## Complex-Analytic Proof of Fourier Transform of Gaussian

Many weeks ago, I wrote a post on how to compute the Fourier transform of the Gaussian function defined by $f(x):=e^{-\pi x^{2}}$ using the methods of ordinary differential equations. Specifically, we showed that

$\widehat{f}(\xi)=\int_{\mathbb{R}}e^{-\pi x^{2}}e^{-2\pi i\xi x}dx=e^{-\pi\xi^{2}},\indent\forall\xi\in\mathbb{R}$

In that post, I promised to later show how to prove the same result by means of complex analysis–Cauchy’s theorem for closed contours. As I keep my promises, though not necessarily with the promptness others might like, I will now give this second proof.

Consider the Fourier transform of the function $f(x):=e^{-ax^{2}}$, for $a > 0$ fixed:

$\displaystyle\widehat{f}(\xi)=\int_{\mathbb{R}}e^{-ax^{2}}e^{-2\pi i \xi x}dx=\int_{\mathbb{R}}e^{-a(x^{2}+2\pi i \xi a^{-1}x)}dx,\indent\forall \xi\in\mathbb{R}$

We can complete the square in the exponent to obtain

$\displaystyle\widehat{f}(\xi)=\int_{\mathbb{R}}e^{-a(x+\pi i \xi a^{-1})^{2}-\pi^{2}\xi^{2}a^{-2}}dx=e^{-\pi^{2}\xi^{2}a^{-1}}\int_{\mathbb{R}}e^{-a(x+i\pi\xi a^{-1})^{2}}dx,\indent\forall\xi\in\mathbb{R}$

Consider the entire function $f(z)=e^{-az^{2}}$, where $a>0$ is a fixed real number. Let $\mathcal{C}_{R}$ denote the positively oriented rectangular contour with vertices $\pm R,\pm R+i\pi\xi a^{-1}$, for $R>0$. By Cauchy’s theorem,

$\displaystyle\int_{\mathcal{C}_{R}}f(z)dz=0$

By the dominated convergence theorem,

$\begin{array}{lcl}\displaystyle\int_{\mathbb{R}}e^{-a(x+i\pi\xi a^{-1})^{2}}dx&=&\displaystyle\lim_{R\rightarrow\infty}\int_{-R}^{R}e^{-a(x+i\pi\xi a^{-1})^{2}}dx\\&=&\displaystyle\lim_{R\rightarrow\infty}\int_{-R}^{R}e^{-a x^{2}}dx+i\int_{0}^{\pi\xi a^{-1}}e^{-a(R+it)^{2}}dx-i\int_{0}^{\pi\xi a^{-1}}e^{-a(-R+it)^{2}}dt\end{array}$

Let’s examine the growth of the second and third integral as $R\rightarrow\infty$.

$\displaystyle\left|\int_{0}^{\pi\xi a^{-1}}e^{-a(\pm R+it)^{2}}idt\right|\leq e^{-aR^{2}}\int_{0}^{\pi\xi a^{-1}}e^{at^{2}}dt\rightarrow 0, R\rightarrow\infty$

It is well-known identity that $\int_{\mathbb{R}}e^{-x^{2}}dx=\sqrt{\pi}$, so by making the change of variable $y=\sqrt{a}x$, we obtain

$\begin{array}{lcl}\displaystyle\lim_{R\rightarrow\infty}\int_{-R}^{R}e^{-ax^{2}}dx=a^{-\frac{1}{2}}\lim_{R\rightarrow\infty}\int_{-\sqrt{a}R}^{\sqrt{a}R}e^{-y^{2}}dy=\displaystyle a^{-\frac{1}{2}}\int_{\mathbb{R}}e^{-y^{2}}dy&=&\sqrt{\dfrac{\pi}{a}}\end{array}$,

where we use the dominated convergence theorem to obtain that penultimate equality. We conclude that

$\displaystyle\widehat{f}(\xi)=\sqrt{\dfrac{\pi}{a}}e^{-\pi^{2}\xi^{2}a^{-1}},\indent\forall \xi\in\mathbb{R}$

In particular, if $a=\pi$, then $f$ is a fixed point of the Fourier transform.