Proving that the Guassian Distribution integral converges to 1

  • #1

Homework Statement



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

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

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.
 
Last edited:

Answers and Replies

  • #2
938
9
There's another double integral trick you can do, which is a little bit easier. When you have

[tex]I = \int dx e^{-x^2} [/tex]
and
[tex] I^2 = \int dx \int dy e^{-x^2 - y^2} [/tex]

do a substitution y = x s, dy = x ds.
 
  • #3
So basically, it ends up like:

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

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

[tex]
y = \frac{(x- \mu )}{2 \sigma} s
[/tex]

[tex]
dy = \frac{(x- \mu )}{2 \sigma} ds
[/tex]

Then I get this:

[tex]
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
[/tex]

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.
 
  • #4
To make it work since the constant is

[tex]\frac{1}{\sigma \sqrt{2 \pi}}[/tex]

I need the integral to equal

[tex]\sigma \sqrt{2 \pi}[/tex]

for it to equal 1
 
  • #5
938
9
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).
 

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