How to invert finding marginals of a rotationally symmetric distribution?

Given

$$f(x) = \int_{-\infty}^{+\infty} g(x^2+y^2) \,\mathrm{d}y,$$

is it possible to reconstruct $g(u)$ analytically? I tried differentiating $f(x)$ but only found

$$f'(x) = \int_{x^2}^{+\infty} \frac{g(u)-g(x^2)}{\sqrt{u-x^2}^3} x \,\mathrm{d}u,$$

which does not seem to be any closer to an inversion formula.

I don't expect $g(x^2+y^2)$ to be a probability density function whenever $f(x)$ is, in fact, it's mainly the counterexamples that I want to study this way (Wigner function in physics). I chose the tag because I believe similar problems have been primarily studied in the theory of probability distributions.

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I think this is relevant: math.stackexchange.com/questions/269998/…. – joriki Mar 6 '13 at 12:57
@joriki: Thanks for pointing me at Radon transform! I guess that answers it. I'll try and see if it helps. – The Vee Mar 6 '13 at 13:01

I believe your problem is solved by the inverse Abel transform. By transforming the integral to $r=\sqrt{x^2+y^2}$ and considering $h(r)=g(r^2)$, you get
$$f(x)=\int_{-\infty}^\infty g\left(x^2+y^2\right)\mathrm dy=2\int_x^\infty\frac{h(r)r}{\sqrt{r^2-x^2}}\mathrm dr\;,$$
$$h(r)=-\frac1\pi\int_r^\infty\frac{f'(x)}{\sqrt{x^2-r^2}}\mathrm dx\;.$$