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  "body": "\n## Statement\n\nLet $f \\colon [0, \\infty) \\to \\C$ be integrable on each interval $[0, T]$.\nAssume that the Laplace integral of $f$ converges at some real number $s_0$, and let $F$ be the [[Laplace transform|05XT5AVJBK]] of $f$.\nThen for every $t \u003e 0$ at which $f$ is continuous,\n$$f(t) = \\lim_{k \\to \\infty} \\frac{(-1)^k}{k!} \\left(\\frac{k}{t}\\right)^{k+1} F^{(k)}\\!\\left(\\frac{k}{t}\\right),$$\nwhere $F^{(k)}$ is the $k$-th derivative of $F$ on the real half-line $s \u003e s_0$.\n\n## Remarks\n\nBy [[differentiation in the transform variable|VEPJN1NK3C]], $\\frac{(-1)^k}{k!} \\left(\\frac{k}{t}\\right)^{k+1} F^{(k)}\\!\\left(\\frac{k}{t}\\right) = \\int_0^\\infty f(u) \\, \\frac{1}{k!} \\left(\\frac{k}{t}\\right)^{k+1} u^k e^{-k u / t} \\, du$.\nThe kernel in this integral is a probability density in $u$ (a gamma density) with mean $t (k+1)/k$ and variance $t^2 (k+1)/k^2$.\nAs $k \\to \\infty$ it concentrates at $u = t$, so the integral tends to $f(t)$ at points of continuity.\nExample: for $f(t) = e^{-t}$, $F(s) = 1/(s + 1)$ and $\\frac{(-1)^k}{k!} F^{(k)}(s) = (s+1)^{-k-1}$, so the expression is $\\left(\\frac{k/t}{k/t + 1}\\right)^{k+1} = \\left(1 + \\frac{t}{k}\\right)^{-k-1} \\to e^{-t}$.\n\n## Explanations\n\n### L1 human\n\nThis formula recovers a function from its Laplace transform using only values of the transform and its derivatives at real points.\nNo complex integration is needed.\nThe price is that one needs derivatives of very high order.\n\n### L2 human\n\nThe formula matters in theory, for example in the theory of semigroups, where it gives the exponential formula for $T(t)$.\nNumerically it is poor: high derivatives amplify small errors, which reflects the [[ill-posedness of real Laplace inversion|TRTN840ZB0]].\nThe method of proof, a kernel that concentrates at one point, is the same as in many approximation theorems.\n",
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