User:Caliburn/Definite Integral to Infinity of Power of x over Hyperbolic Sine of a x/test

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
Let $\mu$ be the Lebesgue measure on $\R$.

For each integer $N \ge 0$, define $f_N : \R \to \R$ by:


 * $\ds \map f u = u^n e^{-\paren {2 N + 1} u}$

We have that:


 * $f_N$ is continuous for each $N \ge 0$.

So, from Continuous Mapping is Measurable:


 * $f_N$ is Borel measurable for each $N \ge 0$.

From Integral of Series of Positive Measurable Functions, we have:


 * $\ds \frac 2 {a^{n + 1} } \int_0^\infty \sum_{N \mathop = 0}^\infty f_N \rd \mu = \frac 2 {a^{n + 1} } \sum_{N \mathop = 0}^\infty \int_0^\infty f_N \rd \mu$

That is, from Lebesgue Integral Coincides with Riemann Integral: (need positive function Lebesgue integrable iff Riemann integrable too)


 * $\ds \frac 2 {a^{n + 1} } \int_0^\infty \sum_{N \mathop = 0}^\infty u^n e^{-\paren {2 N + 1} u} \rd x = \frac 2 {a^{n + 1} } \sum_{N \mathop = 0}^\infty \int_0^\infty u^n e^{-\paren {2 N + 1} u} \rd x$

We have:

So:

giving:


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