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  "body": "\n## Statement\n\nLet $f \\colon [0, \\infty) \\to \\R$ be [[continuous|0P27W9RW0G]] and assume that the limit $f(\\infty) = \\lim_{x \\to \\infty} f(x)$ exists in $\\R$.\nLet $a, b \u003e 0$.\nThen the [[improper integral|0VCTBJMEJF]] below converges and\n$$\\int_0^\\infty \\frac{f(ax) - f(bx)}{x}\\, dx = \\bigl(f(\\infty) - f(0)\\bigr) \\ln \\frac{a}{b},$$\nwhere $\\ln$ is the [[natural logarithm|0044CPHJBV]].\n\n## Remarks\n\nFor $0 \u003c \\varepsilon \u003c R$, the substitutions $t = ax$ and $t = bx$ give\n$$\\int_\\varepsilon^R \\frac{f(ax) - f(bx)}{x}\\, dx = \\int_{a\\varepsilon}^{b\\varepsilon} \\frac{f(t)}{t}\\, dt - \\int_{aR}^{bR} \\frac{f(t)}{t}\\, dt.$$\nAs $\\varepsilon \\to 0$ the first integral tends to $f(0) \\ln(b/a)$ by continuity at $0$, and as $R \\to \\infty$ the second tends to $f(\\infty) \\ln(b/a)$.\nExample: with $f(x) = e^{-x}$ one gets $\\int_0^\\infty \\frac{e^{-ax} - e^{-bx}}{x}\\, dx = \\ln \\frac{b}{a}$.\nThe limit at infinity matters: for $f(x) = \\sin x$ the right side has no meaning, although the integral still converges.\n\n## Explanations\n\n### L1 human\n\nThe formula evaluates a family of integrals that look hard but depend only on the values of $f$ at $0$ and at infinity.\nComparing $f$ at two different scales $ax$ and $bx$ leaves only the behaviour at the two ends, weighted by $\\ln(a/b)$.\n\n### L2 human\n\nFrullani integrals appear in the computation of integral representations of logarithms, such as $\\ln(b/a) = \\int_0^\\infty \\frac{e^{-ax} - e^{-bx}}{x}\\, dx$.\nThe proof is a change of variables in an [[improper integral|0VCTBJMEJF]] together with continuity at the two ends.\nVariants replace continuity by local integrability with suitable mean values of $f$ near $0$ and near infinity.\n",
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