Definite Integral to Infinity of Power of x over Hyperbolic Sine of a x

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Theorem

$\displaystyle \int_0^\infty \frac {x^n} {\sinh a x} \rd x = \frac {2^{n + 1} - 1} {2^n a^{n + 1} } \map \Gamma {n + 1} \map \zeta {n + 1}$

where:

$a$ and $n$ are positive real numbers
$\Gamma$ denotes the gamma function
$\zeta$ denotes the Riemann zeta function.


Proof

\(\displaystyle \int_0^\infty \frac {x^n} {\sinh a x} \rd x\) \(=\) \(\displaystyle \frac 1 a \int_0^\infty \frac {\paren {\frac u a}^n} {\sinh u} \rd u\) substituting $u = a x$
\(\displaystyle \) \(=\) \(\displaystyle \frac 2 {a^{n + 1} } \int_0^\infty \frac {u^n e^{-u} } {1 - e^{-2 u} } \rd u\) Definition of Hyperbolic Sine
\(\displaystyle \) \(=\) \(\displaystyle \frac 2 {a^{n + 1} } \int_0^\infty u^n e^{-u} \paren {\sum_{N \mathop = 0}^\infty \paren {e^{-2 u} }^N} \rd u\) Sum of Infinite Geometric Progression
\(\displaystyle \) \(=\) \(\displaystyle \frac 2 {a^{n + 1} } \sum_{N \mathop = 0}^\infty \int_0^\infty u^n e^{-\paren {2 N + 1} u} \rd u\) Fubini's Theorem
\(\displaystyle \) \(=\) \(\displaystyle \frac 2 {a^{n + 1} } \map \Gamma {n + 1} \sum_{N \mathop = 0}^\infty \frac 1 {\paren {2 N + 1}^{n + 1} }\) Laplace Transform of Real Power

We have:

\(\displaystyle \sum_{N \mathop = 1}^\infty \frac 1 {N^{n + 1} }\) \(=\) \(\displaystyle \map \zeta {n + 1}\) Definition of Riemann Zeta Function
\(\displaystyle \) \(=\) \(\displaystyle \sum_{N \mathop = 1}^\infty \frac 1 {\paren {2 N}^{n + 1} } + \sum_{N \mathop = 0}^\infty \frac 1 {\paren {2 N + 1}^{n + 1} }\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {2^{n + 1} } \sum_{N \mathop = 1}^\infty \frac 1 {N^{n + 1} } + \sum_{N \mathop = 0}^\infty \frac 1 {\paren {2 N + 1}^{n + 1} }\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {2^{n + 1} } \map \zeta {n + 1} + \sum_{N \mathop = 0}^\infty \frac 1 {\paren {2 N + 1}^{n + 1} }\) Definition of Riemann Zeta Function

So:

\(\displaystyle \sum_{N \mathop = 0}^\infty \frac 1 {\paren {2 N + 1}^{n + 1} }\) \(=\) \(\displaystyle \paren {1 - \frac 1 {2^{n + 1} } } \map \zeta {n + 1}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {2^{n + 1} - 1} {2^{n + 1} } \map \zeta {n + 1}\)

giving:

$\displaystyle \int_0^\infty \frac {x^n} {\sinh a x} \rd x = \frac {2^{n + 1} - 1} {2^n a^{n + 1} } \map \Gamma {n + 1} \map \zeta {n + 1}$

$\blacksquare$


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