Fourier Transform of Gaussian Function

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


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

Then:


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

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

Proof
By the definition of a Fourier transform

Taking the derivative with respect to $\zeta$, we have:

Integrating by parts, we have:

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

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

We now have the following:

We solve the differential equation as follows:

We can solve for A by setting $\zeta = 0$

Therefore:


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