I really need to solve this integral using Euler integrals (Gamma & beta functions).
$$I=\int_0^\infty \frac{\sinh^2 bx }{\sinh^2 cx} dx $$
I have tried all that I could, but…
$$I=\int_0^\infty
\left(1 – e^{-2bx} \right)^2
(1 – e^{-2cx})^{-2}
e^{-2(c-b)x}dx = \int_0^\infty \frac{\exp(2bx)+\exp(-2bx)-2}{\left[\exp(cx)-\exp(-cx) \right]^2}dx
$$
After $u=\exp(-2cx)$ i finally get
$$ \frac{1}{2c} \int_0^1 \frac{u^{p-1}-2u^{-1}+u^{p+1} }{(1-u)^2}
$$
(I'm not sure it is correct). I don't know how to create a Beta-function from it. The answer given by book is $$\frac{c-\pi b \cot(b\pi/c)}{2c^2} $$
but I can't get it. Please, help me. Thanks
Best Answer
It is rather related to the digamma function $\psi(a)=\Gamma'(a)/\Gamma(a)$, via the integral $$\int_0^1\frac{x^{a-1}-x^{b-1}}{1-x}\,dx=\psi(b)-\psi(a)\qquad(a,b>0)$$ (here is my answer showing this). Your integral, after your substitution $u=e^{-2cx}$, is really $$I=\int_0^1\left(\frac{u^{-r/2}-u^{r/2}}{u^{-1/2}-u^{1/2}}\right)^2\frac{du}{2cu}=\frac{1}{2c}\int_0^1\frac{u^r+u^{-r}-2}{(1-u)^2}\,du,$$ where $r=b/c$ (and we suppose $|r|<1$). Now we use $\dfrac{1}{(1-u)^2}=\dfrac{d}{du}\dfrac{u}{1-u}$: \begin{align*}2cI&=\underbrace{(u^r+u^{-r}-2)\frac{u}{1-u}\Bigg|_0^1}_{=0}-r\int_0^1\frac{u^r-u^{-r}}{1-u}\,du\\&=r\big(\psi(1+r)-\psi(1-r)\big)=r\frac{d}{dr}\log\big(\Gamma(1+r)\Gamma(1-r)\big)\\&=r\frac{d}{dr}\log\frac{\pi r}{\sin\pi r}=\bbox[5pt,border:2pt solid]{1-\pi r\cot\pi r}\end{align*} as expected. [An alternative to Euler integrals is to use contour integration.]