Partial fraction expansion of $1/(e^x-1)$

Question

I'm reading a paper by Riesel and Gohl; in it they say that "partial fraction expansion" of $1/(e^x-1)$ is $$\frac{1}{e^x-1}=\frac{1}{x}-\frac{1}{2}+2x \sum_{k=1}^{\infty} \frac{1}{x^2+4\pi^2k^2}$$ I tried using Euler's formula as $2\pi k$ are the roots of sine, but it leads nowhere. As the only root of $e^x-1$ is at $x=0$ and I fail to see how I can get this result with partial fractions. The Taylor expansion gives the first two terms correctly, but the rest is a power series involving Bernoulli numbers. So ho do I use partial fractions to expand $1/(e^x-1)$?

Answer 1

Consider $f(z)=\frac{1}{e^z-1}$ as a meromorphic function on the complex plane and apply the residue theorem. $f(z)$ has simple poles for $z\in 2\pi i\mathbb{Z}$.

Answer 2

$~\quad~$ Alternately, start by differentiating the natural logarithm of Euler's infinite product for the sine function, and then use the well-known relations between hyperbolic and trigonometric functions to establish the desired result.