## Neat Integrations: $$\int_{-\infty}^{\infty}e^{-x^2}$$, and Fourier Transform of the Gaussian Curve

My physics prof once said to our class: as physicists, the two things we can do are Taylor expand and Fourier Transform. Indeed, not only is the Fourier Transform a beautiful piece of mathematics, but also an indispensible tool in modern physics. Today, let’s examine the Gaussian Function that we all know and love and its Fourier Transform, which gives rise to the clever neat integral of $$\int_{-\infty}^{\infty}e^{-x^2}$$ (The Gaussian Integral).

Depending on the field you are in, or the books you consult, you will probably see a handful of different forms of the Fourier Transform. This article concerns itself with Fourier Transform of the following form:

$$\mathscr{F}f(x)=\hat{f}(\kappa)=\int_{-\infty}^{\infty}\frac{d\kappa}{2\pi}e^{-i\kappa x}f(x)$$
$$\mathscr{F}^{-1}\hat{f}(\kappa)=f(x)=\int_{-\infty}^{\infty}\frac{d\kappa}{2\pi}e^{i\kappa x}\hat f(x)$$

We know the Gaussian function at 0 takes the following form:

$f(t)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{t^2}{2\sigma^2}}$

It is simple enough to rewrite it in Fourier basis:

$\hat{f}(\omega)=\int_{-\infty}^{\infty}e^{-i\omega t}[ \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{t^2}{2\sigma^2}} ] dt$

Which we can then extract a factor of $$\frac{1}{\sigma\sqrt{2\pi}}$$

$\frac{1}{\sigma\sqrt{2\pi}} \int_{-\infty}^{\infty} \exp \left (-i\omega t -\frac{t^2}{2\sigma^2}\right )dt$

By means of completing the square, we can get:

$\frac{1}{\sigma\sqrt{2\pi}} \int_{-\infty}^{\infty} \exp\left[ \left ( \frac{t}{\sigma\sqrt{2}}+\frac{i\omega t}{\sqrt{2}}-\frac{i\omega \sigma}{\sqrt{2}}\right )^2\right ]dt$

Extracting the part of the exponent where there is no \$$$t$$ dependence gives:

$\frac{1}{\sigma\sqrt{2\pi}} e^{\frac{-\omega^2\sigma^2}{2}} \int_{-\infty}^{\infty} \exp\left[ \left ( \frac{t}{\sigma\sqrt{2}}+\frac{i\omega t}{\sqrt{2}}\right )^2\right ]dt$

We can now integrate by means of substitution:

$u=\frac{t}{\sigma\sqrt{2}}+\frac{i\omega\sigma}{\sqrt{2}}$

$du=\frac{1}{\sigma\sqrt{2}}dt$

$\frac{1}{\sigma\sqrt{2\pi}} e^{\frac{-\omega^2\sigma^2}{2}} \int_{-\infty}^{\infty} e^{-u^2} \sigma\sqrt{2}du$

This is the moment we have all been waiting for, the integral of

$\int e^{-x^2}dx$ is famously famously non-integratable. It is in fact given a name as the error function. But we shall attempt to solve it nonetheless. (p.s. we will solve it, and similar to many other seemingly complicated maths problems, $$\pi$$ magically appears).

Let us define the integral to be:

$I=\int_{-\infty}^{\infty}dx e^{-x^2}$

The following steps I have chosen to take are, in my opinion, sound, but for those of you that are prescriptive mathematicians, kindly look away.

$\int_{-\infty}^{\infty}e^{-x^2} dx \int_{-\infty}^{\infty}e^{-y^2} dy = I^2$

For our purpose, the integral of the function with regards to $$x$$ and $$y$$ are identical. And since they involve different variables, we can combine them into a double integral.

$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \exp [-x^2-y^2] dx dy=I^2$

We shall now engage in what I would call mathematical gymnastics, just for fun, let us now switch to polar coordinates! (It’s actually not just for fun, there is a reason behind this madness.)

$\int_0^{2\pi} \int_{0}^{\infty}dr \exp [-r^2]d\phi=I^2$

It is quite clear that our integral lacks any $$\phi$$ dependence, so we can drop the integral.

$\int_{0}^{\infty}dr \exp [-r^2] =I^2$

We may now integrate by means of substitution:

$z=r^2$

$dz=2rdr$

$\int_{0}^{\infty}\frac{1}{2r}e^{-z}dz$

An integral which we know to converge to $$\pi$$. Thus:

$I^2=2\pi \frac{1}{2}$

And lo and behold:

$I=\sqrt{\pi}$

Plugging it back will give us the Fourier transform of the Gaussian curve:

$\hat{f}(\omega)=\frac{\sqrt{\pi}}{\pi}\exp \left [ \frac{-\omega^2\sigma^2}{2}\right ]$

Albeit this wasn’t much of an article, but rather an exhausting mathematical excercise, nevertheless we encountered some interesting stuff along the way. I think this problem best illustrates the proverb that differentiation is a science, where as integration is an art. We will end with a mesmerising picture. I have included the complex plot for the Fourier transform we arrived at below. Feel free to play with the $$\sigma$$ values, you get some interesting graphs.

I am thinking of something along the lines of solution to the dampened harmonic oscillator with a bit of inhomogeneity for the next article. Let me know if you spotted any mistakes or if you have any suggestions. You can use regular in-text and display $$\LaTeX$$ delimiters in the comment (Not dollar signs, I can’t seem to get them to work with MathJax.)