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  "body": "\n## Definition\n\nLet $n \\ge 1$ and let $g \\colon \\R^n \\to \\C$ be a Lebesgue [[integrable function|X7GTF3D98F]].\nThe *inverse Fourier transform* of $g$ is the function $\\check g \\colon \\R^n \\to \\C$ given by the [[Lebesgue integral|DWJF7ABPR1]]\n$$\\check g(x) = \\int_{\\R^n} g(\\xi) \\, e^{2\\pi i \\, x \\cdot \\xi} \\, d\\xi, \\qquad x \\in \\R^n,$$\nwhere $x \\cdot \\xi = x_1 \\xi_1 + \\cdots + x_n \\xi_n$.\nIt is also written $\\mathcal{F}^{-1} g$.\n\n## Remarks\n\nThis matches the convention $\\hat f(\\xi) = \\int f(x) e^{-2\\pi i x \\cdot \\xi} \\, dx$ for the [[Fourier transform|DPVM34ADZV]].\nSo $\\check g(x) = \\hat g(-x)$, and $\\check g$ is continuous, bounded by $\\int \\abs{g}$, and tends to $0$ at infinity.\nWith the convention $\\hat f(\\xi) = \\int f(x) e^{-i x \\cdot \\xi} \\, dx$ the inverse transform carries the factor $(2\\pi)^{-n}$.\n\nExample: for $n = 1$ and $g = \\mathbf{1}_{[-1/2, 1/2]}$,\n$$\\check g(x) = \\int_{-1/2}^{1/2} e^{2\\pi i x \\xi} \\, d\\xi = \\frac{\\sin(\\pi x)}{\\pi x} \\quad (x \\ne 0), \\qquad \\check g(0) = 1.$$\n\nNon-example: the constant function $g = 1$ on $\\R$ is not integrable, and the integral $\\int_\\R e^{2\\pi i x \\xi} \\, d\\xi$ diverges for every $x$.\nIts inverse Fourier transform exists only in the sense of tempered distributions, where it is the Dirac delta at $0$.\n\nThe name is justified by the [[Fourier inversion theorem|T4Z8YQZKZV]]: if $f$ and $\\hat f$ are integrable, then $(\\hat f)^\\vee = f$ almost everywhere.\n",
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