Integrating exp(-x^2)

Date: 11/09/2005 at 21:43:59
From: Pablo
Subject: integrating exp(-x^2)

I know that integral exp(-x^2) can be evaluated from negative infinity
to positive infinity and gives the result sqrt(pi), but it seems like
all the ways of doing this involve making it a double integral and
using polar coordinates, I was wondering if there is any way to do it
without making it a double integral.

I have tried shifting into polar coordinates with x = rcos(theta) and
then doing another substitution of u = cos^2(theta) but it just gets
really messy.  Maybe it can't be done but I would like a better
explanation for why you can shift it into a double integral, I mean it
makes intuitive sense but I haven't seen any rigorous proof or
reasoning as to why it's valid.

Date: 11/11/2005 at 01:46:22
From: Doctor Minter
Subject: Re: integrating exp(-x^2)

Hi Pablo!

I don't believe that there is any way to integrate this function 
without a change of variables, and thus a double-integral, but there 
is a way to do it with a simpler "u substitution."

Start with defining I as the value of the integral of e^(-x^2) dx 
from zero to infinity.  

Now square both sides of the equation, but instead of simply squaring 
the integral value, multiply it by the same integral, except replace 
x and dx with y and dy, respectively.  Remember that for a definite 
integral, you get a value that is independent of the index (variable 
name) that you choose.  So integrating e^(-x^2) dx is exactly the 
same as integrating e^(-y^2) dy over the same limits of integration.

So now we have

  I^2 = {INT [e^(-x^2) dx]}*{INT [e^(-y^2) dy]}, 

both from -infinity to +infinity, where INT is a single integral.

Which, by the properties of double integrals and the properties of 
exponential functions, is equivalent to

  I^2 = DOUBLE-INT {e^[-(x^2 + y^2)] dx dy, 

both from -infinity to +infinity, where DOUBLE-INT is a double integral.

We now change to polar coordinates, where 

  x^2 + y^2 = r^2, and

dx dy = r dr dT, by the Jacobian Transformation, where T is 
substituted for the usual theta for convenience.

Using these values in the double-integral formula, we now have

  I^2 = DOUBLE-INT [r e^(-r^2) dr dT].  
  
The new limits of integration are: zero to infinity for dr, and zero 
to 2*Pi for theta.  

Since the integrand has no theta dependence, we can replace INT(dT) 
by the difference between its limits, which is 2*Pi.  We can pull Pi 
out of the integral because it is a constant, but leave the 2 inside 
the dr integral for reasons that will be clear shortly.  We are left 
with

  I^2 = Pi*INT[ 2r e^(-r^2) dr ].

Now for the u substitution.  

  Let  u = r^2,
  so  du = 2r dr

  u = 0 when r = 0

and

  u = infinity when r = infinity, 

so the limits of integration do not change.

Our formula is now  

  I^2 = Pi*INT[ e^(-u) du ]

Integrating this is straightforward, and we are left with

  I^2 = Pi*[-e^(-u)], evaluated from zero to infinity.

This leaves us with

  I^2 = Pi*[0 - (-1)], which is also

  I^2 = Pi

Taking the square root of both sides says that the value of the 
original integral is I = sqrt(Pi).

I hope this has made things seem a little easier.  Please feel free 
to write again if you need any further assistance, or if you have any 
other questions.  Thanks for using Dr. Math!

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