Definite Integral to Infinity of Sine of m x over Exponential of 2 Pi x minus One

Theorem

 * $\ds \int_0^\infty \frac {\sin m x} {e^{2 \pi x} - 1} \rd x = \frac 1 4 \coth \frac m 2 - \frac 1 {2 m}$

where:
 * $m$ is a non-zero real number
 * $\coth$ is the hyperbolic cotangent function.

Proof
We have:

By Mittag-Leffler Expansion for Hyperbolic Cotangent Function we have:


 * $\ds \sum_{k \mathop = 1}^\infty \frac z {k^2 + z^2} = \frac \pi 2 \map \coth {\pi z} - \frac 1 {2 z}$

for all $z \in \C$ where $z$ is not an integer multiple of $i$.

Since all variables concerned in this instance all real-valued, we can apply this identity.

Setting $z = \dfrac m {2 \pi}$ in the above we obtain:

So:


 * $\ds \int_0^\infty \frac {\sin m x} {e^{2 \pi x} - 1} \rd x = \frac 1 {2 \pi} \paren {\frac \pi 2 \coth \frac m 2 - \frac \pi m} = \frac 1 4 \coth \frac m 2 - \frac 1 {2 m}$

as required.