3D Sphericall Symmetric Analytical Convolution

Question

I am trying to analytically convolve two spherically symmetric functions in 3D spherical coordinates, a 3D Gaussian and a "box" (really a radial step function). Numerical convolution yields a radially symmetric erfc function. I am unable to reproduce this analytically, however. Let $f$ be our step function and $g$ our Gaussian function.

$$ g(\vec{r}) = \begin{cases} 1 & r - R_p \le 0 \\ 0 & r - R_p > 0 \end{cases}\\f(\vec{r}) = a \exp\left(\frac{-r^2}{2\sigma^2}\right) \\ $$ where $R_p > 0$, $r$ is the scalar magnitude of the $\vec{r}$, and $a$ normalizes $f$ such that the area under $f$ is 1.

There are spherically symmetric analytic convolutions is derived here: http://www.doi.org/10.1364/JOSAA.27.002144 "Operational and convolution properties of three-dimensional Fourier transforms in spherical polar coordinates". The relevant equations (48) and (49) from the paper are reproduced below, respectively.

\begin{equation}\tag{48} f(\vec{r}) * g(\vec{r}) = f(r) * * *g(r) = \int_0^\infty g(r_0) \Phi_{3D}(r-r_0) r_0 dr_0 \end{equation}

\begin{align} \tag{49} \Phi_{3D}(r-r_0) & = \int_0^{2\pi} \int_0^{\pi} f(\vec{r} - \vec{r}_0) \sin(\psi_{r_0}) d\psi_{r_0} d \theta_{r_0} \end{align}

I have tried just naively treating $f$ as spherically symmetric and simplifying $\Phi_{3D}$ to $$\Phi_{3D} = 4\pi f(\vec{r} - \vec{r}_0) $$ though this method does not yield the correct answer (erfc($r$)) and the author even notes this is correct, saying

the function $f$ is shifted from $\vec{r}$ to $\vec{r}_0$ (destroying spherical symmetry), and then the resulting shifted function is integrated over all angular variables. The unshifted function $g$ is still spherically symmetric and unaffected by the integration over angular variables so that the final result is given by Eq. (48).

What is the proper method to solve this analytical convolution? I am unsure how to properly change the bounds of integration for the angular integration as a function of $R_p$.

Answer 1

There is a misprint in that paper. The equation (48) should be $$f({\vec r})*g({\vec r}) = f(r)***g(r)=\int\limits_0^\infty g(r_0)\Phi_{3D}(r-r_0)r_0^2dr_0$$

You can check it with the equation (47), and verify it by taking both $f$ and $g$ to be Gaussians.

The equation (49) in that paper is $$\Phi_{3D}(r-r_0)=\int\limits_0^{2\pi}\int\limits_0^{\pi}f({\vec r}-{\vec r}_0)\sin(\psi_{r_0})d\psi_{r_0}d\theta_{r_0}=8\int\limits_0^\infty f(u)S^{0,0}_0(u,r,r_0)u^2du$$ where $$S^{0,0}_0(u,r,r_0)=\int\limits_0^\infty j_0(\rho u)j_0(\rho r_0)j_0(\rho r)\rho^2d\rho$$$$=\left(\frac{\pi}{2}\right)^{3/2}\frac{1}{\sqrt{ur_0r}}\int\limits_0^\infty J_{1/2}(\rho u)J_{1/2}(\rho r_0)J_{1/2}(\rho r)\rho^{1/2}d\rho$$ because $$j_n(z)=\sqrt{\frac{\pi}{2z}}J_{n+1/2}(z)$$ We have $$f***g=\frac{(2\pi)^{3/2}}{\sqrt{r}}\int\limits_0^\infty dr_0\int\limits_0^\infty d\rho\int\limits_0^\infty du\;\; J_{1/2}(\rho r)\rho^{1/2} \;\; g(r_0)J_{1/2}(\rho r_0)r_0^{3/2} \;\; f(u)J_{1/2}(\rho u)u^{3/2}$$ Substituting $f(u)=\frac{1}{\sigma^3(2\pi)^{3/2}}\exp[-u^2/(2\sigma^2)]$, and taking the integral over $u$, we get $$f***g=\frac{1}{\sqrt{r}}\int\limits_0^\infty dr_0\int\limits_0^\infty d\rho \;\; J_{1/2}(\rho r) e^{-\rho^2\sigma^2/2} \rho \;\; g(r_0)J_{1/2}(\rho r_0)r_0^{3/2}$$ Next, take the integral over $r_0$: $$f***g=\sqrt{\frac{2}{\pi r}}\int\limits_0^\infty d\rho\, J_{1/2}(\rho r) e^{-\rho^2 \sigma^2/2}\left(\frac{\sin(\rho R_p)}{\rho^{3/2}}-\frac{R_p\cos(\rho R_p)}{\rho^{1/2}}\right)$$ Finally, $$f***g=\frac{1}{\sqrt{2\pi}}\frac{\sigma}{r}e^{-\frac{(r+R_p)^2}{2\sigma^2}}\left[1-e^{-\frac{2rR_p}{\sigma^2}}\right]+\frac{1}{2}\left[\text{erf}\left(\frac{r-R_p}{\sigma\sqrt{2}}\right)-\text{erf}\left(\frac{r+R_p}{\sigma\sqrt{2}}\right)\right]$$

Checking it with Fourier transforms (see this link, also this):

$$F(w) = e^{-w^2\sigma^2/2}, \;\;\;\;G(w)=\frac{4\pi}{w^3}\left[\sin(R_pw)-R_pw\cos(R_pw)\right]$$

Multiplying them, and taking the inverse Fourier transform

$$f***g=\frac{1}{\pi}\int\limits_0^\pi d\theta\int\limits_0^\infty dr\;\frac{e^{-w^2\sigma^2/2}}{w}\left[\sin(R_pw)-R_pw\cos(R_pw)\right]\sin(\theta)e^{iwr\cos(\theta)}$$

Taking this integral gives the same expression only with the opposite sign. I am not sure where I lost the sign.