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  "body": "\n## Statement\n\nLet $f, g : [0, \\infty) \\to \\mathbb{C}$ be piecewise continuous, and let $M, a$ be real constants with $|f(t)| \\le M e^{a t}$ and $|g(t)| \\le M e^{a t}$ for all $t \\ge 0$.\nLet the convolution $f * g : [0, \\infty) \\to \\mathbb{C}$ be\n$$(f * g)(t) = \\int_0^t f(\\tau)\\, g(t - \\tau)\\, d\\tau.$$\nLet $F$ and $G$ be the [[Laplace transforms|05XT5AVJBK]] of $f$ and $g$.\nThen for each $s \\in \\mathbb{C}$ with $\\operatorname{Re} s \u003e a$, the Laplace transform of $f * g$ converges at $s$, and $\\mathcal{L}\\{f * g\\}(s) = F(s)\\, G(s)$.\n\n## Remarks\n\nThe proof writes $F(s) G(s)$ as a double integral over the quadrant $\\tau, u \\ge 0$, and changes variables to $t = \\tau + u$. The absolute convergence for $\\operatorname{Re} s \u003e a$ justifies the change of the order of integration.\nThe convolution is commutative: $f * g = g * f$.\nExample: $\\mathcal{L}^{-1}\\left\\{\\frac{1}{s^2 (s - 1)}\\right\\}$ is the convolution of $t$ and $e^t$, that is $e^t - 1 - t$.\nIn [[solving initial value problems|0K3HFZJK5Q]], the solution with zero initial values is the convolution of the forcing term with the impulse response.\n\n## Explanations\n\n### L1 human\n\nMultiplying two transforms corresponds to a special kind of product of the functions, the convolution.\nThe convolution mixes the past values of one function with the other, as a memory of earlier inputs.\n",
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        "locator": "Section 6.3, Theorem 6.3.1",
        "note": "J. Lebl, Notes on Diffy Qs. The theorem states the result for functions of exponential order.",
        "url": "https://www.jirka.org/diffyqs/html/convolution_section.html"
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        "locator": "Properties and theorems",
        "url": "https://en.wikipedia.org/wiki/Laplace_transform"
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