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  "body": "\n## Statement\n\nLet $f \\colon [0, \\infty) \\to \\C$ be integrable on each interval $[0, T]$, and let $F$ be its [[Laplace transform|05XT5AVJBK]].\nLet $a$ be a real number such that $\\int_0^\\infty \\abs{f(t)} e^{-\\sigma t} \\, dt \u003c \\infty$ for every $\\sigma \u003e a$.\nThen $F$ is [[holomorphic|0DG0Z72CQD]] on the half-plane $\\operatorname{Re} s \u003e a$, and for every integer $n \\ge 1$ and every $s$ with $\\operatorname{Re} s \u003e a$,\n$$F^{(n)}(s) = \\int_0^\\infty (-t)^n f(t) \\, e^{-s t} \\, dt = \\mathcal{L}\\{(-t)^n f(t)\\}(s).$$\n\n## Remarks\n\nThe hypothesis says that the Laplace integral converges absolutely to the right of $a$; the least such $a$ is the [[abscissa of absolute convergence|YCSH2WK69M]].\nFor $\\operatorname{Re} s \\ge \\sigma \u003e \\sigma' \u003e a$, one has $t^n e^{-\\sigma t} \\le C e^{-\\sigma' t}$, so $t^n \\abs{f(t)} e^{-\\sigma t}$ is integrable.\nThis domination justifies differentiation under the integral sign.\nExample: from $\\mathcal{L}\\{e^{b t}\\}(s) = 1/(s - b)$ one gets $\\mathcal{L}\\{t e^{b t}\\}(s) = 1/(s - b)^2$.\nFrom $\\mathcal{L}\\{\\sin \\omega t\\}(s) = \\omega/(s^2 + \\omega^2)$ one gets $\\mathcal{L}\\{t \\sin \\omega t\\}(s) = 2 \\omega s / (s^2 + \\omega^2)^2$.\n\n## Explanations\n\n### L1 human\n\nMultiplying a function by $-t$ corresponds to differentiating its transform.\nSo the transform of $t \\cdot f(t)$ is minus the slope of the transform of $f$.\n\n### L2 human\n\nThis rule produces the transforms of $t^n e^{b t}$, $t \\cos \\omega t$ and similar functions, which appear as resonant solutions of linear equations.\nIt also shows that a Laplace transform is infinitely differentiable, and in fact holomorphic, in its half-plane of convergence.\nThe reverse operation is [[integration in the transform variable|GXX2WPVRJ2]].\n",
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