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  "body": "\n## Definition\n\nLet $f \\colon \\R \\to \\C$ be measurable.\nIts two-sided Laplace transform is $\\mathcal{B}f(s) = \\int_{\\R} f(x) e^{-sx} \\, dx$, for $s \\in \\C$ at which $x \\mapsto f(x) e^{-sx}$ is [[integrable|X7GTF3D98F]].\nLet $\\sigma \\in \\R$ be such that $x \\mapsto e^{-\\sigma x} f(x)$ is integrable.\nThe *restriction of $\\mathcal{B}f$ to the line $\\operatorname{Re} s = \\sigma$* is the function $\\xi \\mapsto \\mathcal{B}f(\\sigma + 2\\pi i \\xi)$ on $\\R$.\nFor $\\sigma = 0$, that is, for $f$ integrable, this is the restriction of $\\mathcal{B}f$ to the imaginary axis, and it equals the [[Fourier transform|DPVM34ADZV]]:\n$$\\hat f(\\xi) = \\int_{\\R} f(x) e^{-2\\pi i x \\xi} \\, dx = \\mathcal{B}f(2\\pi i \\xi), \\qquad \\xi \\in \\R.$$\n\n## Remarks\n\nIn general $\\mathcal{B}f(\\sigma + 2\\pi i \\xi) = \\widehat{f_\\sigma}(\\xi)$ with $f_\\sigma(x) = e^{-\\sigma x} f(x)$, by writing $e^{-(\\sigma + 2\\pi i \\xi) x} = e^{-\\sigma x} e^{-2\\pi i x \\xi}$.\nExample: $f(x) = e^{-\\abs{x}}$ is integrable, and $\\mathcal{B}f(s) = \\frac{1}{1 + s} + \\frac{1}{1 - s} = \\frac{2}{1 - s^2}$ for $\\abs{\\operatorname{Re} s} \u003c 1$.\nAt $s = 2\\pi i \\xi$ this gives $\\hat f(\\xi) = 2/(1 + 4\\pi^2 \\xi^2)$, the known Fourier transform of $e^{-\\abs{x}}$.\nNon-example: $f(x) = e^{x}$ for $x \\ge 0$ and $f(x) = 0$ for $x \u003c 0$ is not integrable, so the imaginary axis is not in its region of convergence.\nHere $\\mathcal{B}f(s) = 1/(s - 1)$ for $\\operatorname{Re} s \u003e 1$, and this function is not the Fourier transform of $f$, which does not exist as an integral.\nWith the convention $\\int f(x) e^{-i \\omega x} \\, dx$, the same relation reads $\\hat f(\\omega) = \\mathcal{B}f(i\\omega)$.\n",
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