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  "body": "\n## Statement\n\nLet $f \\colon [0, \\infty) \\to \\C$ be piecewise continuous and of [[exponential order|TKZAV3KR86]] $a$, that is, $\\abs{f(t)} \\le M e^{a t}$ for all $t \\ge 0$ and some constants $M$ and $a$.\nLet $F$ be the [[Laplace transform|05XT5AVJBK]] of $f$, and let $g(t) = \\int_0^t f(\\tau) \\, d\\tau$.\nThen for every complex $s$ with $\\operatorname{Re} s \u003e \\max(a, 0)$,\n$$\\mathcal{L}\\{g\\}(s) = \\frac{F(s)}{s}.$$\n\n## Remarks\n\nThe function $g$ is continuous, $g(0) = 0$, and $g' = f$ at each point where $f$ is continuous.\nThe bound on $f$ gives a bound $\\abs{g(t)} \\le M' e^{b t}$ for every $b \u003e \\max(a, 0)$, so $\\mathcal{L}\\{g\\}$ exists on $\\operatorname{Re} s \u003e \\max(a, 0)$.\nThe [[rule for derivatives|0K090GVHVN]] then gives $F(s) = s \\, \\mathcal{L}\\{g\\}(s) - g(0) = s \\, \\mathcal{L}\\{g\\}(s)$.\nExample: with $f = 1$, one gets $\\mathcal{L}\\{t\\}(s) = 1 / s^2$; repeating gives $\\mathcal{L}\\{t^n / n!\\}(s) = 1 / s^{n + 1}$.\n\n## Explanations\n\n### L1 human\n\nIntegrating in time corresponds to dividing by $s$ in the transform.\nIt is the opposite of differentiation, which corresponds to multiplying by $s$.\n\n### L2 human\n\nRead backwards, the rule inverts transforms of the form $F(s) / s$: the answer is the running integral of $\\mathcal{L}^{-1}\\{F\\}$.\nFor example $1 / (s (s^2 + 1))$ inverts to $\\int_0^t \\sin \\tau \\, d\\tau = 1 - \\cos t$.\nIt is a special case of the [[convolution theorem|0JNMJEDD93]], since $g = f * 1$ and $\\mathcal{L}\\{1\\} = 1/s$.\n",
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