Definite Integral to Infinity of Hyperbolic Sine of a x over Exponential of b x plus One

Theorem

 * $\displaystyle \int_0^\infty \frac {\sinh a x} {e^{b x} + 1} \rd x = \frac \pi {2 b} \csc \frac {a \pi} b - \frac 1 {2 a}$

where:
 * $a$ and $b$ are positive real numbers with $b > a$
 * $\csc$ is the cosecant function.

Proof
Note that as $b > a$, we have that $a - b < 0$.

As $b > 0$, we therefore have $a - \paren {n + 1} b < 0$ for all positive integer $n$.

We also have that as $a + \paren {n + 1} b > 0$, that $-\paren {a + \paren {n + 1} b} < 0$.

So, by Exponential Tends to Zero and Infinity:


 * $\displaystyle \lim_{x \mathop \to \infty} \frac {e^{\paren {a - \paren {n + 1} b} x} } {a - \paren {n + 1} b} = 0$

and:


 * $\displaystyle \lim_{x \mathop \to \infty} \frac {e^{-\paren {a + \paren {n + 1} b} x} } {a + \paren {n + 1} b} = 0$

We therefore have:

We have by Mittag-Leffler Expansion for Cosecant Function:


 * $\displaystyle \pi \map \csc {\pi z} = \frac 1 z + 2 \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac z {z^2 - n^2}$

Setting $z = \dfrac a b$ we have:


 * $\displaystyle \pi \map \csc {\frac {\pi a} b} = \frac b a + 2 \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {\paren {\frac a b} } {\paren {\frac a b}^2 - n^2}$

Rearranging gives:


 * $\displaystyle \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {\paren {\frac a b} } {\paren {\frac a b}^2 - n^2} = \frac \pi 2 \map \csc {\frac {\pi a} b} - \frac b {2 a}$

Therefore: