The Fourier Transform of a Gaussian:

The definition of the Fourier transform is:

$$F(s) \equiv \int_{-\infty}^\infty f(x) \exp ( - i 2 \pi s x) d x$$

and the general Gaussian function is:

$$f(x) = {1 \over \sqrt{2 \pi }\sigma} \exp\biggl( - {x^2 \over 2 \sigma^2}\biggr)~.$$

It is convenient to consider the particular Gaussian function $f(x) = \exp ( - \pi x^2)$, for which $\sigma^2 = 1 / (2 \pi)$. 
$$F(s) = \int_{-\infty}^\infty \exp ( - \pi x^2) \exp ( -i 2 \pi s x) d x$$
$$F(s) = \int_{-\infty}^\infty \exp [ - \pi (x^2 + i 2 s x + s^2 - s^2) ] dx$$ $$F(s) = \exp ( - \pi s^2) \int_{-\infty + i s}^{\infty + i s} \exp [ - \pi (x + i s)^2 ] d(x + is)$$
$$F(s) = \exp ( - \pi s^2)\int_{-\infty}^\infty \exp ( - \pi x^2) dx$$
To evaluate this one-dimensional integral, we first break it into the product of two integrals and change the dummy variable in the second from $x$ to $y$ to suggest the two cartesian coordinates in a plane:
$$F(s) = \exp (- \pi s^2) \biggl [ \int_{-\infty}^\infty \exp ( - \pi x^2) dx \int_{-\infty}^\infty \exp ( - \pi y^2) dy \biggl]^{1/2}$$
$$F(s) = \exp (- \pi s^2) \biggl [ \int_{-\infty}^\infty \int_{-\infty}^{\infty} \exp [ - \pi (x^2 +y^2)] dx dy \biggr]^{1/2}$$
The clever trick is to introduce polar coordinates $r, \theta$ so $r^2 = x^2 + y^2$ and $dx dy = r dr d\theta$.  Then
$$F(s) = \exp (- \pi s^2) \biggl [ \int_{r = 0}^\infty \int_{\theta = 0}^{2\pi} \exp [ - \pi (r^2)] r dr d \theta \biggr]^{1/2}$$
and let $u \equiv \pi r^2$ so $du = 2 \pi r dr$:
$$F(s) = \exp (- \pi s^2) \biggl [ 2 \pi \int_{u = 0}^\infty \exp ( -u) {du \over 2 \pi} \biggr]^{1/2}$$
$$F(s) = \exp (- \pi s^2) \biggl [ -e^{-u} \big|_0^\infty \biggr]^{1/2}$$ $$F(s) = \exp (- \pi s^2)$$
Thus the Fourier transform of the Gaussian $f(x) = \exp (- \pi x^2)$ is the Gaussian $F(s) = \exp (- \pi s^2)$.