Other ways to compute this integral? The following (improper) integral comes up in exercise 2.27 in Folland (see this other question): $$I = \int_0^\infty \frac{a}{e^{ax}-1} - \frac{b}{e^{bx}-1}\,dx.$$
I computed it as follows. 
An antiderivative for $a(e^{ax}-1)^{-1}$ is $\log(1-e^{-ax})$, found by substituting $u = e^{ax}-1$ and noting that $1/u(u+1) = 1/u - 1/(u+1)$.
Therefore, $$\int_{\varepsilon}^R \frac{a}{e^{ax}-1} - \frac{b}{e^{bx}-1}\,dx = \log\left(\frac{1-e^{-aR}}{1-e^{-bR}}\right) + \log\left(\frac{1-e^{-b\varepsilon}}{1-e^{-a\varepsilon}}\right).$$
The first term goes to $\log(1) = 0$ as $R\to\infty$. 
For the second term, we have 
$$\lim_{\varepsilon\to 0^+} \log\left(\frac{1-e^{-b\varepsilon}}{1-e^{-a\varepsilon}}\right) = \log \lim_{\varepsilon\to 0^+} \frac{1-e^{-b\varepsilon}}{1-e^{-a\varepsilon}},$$
which looks like $0/0$. Applying l'Hospital's rule, 
we get $$\lim_{\varepsilon\to 0^+} \frac{1-e^{-b\varepsilon}}{1-e^{-a\varepsilon}} = \lim_{\varepsilon\to 0^+} \frac{be^{-b\varepsilon}}{ae^{-a\varepsilon}} = \frac{b}{a}$$
so $I = \log(b/a)$. 
What are some other ways to compute this integral? Perhaps there is a method incorporating Frullani's theorem?

Here's a nice solution. 
If we let $$f(x) = \frac{x}{e^x-1} = \frac{1}{1+\frac{x}{2}+\frac{x^2}{6}+\dotsb}$$
then $f(0) = 1$ and $f(\infty) = 0$ and $$\frac{f(ax) - f(bx)}{x} = \frac{a}{e^{ax}-1} - \frac{b}{e^{bx}-1},$$
and now apply Frullani's theorem.
