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  "body": "\n## Statement\n\nLet $f \\colon [0, \\infty) \\to \\C$ be piecewise continuous and of [[exponential order|TKZAV3KR86]] $a$, that is, $\\abs{f(t)} \\le M e^{a t}$ for all $t \\ge 0$.\nAssume that $\\int_0^1 \\abs{f(t)} / t \\, dt \u003c \\infty$, which holds for example when $\\lim_{t \\to 0^+} f(t)/t$ exists.\nLet $F$ be the [[Laplace transform|05XT5AVJBK]] of $f$.\nThen for every real $s \u003e a$,\n$$\\mathcal{L}\\left\\{\\frac{f(t)}{t}\\right\\}(s) = \\int_s^\\infty F(\\sigma) \\, d\\sigma,$$\nwhere the integral on the right is taken along the real axis and converges.\n\n## Remarks\n\nWrite $\\int_s^\\infty F(\\sigma) \\, d\\sigma = \\int_s^\\infty \\int_0^\\infty f(t) e^{-\\sigma t} \\, dt \\, d\\sigma$ and swap the integrals with [[Fubini's theorem|0R0T1VWH5X]].\nThe swap is allowed because $\\int_0^\\infty \\abs{f(t)} e^{-s t} / t \\, dt \u003c \\infty$ under the hypotheses.\nThe inner integral is $\\int_s^\\infty e^{-\\sigma t} \\, d\\sigma = e^{-s t} / t$.\nExample: with $f(t) = \\sin t$, one gets $\\mathcal{L}\\{\\sin t / t\\}(s) = \\int_s^\\infty d\\sigma / (\\sigma^2 + 1) = \\pi/2 - \\arctan s$.\nThe hypothesis at $0$ is needed: for $f = 1$, the function $1/t$ has no Laplace transform, and $\\int_s^\\infty d\\sigma / \\sigma$ diverges.\n\n## Explanations\n\n### L1 human\n\nDividing a function by $t$ corresponds to integrating its transform from $s$ out to infinity.\nIt is the reverse of the rule that multiplying by $t$ corresponds to differentiating the transform.\n\n### L2 human\n\nThe rule gives transforms of functions such as $\\sin t / t$ and $(1 - e^{-t})/t$ that are hard to compute directly.\nLetting $s \\to 0^+$ in the example, with a limit argument, recovers the Dirichlet integral $\\int_0^\\infty \\sin t / t \\, dt = \\pi / 2$.\nCompare [[differentiation in the transform variable|VEPJN1NK3C]].\n",
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  "name": "Integration in the transform variable",
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