Indefinite Integral

Date: 6/1/96 at 0:48:9
From: Anonymous
Subject: Integrals

I just want to find the indefinite integral:

(
| e^(x*x) dx    "e to the x squared"
)

AND:

(
| e^(-x*x) dx "negative x squared"
)

Thank you.

Date: 6/1/96 at 8:11:43
From: Doctor Anthony
Subject: Re: Integrals

The integral of e^(x^2) cannot be done by elementary methods.  One way 
of proceeding is to write I^2 = [INT(e^(x^2)).dx]^2
                              = [INT(e^(x^2)).dx][INT(e^(y^2)).dy]
which can be written as a double integral over a region of the xy 
plane.

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

Now change to polar coordinates, remembering that an elementary unit 
of area dx.dy becomes an elementary unit of area r.d(theta).dr  This 
is known as the Jacobian of the transformation if you want to look up 
change of variables in multiple integrals.  

Also x^2+y^2 = r^2

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

From here we can integrate first with respect to r.

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

Put r^2 = u, so 2r.dr = du and r.dr = (1/2)du

 INT[e^(r^2).r.dr] = (1/2)INT[e^(u).du]
                   = (1/2)e^(u)
                   = (1/2)e^(r^2) + const.
                   = f(r)

Now if we had limits for r we could evaluate this expression, but in 
any case treat r as a constant for the integration with respect to 
theta.

Returning to the main integral

 I^2 = f(r).INT[d(theta)]

     = f(r)[theta + const.]

And finally take the square root

 I = sqrt{f(r)[theta+const.]}


In the case of integrating e^(-x^2) we proceed in exactly the same 
manner. The only difference is that f(r) would be (-1/2)e^(-r^2) + 
const.  

In this case if we are integrating x from 0 to infinity, the region of 
integration is the positive quadrant of the xy plane, and r is from 0 
to infinity while theta is from 0 to pi/2.

We would have f(r) = (-1/2)[0-1] = 1/2

and  I^2 = (1/2)[pi/2 - 0]
         = pi/4

Giving I = (1/2)sqrt(pi)

This last integral is, of course, used in evaluating areas under the 
Normal distribution curve. 

-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/