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  "body": "\n## Statement\n\nLet $f : [0, \\infty) \\to \\mathbb{C}$ be continuously differentiable, and let $M, a$ be real constants with $|f(t)| \\le M e^{a t}$ for all $t \\ge 0$.\nLet $F$ be the [[Laplace transform|05XT5AVJBK]] of $f$.\nThen for each $s \\in \\mathbb{C}$ with $\\operatorname{Re} s \u003e a$, the Laplace transform of $f'$ converges at $s$ and\n$$\\mathcal{L}\\{f'\\}(s) = s F(s) - f(0).$$\n\n## Remarks\n\nProof: integration by parts gives $\\int_0^T f'(t) e^{-st}\\, dt = f(T) e^{-sT} - f(0) + s \\int_0^T f(t) e^{-st}\\, dt$. The term $f(T) e^{-sT}$ tends to $0$ for $\\operatorname{Re} s \u003e a$.\nApplied twice, under the same hypotheses on $f$ and $f'$: $\\mathcal{L}\\{f''\\}(s) = s^2 F(s) - s f(0) - f'(0)$.\nThis rule is the basis of [[solving initial value problems with the Laplace transform|0K3HFZJK5Q]]. The initial values enter the transformed equation directly.\n\n## Explanations\n\n### L1 human\n\nDifferentiation in time becomes multiplication by $s$ after the transform, with a correction by the initial value.\nSo a differential equation becomes an algebraic equation for $F$.\n",
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