Use of Symmetry Arguments I am trying to solve the problem
$$\int d^3x\,e^{i{\bf a}\cdot{\bf x}}e^{-br^2}$$
using symmetry arguments. Could someone direct me to a similar question or guide me through a similar problem so I can learn the method to solving problems like this?
edit:
Here are the steps I've taken:
$$\int d^3x\,e^{i{\bf a}\cdot{\bf x}}e^{-br^2}$$
$$\int_V r^2\sin\theta\, e^{iar\sin^2\theta\cos^2\phi}e^{-br^2}dr\,d\theta\, d\phi,$$
and I think I can make the argument that $\theta = \pi /2$ and $\phi = 0$, but I don't know if or how that will affect the integral.
edit 2:
After receiving advice from Daniel and Cosmas, I see now why spherical coordinates are not the correct move. Following the steps detailed by Cosmas,
$$\int_0^ze^{iaz-bz^2}dz\int_0^xe^{-bx^2}dx\int_0^ye^{-by^2}dy$$
$$= \int_0^ze^{iaz-bz^2}dz\left( \frac{\pi}{4b}erf(\sqrt{b}x)erf(\sqrt{b}y)\right)$$
$$=\left( \frac{\sqrt{\pi}e^{-\frac{a^2}{4b}}erf(\sqrt{b}z-\frac{ia}{2\sqrt{b}})}{2\sqrt{b}} \right)\left( \frac{\pi}{4b}erf(\sqrt{b}x)erf(\sqrt{b}y)\right),$$
which I assume you would solve further given more information on the values of x, y, and z, since small valued and large valued error functions have different expansions. Cosmas recommended removing the b term immediately but I did it in parts using u substitution.
 A: This is the product of the Fourier transforms of three Gaussians. But, of course, with substantial symmetry, so it would be a waste to perform all three.
First, by redefining $\sqrt{b} ~{\mathbf x}\mapsto {\mathbf x}$, scale b out of the problem,
$$\int d^3x\,e^{i{\bf a}\cdot{\bf x}}e^{-br^2}= b^{-3/2} \int d^3x\,e^{i{{\bf a}\over \sqrt{b}}\cdot{\bf x}}e^{-r^2}.$$
Your then note that, since you integrate over all directions, your may choose your z -axis to be in the direction of a, that is ${\mathbf a}= a\hat {\mathbf z} $,
$$ b^{-3/2} \int d^3x\,e^{i{{\bf a}\over \sqrt{b}}\cdot{\bf x}}e^{-r^2}= b^{-3/2} \int\!\! dz~ e^{i{ a\over \sqrt{b}} z}e^{-z^2}\int\!\! dx dy ~~e^{-x^2-y^2}.$$
So you have two Gaussians and the Fourier transform of a Gaussian, which is also a Gaussian with its width related inversely to the original width.  You may finish up.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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$\ds{\bbox[5px,#ffd]{\int\dd^{3}x\,\expo{\ic{\bf a}\cdot{\bf x}}
\expo{-br^{2}}}:\ {\Large ?}}$.

\begin{align}
&\bbox[5px,#ffd]{\int\dd^{3}x\,\expo{\ic{\bf a}\cdot{\bf x}}
\expo{-br^{2}}} =
\int_{0}^{\infty}\expo{-br^{2}}
\pars{\int_{\Omega}\expo{\ic{\bf a}\cdot{\bf x}}{\dd\Omega \over 4\pi}}
4\pi r^{2}\,\dd r
\\[5mm] = &\
\int_{0}^{\infty}\expo{-br^{2}}
\bracks{\sin\pars{ar} \over ar}4\pi r^{2}\,\dd r
\,\,\,\stackrel{\large ar\ \mapsto\ r}{=}\,\,\,
\left.{2\pi \over a^{3}}\int_{-\infty}^{\infty}\expo{-\mu r^{2}}
\bracks{\sin\pars{r} \over r}r^{2}\,\dd r
\,\right\vert_{\large\ \mu\ \equiv\ b/a^{2}}\label{1}\tag{1}
\\[5mm] = &\
{2\pi \over a^{3}}\int_{0}^{\infty}\expo{-\mu r^{2}}
\bracks{{1 \over 2}\int_{-1}^{1}\expo{\ic kr}\dd k}r^{2}\,\dd r
\\[5mm] = &\
{\pi \over a^{3}}\int_{-1}^{1}\int_{-\infty}^{\infty}
\exp\pars{-\mu\bracks{r - {k \over 2\mu}\,\ic}^{2} - {k^{2} \over 4\mu}}
r^{2}\,\dd r\,\dd k
\\[5mm] = &\
{\pi \over a^{3}}\int_{-1}^{1}\exp\pars{-\,{k^{2} \over 4\mu}}
\int_{-\infty}^{\infty}
\exp\pars{-\mu r^{2}}
\pars{r^{2} + {k \over \mu}\,r\,\ic - {k^{2} \over 4\mu^{2}}}\,\dd r\,\dd k
\\[5mm] = &\
{\pi \over a^{3}}\int_{-1}^{1}\exp\pars{-\,{k^{2} \over 4\mu}}
\bracks{%
\underbrace{\int_{-\infty}^{\infty}\exp\pars{-\mu r^{2}}r^{2}\,\dd r}
_{\ds{\root{\pi} \over 2\mu^{3/2}}}\ -\
{k^{2} \over 4\mu^{2}}\
\underbrace{\int_{-\infty}^{\infty}\exp\pars{-\mu r^{2}}\,\dd r}
_{\ds{\root{\pi} \over \mu^{1/2}}}}\,\dd k
\\[5mm] = &\
{\pi \over a^{3}}{\root{\pi} \over 2\mu^{3/2}}\
\underbrace{\int_{-1}^{1}\exp\pars{-\,{k^{2} \over 4\mu}}
\pars{1 - {k^{2} \over 4\mu}}\dd k}
_{\ds{\exp\pars{-\,{1 \over 4\mu}} + \root{\pi\mu}\,\mrm{erf}\pars{1 \over 2\root{\mu}}}}
\\[5mm] = &\
\bbx{{1 \over 2}\pars{\pi \over b}^{3/2}\exp\pars{-\,{a^{2} \over 4b}} +
{\pi^{2} \over 2ab}\,\mrm{erf}\pars{a \over 2\root{b}}}
\\ &
\end{align}
