# How do I find the inverse Fourier transform of a function that is separable into a radial and an angular part?

I need to take the inverse Fourier transform of a function that is initially specified in spherical coordinates: $f(r, \theta, \phi) = \int_{R^3}F(k, \Theta,\Phi)e^{i\vec{k}\vec{r}}k^2sin(\Theta)dkd\Theta d\Phi$.

My function is separable into a radial and an angular part: $F(k,\Theta,\Phi)=R(k)A(\Theta,\Phi)$.

As an answer to a previous post: How do I find the Fourier transform of a function that is separable into a radial and an angular part? joriki tipped me (thanks!) on how I could expand the plane wave term into spherical waves:

$e^{i\vec{k}\vec{r}}=4\pi\sum_{l=0}^{\infty}i^lj_l(kr)\sum_{m=-l}^{+l}Y_{l,m}^*(\theta, \phi)Y_{l,m}(\Theta, \Phi)$

where $j_l$ is the spherical Bessel function and $Y_{l,m}$ is the Spherical Harmonic of degree l and order m.

Plugging this expression into the inverse Fourier transform I get: $f(r, \theta,\phi)=4\pi\sum_{l=0}^{\infty}i^l[\int_{0}^{\infty}R(k)j_l(kr)k^2dk][\sum_{m=-l}^{+l}Y_{l,m}^*(\theta, \phi)\int_0^{\pi}\int_0^{2\pi}A(\Theta, \Phi)Y_{l,m}(\Theta, \Phi)sin(\Theta)d\Theta d\Phi]$

I guess the last two integrals equals the coefficients in a real spherical harmonic expansion of A (?): $A(\Theta,\Phi)=\sum_{l=0}^{\infty}\sum_{m=-l}^{+l}A_{l,m}Y_{l,m}$, where $A_{l,m}=\int_0^{\pi}\int_0^{2\pi}A(\Theta, \Phi)Y_{l,m}(\Theta, \Phi)sin(\Theta)d\Theta d\Phi$

Therefore: $f(r, \theta,\phi)=4\pi\sum_{l=0}^{\infty}i^l[\int_{0}^{\infty}R(k)j_l(kr)k^2dk][\sum_{m=-l}^{+l}A_{l,m}Y_{l,m}^*(\theta, \phi)]$

It seems the tricky part is the $\int_{0}^{\infty}R(k)j_l(kr)k^2dk$ term. Any ideas on how to proceed with this?