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  "body": "\n## Example\n\nLet $f_0 \\colon (0, \\infty) \\to \\C$ satisfy $\\int_0^\\infty \\abs{f_0(r)}\\, r\\, dr \u003c \\infty$, and let $f(x) = f_0(\\abs{x})$ on $\\R^2$, an integrable radial function.\nWith the [[Fourier transform|DPVM34ADZV]] $\\hat f(\\xi) = \\int_{\\R^2} f(x) e^{-2\\pi i x \\cdot \\xi}\\, dx$, one has for $\\rho = \\abs{\\xi}$\n$$\\hat f(\\xi) = 2\\pi \\int_0^\\infty f_0(r)\\, J_0(2\\pi r \\rho)\\, r\\, dr,$$\nwhich is $2\\pi$ times the [[Hankel transform|2P9YWDGKC4]] of order $0$ of $f_0$ at $k = 2\\pi\\rho$.\nCheck: rotate so that $\\xi$ lies along the first axis.\nIn polar coordinates $x = (r\\cos\\theta, r\\sin\\theta)$ one gets\n$$\\hat f(\\xi) = \\int_0^\\infty f_0(r) \\Big( \\int_0^{2\\pi} e^{-2\\pi i r \\rho \\cos\\theta}\\, d\\theta \\Big) r\\, dr.$$\nThe inner integral is $2\\pi J_0(2\\pi r\\rho)$ by Bessel's integral $J_0(z) = \\frac{1}{2\\pi} \\int_0^{2\\pi} e^{-iz\\cos\\theta}\\, d\\theta$, where $J_0$ is the [[Bessel function|CZCMRDWDYW]] of order $0$.\nThis is the case $n = 2$ of the [[Fourier transform of radial functions|3CSEGRAD75]].\n\nFor the Gaussian $f_0(r) = e^{-\\pi r^2}$, the known integral $\\int_0^\\infty e^{-a r^2} J_0(kr)\\, r\\, dr = \\frac{1}{2a} e^{-k^2/(4a)}$ with $a = \\pi$ and $k = 2\\pi\\rho$ gives $\\hat f(\\xi) = 2\\pi \\cdot \\frac{1}{2\\pi} e^{-\\pi \\rho^2} = e^{-\\pi\\abs{\\xi}^2}$, as expected for $e^{-\\pi \\abs{x}^2}$.\n\n## Explanations\n\n### L1 human\n\nA function in the plane that only depends on the distance to the origin has a Fourier transform with the same symmetry.\nIts two-dimensional Fourier transform then reduces to a one-dimensional integral over the radius, with the Bessel function $J_0$ as kernel.\nThat integral is the Hankel transform of order $0$, which is why Hankel transforms appear in every problem with circular symmetry.\n",
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