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  "body": "\n## Statement\n\nLet $f \\colon [0, \\infty) \\to \\C$ be integrable on each interval $[0, T]$, and let $\\gamma$ be a real number with $\\int_0^\\infty \\abs{f(t)} e^{-\\gamma t} \\, dt \u003c \\infty$.\nLet $F$ be the [[Laplace transform|05XT5AVJBK]] of $f$.\nLet $t \u003e 0$, and assume that $f$ has [[bounded variation|F34S6P61SW]] on some interval around $t$.\nThen the [[Bromwich integral|9WRE8KEJJX]] of $F$ along the line $\\operatorname{Re} s = \\gamma$, that is, the symmetric limit below, exists at $t$, and\n$$\\frac{1}{2 \\pi i} \\lim_{T \\to \\infty} \\int_{\\gamma - i T}^{\\gamma + i T} e^{s t} F(s) \\, ds = \\frac{f(t^+) + f(t^-)}{2}.$$\nIn particular the integral equals $f(t)$ at each $t \u003e 0$ where $f$ is continuous.\nFor every $t \u003c 0$ the same integral exists and equals $0$.\n\n## Remarks\n\nPut $g(\\tau) = f(\\tau) e^{-\\gamma \\tau}$ for $\\tau \\ge 0$ and $g(\\tau) = 0$ for $\\tau \u003c 0$; then $g$ is integrable on $\\R$.\nOn the line $s = \\gamma + i y$, one has $F(\\gamma + i y) = \\hat{g}(y) = \\int_\\R g(\\tau) e^{-i y \\tau} \\, d\\tau$, and the Bromwich integral is $e^{\\gamma t} \\cdot \\frac{1}{2\\pi} \\lim_{T \\to \\infty} \\int_{-T}^{T} \\hat{g}(y) e^{i y t} \\, dy$.\nSo the statement is the [[Fourier inversion theorem|T4Z8YQZKZV]] in its Dirichlet–Jordan form, applied to $g$.\nExample: for $f(t) = e^{b t}$ with $b$ real and $\\gamma \u003e b$, $F(s) = 1/(s - b)$, and the residue of $e^{s t}/(s - b)$ at $b$ gives back $e^{b t}$ for $t \u003e 0$.\nAt a jump of $f$ the integral gives the midpoint of the jump, not either one-sided value.\n\n## Explanations\n\n### L1 human\n\nIf you know the Laplace transform $F$ of a function, this formula gives the function back.\nYou integrate $e^{s t} F(s)$ along a vertical line far enough to the right in the complex plane.\nAt a jump, the formula returns the average of the values on the two sides.\n\n### L2 human\n\nThe formula shows that the Laplace transform is a Fourier transform in disguise: along a vertical line it is the Fourier transform of $f(t) e^{-\\gamma t}$.\nIn practice one rarely integrates along the line directly; one deforms the contour and uses residues, as in [[residue calculus for Laplace inversion|E14W2V09HM]].\nThe same formula, after the change of variables $x = e^{-t}$, gives the inversion theorem for the Mellin transform.\n",
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