Frullani integral
/ froo-LAH-nee /
The Frullani integral is a small marvel: a whole class of definite integrals over [0, infinity] that all collapse to the same simple logarithmic answer, no matter what well-behaved function you feed in. It handles integrals of the form [f(ax) - f(bx)] / x dx, where the same function f is sampled at two different scalings a and b of the variable. The astonishing thing is that the answer barely depends on f at all — only on its endpoint behavior.
When f is continuous on [0, infinity) with a finite limit at 0 and a finite limit at infinity, the integral from 0 to infinity of [f(ax) - f(bx)] / x dx equals [f(0) - f(infinity)] times ln(b/a). You can feel why: write the difference as an integral of f' along x from ax to bx and swap the order of integration, and the 1/x weight turns the gap between the two scalings into the logarithm ln(b/a), while the values of f at the two ends supply the prefactor. The interior shape of f washes out; only the two boundary values and the ratio of scales matter.
Frullani integrals are a favorite of integral tables and competition problems, and they appear when a physical quantity is built from the difference of the same response at two frequencies or two length scales. They are also a clean illustration of a recurring theme in advanced integration — that a global structural feature (here, the difference-of-scalings shape) can determine an integral with almost no detailed computation. The hypotheses matter, though: if f has no finite limit at infinity, the basic formula must be adjusted or may fail outright.
integral from 0 to infinity of (e^(-ax) - e^(-bx)) / x dx, with f(x) = e^(-x), f(0) = 1, f(infinity) = 0, equals (1 - 0) * ln(b/a) = ln(b/a).
Only the boundary values of f and the ratio b/a survive — the function's middle is irrelevant.
The clean formula needs both endpoint limits to exist and be finite. If f(infinity) does not exist, you cannot just plug in zero — the standard result no longer applies and the integral may even diverge.