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  "body": "\n## Definition\n\nLet $\\nu \\ge -\\frac12$, and let $J_\\nu$ be the [[Bessel function of the first kind|CZCMRDWDYW]] of order $\\nu$.\nLet $f \\colon (0, \\infty) \\to \\C$ be measurable with $\\int_0^\\infty \\abs{f(r)}\\, r^{1/2}\\, dr \u003c \\infty$.\nThe *Hankel transform of order $\\nu$* of $f$ is the function on $(0, \\infty)$ given by the [[Lebesgue integral|DWJF7ABPR1]]\n$$F_\\nu(k) = \\int_0^\\infty f(r)\\, J_\\nu(kr)\\, r\\, dr.$$\n\n## Remarks\n\nThe integral converges absolutely, because $\\sqrt{z}\\, \\abs{J_\\nu(z)}$ is bounded on $(0, \\infty)$ for $\\nu \\ge -\\frac12$, so $\\abs{J_\\nu(kr)}\\, r \\le C k^{-1/2} r^{1/2}$.\nThe inverse transform has the same form: under suitable conditions $f(r) = \\int_0^\\infty F_\\nu(k)\\, J_\\nu(kr)\\, k\\, dk$.\nOther normalizations use the kernel $J_\\nu(kr) \\sqrt{kr}$ with $dr$ instead of $r\\, dr$.\nThrough the [[Fourier transform|DPVM34ADZV]] in the plane, the order-$0$ transform describes radial functions: if $g(x) = f(\\abs{x})$ on $\\R^2$, then $\\int_{\\R^2} g(x) e^{-i x \\cdot \\xi}\\, dx = 2\\pi F_0(\\abs{\\xi})$.\n\nExample: $\\nu = 0$ and $f(r) = e^{-r^2/2}$.\nThe plane Gaussian $e^{-\\abs{x}^2/2}$ has Fourier transform $2\\pi e^{-\\abs{\\xi}^2/2}$ with this kernel, so $F_0(k) = e^{-k^2/2}$.\n\nNon-example: $f(r) = 1$ fails the integrability condition, and $\\int_0^R J_0(kr)\\, r\\, dr = \\frac{R}{k} J_1(kR)$ has no limit as $R \\to \\infty$, since $J_1(kR)$ decays only like $R^{-1/2}$.\n",
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