Abel inversion formula

Remarks

Proof idea: insert the formula for $f$ into $F$ and exchange the order of integration. This gives $F(x) = \int_0^x \varphi(s) \left( \int_s^x (x - t)^{\alpha - 1} (t - s)^{-\alpha} \, dt \right) ds$. The substitution $t = s + u(x - s)$ turns the inner integral into the Beta function value $B(\alpha, 1 - \alpha) = \Gamma(\alpha) \Gamma(1 - \alpha) = \pi / \sin(\pi \alpha)$, which does not depend on $x$ or $s$. So $F(x) = \frac{\pi}{\sin(\pi \alpha)} \int_0^x \varphi(s) \, ds$, and the fundamental theorem of calculus gives the formula.

Example: 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. With $f(x) = 2\sqrt{x}$, the inner integral is $2 x B(\tfrac32, \tfrac12) = \pi x$, so $\varphi = 1$; indeed $\int_0^x (x - t)^{-1/2} \, dt = 2\sqrt{x}$.

The statement starts from a solution $\varphi$; it does not say that every $f$ arises. A given $f$ is of this form for a continuous $\varphi$ only if $F$ is continuously differentiable; for instance $f$ must vanish at $0$. If $f$ is continuously differentiable with $f(0) = 0$, then the solution can also be written $\varphi(x) = \frac{\sin(\pi \alpha)}{\pi} \int_0^x \frac{f'(t)}{(x - t)^{1 - \alpha}} \, dt$.

Going deeper

For a student

In the language of fractional calculus, the equation says $\Gamma(1 - \alpha) \, I^{1 - \alpha} \varphi = f$, where $I^\beta$ is the Riemann–Liouville integral of order $\beta$. The 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 shows that this agrees with the statement. It solves the tautochrone problem, and its variant for the Abel transform is the basis of Abel inversion in tomography of radially symmetric objects.