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Find $$\int_{0}^{\infty}{f(x)-f(2x)\over x}\, \mathrm{d}x$$ if $f\in C([0,\infty])$ and $\lim\limits_{x\to \infty}{f(x)=L}$.

I tried denoting $\displaystyle \int{f(x)\over x}dx=F(x)$, but I don't know what to do with $F(\infty)-F(0)$. I also tried somehow to use l'Hopital's rule but don't seem to come by a plausible justification for it, nor do I arrive at a clear expression. I would really appreciate your help in this.

Travis Willse
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Meitar
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2 Answers2

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If one assumes that $f$ is sufficiently well-behaved (and in particular, differentiable), one can solve this by differentiating under the integral sign (this is sometimes called Feynman's trick).

Define $$I(a) := \int_0^{\infty} \frac{f(x) - f(a x)}{x} \,dx.$$ Differentiating with respect to $a$ (and justifying passing the derivative sign through the improper integral) gives (for $a > 0)$) that \begin{align} I'(a) &= \frac{d}{da} \int_0^{\infty} \frac{f(x) - f(a x)}{x} \,dx \\ &= \int_0^{\infty} \frac{d}{da} \left(\frac{f(x) - f(a x)}{x}\right) dx \\ &= -\int_0^{\infty} f'(ax) \,dx \\ &= -\left.\frac{1}{a} f(ax)\right\vert_0^{\infty} \\ &= \frac{1}{a}(f(0) - L) . \end{align} By the F.T.C., integrating w.r.t. $a$ gives $$I(2) - I(1) = \int_1^2 \frac{1}{a}(f(0) = L)\, da = (f(0) - L) \log a \vert_1^2 = (f(0) - L) \log 2.$$ On the other hand, $I(2)$ is the integral we want to evaluate and $$I(1) = \int_0^{\infty} \frac{f(x) - f(x)}{x} dx = 0.$$ Putting this all together gives, as desired, $$\color{#bf0000}{\int_0^{\infty} \frac{f(x) - f(2 x)}{x} \,dx = (f(0) - L) \log 2}.$$

Here's a different way to present essentially the same argument: The integrand is a difference of the values of a particular expression at two points, which is precisely the form of one side of the F.T.C. Reverse-engineering gives $$\frac{f(x) - f(2 x)}{x} = -\int_1^2 f'(y x) \,dy.$$ Justifying a reversal of the order of integration again gives \begin{align} \color{#bf0000}{\int_0^{\infty} \frac{f(x) - f(2 x)}{x} dx} &= \int_0^{\infty} \left[-\int_1^2 f'(y x) \,dy\right] dx \\ &= -\int_1^2 \int_0^{\infty} f'(y x) \,dx \,dy \\ &= -\int_1^2 \left.\frac{1}{y} f(y x)\right\vert_0^{\infty} dy \\ &= -(L - f(0)) \int_1^2 \frac{dy}{y} \\ &\color{#bf0000}{= (f(0) - L) \log 2} . \end{align}

More generally, the same argument gives (for $f$ satisfying the same conditions as in the question) that $$\int_0^{\infty} \frac{f(bx) - f(ax)}{x} dx = (L - f(0)) \log \frac{b}{a}.$$

Travis Willse
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  • I believe the general formula in the last line of the display (or, at least, one very similar to it) has a name, but I could neither recall nor find it. I'd be grateful if someone could remind me. – Travis Willse Jul 01 '15 at 18:41
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    I believe the general formula is Frullani's. – kobe Jul 01 '15 at 18:45
  • @kobe Yes, that's it, thank you! – Travis Willse Jul 01 '15 at 18:49
  • @Travis Now how do you justify differentiating under the integral sign? Or reversing the order of integration? – David C. Ullrich Jul 01 '15 at 20:54
  • @Travis Or for that matter, how do you even show $I(\alpha)$ exists in the first place? – David C. Ullrich Jul 01 '15 at 21:25
  • @DavidC.Ullrich You're right that this probably requires some niceness condition on $f$. For reversing the order of integration, for example, I think $f' \in L^1([0, \infty))$ should suffice. – Travis Willse Jul 02 '15 at 04:42
  • @Travis Yes, that's enough for the Fubini. Not clear to me that that's enough for the differentiation under the integral sign. I guess I'm curious why we like these arguments better than the other one. The other one proves much more, and also seems much simpler. (This one isn't even a proof yet, until we add those "justifications". The justifications are really not optional, this is the sort of improper integral situation where things can go wrong. As we see from the guy who said the answer is $\int_0^\infty f(t)/t - \int_0^\infty f(t)/t=0$; that would be fine if we could "justify" it...) – David C. Ullrich Jul 02 '15 at 16:03
  • I don't know "we" is (is that in particular a complaint about upvote counts?), but I don't claim that the argument in my answer is better or otherwise preferable that yours, and I won't speculate on the preferences of others. – Travis Willse Jul 04 '15 at 07:30
  • Also, like many other answers on this site, this answer is not intended to be a detailed, self-contained proof (the hedge at the beginning of the proof is supposed to emphasize this); rather, I wrote it to expose an (in my view) interesting connection with a circle of ideas that's interesting but underexposed. Of course, I mention the justifications precisely to point out what is needed to make this a complete, detailed proof---if they were optional, I wouldn't have bothered to mention them in the first place. – Travis Willse Jul 04 '15 at 07:34
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There are possible singularities at the origin and at infinity; by definition the integral is the limit of $\int_\delta^A$ as $\delta\to0$ and $A\to\infty$. You have $$\begin{eqnarray*} \int_\delta^A\frac{f(t)}{t}-\int_\delta^A\frac{f(2t)}{t}&=\int_\delta^A\frac{f(t)}{t}-\int_{2\delta}^{2A}\frac{f(t)}t \\&=\int_{\delta}^{2\delta}\frac{f(t)}t-\int_{A}^{2A}\frac{f(t)}t \\&=\int_1^2\frac{f(\delta t)}{t}-\int_1^2\frac{f(At)}{t} \\&\to(f(0)-L)\log(2), \end{eqnarray*}$$ by uniform convergence on $[1,2]$.

David C. Ullrich
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