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  "body": "\n## Definition\n\nLet $f, g \\colon [0, \\infty) \\to \\C$ be locally integrable, that is, Lebesgue integrable on $[0, T]$ for every $T \u003e 0$.\nThe *convolution* of $f$ and $g$ is the function $f * g$ on $[0, \\infty)$ given by\n$$(f * g)(t) = \\int_0^t f(\\tau) \\, g(t - \\tau) \\, d\\tau$$\nat each $t \\ge 0$ at which the [[Lebesgue integral|DWJF7ABPR1]] converges absolutely.\n\n## Remarks\n\nThe integral converges for almost every $t$, and $f * g$ is again locally integrable.\nIf $f$ and $g$ are continuous, then $f * g$ is defined everywhere and continuous, with $(f * g)(0) = 0$.\nExtending $f$ and $g$ by $0$ on $(-\\infty, 0)$ turns this into the [[convolution on the line|ERP5737VZ1]].\n\nExample: for integers $m, n \\ge 0$, the functions $t^m$ and $t^n$ give\n$$(t^m * t^n)(t) = \\int_0^t \\tau^m (t - \\tau)^n \\, d\\tau = \\frac{m! \\, n!}{(m + n + 1)!} \\, t^{m + n + 1},$$\nby the Beta integral; for $m = n = 0$ this is $1 * 1 = t$.\n\nNon-example: $f(t) = 1/t$ for $t \u003e 0$ is not integrable on $[0, 1]$, so it is not locally integrable on $[0, \\infty)$.\nIndeed $(f * 1)(t) = \\int_0^t d\\tau / \\tau$ diverges for every $t \u003e 0$.\n",
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