Definite Integral to Infinity of Sine of m x over Exponential of 2 Pi x minus One
Jump to navigation
Jump to search
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
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 15$: Definite Integrals involving Exponential Functions: $15.83$