Jach's personal blog

(Largely containing a mind-dump to myselves: past, present, and future)
Current favorite quote: "Supposedly smart people are weirdly ignorant of Bayes' Rule." William B Vogt, 2010

A new favorite equation?

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

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:

[math]\int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi}[/math]

Whaaaat? The proof is fascinating.


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

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:

[math]x^2 + y^2 = r^2[/math]

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

[math]\int_0^\tau \int_0^{\infty} f(r, \theta) r\ dr\ d\theta[/math]
[math]Note\ \tau = 2\pi[/math]

(Tau for the win!)

So we have

[math]\int_0^\tau \int_0^{\infty} e^{-r^2} r\ dr\ d\theta[/math]

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

[math]\frac{1}{2} \int_0^{\infty} e^{-r^2} 2r\ dr = -\frac{1}{2} e^{-r^2} |_0^{\infty} = \frac{1}{2}[/math]

Evaluating the outer integral becomes trivial.

[math]\int_0^\tau \frac{1}{2} d\theta = \frac{1}{2} \tau = \pi[/math]

And since we were solving for I2,

[math]I = \sqrt{\pi} = \int_{-\infty}^{\infty} e^{-x^2} dx[/math]

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

[math]f(x) = e^{-\pi x^2}[/math]
[math]\int_{-\infty}^{\infty} e^{-\pi x^2} dx = 1[/math]

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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