Fourier Transform of Gaussian Function

Theorem
Let $\map f x$ be defined as $\sqrt \pi$ times the Gaussian probability density function where $\mu = 0$ and $\sigma = \dfrac {\sqrt 2} 2$:


 * $\map f x = e^{-x^2}$

Then:


 * $\map {\hat f} s = \sqrt \pi e^{-\paren {\pi s }^2}$

where $\map {\hat f} s$ is the Fourier transform of $\map f x$.

Proof
By the definition of a Fourier transform:

Taking the derivative with respect to $s$, we have:

Integrating by parts, we have:

Let $u = e^{-2 \pi i x s}$ and $\rd v = x e^{-x^2} \rd x$

Then:
 * $\rd u = -2 \pi i s e^{-2 \pi i x s} \rd x$ and $v = -\dfrac 1 2 e^{-x^2}$

Hence:

We now have the following:

We solve for $A$ by setting $s = 0$:

Therefore:


 * $\map {\hat f} s = \sqrt \pi e^{-\paren {\pi s}^2}$