1. ## very strong integral

find:

$\large \int_{0}^{\infty }\frac{\sinh (ax)}{e^{bx}+1}dx{\color{red} }$

2. Start by writing sinh(ax) as $\displaystyle \frac{e^{ax}- e^{-ax}}{2}$.

3. Note that

$\displaystyle \frac{ \sinh(ax)}{e^{bx} + 1} = \frac{ e^{-bx} \sinh(ax) }{ 1 + e^{-bx}}$

$\displaystyle = \sum_{k=1}^{\infty} (-1)^{k+1} \sinh(ax) e^{-bkx}$

Now this strong integral ,

$\displaystyle \int_0^{\infty} \frac{ \sinh(ax)}{e^{bx} + 1}~dx$

$\displaystyle = \int_0^{\infty} \sum_{k=1}^{\infty} (-1)^{k+1} \sinh(ax) e^{-bkx} ~dx$

Recall the Laplace Transform of $\displaystyle \sinh(ax)$ which is $\displaystyle \frac{a}{s^2 - a^2}$

but this time $\displaystyle s = bk$ ( we assume $\displaystyle b > a$ )

This strong integral are therefore weakened by this Laplace Transform

$\displaystyle = \sum_{k=1}^{\infty} (-1)^{k+1} \frac{ a }{ (bk)^2 - s^2 }$

$\displaystyle = \sum_{k=1}^{\infty} (-1)^{k+1} \frac{1}{b} \cdot \frac{ a/b}{ k^2 - (a/b)^2}$

To totally defeat this strong integral , we must recall the infinite product for tagent .

$\displaystyle \tan(\frac{\pi x}{2}) = \frac{\pi x}{2} \prod_{k=1}^{\infty} \left( 1- \frac{x^2}{ (2k)^2} \right ) \left ( \prod_{k=1}^{\infty} [ 1 - \frac{ x^2}{ ( 2k-1)^2}] \right )^{-1}$

Then take logarithmic derivative ,

$\displaystyle \pi \csc(\pi x ) = \frac{1}{x} + \sum_{k=1}^{\infty} (-1)^{k+1} \frac{ 2x }{ k^2 - x^2 }$

$\displaystyle \sum_{k=1}^{\infty} (-1)^{k+1} \frac{ x }{ k^2 - x^2 } = \frac{1}{2} \left ( \pi \csc(\pi x) - \frac{1}{x} \right )$

Sub. $\displaystyle x = a/b$

The integral $\displaystyle = \frac{\pi}{2b} \csc(\frac{\pi a}{b}) - \frac{1}{2a}$