If $f$ is a radial function and $f(x)=f_0(|x|)$, I want to show $$\int_{\mathbb{R}^d}f(u)e^{-2\pi ix \cdot u}du=\int_{0}^{\infty}f_0(s)\left(\int_{\mathbb{S}^{d-1}} e^{-2\pi i |x| s(x' , u')}du'\right)s^{d-1}ds.$$
My trying is taking $s=|u|$ and applying the spherical coordinate transformation. I don't know where is the $|x|s$ in the power of $e$ in the RHS comes from. Besides, where should I use the function $f$ is radial in the calculation?
Is there any special integral formulas about radial functions used here?
Thanks a lot!
The general form (I think that it is in Folland's blue book) of the change of variables is:
$$\int_{\mathbb{R^d}}g(x)dx = \int_0^\infty \left( \int_{S^{d-1}} g(r\omega) d\omega \right) r^{d-1} dr = \int_{S^{d-1}} \left( \int_0^\infty g(r\omega) r^{d-1}dr \right) d\omega,$$ where $\omega$ denotes a direction on the sphere and $d\omega$ denotes the sphere surface measure, and $r$ denotes the radius. (Note: the $r^{d-1}$ is a scaling factor from integrating on spheres of radius $r$.) So, $r\omega$ denotes an arbitrary point in $\mathbb{R}^d$. In particular, for $y \in \mathbb{R}^d$ nonzero, $y = |y|\frac{y}{|y|} = r\omega$.
Keep in mind. If $f$ is radial, then $f(r\omega)$ only depends on $r$.