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

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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:

\(\ds \int_0^\infty \frac {\sin m x} {e^{2 \pi x} - 1} \rd x\) \(=\) \(\ds \int_0^\infty \frac {e^{-2 \pi x} \sin m x} {1 - e^{-2 \pi x} } \rd x\)
\(\ds \) \(=\) \(\ds \int_0^\infty e^{-2 \pi x} \sin m x \paren {\sum_{k = 0}^\infty e^{-2 \pi k x} } \rd x\) Sum of Infinite Geometric Sequence
\(\ds \) \(=\) \(\ds \sum_{k = 0}^\infty \paren {\int_0^\infty e^{-x \paren {2 \pi \paren {1 + k} } } \sin m x \rd x}\)
\(\ds \) \(=\) \(\ds \sum_{k = 0}^\infty \frac m {\paren {2 \pi \paren {1 + k} }^2 + m^2}\) Definite Integral to Infinity of $e^{-a x} \sin b x$
\(\ds \) \(=\) \(\ds \frac 1 {2 \pi} \sum_{k = 0}^\infty \frac {\frac m {2 \pi} } {\paren {k + 1}^2 + \paren {\frac m {2 \pi} }^2}\) dividing by $\paren {2 \pi}^2$
\(\ds \) \(=\) \(\ds \frac 1 {2 \pi} \sum_{k = 1}^\infty \frac {\frac m {2 \pi} } {k^2 + \paren {\frac m {2 \pi} }^2}\) shifting the index

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:

\(\ds \sum_{k = 1}^\infty \frac {\frac m {2 \pi} } {k^2 + \paren {\frac m {2 \pi} }^2}\) \(=\) \(\ds \frac \pi 2 \map \coth {\pi \times \frac m {2 \pi} } - \frac 1 {2 \times \frac m {2 \pi} }\)
\(\ds \) \(=\) \(\ds \frac \pi 2 \coth \frac m 2 - \frac \pi m\)

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.

$\blacksquare$


Sources

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