# A new favorite equation?

In my differential equations class, we noted the function:
$f(x) = e^{-x^2}$

and how it looks really neat when plotted, but in a normal calculus course you'll probably be told that you can't integrate it. We then noted that:

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

Whaaaat? The proof is fascinating.

$\begin{eqnarray} let\ I &= \int_{-\infty}^{\infty} e^{-x^2} dx = \int_{-\infty}^{\infty} e^{-y^2} dy \ with\ x=y \\ I^2 =& (\int_{-\infty}^{\infty} e^{-x^2} dx) * (\int_{-\infty}^{\infty} e^{-y^2} dy) \\ =& \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(x^2 + y^2)} dx dy \end{eqnarray}$

Notice we're integrating over the entire x-y plane (with x = y), we can therefore switch to Polar coordinates and integrate over the entire polar plane. (Proof that the x = y invariance holds is left to the reader.) We note the conversion formula:

$x^2 + y^2 = r^2$

The typical formula for integrating all over the polar plane is:

$\int_0^\tau \int_0^{\infty} f(r, \theta) r\ dr\ d\theta$
$Note\ \tau = 2\pi$

(Tau for the win!)

So we have

$\int_0^\tau \int_0^{\infty} e^{-r^2} r\ dr\ d\theta$

Evaluating the inner integral (we add a factor of two inside by pulling a factor of 1/2 outside):

$\frac{1}{2} \int_0^{\infty} e^{-r^2} 2r\ dr = -\frac{1}{2} e^{-r^2} |_0^{\infty} = \frac{1}{2}$

Evaluating the outer integral becomes trivial.

$\int_0^\tau \frac{1}{2} d\theta = \frac{1}{2} \tau = \pi$

And since we were solving for I2,

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

Isn't that neat? Another neat property is that if we normalize it so that the integral is 1, we get

$f(x) = e^{-\pi x^2}$
$\int_{-\infty}^{\infty} e^{-\pi x^2} dx = 1$

which has the interesting property that its Fourier Transform is equal to itself. (Which is why you would choose that normalization.) The proved equation may become my favorite, upsetting Euler's...

#### Posted on 2011-04-05 by Jach

Tags: math

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