Show $\int_0^\infty \frac{\cos a x-\cos b x}{\sinh \beta x}\frac{dx}{x}=\log\big( \frac{\cosh \frac{b\pi}{2 \beta}}{\cosh \frac{a\pi}{2\beta}}\big)$

Question

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 1

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

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

Answer 2

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}

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

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}