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  "body": "\n## Example\n\nLet $\\operatorname{erf}(x) = \\frac{2}{\\sqrt{\\pi}} \\int_0^x e^{-u^2}\\, du$ be the [[error function|1EG3DWPNZN]].\nThe [[Laplace transform|05XT5AVJBK]] of $f(t) = \\operatorname{erf}(\\sqrt{t})$ is\n$$\\mathcal{L}\\{\\operatorname{erf}(\\sqrt{t})\\}(s) = \\frac{1}{s \\sqrt{s + 1}} \\quad (s \u003e 0).$$\nCheck: $f(0) = 0$ and, for $t \u003e 0$, $f'(t) = \\frac{2}{\\sqrt{\\pi}} e^{-t} \\cdot \\frac{1}{2\\sqrt{t}} = \\frac{e^{-t}}{\\sqrt{\\pi t}}$.\nBy the [[transform of a power|RDFYHS2Q55]] with $\\alpha = -1/2$ and the shift $s \\mapsto s + 1$ caused by the factor $e^{-t}$,\n$$\\mathcal{L}\\{f'\\}(s) = \\frac{1}{\\sqrt{\\pi}} \\cdot \\frac{\\Gamma(1/2)}{\\sqrt{s + 1}} = \\frac{1}{\\sqrt{s + 1}},$$\nusing $\\Gamma(1/2) = \\sqrt{\\pi}$ for the [[gamma function|0V3FZMVB4K]].\nSince $f(t) = \\int_0^t f'(u)\\, du$, its transform is that of $f'$ divided by $s$, which gives $1/(s\\sqrt{s + 1})$.\nThe formula extends to $\\operatorname{Re} s \u003e 0$ with the principal square root.\n\n## Explanations\n\n### L1 human\n\nThe error function measures the area under the bell curve, and it appears in heat flow and probability.\nIts transform, with argument $\\sqrt{t}$, is a simple algebraic expression in $s$.\nThis is why Laplace transforms solve heat conduction problems: the hard function erf on the time side becomes a square root on the transform side.\n",
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