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### 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].

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/
```
Associated Topics:
College Calculus

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