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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 $f$.\nAssume that the one-sided [[limit|0TXTSEMHR5]] $f(0^+) = \\lim_{t \\to 0^+} f(t)$ exists and is finite.\nThen\n$$\\lim_{s \\to +\\infty} s F(s) = f(0^+),$$\nwhere $s$ tends to $+\\infty$ through real values.\n\n## Remarks\n\nSince $s \\int_0^\\infty e^{-s t} \\, dt = 1$, the difference $s F(s) - f(0^+)$ is $s \\int_0^\\infty e^{-s t} (f(t) - f(0^+)) \\, dt$.\nThe weight $s e^{-s t}$ has total mass $1$ and concentrates near $t = 0$ as $s \\to \\infty$.\nOn $[0, \\delta]$ the integrand is small by the existence of $f(0^+)$; on $[\\delta, \\infty)$ the contribution tends to $0$ by the convergence at $s_0$.\nExample: for $f(t) = \\cos t$, $s F(s) = s^2 / (s^2 + 1) \\to 1 = f(0)$.\nThe finiteness of $f(0^+)$ matters: for $f(t) = t^{-1/2}$ one has $F(s) = \\sqrt{\\pi / s}$, so $s F(s) = \\sqrt{\\pi s} \\to +\\infty$.\n\n## Explanations\n\n### L1 human\n\nThe behaviour of the transform for very large $s$ tells the starting value of the function.\nOne can read $f(0^+)$ from $F$ without inverting the transform.\n\n### L2 human\n\nIn control and circuit theory one uses it to check the value of a response just after a switch is closed.\nIt is the simplest Abelian theorem for the Laplace transform: behaviour of $f$ near $0$ gives behaviour of $F$ near $\\infty$.\nIts companion is the [[final value theorem|BDHADVMJ8G]], which links $f$ near $\\infty$ with $F$ near $0$.\n",
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