I'm fairly sure that transformation:
$$ h(y) = \int_0^\infty g(t) t^2 \frac{\sin(yt)}{yt} dt $$
inverts the 3D radial "Fourier transform" (it's usually just called the Fourier transform in physics, even though its not periodic) given by
$$ g(t) = \int_0^\infty f(x) x^2 \frac{\sin(xt)}{xt} $$
A very handwavy demonstration of this is to consider the sequential application
$$ h(y) = \int_{t=0}^{t=\infty} \int_{x=0}^{x=\infty} f(x) x^2 \frac{\sin(xt)}{xt} dx\; t^2 \frac{\sin(yt)}{yt} dt $$
and then (improperly) swap the order of integration, to give
$$ \int_{x=0}^{x=\infty} f(x) \int_{t=0}^{t=\infty} x^2 t^2 \frac{\sin(xt)}{xt} \frac{\sin(yt)}{yt} dt \; dx $$
The inner integral can be then written as
$$ \frac{x}{y} \int_{t=0}^{t=\infty} \sin(xt)\sin(yt) dt $$
If $x=y$ the leading term will be equal to one, and the integral will be infinite. When $x\neq y$ it will be zero, just like a delta function. So it seems to be intuitively correct, but this is clearly not a correct way of doing this formally.
What's the right way?