---
tag: A58Z1DQ593
kind: statement
role: theorem
name: Frullani integral
msc: [26A42]
uses: [0P27W9RW0G, 0VCTBJMEJF, 0044CPHJBV]
about: [0VCTBJMEJF]
sources:
  - url: https://en.wikipedia.org/wiki/Frullani_integral
same_as:
  wikidata: Q1470953
  wikipedia: Frullani integral
---

## Statement

Let $f \colon [0, \infty) \to \R$ be [[continuous|0P27W9RW0G]] and assume that the limit $f(\infty) = \lim_{x \to \infty} f(x)$ exists in $\R$.
Let $a, b > 0$.
Then the [[improper integral|0VCTBJMEJF]] below converges and
$$\int_0^\infty \frac{f(ax) - f(bx)}{x}\, dx = \bigl(f(\infty) - f(0)\bigr) \ln \frac{a}{b},$$
where $\ln$ is the [[natural logarithm|0044CPHJBV]].

## Remarks

For $0 < \varepsilon < R$, the substitutions $t = ax$ and $t = bx$ give
$$\int_\varepsilon^R \frac{f(ax) - f(bx)}{x}\, dx = \int_{a\varepsilon}^{b\varepsilon} \frac{f(t)}{t}\, dt - \int_{aR}^{bR} \frac{f(t)}{t}\, dt.$$
As $\varepsilon \to 0$ the first integral tends to $f(0) \ln(b/a)$ by continuity at $0$, and as $R \to \infty$ the second tends to $f(\infty) \ln(b/a)$.
Example: with $f(x) = e^{-x}$ one gets $\int_0^\infty \frac{e^{-ax} - e^{-bx}}{x}\, dx = \ln \frac{b}{a}$.
The limit at infinity matters: for $f(x) = \sin x$ the right side has no meaning, although the integral still converges.

## Explanations

### L1 human

The formula evaluates a family of integrals that look hard but depend only on the values of $f$ at $0$ and at infinity.
Comparing $f$ at two different scales $ax$ and $bx$ leaves only the behaviour at the two ends, weighted by $\ln(a/b)$.

### L2 human

Frullani integrals appear in the computation of integral representations of logarithms, such as $\ln(b/a) = \int_0^\infty \frac{e^{-ax} - e^{-bx}}{x}\, dx$.
The proof is a change of variables in an [[improper integral|0VCTBJMEJF]] together with continuity at the two ends.
Variants replace continuity by local integrability with suitable mean values of $f$ near $0$ and near infinity.