Abel transform

Remarks

The integrand has an integrable singularity at $r = y$, of order $(r - y)^{-1/2}$, when $f$ is bounded near $y$ and $y > 0$. If $f$ is continuously differentiable and decays fast enough at infinity, the transform can be inverted by $$f(r) = -\frac{1}{\pi} \int_r^\infty \frac{F'(y)}{\sqrt{y^2 - r^2}}\, dy.$$

Example: $f(r) = e^{-r^2}$. The substitution $s = r^2 - y^2$, $ds = 2r\, dr$, gives $F(y) = \int_0^\infty \frac{e^{-y^2 - s}}{\sqrt s}\, ds = e^{-y^2}\, \Gamma(1/2) = \sqrt{\pi}\, e^{-y^2}$.

Non-example: $f(r) = 1$ for all $r$. Then $\frac{r}{\sqrt{r^2 - y^2}} \ge 1$ for $r > y$, so the integral diverges for every $y$ and $f$ has no Abel transform.