I am interested in finding the inverse of something that looks like an abel transform of a spherically symmetric function, only with integrals over planes rather than lines. Ie, rather than finding $A^{-1}$ where $A$ behaves like so:
$$A[f](x) = \int_{-\infty}^\infty f\left(\sqrt{x^2 + y^2}\right) dy$$
I would like to find $B^{-1}$ such that
$$B[f](x) = \int_{-\infty}^\infty \int_{-\infty}^\infty f\left(\sqrt{x^2 + y^2 + z^2}\right) dy dz$$
Or equivalently
$$B[f](x) = 2 \pi \int_0^\infty f\left(\sqrt{x^2 + s^2}\right) s ds$$
Is it possible to derive a simple inverse formula for this transform, or better still write it in terms of other well established transforms and inverses? Thanks!
So I thought about this a bit more and realized it's fairly straightforward:
$$B[f](x) = 2 \pi \int_0^\infty f\left(\sqrt{x^2 + s^2}\right) s ds$$ $$= 2 \pi \int_x^\infty f\left(r\right) r dr$$ where $r^2 = x^2 + s^2$. But then by the fundamental theorem of calculus:
$$\frac{d}{dx}B[f](x) = -2 \pi f(x)x$$
And so
$$B^{-1} = -\frac{1}{2\pi x}\frac{d}{dx}$$