Bessel Functions and Fourier Transform of Radial Functions

Definition of Bessel Functions

There are several ways to define Bessel functions. The one given by (Grafakos, 2008) Appendix B is as follows,

$$J_v(t) = \frac{(t/2)^v}{\Gamma(v+1/2)\Gamma(1/2)} \int_{-1}^{1} e^{its}(1-s^2)^v\frac{ds}{\sqrt{1-s^2}}\,.\tag{1}$$

Here $v$ is any complex number with real part greater than $1/2$. The second definition is that $J_v(t)$ is the solution of the ODE,

$$t^2\frac{d^2y}{dt^2}+t\frac{dy}{dt}+(t^2-v^2)y=0\,.$$

The third one is the original Bessel’s definition, which only works for integer values of $v$.

$$J_v(t)=\frac{1}{2\pi}\int_{-\pi}^\pi e^{i(t\sin\tau-v\tau)}d\tau\,.\tag{2}$$

Now we show that these three definitions are equivalent. First, applying integration by parts, we can show that

$$\begin{align} & 2\left(v+\frac{1}{2}\right)\int_{-1}^1ise^{its}(1-s^2)^v\frac{ds}{\sqrt{1-s^2}} \\ & = -i\int_{-1}^1 e^{its}d(1-s^2)^{v+1/2} \\ & = \int_{-1}^1te^{its}(1-s^2)^{v+1/2}ds \\ & = t\int_{-1}^1e^{its}(1-s^2)^{v+1}\frac{ds}{\sqrt{1-s^2}}\,. \end{align}$$

With the result above, to show that $J_v(t)$ is the solution of the differential equation, we only need to plug it into the equation and check. Note that the second order differential equation has two independent solutions, $J_v(t)$ is only one of them.

The simple derivation also implies the following identity,

$$\frac{d}{dt}\left(t^{-v}J_v(t)\right)=-t^{-v}J_{v+1}(t)\,.\tag{3}$$

With this identity, we can show the definition for integer $v$. It is proved by induction. First when $v=0$,

$$\begin{equation} J_0(t) = \frac{1}{\Gamma^2(1/2)} \int_{-1}^1 e^{its}\frac{ds}{\sqrt{1-s^2}}\,. \end{equation}$$

Make the variable substitution $s=\sin(\tau)$ and use the fact $\Gamma(1/2)=\sqrt{\pi}$. We get the base case. Then we verify that (2) also satisfies the recurrence relation (3). The key is still integration by parts and the famous Euler’s identity,

$$\begin{equation} e^{i\tau}=\cos\tau+i\sin\tau\,. \end{equation}$$

Fourier Transform of Measure on Unit Sphere

Denote the uniform measure over the unit sphere $S^{n-1}$ by $\sigma_{n-1}$. Its Fourier transform

$$\begin{equation} \mathcal{F}\sigma_n(t) = \int_{S^{n}}e^{-2\pi it\cdot x}d\sigma_n(x) \end{equation}$$

can be evaluated in the following way. First, for any fixed $t$, we consider the cylinder coordinates with axis along the direction of $t$, $(\theta,s)$. $\theta$ is the $n-1$ spherical coordinates and $s$ is the Cartesian coordinate ranging from $[-1,1]$. Then $t\cdot x=\vert t\vert s$, and the integral becomes

$$\begin{align} \mathcal{F}\sigma_n(t) & =\int_{-1}^1\int_{\sqrt{1-s^2}S^{n-1}} e^{-2\pi i\vert t\vert s} d\sigma_{n-1}(\theta)\frac{ds}{\sqrt{1-s^2}} \\ & = A_{n-1}\int_{-1}^1 e^{-2\pi i\vert t\vert s} (1-s^2)^{(n-1)/2}\frac{ds}{\sqrt{1-s^2}} \\ & = \frac{n\Gamma^{n}(1/2)J_{(n-1)/2}(2\pi\vert t\vert)\Gamma(n/2)\Gamma(1/2)} {\Gamma(n/2+1)(\pi\vert t\vert)^{(n-1)/2}} \\ & = \frac{2\pi J_{(n-1)/2}(2\pi\vert t\vert)} {(\vert t\vert)^{(n-1)/2}}\,. \end{align}$$

Here $A_{n-1}$ is the surface area of $n-1$ sphere. With this formula we can further get the Fourier transform of any radial functions $f(x)=f_0(\vert x\vert)$ as follows,

$$\begin{equation} \mathcal{F}f(t)=\frac{2\pi}{\vert t\vert^{(n-2)/2}} \int_0^\infty f_0(r)J_{n/2-1}(2\pi r\vert t\vert)r^{n/2}dr\,. \end{equation}$$

Reference

1. Grafakos, L. (2008). Classical Fourier Analysis. Springer New York.