Other ways to compute this integral?

Question

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?

Answer 1

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.

Answer 2

Well, I guess the Frullani way really just involves expressing the integrand as an integral over its derivative and then switching the order of integration. So, consider

$$f(u) = \frac{u}{e^{u x}-1}$$

$$f'(u) = -\frac{u \, x \, e^{u x}}{(e^{u x}-1)^2} + \frac1{e^{u x}-1}$$

$$\int_b^a du \, f'(u) = \frac{a}{e^{a x}-1} - \frac{b}{e^{b x}-1} $$

Thus, we assert that

$$\begin{align}\int_0^{\infty} dx \, \left (\frac{a}{e^{a x}-1} - \frac{b}{e^{b x}-1} \right ) &= \int_0^{\infty} dx \, \int_b^a du \, \left [-\frac{u \, x \, e^{u x}}{(e^{u x}-1)^2} + \frac1{e^{u x}-1} \right ] \\ &= \int_b^a du \, \int_0^{\infty} dx \,\left [-\frac{u \, x \, e^{u x}}{(e^{u x}-1)^2} + \frac1{e^{u x}-1} \right ] \end{align}$$

We assert this because we know (ahem) that each integral is finite by itself. (I know this reasoning allowing us to switch the order of integration can be improved upon but I want to stress the mechanics of the computation for now.)

Now we must do the inner integral. The funny thing is that the integrand is completely symmetric in $u$ and $x$ so that we may simply write down the antiderivative as $f$, but now seen as a function of $x$ rather than $u$. Thus,

$$\int_0^{\infty} dx \,\left [-\frac{u \, x \, e^{u x}}{(e^{u x}-1)^2} + \frac1{e^{u x}-1} \right ] = \left [\frac{x}{e^{u x}-1} \right ]_0^{\infty} = -\frac1{u}$$

Thus, the integral we seek is

$$-\int_b^a \frac{du}{u} = \log{\frac{b}{a}}$$