Abel transform as the projection of a radial function

Remarks

Since $g(x, y)$ is even in $x$, the line integral is $2 \int_0^\infty f\big(\sqrt{x^2 + y^2}\big)\, dx$. For $y \ne 0$, the substitution $r = \sqrt{x^2 + y^2}$, with $dx = \frac{r\, dr}{\sqrt{r^2 - y^2}}$, maps $x \in (0, \infty)$ onto $r \in (\abs{y}, \infty)$ and gives the formula; for $y = 0$ it is immediate. The absolute convergence follows the same way: $\frac{r}{\sqrt{r^2 - y^2}}$ is integrable near $r = \abs{y}$ and tends to $1$ as $r \to \infty$, while $f$ is bounded near $\abs{y}$ and integrable. By rotation invariance, the integral of $g$ along any line at distance $\abs{y}$ from the origin is also $F(\abs{y})$; this is the Radon transform of $g$.

Example: for $f(r) = e^{-r^2}$ the line integral is $\int_\R e^{-x^2 - y^2}\, dx = \sqrt{\pi}\, e^{-y^2}$, which is the Abel transform of $f$ at $\abs{y}$.

Going deeper

For a student

The statement identifies the Abel transform with the two-dimensional Radon transform restricted to radial functions. Together with the projection-slice theorem, it shows that the Fourier transform of the Abel transform of $f$ is the Hankel transform of order $0$ of $f$, up to constants. This chain is used in tomography of axially symmetric objects, where only one projection is needed.