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  "body": "\n## Example\n\nLet $\\mu$ be a [[probability measure|059TJN0NZ1]] on $\\R$, for example the law of a real random variable $X$.\nIts [[two-sided Laplace transform|6J2AGD0BYY]] is\n$$\\mathcal{B}\\mu(s) = \\int_{\\R} e^{-sx} \\, d\\mu(x) = E\\left[e^{-sX}\\right],$$\nfor the complex $s$ at which the integral converges absolutely.\nSince $\\abs{e^{-sx}} = e^{-x \\operatorname{Re} s}$, convergence depends only on $\\operatorname{Re} s$, and it always holds on the line $\\operatorname{Re} s = 0$, where the integrand has modulus $1$.\nFor real $t$, the [[moment-generating function|0WQVXP7PDB]] is $M_X(t) = \\mathcal{B}\\mu(-t)$.\n\nTake $\\mu$ the standard [[normal distribution|05AQE0QR0D]], with density $(2\\pi)^{-1/2} e^{-x^2/2}$.\nCompleting the square gives $-sx - x^2/2 = -(x + s)^2/2 + s^2/2$.\nSo\n$$\\mathcal{B}\\mu(s) = e^{s^2/2} \\cdot \\frac{1}{\\sqrt{2\\pi}} \\int_{\\R} e^{-(x+s)^2/2} \\, dx = e^{s^2/2}$$\nfor every $s \\in \\C$, where the last integral equals $\\sqrt{2\\pi}$ by a shift of the line of integration.\nHere the strip of convergence is the whole plane.\n\nTake $\\mu$ the [[exponential distribution|0E99F51GVB]] with rate $\\lambda \u003e 0$, with density $\\lambda e^{-\\lambda x}$ on $[0, \\infty)$.\nThen $\\int_0^\\infty e^{-sx} \\lambda e^{-\\lambda x} \\, dx = \\lambda / (\\lambda + s)$ when $\\operatorname{Re} s \u003e -\\lambda$, and the integral diverges when $\\operatorname{Re} s \\le -\\lambda$.\nHere the strip of convergence is the half-plane $\\operatorname{Re} s \u003e -\\lambda$.\n\n## Explanations\n\n### L1 human\n\nA probability law can be turned into a function of a complex variable by averaging the exponential $e^{-sX}$.\nThis is the moment-generating function read with the opposite sign and allowed to take complex arguments.\nFor the bell curve the answer is the very simple function $e^{s^2/2}$.\nFor the exponential law the answer $\\lambda / (\\lambda + s)$ only makes sense to the right of the line $\\operatorname{Re} s = -\\lambda$.\nThe reason is that the tail of this law decays only like $e^{-\\lambda x}$.\n",
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