How to do integral $ \int_0^{\infty} \left(\frac{a}{(e^{ax}-1)}-\frac{b}{(e^{bx}-1)}\right)\mathrm{d}x$? I met this integral: 
$$
\int_0^{\infty} \left(\frac{a}{(e^{ax}-1)}-\frac{b}{(e^{bx}-1)}\right)\mathrm{d}x, \text{ where } 
 \,\,0<a<b .
$$
It is a problem that showed up on my final exam of the lesson Mathematics I. Would anyone help me calculate this integral? Thanks.
 A: Here is an hint for solving $\displaystyle\int \frac{a}{e^{ax}-1}dx $.
Let $e^{ax}-1=u $. This substitution will give us the integral $$\int\frac{1}{u(u+1)}du=\int\left(\frac1u-\frac{1}{(u+1)}\right)du.$$ Can you take it from here?
A: Hints:
0) Prove the integral to be absolutely convergent.
1) Integrate from $\epsilon >0$ instead and prove that you may split the integral.
2) Change variables so that integrands are the same.
3) See what happens and remember $e^x=1+x+o(x)$ at $x \rightarrow 0$. 
A: Another hint for solving $\int \frac{a}{e^{ax}-1} dx$
$$\int \frac{a}{e^{ax}-1} dx = \int \frac{a(e^{ax}+1)}{e^{2ax}-1} dx = \frac{1}{2}\int \frac{a}{e^{ax}-1} dx + \int \frac{a e^{ax}}{e^{2ax}-1} dx$$
Reagrouping terms and a change of variable s.t. $u = e^{ax}$ and $du = a e^{ax}dx$, leads to:
$$\frac{1}{2}\int \frac{a}{e^{ax}-1} dx = \int \frac{1}{u^2-1} du= \frac{1}{2} \left( \int \frac{1}{u-1} du - \int \frac{1}{u+1} du\right)$$
$$
\boxed{
\int \frac{a}{e^{ax}-1} dx = \ln\left( e^{2ax} - 1\right) - \ln\left( e^{2ax} + 1\right)
}$$
