I am trying to evaluate the integral
$$
\int_0^\infty \frac{\sin x}{x}\frac{\sinh(ax)}{\sinh(bx)}e^{-cx}\,dx
$$
with $c>0$ and $0<a<b$.
I have tried to look for a suitable contour, or to differentiate under integral sign with respect to $c$ (as I would do without the sinh's), but without success.
Thank you in advance for any hint!
Best Answer
Let
$$ I(a)= \int_{-\infty}^\infty \frac{\sinh(tx)\sinh(ax)\cosh(cx)}{(tx)\sinh(bx)}\,dx $$ Then \begin{align} I’(a) & = \int_{-\infty}^\infty \frac{\sinh(tx)\cosh(ax)\cosh(cx)}{t\sinh(bx)}\,dx\\ &= \frac1{4t}[J(a+c+t)-J(a-c-t) +J(a-c+t)- J(a+c-t)] \end{align} where $$J(p) = \int_{-\infty}^\infty \frac{\sinh(px)}{\sinh(bx)}\,dx =\frac\pi b \tan\frac{\pi p}{2b} $$ (obtained with contour integration.) Apply $\int_0^a \tan s\>ds=\ln \sec a$ to get
$$I(a) = \int_0^a I’(s)ds=\frac1{2t} \ln \frac{\sec\frac{\pi(a+c+t)}{2b}} {\sec\frac{\pi(a+c-t)}{2b}} + \frac1{2t} \ln \frac{\sec\frac{\pi(a-c+t)}{2b} } {\sec\frac{\pi(a-c-t)}{2b} } $$
Finally, with $\frac{\sec(u+i v)} {\sec(u-i v)}=\frac{1+i \tan u\tanh v}{1-i \tan u\tanh v} $
\begin{align} & \int_0^\infty \frac{\sin x\sinh(ax)}{x\sinh(bx)}e^{-cx}\,dx =\frac12 I(a)|_{t=i}\\=& \frac12\tan^{-1}\left( \tan\frac{\pi(a+c)}{2b}\tanh\frac\pi{2b} \right) + \frac12\tan^{-1}\left( \tan\frac{\pi(a-c)}{2b}\tanh\frac\pi{2b} \right) \end{align}