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

## 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:

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

$\blacksquare$