Associated Topics || Dr. Math Home || Search Dr. Math

### Algebraic Integration of Standard Normal Distribution Function

```
Date: 02/11/98 at 05:00:54
From: julian keites
Subject: Algebraic integration of standard normal distribution
function

I've tried inspection and partial integration and don't seem to get
the right results - I wonder if this is a function whose integral
cannot be solved algebraically? Is numeric integration the only option
or is there a known approximation?

f(Z)=(1/(srqt(2*PI))*e^((-Z^2)/2)
```

```
Date: 02/11/98 at 16:01:22
From: Doctor Anthony
Subject: Re: Algebraic integration of standard normal distribution
function

You need special techniques for handling this integral. Below is the
method of integrating under the complete curve. Areas at intermediate
values of z will be even more difficult.

INTEGRATING THE NORMAL CURVE

I will carry out the integral from 0 to infinity, and doubling the
result (since the graph is symmetrical about the y axis) will give the
total area from -infinity to +infinity.

Let  I = INT(0 to infinity)[e^(-x^2/2).dx]

This cannot be evaluated using elementary methods, so we proceed as
follows:

I^2 = [INT(0 to infinity)(e^(-x^2/2).dx]^2

= INT(0 to infinity){e^(-x^2/2)dx) INT(0 to infinity)(e^(-y^2/2).dy)

= INT(dx INT[e^(-(x^2+y^2)/2).dy]   all integrals are from 0 to
infinity.

= INT.INT[e^(-(x^2+y^2)/2).dx.dy]

where the region of integration is the whole of the positive quadrant
of the xy plane.

If we transform to polar coordinates, x^2+y^2 = r^2, also the element
of area dx.dy is now given in polar coordinates by element of area
r.d(theta).dr  The limits of integration will be 0 to infinity for r,
and 0 to pi/2 for theta.

So our integral now becomes

I^2 = INT.INT[e^(-r^2/2)r.dr.d(theta)]

= INT[d(theta)INT[e^(-r^2/2)r.dr]

make the substitution  r^2/2 = u    then r.dr = du and the inner
integral becomes

INT[e^(-u).du]   for u from 0 to infinity.

= - e^(-u)  = -(0 - 1) = 1

So now we have

I^2 = INT[d(theta)] from 0 to pi/2

=  [theta] from 0 to pi/2

=  pi/2

and so I = sqrt(pi/2)

and   2I = sqrt(2pi)  giving area from -infinity to +infinity.

-Doctor Anthony,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Calculus

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/