Integrating the Standard Distribution Curve

Date: 06/25/99 at 20:39:31
From: John
Subject: Integrating the standard distribution curve

Does anyone here know how to integrate the standard distribution curve 
for statistics?

I need to find the integral from infinity to negative infinity of

     (e^((-x^2)/2))/((2pi)^(1/2))

This is a self-interest project. I may not even understand what the 
explanation means, but please write down everything needed to prove 
the answer as if I do. I will find some way to understand it.
 
Thank you

Date: 06/26/99 at 07:17:12
From: Doctor Jerry
Subject: Re: Integrating the standard distribution curve

Hi John,

It only takes a change of variable to relate the above to the integral

     I = int[0,oo,e^(-x^2)*dx].

I'll work with the latter. One can write

     I^2 = int[0,oo,e^(-x^2)*dx]*int[0,oo,e^(-x^2)*dx]

         = int[0,oo,int[0,oo,e^(-x^2)*e^(-y^2)*dx]*dy]

         = int[0,oo,int[0,oo,e^(-(x^2+y^2))*dx]*dy]

Now make a change of variable, to polar coordinates. This gives

     I^2 = int[0,pi/2,int[0,oo,r*e^(-r^2)*dr]*dt]

The inner integral is easy and the outer one is trivial.

This is the stardard way of evaluating I.

- Doctor Jerry, The Math Forum
  http://mathforum.org/dr.math/   

Date: 06/26/99 at 18:49:59
From: NYSE DOW JONES
Subject: Re: Integrating the standard distribution curve

Doctor Jerry,

Thanks for doing this, but could you please tell me what "0,oo" is? 
Could you please explain this to me as if this is a 20-point final 
exam question and you need to explain to your professor as clearly as 
possible? In other words, would you please show me more steps in 
between so that it is more comfortable for a person who only knows 
first year calculus to dig through. That way, I can look in books and 
figure out all this by myself more easily.

Thanks

Date: 06/27/99 at 07:58:19
From: Doctor Jerry
Subject: Re: Integrating the standard distribution curve

Hi John,

Short of trying to type out a complete course on multi-dimensional 
integration, I'll try to fill in some details.

The "oo" is meant to suggest the infinity symbol. So, when I wrote

     I = int[0,oo,e^(-x^2)*dx]

I meant the integral, from 0 to infinity, of e^(-x^2). Of course, this 
is shorthand for the limit as a -> oo of int[0,a,e^(-x^2)*dx].

Here's a copy of what I wrote, with some added comments.

     I^2 = int[0,oo,e^(-x^2)*dx]*int[0,oo,e^(-x^2)*dx]

The above, a key step, assumes that the improper integral 

     Int[0,oo,e^(-x^2)*dx]

converges to a finite number, called I. This is not hard to prove. The 
guts of the argument are: for a>1,

     int[0,a,e^(-x^2)*dx] = int[0,1,e^(-x^2)*dx]+int[1,a,e^(-x^2)*dx]
                          < 1+int[1,a,e^(-x)*dx] = 1+1/e-1/e^a.

In the second step

         = int[0,oo,int[0,oo,e^(-x^2)*e^(-y^2)*dx]*dy]

we take advantage of the fact that

     int[0,oo,e^(-x^2)*dx] = int[0,oo,e^(-y^2)*dy]

and the latter can be taken inside the other factor of the product, 
because it doesn't contain x.

The next step,

         = int[0,oo,int[0,oo,e^(-(x^2+y^2))*dx]*dy]

is clear.

Now we change variable, specifically,

     X = r*cos(t)
     y = r*sin(t),

At this point, you need to calculate the Jacobian of this change of 
variable. You need to look this up. It amounts to replacing dx*dy by 
r*dr*dt.

And the region of integration changes from the first quadrant, that 
is, x from 0 to oo and y from 0 to oo, to r from 0 to oo and t from 0 
to pi/2.

     I^2 = int[0,pi/2,int[0,oo,r*e^(-r^2)*dr]*dt]
 
The rest is one-dimensional integration (first-year calculus).

I hope this helps a bit.

- Doctor Jerry, The Math Forum
  http://mathforum.org/dr.math/