansion due to Mittag-Leffler:

`\displaystyle \pi \coth(\pi z) = \frac1z + 2 \sum_(n=1)^\infty \frac z{z^2+n^2}`

The proof of this is easy to follow. (I'll try to include a link)

With some simple rearrangement, we get the desired result.

`\displaystyle \sum_(n=1)^\infty \frac1{n^2 + x^2} = -\frac1{2x^2} + (\pi \coth(\pi x))/(2x) \\\\ ~~~~~~~~ = -\frac1{2x^2} + \frac\pi{2x} + \frac\pi{x(e^(2\pi x)-1)}`