For a simple power law function $$f(\mathbf{r}) = \frac{1}{r^\alpha}\,,$$ it is well-known that its Fourier transform in $d$-dimensions is given by $$\hat{f}(\mathbf{k}) = \frac{1}{k^{d-\alpha}}\,.$$.
Now consider the function $$f(\mathbf{r}) = \frac{1}{r^\alpha} g(\Omega)$$
where $g(\Omega)$ is some dimensionless function of the angular coordinates. By simple dimensional analysis, it seems it must be the case that the Fourier transform is given by $$\hat{f}(\mathbf{k}) = \frac{1}{k^{d-\alpha}} \hat{g}(\Omega_k)$$ for some function $\hat{g}(\Omega_k)$ of the angular coordinates in momentum space. Is there an explicit formula that relates $g(\Omega)$ to $\hat{g}(\Omega_k)$? I have been unable to derive such a relation.