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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 [[limit|0TXTSEMHR5]] $f(\\infty) = \\lim_{t \\to \\infty} f(t)$ exists and is finite.\nThen the Laplace integral $F(s)$ of $f$ converges for every real $s \u003e 0$, and\n$$\\lim_{s \\to 0^+} s F(s) = f(\\infty),$$\nwhere $s$ tends to $0$ through positive real values.\n\n## Remarks\n\nSince $f$ has a finite limit, it is bounded for large $t$, so $F(s)$ converges for $s \u003e 0$.\nSince $s \\int_0^\\infty e^{-s t} \\, dt = 1$, the difference $s F(s) - f(\\infty)$ equals $s \\int_0^\\infty e^{-s t} (f(t) - f(\\infty)) \\, dt$.\nAs $s \\to 0^+$ the weight $s e^{-s t}$ spreads its mass over large $t$, where $f(t) - f(\\infty)$ is small, and the part over a fixed interval $[0, T]$ tends to $0$.\nExample: for $f(t) = 1 - e^{-t}$, $s F(s) = 1 - s/(s + 1) \\to 1 = f(\\infty)$.\nThe hypothesis that $f(\\infty)$ exists is needed: for $f(t) = \\sin t$, $s F(s) = s / (s^2 + 1) \\to 0$, but $\\sin t$ has no limit.\n\n## Explanations\n\n### L1 human\n\nThe behaviour of the transform for small $s$ tells the long-run value of the function, if that value exists.\nOne can read the steady state from $F$ without inverting the transform.\n\n### L2 human\n\nIn control theory one uses it to find the steady-state error of a stable system.\nFor a rational $F$, the limit $f(\\infty)$ exists when all poles of $s F(s)$ lie in the open left half-plane; then the theorem applies.\nIf $s F(s)$ has a pole on the imaginary axis or to its right, the formula can give a wrong answer, as the example $\\sin t$ shows.\nThe converse direction, from $F$ to $f$, needs extra hypotheses and belongs to the [[Tauberian theorems|NV54MWKT6T]].\nIts companion is the [[initial value theorem|KZPVHDKV9E]].\n",
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