Integrating $\int_{-\infty}^{+\infty}{\frac{\sin(ax)\sinh(bx)}{(c+\cosh(bx))^2}dx}$

Question

I want to integrate this expression :

$$I = \int_{-\infty}^{+\infty}{\frac{\sin(ax)\sinh(bx)}{(c+\cosh(bx))^2}dx}$$ where $a>0$, $b>0$ and $c<0$

Answer 1

$$I(a,b,c) = \int_{-\infty}^{+\infty}\frac{\sin(ax)\sinh(bx)}{(c+\cosh(bx))^2}dx$$ $I(a,b,c)$ diverges for arbitrary $a,b>0$ and $c\leqslant -1$ (except for a very specific case of $c=-\cosh\bigl(\frac{b}{a}\pi k\bigr), k=\pm1, \pm2 \,..\,$, when the integral can be evaluated in the principal value sense and is equal to zero).

Therefore, we consider the case $c\in(-1;0)$. Integrating by part $$I=-\frac{1}{b}\frac{\sin (ax)}{c+\cosh (bx)}|_{-\infty}^{\infty}+\frac{a}{b}\int_{-\infty}^{+\infty}\frac{\cos (ax)}{c+\cosh(bx)}dx=\frac{a}{b^2}\Re\int_{-\infty}^{+\infty}\frac{e^{i\frac{a}{b}t}}{c+\cosh t}dt$$

Denoting $\alpha=\frac{a}{b}\,$ and $\,c=-\cos\beta$, $\,\,\beta\in(0;\frac{\pi}{2})$ we get $$I=\frac{a}{b^2}\Re\int_{-\infty}^{+\infty}\frac{e^{i\alpha x}}{\cosh x-\cos\beta}dx=\frac{a}{b^2}\,\Re \,J(\alpha, \beta)$$ Let's evaluate $J(\alpha, \beta)=\int_{-\infty}^{+\infty}\frac{e^{i\alpha x}}{\cosh x-\cos\beta}dx$. We choose the rectangular contour in the complex plane:

Due to periodicity of $\cosh(x+2\pi i)=\cosh x$ we get $$\oint=J(\alpha, \beta)+I_1-J(\alpha, \beta)e^{-2\pi\alpha}+I_2=2\pi i \sum Res \frac{e^{i\alpha x}}{\cosh x-\cos\beta}$$ It can be shown that integral along path $1$ and $2$ $\to 0$ as $R\to\infty$. We also have a couple of simple poles inside the contour at $x=i\beta$ and $x=2\pi i-i\beta$.

Near $x=i\beta$ we have $$\cosh(i\beta+\epsilon)-\cos\beta = \frac{1}{2}\Bigl(e^{i\beta+\epsilon}+e^{-i\beta-\epsilon}-e^{i\beta}-e^{-i\beta}\Bigr)\to-i\epsilon\sin\beta\,\, \text{at}\,\,\epsilon\to0$$ In the same way we evaluate the residual at $x=2\pi i-i\beta$. $$J(\alpha, \beta)(1-e^{-2\pi\alpha})=2\pi i\Bigl(\frac{e^{-\alpha\beta}}{i\sin \beta}+\frac{e^{\alpha\beta-2\pi\alpha}}{-i\sin \beta}\Bigr)=\frac{4\pi}{\sin \beta}e^{-\pi\alpha}\sinh\alpha(\pi-\beta)$$ $$J(\alpha, \beta)=\frac{2\pi\sinh\alpha(\pi-\beta)}{\sin \beta\,\sinh\alpha\pi}$$ Quick check: at $\alpha\to0$ and $\beta\to0$ $J\to\frac{2\pi}{\sin\beta}\sim\frac{2\pi}{\beta}$; on the other hand

$J(\alpha=0, \beta\to 0)=\int_{-\infty}^{+\infty}\frac{1}{\cosh x-\cos\beta}dx\sim 2\int_{-\infty}^{+\infty}\frac{dx}{e^x+e^{-x}-2\bigl(1-\beta^2/2\bigr)}\sim2\int_{-\infty}^{+\infty}\frac{dx}{x^2+\beta^2}\sim\frac{2\pi}{\beta}$ $$I(a,b,c)=\frac{2\pi a}{\,b^2}\,\frac{\sinh\Bigl(\frac{a}{b}(\pi-\arccos|c|)\Bigr)}{\sqrt{1-|c|^2}\,\sinh\bigl(\frac{a}{b}\pi\bigr)},\,\,\, a,b>0 \,\,\text{and} \,\,c\in(-1,0)$$ Just in case, for information

1. $c\in(0,1) \,\,I=\frac{2\pi a}{\,b^2}\,\frac{\sinh\bigl(\frac{a}{b}\arccos c\bigr)}{\sqrt{1-c^2}\,\sinh\bigl(\frac{a}{b}\pi\bigr)}$

2. $c>1\, \,\,I=\frac{2\pi a}{\,b^2}\,\frac{\sin\bigl(\frac{a}{b}\cosh^{-1} c\bigr)}{\sqrt{c^2-1}\,\sinh\bigl(\frac{a}{b}\pi\bigr)}$

3. $c=1\,\,\, I=\frac{2\pi a^2}{\,b^3}\,\frac{1}{\sinh\bigl(\frac{a}{b}\pi\bigr)}$

4. $c=0\,\,\,I=\frac{2\pi a}{\,b^2}\,\frac{\sinh\bigl(\frac{\pi a}{2b}\bigr)}{\sinh\bigl(\frac{a}{b}\pi\bigr)}=\frac{\pi a}{\,b^2}\,\frac{1}{\cosh\bigl(\frac{\pi a}{2b}\bigr)}$