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  "body": "\n## Definition\n\nLet $f \\colon [0, \\infty) \\to \\C$ be continuous, and assume that $f$ is of [[exponential order|TKZAV3KR86]]: there are constants $M \u003e 0$ and $c \\ge 0$ with $\\abs{f(t)} \\le M e^{ct}$ for all $t \\ge 0$.\nThe *Sumudu transform* of $f$ is the function $G$ given by\n$$G(u) = \\int_0^\\infty f(ut)\\, e^{-t}\\, dt$$\nfor $0 \\le u \u003c 1/c$, and for all $u \\ge 0$ when $c = 0$.\n\n## Remarks\n\nThe integral converges absolutely, since $\\abs{f(ut)} e^{-t} \\le M e^{-(1 - cu) t}$ and $cu \u003c 1$.\nAt $u = 0$ one gets $G(0) = f(0)$.\nFor $u \u003e 0$ the substitution $s = ut$ gives $G(u) = \\frac{1}{u} \\int_0^\\infty f(s) e^{-s/u}\\, ds = \\frac{1}{u} F\\big(\\frac{1}{u}\\big)$, where $F$ is the [[Laplace transform|0V49JEZW06]] of $f$.\nSo every rule for the Laplace transform has a Sumudu counterpart; for example, the transform of $f'$ is $\\frac{G(u) - f(0)}{u}$ when $f'$ is continuous and of exponential order.\n\nExample: $f(t) = e^{at}$ with $a \\in \\R$.\nThen $G(u) = \\int_0^\\infty e^{-(1 - au) t}\\, dt = \\frac{1}{1 - au}$ for $0 \\le u$ and $au \u003c 1$.\nAlso $f(t) = t^n$ gives $G(u) = u^n \\int_0^\\infty t^n e^{-t}\\, dt = n!\\, u^n$.\n\nNon-example: $f(t) = e^{t^2}$ is not of exponential order, and $\\int_0^\\infty e^{u^2 t^2 - t}\\, dt$ diverges for every $u \u003e 0$.\n",
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