Fourier transform of radial functions

Remarks

The transform commutes with rotations: if $R$ is an orthogonal matrix, then the transform of $f \circ R$ is $\hat f \circ R$. So the transform of a radial function is radial. To compute it, write the integral in polar coordinates $x = r \omega$ and use the formula $\int_{S^{n-1}} e^{-2\pi i r \rho\, \omega \cdot e_1}\, d\sigma(\omega) = 2\pi (r\rho)^{1 - n/2} J_{n/2-1}(2\pi r \rho)$ for the surface measure $\sigma$ on the unit sphere. The integral converges absolutely, because $z^{-\nu} J_\nu(z)$ is bounded on $(0, \infty)$ for $\nu \ge -\frac12$ and the hypothesis on $f_0$ holds.

Example: for $n = 3$, $J_{1/2}(z) = \sqrt{2/(\pi z)}\, \sin z$, and the formula becomes $$\hat f(\xi) = \frac{2}{\rho} \int_0^\infty f_0(r)\, r \sin(2\pi r \rho)\, dr.$$ For $n = 1$ the same formula, with $J_{-1/2}(z) = \sqrt{2/(\pi z)} \cos z$, reduces to the cosine transform of an even function.

Up to the factors of $\rho$ and $r$, the one-dimensional integral is a Hankel transform of order $n/2 - 1$.

Going deeper

For a student

This reduction is the main tool for radial problems on $\R^n$. The transforms of the indicator of a ball, of the surface measure on a sphere and of the Bochner–Riesz multipliers are all computed this way. The decay of Bessel functions, $J_\nu(z) = O(z^{-1/2})$ as $z \to \infty$, has a direct consequence. The Fourier transform of the surface measure on a sphere decays like $\abs{\xi}^{-(n-1)/2}$, a fact at the heart of restriction theory.