Show $\int_0^\infty \frac{\cos a x-\cos b x}{\sinh \beta x}\frac{dx}{x}=\log\big( \frac{\cosh \frac{b\pi}{2 \beta}}{\cosh \frac{a\pi}{2\beta}}\big)$ 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
 A: Consider the integral
\begin{align}
\int_{a}^{b} \sin(\mu x) \ d\mu = \left[ \frac{\cos(x \mu)}{x} \right]_{a}^{b} = 
\frac{\cos(ax) - \cos(bx)}{x}.
\end{align}
The integral to calculate is
\begin{align}
I = \int_{0}^{\infty} \frac{\cos(ax) - \cos(bx)}{x \ \sinh(\beta x)} \ dx
\end{align}
and can be seen to be
\begin{align}
I &= \int_{0}^{\infty} \frac{\cos(ax) - \cos(bx)}{x \ \sinh(\beta x)} \ dx \\
&= \int_{a}^{b} \left( \int_{0}^{\infty} \frac{\sin(\mu x)}{\sinh(\beta x)} \ dx \right) \ d\mu.
\end{align}
The inner integral can be quickly calculated by using the known integral
\begin{align}
\int_{0}^{\infty} \frac{\sinh(\alpha x)}{\sinh(\beta x)} \ dx = \frac{\pi}{2 \beta} \ 
\tan\left( \frac{\alpha \pi}{2 \beta} \right).
\end{align}
By letting $\alpha = i \mu$ this becomes
\begin{align}
\int_{0}^{\infty} \frac{\sin(\mu x)}{\sinh(\beta x)} \ dx = \frac{\pi}{2 \beta} \ 
\tanh\left( \frac{\mu \pi}{2 \beta} \right)
\end{align}
and leads to
\begin{align}
I &= \frac{\pi}{2 \beta} \ \int_{a}^{b} \tanh\left( \frac{\pi \mu}{2 \beta} \right) \ d\mu \\
&= \int_{(a\pi/2\beta)}^{(b\pi/2\beta)} \tanh(x) \ dx \\
&= \left[ \ln(\cosh(x)) \right]_{(a\pi/2\beta)}^{(b\pi/2\beta)}
\end{align}
which can be restated as
\begin{align}
\int_{0}^{\infty} \frac{\cos(ax) - \cos(bx)}{x \ \sinh(\beta x)} \ dx = \ln\left( \frac{\cosh
\left( \frac{b \pi}{2\beta} \right)}{\cosh\left(\frac{a\pi}{2\beta}\right)}\right).
\end{align}
A: 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 $a, b \in \mathbb{R}$,
$$ \begin{align} \int_{0}^{\infty} \frac{\cos ax - \cos bx}{\sinh \beta x} \frac{dx}{x} &= \int_{0}^{\infty} \int_{a}^{b} \frac{\sin xt}{\sinh \beta x} \, dt \, dx= \int_{a}^{b} \int_{0}^{\infty} \frac{\sin tx}{\sinh \beta x} \, dx \, dt \\ &= \frac{\pi}{2 \beta}\int_{a}^{b} \tanh \left(\frac{\pi t}{2 \beta} \right) \, dt = \int_{\frac{\pi a}{2 \beta}}^{\frac{\pi b}{2 \beta}} \tanh u \, du \\ &= \log\left(\cosh  \frac{\pi b}{2 \beta}\right) - \log \left( \cosh \frac{\pi a}{2 \beta}\right) \\ &= \log \left( \frac{\cosh \frac{\pi b}{2 \beta}}{\cosh \frac{\pi a}{2 \beta}}\right)\end{align} $$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Lets $\ds{\mu \equiv {a \over 2\beta}\,,\ \nu \equiv {b \over 2\beta}}$:

\begin{align}
\mc{I} & \equiv
\int_{0}^{\infty}{\cos\pars{ax} - \cos\pars{bx} \over \sinh\pars{\beta x}}
\,{\dd x \over x} =
\int_{0}^{\infty}{\cos\pars{2\mu x} - \cos\pars{2\nu x} \over \sinh\pars{x}}
\,{\dd x \over x}
\\[5mm] & =
2\,\Re\int_{0}^{\infty}
{\expo{-2\ic\mu x} - \expo{-2\ic\nu x} \over 1 - \expo{-2x}}\expo{-x}
{\dd x \over x}
\,\,\,\stackrel{\expo{-2x}\ =\ t}{=}\,\,\,
2\,\Re\int_{1}^{0}{t^{\ic\mu} - t^{\ic\nu} \over 1 - t}\,t^{1/2}\,
{-\dd t/\pars{2t} \over \ln\pars{t}/\pars{-2}}
\\[5mm] & =
-2\,\Re\int_{0}^{1}{t^{-1/2 + \ic\mu} - t^{-1/2 + \ic\nu} \over 1 - t}\,
{\dd t \over \ln\pars{t}} =
2\,\Re\int_{0}^{1}{t^{-1/2 + \ic\mu} - t^{-1/2 + \ic\nu} \over 1 - t}
\int_{0}^{\infty}t^{\xi}\,\dd\xi\,\dd t
\\[5mm] & =
2\,\Re\int_{0}^{\infty}\bracks{%
\int_{0}^{1}{1 - t^{\xi -1/2 + \ic\nu} \over 1 -t}\,\dd t -
\int_{0}^{1}{1 - t^{\xi -1/2 + \ic\mu} \over 1 -t}\,\dd t}\dd\xi
\\[5mm] & =
2\,\Re\int_{0}^{\infty}\bracks{%
\Psi\pars{\xi + {1 \over 2} + \ic\nu} - \Psi\pars{\xi + {1 \over 2} + \ic\mu}}
\dd\xi =
\left.2\,\Re
\ln\pars{\Gamma\pars{\xi + 1/2 + \ic\nu} \over \Gamma\pars{\xi + 1/2 + \ic\mu}}
\right\vert_{\ \xi\ =\ 0}^{\ \xi\ \to\ \infty}
\\[5mm] & =
2\,\Re\ln\pars{\Gamma\pars{1/2 + \ic\mu} \over \Gamma\pars{1/2 + \ic\nu}} =
\ln\pars{\verts{\Gamma\pars{{1 \over 2} + \ic\mu}}^{2}} -
\ln\pars{\verts{\Gamma\pars{{1 \over 2} + \ic\nu}}^{2}}
\\[5mm] & =
\ln\pars{\pi \over \cosh\pars{\pi\mu}} -
\ln\pars{\pi \over \cosh\pars{\pi\nu}} =
\ln\pars{\cosh\pars{\pi\nu} \over \cosh\pars{\pi\mu}} =
\bbx{\ds{\ln\pars{\cosh\pars{\pi b \over 2\beta} \over
\cosh\pars{\pi a \over 2\beta}}}}
\end{align}
