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.

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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:

$$\int\int\int{e^{-i(r_1q_1+r_2q_2+r_3q_3)}}e^{-a^2(r_1^2+r_2^2+r_3^2)}dr_1dr_2dr_3$$

This equals to the product of evaluating each integral separately: $$\int{e^{-ir_1q_1-a^2r_1^2}}dr_1\int{e^{-ir_2q_2-a^2r_2^2}}dr_2\int{e^{-ir_3q_3-a^2r_3^2}}dr_3$$

Hence, you got three Fourier transform of a Gaussian.

A Fourier transform of each Gaussian is given by: $$\int_{-\infty}^\infty{e^{-irq-a^2r^2}}dr=\sqrt{\frac{\pi}{a}}e^{-\frac{q^2}{4a}}$$

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:

$$\int\int\int{e^{-i(r_1q_1+r_2q_2+r_3q_3)}}e^{-a^2(r_1^2+r_2^2+r_3^2)}dr_1dr_2dr_3=(\frac{\pi}{a})^{1.5}e^{-\frac{q_1^2+q_2^2+q_3^2}{4a}}$$

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Many thanks, Gil –  Glen Feb 18 at 14:53

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 r^2} j_0(q r) \\ &= \frac{4 \pi}{q} \int_0^{\infty} dr \: r e^{-a r^2} \sin{q r} \\ &= \sqrt{\frac{\pi^3}{a^3}} e^{-\frac{q^2}{4 a}} \end{align}

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Many thanks, rigordonma –  Glen Feb 18 at 14:56