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  "body": "\n## Definition\n\nLet $T$ be a distribution on $\\R$ with support in $[0, \\infty)$.\nLet $\\sigma_0 \\in \\R$ be such that $e^{-\\sigma_0 t} T$ is a [[tempered distribution|AES2T5JPFP]].\nFix a smooth function $\\chi \\colon \\R \\to [0, 1]$ with $\\chi = 1$ on $[-1/2, \\infty)$ and $\\chi = 0$ on $(-\\infty, -1]$.\nThe *Laplace transform* of $T$ is the function defined for $\\operatorname{Re} s \u003e \\sigma_0$ by\n$$(\\mathcal{L} T)(s) = \\left\\langle e^{-\\sigma_0 t} T, \\ \\chi(t) \\, e^{-(s - \\sigma_0) t} \\right\\rangle.$$\n\n## Remarks\n\nFor $\\operatorname{Re} s \u003e \\sigma_0$, the function $\\chi(t) e^{-(s - \\sigma_0) t}$ lies in the [[Schwartz space|CG9JEATW3Q]], so the pairing makes sense.\nThe value does not depend on the choice of $\\chi$ or of $\\sigma_0$, because $T$ is supported in $[0, \\infty)$; formally it is $\\langle T, e^{-st} \\rangle$.\n$\\mathcal{L} T$ is holomorphic on $\\operatorname{Re} s \u003e \\sigma_0$.\nFor a locally integrable function $f$ of exponential order vanishing on $(-\\infty, 0)$, it agrees with the ordinary [[Laplace transform|0V49JEZW06]] $\\int_0^\\infty e^{-st} f(t) \\, dt$.\nFor all such $T$, $\\mathcal{L}(T') = s \\, \\mathcal{L} T$, and the transform of a convolution is the product of the transforms.\n\nExample: the [[Dirac delta|0KPHENZ5S7]] $\\delta$ has support $\\set{0}$ and is tempered, so $\\sigma_0 = 0$ works, and $\\mathcal{L}\\delta(s) = \\chi(0) e^{0} = 1$.\nLikewise $\\mathcal{L}\\delta'(s) = -\\frac{d}{dt}\\left(\\chi(t) e^{-st}\\right)\\big|_{t=0} = s$.\nWith the [[Heaviside step function|09C4JP7Y0M]] $H$, $\\mathcal{L} H = 1/s$ and $H' = \\delta$, consistent with $\\mathcal{L}(H') = s \\cdot \\frac{1}{s} = 1$.\n\nNon-example: $T = H(t) e^{t^2}$ is a distribution supported in $[0, \\infty)$, but the positive function $e^{-\\sigma_0 t} e^{t^2}$ grows faster than any polynomial. Hence $e^{-\\sigma_0 t} T$ is tempered for no $\\sigma_0$, and $T$ has no Laplace transform.\n",
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