3
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Perhaps use residues? This method already seems fairly painless... $\endgroup$ – Potato Oct 24 '15 at 0:16
  • $\begingroup$ @hermes There is no typo; moreover, your integral is divergent. $\endgroup$ – Unit Oct 24 '15 at 0:17
  • $\begingroup$ @Potato Perhaps I should ask simply for other and not quicker ways. My solution feels gritty to me. $\endgroup$ – Unit Oct 24 '15 at 0:25
  • $\begingroup$ @Unit You did the straightforward thing. I don't see anything wrong with that aesthetically. $\endgroup$ – Potato Oct 24 '15 at 0:27
  • $\begingroup$ @Potato Perhaps you would be interested in my answer below. $\endgroup$ – Unit Oct 24 '15 at 12:07
2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ 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 $\int_a^b\int_0^\infty f'(xt)\,dx\,dt = f(xt)|_{x=0}^\infty \int_a^b dt/t$! $\endgroup$ – Unit Oct 24 '15 at 0:55
  • $\begingroup$ 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 $f(0)-f(\infty)$. $\endgroup$ – Unit Oct 24 '15 at 1:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.