# 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?

• Perhaps use residues? This method already seems fairly painless... – Potato Oct 24 '15 at 0:16
• @hermes There is no typo; moreover, your integral is divergent. – Unit Oct 24 '15 at 0:17
• @Potato Perhaps I should ask simply for other and not quicker ways. My solution feels gritty to me. – Unit Oct 24 '15 at 0:25
• @Unit You did the straightforward thing. I don't see anything wrong with that aesthetically. – Potato Oct 24 '15 at 0:27
• @Potato Perhaps you would be interested in my answer below. – Unit Oct 24 '15 at 12:07

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.

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

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

• This is nice! But it's really just a proof of Frullani's theorem in disguise, as explained by your first sentence! For if we have some $C^1$ function $f$ then $\frac{\partial}{\partial t} f(xt)/x = \frac{\partial}{\partial x} f(xt)/t = f'(xt)$, then $\int_b^a f'(xt) = (f(ax)-f(bx))/x$ and $\int_a^b\int_0^\infty f'(xt)\,dx\,dt = f(xt)|_{x=0}^\infty \int_a^b dt/t$! – Unit Oct 24 '15 at 0:55
• Took too long editing my comment; the last integral should be from $b$ to $a$ and this is what changes $f(\infty)-f(0)$ to $f(0)-f(\infty)$. – Unit Oct 24 '15 at 1:02