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  "body": "\n## Statement\n\nLet $f, g \\colon [0, \\infty) \\to \\C$ be integrable on each interval $[0, T]$ with $T \u003e 0$, and let $\\alpha, \\beta \\in \\C$.\nLet $s \\in \\C$ be such that the [[Laplace integrals|05XT5AVJBK]] $F(s) = \\int_0^\\infty f(t) e^{-st}\\, dt$ and $G(s) = \\int_0^\\infty g(t) e^{-st}\\, dt$ both converge.\nThen the Laplace integral of $\\alpha f + \\beta g$ converges at $s$, and\n$$\\mathcal{L}\\{\\alpha f + \\beta g\\}(s) = \\alpha F(s) + \\beta G(s).$$\n\n## Remarks\n\nBy [[linearity of the integral|080RWVVSB6]], $\\int_0^T (\\alpha f + \\beta g) e^{-st}\\, dt = \\alpha \\int_0^T f e^{-st}\\, dt + \\beta \\int_0^T g e^{-st}\\, dt$ for each $T$, and the limit of a sum is the sum of the limits.\nSo the identity holds on the half-plane $\\operatorname{Re} s \u003e \\max(\\sigma_c(f), \\sigma_c(g))$, where $\\sigma_c$ is the [[abscissa of convergence|FTHFSJ30WJ]].\nExample: $\\cosh t = \\frac{1}{2}(e^t + e^{-t})$ has transform $\\frac{1}{2}\\left(\\frac{1}{s - 1} + \\frac{1}{s + 1}\\right) = \\frac{s}{s^2 - 1}$ for $\\operatorname{Re} s \u003e 1$.\nThe transform of $\\alpha f + \\beta g$ may converge on a larger half-plane than those of $f$ and $g$: for $f(t) = 1 + e^t$ and $g(t) = -e^t$, both have abscissa $1$, while $f + g = 1$ has abscissa $0$.\n\n## Explanations\n\n### L1 human\n\nThe transform of a sum is the sum of the transforms, and constants can be pulled out.\nThis is what lets one use a table: split a function into simple pieces, transform each piece, and add.\nIt also works backwards, when a transform is split into partial fractions and each fraction is inverted separately.\n\n### L2 human\n\nLinearity is the reason the Laplace transform fits linear differential equations with constant coefficients: together with the rule for derivatives, it turns the equation into an algebraic equation for $F$.\nIn operator language, $\\mathcal{L}$ is a linear map from a space of functions (for instance those of exponential order) to a space of holomorphic functions on half-planes.\nCombined with Lerch's uniqueness theorem, linearity also makes the inverse transform linear.\n",
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  "name": "Linearity of the Laplace transform",
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