3D Fourier transform I don't know how to evaluate an integral of the form 
$$\int d^3 r \exp(-i \vec r\cdot\vec q)\exp(-a^2 r^2)$$
where $a\in \mathbb R$.
Could anyone please teach me how to do this integral? 
Many thanks.
 A: Note that this integral is separable.
Assuming $\vec{r}=(r_1,r_2,r_3)$ and by $r^2$ you mean $\vec{r} \cdot \vec{r}=r_1^2+r_2^2+r_3^2$ and $\vec{q}=(q_1,q_2,q_3)$ the integral can be reformulated as:
\begin{equation}
\int\int\int{e^{-i(r_1q_1+r_2q_2+r_3q_3)}}e^{-a^2(r_1^2+r_2^2+r_3^2)}dr_1dr_2dr_3
\end{equation}
This equals to the product of evaluating each integral separately:
\begin{equation}
\int{e^{-ir_1q_1-a^2r_1^2}}dr_1\int{e^{-ir_2q_2-a^2r_2^2}}dr_2\int{e^{-ir_3q_3-a^2r_3^2}}dr_3
\end{equation}
Hence, you got three Fourier transform of a Gaussian.
A Fourier transform of each Gaussian is given by:
\begin{equation}
\int_{-\infty}^\infty{e^{-irq-a^2r^2}}dr=\sqrt{\frac{\pi}{a}}e^{-\frac{q^2}{4a}}
\end{equation}
So your integral all in all is equal to a product of those three Gaussians in $(q_1,q_2,q_3)$. Assuming I didn't make any mistake along the way:
\begin{equation}
\int\int\int{e^{-i(r_1q_1+r_2q_2+r_3q_3)}}e^{-a^2(r_1^2+r_2^2+r_3^2)}dr_1dr_2dr_3=(\frac{\pi}{a})^{1.5}e^{-\frac{q_1^2+q_2^2+q_3^2}{4a}}
\end{equation}
A: Use spherical coordinates: $x = r \sin{\theta} \cos{\phi}$, $y = r \sin{\theta} \sin{\phi}$, $z = r \cos{\theta}$, and
$$\mathrm{d^3}\vec{r} = r^2 \sin{\theta} \, dr \, d \theta \, d \phi $$
Assume a plane wave expansion of the form
$$e^{-i \vec r\cdot\vec q} = e^{-i q r \cos{\theta}} = \sum_{l=0}^\infty\mathrm i^l(2l+1)j_l(q r)P_l(\cos\theta)$$
where $j_0$ is a spherical Bessel function and $P_l$ is a Legendre polynomial.
Note that, as the integral is over a spherically symmetric function, the only term in the sum that does not vanish is the $l=0$ term.  Integrating over $\theta$ and $\phi$ produces a factor of $4 \pi$. We then have
$$\begin{align}\int_{\mathbb{R}^3} d^3 r \exp(-i \vec r\cdot\vec q)\exp(-a^2 r^2) &= 4 \pi \int_0^{\infty} dr \: r^2 e^{-a^2 r^2} j_0(q r) \\ &= \frac{4 \pi}{q} \int_0^{\infty} dr \: r e^{-a^2 r^2} \sin{q r} \\ &= \frac{\sqrt{\pi^3}}{a^3} e^{-\frac{q^2}{4 a^2}} \end{align}$$
A: The Fourier Transform of each Gaussian should be(referring to http://mathworld.wolfram.com/FourierTransformGaussian.html)
$$
   \int e^{−irq}e^{−a^2r^2} dr = \sqrt{\frac{\pi}{a^2}}e^{-\frac{q^2}{4a^2}}
$$
The 3d Fourier Transform is
$$
   (\frac{\pi}{a^2})^{3/2}e^{-\frac{q_1^2+q_2^2+q_3^2}{4a^2}}
$$
