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  "body": "\n## Statement\n\nLet $f \\colon [0, \\infty) \\to \\C$ be measurable, and assume that $\\int_0^\\infty \\abs{f(t)} e^{-\\sigma_0 t}\\, dt \u003c \\infty$ for some real $\\sigma_0$.\nLet $F(s) = \\int_0^\\infty f(t) e^{-st}\\, dt$ be its [[Laplace transform|05XT5AVJBK]], defined for $\\operatorname{Re} s \\ge \\sigma_0$.\nThen\n$$\\sup_{\\operatorname{Re} s = \\sigma} \\abs{F(s)} \\to 0 \\quad (\\sigma \\to +\\infty).$$\nThat is, $F(s) \\to 0$ as $\\operatorname{Re} s \\to +\\infty$, uniformly in $\\operatorname{Im} s$.\n\n## Remarks\n\nFor $\\operatorname{Re} s = \\sigma \\ge \\sigma_0$, $\\abs{F(s)} \\le \\int_0^\\infty \\abs{f(t)} e^{-\\sigma t}\\, dt$.\nThe integrand decreases to $0$ for each $t \u003e 0$ as $\\sigma \\to \\infty$ and is bounded by $\\abs{f(t)} e^{-\\sigma_0 t}$, so the right side tends to $0$ by [[dominated convergence|BYNV84PF86]].\nThe hypothesis says that $\\sigma_0$ lies to the right of the [[abscissa of absolute convergence|YCSH2WK69M]], or on it.\nExample: $\\mathcal{L}\\{e^{at}\\}(s) = 1/(s - a)$ tends to $0$ as $\\operatorname{Re} s \\to \\infty$.\nConsequence: functions such as $1$, $s$, $e^{s}$ or a rational function $P/Q$ with $P \\ne 0$ and $\\deg P \\ge \\deg Q$ are not Laplace transforms of any such $f$.\nThe constant $1$ is the transform of the Dirac delta, which is a measure and not a function.\n\n## Explanations\n\n### L1 human\n\nAs $\\operatorname{Re} s$ grows, the weight $e^{-st}$ crushes everything except what happens at times very close to $0$.\nAn ordinary function has no mass concentrated at a single instant, so the transform dies out.\nThis gives a quick test: if a candidate $F(s)$ does not go to zero, it cannot be the transform of a function.\n\n### L2 human\n\nThe test is used in partial fraction inversion: only proper rational functions, with numerator degree less than denominator degree, are inverted to ordinary functions.\nImproper parts correspond to the Dirac delta and its derivatives, which belong to the Laplace transform of distributions.\nThe initial value theorem refines the statement: if $f(t) \\to c$ as $t \\to 0^+$, then $sF(s) \\to c$, so $F(s)$ behaves like $c/s$.\n",
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