Proving that the Guassian Distribution integral converges to 1

[nickclarson]

Homework Statement

Prove that the integral of the Guassian Distribution converges to 1:

$$\int_{- \infty}^{\infty} \frac{1}{\sigma \sqrt{2 \pi}} e^{- \frac{(x- \mu )^2}{2 \sigma ^2}} dx = 1$$

Homework Equations

none

The Attempt at a Solution

So I get that I can pull the constants out for the first step, but I have no clue where to go from there. I looked online for a solution and read about having to convert it to polar or something to solve it so now I am really confused. I thought it might work with a simple substitution, one that I just can't think of.

Any help greatly appreciated.

 

[clamtrox]

There's another double integral trick you can do, which is a little bit easier. When you have

$$I = \int dx e^{-x^2}$$
and
$$I^2 = \int dx \int dy e^{-x^2 - y^2}$$

do a substitution y = x s, dy = x ds.

 

[nickclarson]

So basically, it ends up like:

$$I^2 = \frac{1}{\sigma \sqrt{\pi}} \int dx \int e^{- \frac{(x- \mu )^2}{2 \sigma ^2} - y^2} dy$$

But now I'm having trouble with the substitution. Would my substitution look like this?

$$y = \frac{(x- \mu )}{2 \sigma} s$$

$$dy = \frac{(x- \mu )}{2 \sigma} ds$$

Then I get this:

$$I^2 = \frac{1}{\sigma \sqrt{\pi}} \int dx \int \frac{(x- \mu )}{2 \sigma} e^{- \frac{(x- \mu )^2}{2 \sigma ^2} - (\frac{(x- \mu )}{2 \sigma} s)^2} ds$$

I feel like I'm doing something horribly wrong and making it more complicated, pardon my ignorance. Thanks for your help so far though, I appreciate it.

 

[nickclarson]

To make it work since the constant is

$$\frac{1}{\sigma \sqrt{2 \pi}}$$

I need the integral to equal

$$\sigma \sqrt{2 \pi}$$

for it to equal 1

 

[clamtrox]

Get rid of all the constants first. Notice that you can shift x-mu to x since you're integrating over the entire real line, and then do a substitution to make exponential of the form e^(-x^2).