Suppose we have a function $\phi(\vec r)$ and we have $\Psi(y,z)$ that is its integral over the $x$ axis, i.e. $$\Psi(y,z) =\int_{-∞}^{+∞} \phi(x,y,z)\,\mathrm dx.$$ Of course there is no general way to recover $\phi$ starting from $\Psi$. Now let's explore the case where $\phi$ is spherically symmetric, so that we can write it as $\phi(r)$. Then, we can find $\Psi(R)$ in this way: $$\Psi(R) =\int_{-∞}^{+∞} \phi(\sqrt{R^2+x^2})\,\mathrm dx,$$ where $R$ is the cylindrical radius. It easy to prove (change $s=\sqrt{R^2+x^2}$) that the above integral is equal to $$\Psi(R) =2\int_R^{+∞} \phi(s) {\left(1+\left(R/s\right)^2\right)^{-1/2}}\,\mathrm ds.$$
The question is: How can I invert $\Psi(R)$ and recover $\phi(r)?$ Unluckily I cannot just make the derivative of $\Psi(R)$ because the variable $R$ appears in the right side both in the integration bounds and in the integrand.