Roger's Equations

The first time I saw the Gaussian Integral, I was mystified. How could an integral that looks like this (see below) be equal to the square root of π? How did they figure that out?

Unfortunately I wasn't shown the relatively straight forward derivation of the solution, which pretty much kills the mystery, until a few years later. I'll provide this derivation below. Start with the left hand side of the expression above. Square it and take the square root. Basically in the first step we are saying the following;

In the same way that this is true in the above, the same trick applies below;

=

=

To solve the integral we switch to polar coordinates where

x = rcosθ
y = rsinθ

x² = r²cos²θ
y² = r²sin²θ
x² + y² = r²cos²θ + r²sin²θ = r²(cos²θ + sin²θ) = r²(1) = r²

dA = dxdy = rdrdθ

so substituting and integrating we get;

Notice above we are making use of e^-∞= 0 and e⁰ = 1.

The π comes from the integration range of 0 to 2π for θ. Not out of thin air like I thought for two years. The gaussian integral is pretty useful, showing up in probability, quantum mechanics, scattering problems, etc. A more generalized form of the equation is;

There are all types of tricks to find the solutions of closely related integrals like taking the derivative or substituting trig functions for the x, so in a way, this trick above allows us to solve a whole range of Integrals exactly, which is very useful.

Thanks to the following sites for their help:

http://www.umich.edu/~chem461/Gaussian%20Integrals.pdf
http://mathworld.wolfram.com/GaussianIntegral.html
http://www.answers.com/topic/gaussian-integral