calculus - Show $int_0^infty frac{cos a x-cos b x}{sinh beta x}frac{dx}{x}=logbig( frac{cosh frac{bpi}{2 beta}}{cosh frac{api}{2beta}}big)$

Hi I am trying to prove this interesting integral
$$
\mathcal{I}:=\int_0^\infty \frac{\cos a x-\cos b x}{\sinh \beta x}\frac{dx}{x}=\log\left( \frac{\cosh \frac{b\pi}{2 \beta}}{\cosh \frac{a\pi}{2\beta}}\right), \qquad Re(\beta)>0.
$$

I was thinking this was possibly a Frullani integral

$$
\int_0^\infty \frac{f(ax)-f(bx)}{x}dx=\big[f(0)-f(\infty)\big]\log \frac{b}{a},
$$
but am unable to get rid of the $\sinh \beta x$ in the denominator. I have tried partial integration and splitting the integral up but ran into convergent issues. How can we solve this integral? Possibly we can go to exponential representation using $2\sinh x=e^x-e^{-x}$, but that didn't help either. Thanks

Answer

First I'm going to evaluate $ \displaystyle\int_{0}^{\infty}\frac{\sin tx}{\sinh \beta x} \, dx \ , \ (t \in \mathbb{R}, \, \text{Re}(\beta) >0) $.

$$ \begin{align}\int_{0}^{\infty} \frac{\sin tx}{\sinh \beta x} \, dx &= 2 \int_{0}^{\infty} \frac{\sin tx}{e^{\beta x}-e^{-\beta x}} \, dx = 2 \int_{0}^{\infty} \sin tx \ \frac{e^{- \beta x}}{1-e^{-2 \beta x}} \, dx \\ &= 2 \int_{0}^{\infty} \sin tx \sum_{n=0}^{\infty} e^{-(2n+1)\beta x} \, dx = 2 \sum_{n=0}^{\infty} \int_{0}^{\infty} \sin tx \ e^{-(2n+1) \beta x} \, dx \\ &= 2 \sum_{n=0}^{\infty} \frac{t}{(2n+1)^{2}\beta^{2} + t^{2}} = \frac{\pi}{2 \beta} \tanh \left( \frac{\pi t}{2\beta}\right) , \end{align} $$

where I used the partial fraction expansion $$\tanh z = \sum_{n=0}^{\infty} \frac{2z}{\left(\frac{(2n+1) \pi}{2}\right)^{2}+z^{2}} $$

Then for $a, b \in \mathbb{R}$,

$$ \begin{align} \int_{0}^{\infty} \frac{\cos ax - \cos bx}{\sinh \beta x} \frac{dx}{x} &= \int_{0}^{\infty} \int_{a}^{b} \frac{\sin xt}{\sinh \beta x} \, dt \, dx= \int_{a}^{b} \int_{0}^{\infty} \frac{\sin tx}{\sinh \beta x} \, dx \, dt \\ &= \frac{\pi}{2 \beta}\int_{a}^{b} \tanh \left(\frac{\pi t}{2 \beta} \right) \, dt = \int_{\frac{\pi a}{2 \beta}}^{\frac{\pi b}{2 \beta}} \tanh u \, du \\ &= \log\left(\cosh \frac{\pi b}{2 \beta}\right) - \log \left( \cosh \frac{\pi a}{2 \beta}\right) \\ &= \log \left( \frac{\cosh \frac{\pi b}{2 \beta}}{\cosh \frac{\pi a}{2 \beta}}\right)\end{align} $$