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  "body": "\n## Statement\n\nLet $b \u003e 0$, let $0 \u003c \\alpha \u003c 1$, and let $\\varphi \\colon [0, b] \\to \\C$ be continuous.\nLet\n$$f(x) = \\int_0^x \\frac{\\varphi(t)}{(x - t)^{\\alpha}} \\, dt, \\quad x \\in [0, b].$$\nThen the function $F(x) = \\int_0^x \\frac{f(t)}{(x - t)^{1 - \\alpha}} \\, dt$ is continuously differentiable on $[0, b]$, and\n$$\\varphi(x) = \\frac{\\sin(\\pi \\alpha)}{\\pi} \\, F'(x) = \\frac{\\sin(\\pi \\alpha)}{\\pi} \\, \\frac{d}{dx} \\int_0^x \\frac{f(t)}{(x - t)^{1 - \\alpha}} \\, dt, \\quad x \\in [0, b].$$\nIn particular, the [[Volterra equation of the first kind|H6ZGEDM4VV]] $\\int_0^x \\varphi(t) (x - t)^{-\\alpha} \\, dt = f(x)$ has at most one continuous solution $\\varphi$.\n\n## Remarks\n\nProof idea: insert the formula for $f$ into $F$ and exchange the order of integration.\nThis gives $F(x) = \\int_0^x \\varphi(s) \\left( \\int_s^x (x - t)^{\\alpha - 1} (t - s)^{-\\alpha} \\, dt \\right) ds$.\nThe substitution $t = s + u(x - s)$ turns the inner integral into the [[Beta function|QHNWVSC6AQ]] value $B(\\alpha, 1 - \\alpha) = \\Gamma(\\alpha) \\Gamma(1 - \\alpha) = \\pi / \\sin(\\pi \\alpha)$, which does not depend on $x$ or $s$.\nSo $F(x) = \\frac{\\pi}{\\sin(\\pi \\alpha)} \\int_0^x \\varphi(s) \\, ds$, and the fundamental theorem of calculus gives the formula.\n\nExample: for $\\alpha = \\tfrac12$ the formula reads $\\varphi(x) = \\frac{1}{\\pi} \\frac{d}{dx} \\int_0^x \\frac{f(t)}{\\sqrt{x - t}} \\, dt$, the inversion of [[Abel's integral equation|G4GWBYTW7B]].\nWith $f(x) = 2\\sqrt{x}$, the inner integral is $2 x B(\\tfrac32, \\tfrac12) = \\pi x$, so $\\varphi = 1$; indeed $\\int_0^x (x - t)^{-1/2} \\, dt = 2\\sqrt{x}$.\n\nThe statement starts from a solution $\\varphi$; it does not say that every $f$ arises.\nA given $f$ is of this form for a continuous $\\varphi$ only if $F$ is continuously differentiable; for instance $f$ must vanish at $0$.\nIf $f$ is continuously differentiable with $f(0) = 0$, then the solution can also be written $\\varphi(x) = \\frac{\\sin(\\pi \\alpha)}{\\pi} \\int_0^x \\frac{f'(t)}{(x - t)^{1 - \\alpha}} \\, dt$.\n\n## Explanations\n\n### L1 human\n\nAbel's equation hides the unknown $\\varphi$ inside an integral with a kernel that blows up at the endpoint.\nThe inversion formula recovers $\\varphi$ by applying a second integral of the same kind, with the complementary exponent $1 - \\alpha$, and then differentiating.\nThe two integrals together amount to one ordinary integration, which the derivative undoes.\n\n### L2 human\n\nIn the language of fractional calculus, the equation says $\\Gamma(1 - \\alpha) \\, I^{1 - \\alpha} \\varphi = f$, where $I^\\beta$ is the [[Riemann–Liouville integral|JTKCZ44NN3]] of order $\\beta$.\nThe formula then reads $\\varphi = \\frac{1}{\\Gamma(1 - \\alpha)} \\frac{d}{dx} I^{\\alpha} f$, using $I^\\alpha I^{1 - \\alpha} = I^1$; the reflection formula $\\Gamma(\\alpha) \\Gamma(1 - \\alpha) = \\pi / \\sin(\\pi \\alpha)$ for the [[Gamma function|0H5SYBSED9]] shows that this agrees with the statement.\nIt solves the [[tautochrone problem|A56P5S5D6Z]], and its variant for the Abel transform is the basis of Abel inversion in tomography of radially symmetric objects.\n",
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