# Convolution of radially symmetric Gaussian and with exponential and power-law

I wish to calculate a 3D convolution between two radially symmetric functions, $$f(\vec r) = \exp(-3 r^2/(4a^2))$$ and $$g(\vec r) = A e^{-b r}r^n$$ where $$n \in \mathbb{Z}$$ and $$r = |\vec r|$$.

Essentially, this corresponds to the integral of the form, $$f(r)\otimes_{3d}g(r)=A\int d^3 r'\;e^{-3(r-r')^2/4a^2}(r')^n e^{-br'}$$ One way to evaluate this integral would be chose the orientation along z-axis and write the integral in spherical polar coordinates,

$$\int_0^\infty dr'(r')^{n+2}e^{-br'}\int_0^\pi d\theta\sin(\theta)\exp{\left[-\frac{3(r^2+r'^2-2rr'\cos(\theta)}{4a^2}\right]}$$

I am stuck at this point on how to evaluate this integral further to obtain an expression for general integer values of $$n \ge 0$$

Another approach worth trying may be using the convolution property of the Fourier Transform.

The 3-D cartesian Fourier Transform: $$F(x, y, z) = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x,y,z) e^{-2\pi i (xu + yv + zw)} \space dx \space dy \space dz$$ when there is 3-D spherical symmetry, can be expressed as: $$F(q) = 4\pi \int_0^\infty f(r)r^2\mathrm{sinc}(2qr)dr$$ with the inverse transform expressed as: $$f(r) = 4\pi \int_0^\infty F(q)q^2\mathrm{sinc}(2rq)dq$$ with $$\mathrm{sinc}(x) \equiv \dfrac{\sin(\pi x)}{\pi x}$$

So starting with the case $$n = 0$$

$$f(r) = e^{-\frac{3}{4\pi a^2}\pi r^2} \quad \text{so} \quad F(q) = \sqrt{\dfrac{4\pi a^2}{3}}e^{-\frac{4\pi a^2}{3}\pi q^2}$$ $$g(r) = Ae^{-br} \quad \text{so} \quad G(q) = A\dfrac{8\pi b}{\left[b^2+(2\pi q)^2\right]^2}$$

(For $$n > 0$$, a derivation of a derivative property for the 3-D Fourier Transform will probably be needed to get $$G(q)$$.)

So now to attempt the convolution for the $$n= 0$$ case: \begin{align*}(f *g)(r) &= \mathscr{F}^{-1}\left\{F(q)G(q)\right\} \\ \\ &= 4\pi \int_0^\infty \sqrt{\dfrac{4\pi a^2}{3}}e^{-\frac{4\pi a^2}{3}\pi q^2} A\dfrac{8\pi b}{\left[b^2+(2\pi q)^2\right]^2} q^2 \mathrm{sinc}(2rq) dq\\ \\ &= \dfrac{32\pi Aab}{r}\sqrt{\dfrac{\pi}{3}}\int_0^\infty e^{-\frac{4\pi a^2}{3}\pi q^2} \dfrac{1}{\left[b^2+(2\pi q)^2\right]^2} q \sin(2\pi rq) dq \end{align*}

And that integral looks a little more tractable than the one in the original question. However it is still not obvious if one can get a closed form solution using substitutions and integration by parts.

Making an attempt at working the original integral in the question

\begin{align*} (f*g)(r) &= \int_0^\infty dr'(r')^{n+2}e^{-br'}\int_0^\pi d\theta\sin(\theta)\exp{\left[-\frac{3(r^2+r'^2-2rr'\cos(\theta))}{4a^2}\right]}\\ \\ &= \exp{\left[-\frac{3}{4a^2}r^2\right]}\int_0^\infty dr'(r')^{n+2}\exp{\left[-\frac{3}{4a^2}r'^2-br'\right]}\int_0^\pi d\theta\sin(\theta)\exp{\left[\frac{6rr'\cos(\theta)}{4a^2}\right]}\\ \\ &= \exp{\left[-\frac{3}{4a^2}r^2\right]}\int_0^\infty dr'(r')^{n+2}\exp{\left[-\frac{3}{4a^2}r'^2-br'\right]}\int_{-1}^{1} du\exp{\left[\frac{6rr'u}{4a^2}\right]}\\ \\ &= \exp{\left[-\frac{3}{4a^2}r^2\right]}\int_0^\infty dr'(r')^{n+2}\exp{\left[-\frac{3}{4a^2}r'^2-br'\right]}\left( \frac{4a^2}{6rr'}\exp{\left[\frac{6rr'u}{4a^2}\right]}\biggr\rvert_{-1}^1 \right)\\ \\ &= \frac{4a^2}{6r} \exp{\left[-\frac{3}{4a^2}r^2\right]}\int_0^\infty dr'(r')^{n+1}\exp{\left[-\frac{3}{4a^2}r'^2-br'\right]}\left[\exp\left(\frac{6r}{4a^2}r'\right)-\exp\left(-\frac{6r}{4a^2}r'\right)\right]\\ \\ &= \frac{4a^2}{6r} \exp{\left[-\frac{3}{4a^2}r^2\right]}\int_0^\infty dr'(r')^{n+1}\left(\exp{\left[-\frac{3}{4a^2}r'^2+\left(-b+\frac{6r}{4a^2}\right)r'\right]}-\exp{\left[-\frac{3}{4a^2}r'^2+\left(-b-\frac{6r}{4a^2}\right)r'\right]}\right)\\ \end{align*}

and then maybe complete the square on the exponents that are quadratic in $$r'$$. I think one winds up with integrals that can actually be expressed in terms of exponentials or error functions, depending on whether $$n$$ is even or odd.