Calculating the definite integral $\int_0^{+∞}\frac{\sinh^2(bx)}{x(e^{ax}+1)}\,\mathrm dx$

Question

So firstly I need to understand for what $a$ and $b$ this integral converges. If $a>2b>1$ integral converges. Then my professor told me to find derivative of numerator $$\frac{d(\sinh^2(bx))}{db}$$ so that I can rid off $x$ in denominator and than make this Euler’s integral (Beta function) and after this integrate again $db$. I tried to do that way but how I make beta function from ${\displaystyle \frac{e^{2bx}}{e^{ax}+1}}$?

Answer 1

In my opinion, I wouldn’t use the beta function to evaluate$$J=\int\limits_0^{\infty}dx\,\frac {e^{2bx}}{1+e^{ax}}$$It just doesn’t seem to fit the integral. Instead, what we can do is divide both the numerator and denominator by $e^{ax}$ to get$$J=\int\limits_0^{\infty}dx\,\frac {e^{2bx-ax}}{1+e^{-ax}}$$Now make the substitution $x\mapsto e^{-ax}$ so the limits change to zero and one. Hence$$\begin{align*}J & =\frac 1a\int\limits_0^1dx\,\frac {x^{-2b/a}}{1+x}\end{align*}$$Now use the geometric series to expand the inner denominator, integrate term wise, and simplify your final result.