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  "body": "\n## Statement\n\nLet $n \\ge 1$ and let $f \\colon \\R^n \\to \\C$ be [[integrable|X7GTF3D98F]].\nLet $\\hat f(\\xi) = \\int_{\\R^n} f(x) e^{-2\\pi i x \\cdot \\xi} \\, dx$ be its [[Fourier transform|DPVM34ADZV]], and assume that $\\hat f$ is also integrable.\nThen\n$$f(x) = \\int_{\\R^n} \\hat f(\\xi) \\, e^{2\\pi i x \\cdot \\xi} \\, d\\xi$$\nfor [[almost every|RARTMA0ZEC]] $x \\in \\R^n$, that is, $f$ equals the [[inverse Fourier transform|JB8QMT8DCH]] of $\\hat f$ almost everywhere.\nIf moreover $f$ is continuous, the equality holds for every $x \\in \\R^n$.\n\n## Remarks\n\nProof idea: insert the Gaussian factor $e^{-\\pi \\varepsilon^2 \\abs{\\xi}^2}$ into the inversion integral.\nWith this factor the double integral converges absolutely.\nSince the Gaussian is its own Fourier transform, Fubini's theorem turns it into the convolution of $f$ with a Gaussian of width $\\varepsilon$ and total mass $1$.\nAs $\\varepsilon \\to 0$, the left side tends to the inversion integral by dominated convergence, since $\\hat f$ is integrable.\nThe convolutions tend to $f$ in $L^1$, and hence almost everywhere along a subsequence.\n\nExample: $f(x) = e^{-\\abs{x}}$ on $\\R$ has $\\hat f(\\xi) = \\frac{2}{1 + 4\\pi^2 \\xi^2}$, which is integrable.\nSince $f$ is continuous, $e^{-\\abs{x}} = \\int_\\R \\frac{2 e^{2\\pi i x \\xi}}{1 + 4\\pi^2 \\xi^2} \\, d\\xi$ for every $x$.\n\nThe integrability of $\\hat f$ is needed for the integral to make sense.\nFor $f = \\mathbf{1}_{[-1/2, 1/2]}$, $\\hat f(\\xi) = \\frac{\\sin \\pi \\xi}{\\pi \\xi}$ is not integrable, so the theorem does not apply as stated.\nThen $f$ is recovered only by the symmetric limits $\\lim_{R \\to \\infty} \\int_{-R}^{R}$, which give $f(x)$ for $\\abs{x} \\ne 1/2$ and $1/2$ at the jumps.\nA consequence of the theorem is uniqueness: if $f$ is integrable and $\\hat f = 0$, then $f = 0$ almost everywhere.\n\n## Explanations\n\n### L1 human\n\nThe Fourier transform breaks a function into waves of all frequencies.\nThe inversion theorem says that adding these waves back together gives the original function.\nSo no information is lost in passing to the frequency side.\n\n### L2 human\n\nThe theorem lets one solve problems on the frequency side, where derivatives become multiplications, and then return.\nIt is the basis of the Fourier method for linear differential equations with constant coefficients.\nOn the Schwartz space the transform is a bijection and inversion holds everywhere without extra hypotheses.\nThe [[Plancherel theorem|HPZHRZ0502]] gives the corresponding statement in $L^2$, where the inverse transform is the adjoint of the transform.\n",
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