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  "body": "\n## Definition\n\nLet $f \\colon [0, \\infty) \\to \\C$ be measurable, and let $y \\ge 0$ be such that $\\int_y^\\infty \\frac{\\abs{f(r)}\\, r}{\\sqrt{r^2 - y^2}}\\, dr \u003c \\infty$.\nThe *Abel transform* of $f$ at $y$ is the [[Lebesgue integral|DWJF7ABPR1]]\n$$F(y) = 2 \\int_y^\\infty \\frac{f(r)\\, r}{\\sqrt{r^2 - y^2}}\\, dr.$$\n\n## Remarks\n\nThe integrand has an integrable singularity at $r = y$, of order $(r - y)^{-1/2}$, when $f$ is bounded near $y$ and $y \u003e 0$.\nIf $f$ is continuously differentiable and decays fast enough at infinity, the transform can be inverted by\n$$f(r) = -\\frac{1}{\\pi} \\int_r^\\infty \\frac{F'(y)}{\\sqrt{y^2 - r^2}}\\, dy.$$\n\nExample: $f(r) = e^{-r^2}$.\nThe substitution $s = r^2 - y^2$, $ds = 2r\\, dr$, gives $F(y) = \\int_0^\\infty \\frac{e^{-y^2 - s}}{\\sqrt s}\\, ds = e^{-y^2}\\, \\Gamma(1/2) = \\sqrt{\\pi}\\, e^{-y^2}$.\n\nNon-example: $f(r) = 1$ for all $r$.\nThen $\\frac{r}{\\sqrt{r^2 - y^2}} \\ge 1$ for $r \u003e y$, so the integral diverges for every $y$ and $f$ has no Abel transform.\n",
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