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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 $c \u003e 0$ be real, and let $s$ be a complex number such that the Laplace integral of $f$ converges at $s / c$.\nThen the Laplace integral of $t \\mapsto f(c t)$ converges at $s$, and\n$$\\mathcal{L}\\{f(c t)\\}(s) = \\frac{1}{c} F\\!\\left(\\frac{s}{c}\\right).$$\n\n## Remarks\n\nThe substitution $\\tau = c t$ turns $\\int_0^\\infty f(c t) e^{-s t} \\, dt$ into $\\frac{1}{c} \\int_0^\\infty f(\\tau) e^{-(s / c) \\tau} \\, d\\tau$.\nExample: from $\\mathcal{L}\\{\\sin t\\}(s) = 1 / (s^2 + 1)$ one gets $\\mathcal{L}\\{\\sin \\omega t\\}(s) = \\frac{1}{\\omega} \\cdot \\frac{1}{(s / \\omega)^2 + 1} = \\frac{\\omega}{s^2 + \\omega^2}$ for $\\omega \u003e 0$.\nThe condition $c \u003e 0$ is needed: for $c \u003c 0$ the function $f(c t)$ uses values of $f$ at negative times.\n\n## Explanations\n\n### L1 human\n\nSpeeding up time by a factor $c$ stretches the transform by the same factor and scales its height by $1/c$.\nFast signals have transforms that are spread out in $s$.\n\n### L2 human\n\nThe rule lets one derive a whole family of transforms from one transform, for example all $\\sin \\omega t$ from $\\sin t$.\nIt also explains how the abscissa of convergence changes: it is multiplied by $c$.\n",
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