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Trying to Integrate f(x) = exp(-ax^2)


Date: 11/23/2001 at 09:07:34
From: Craig Robinson
Subject: Integration

How would you go about integrating a function of the form 
f(x) = exp(-ax^2)?


Date: 11/23/2001 at 09:32:53
From: Doctor Jubal
Subject: Re: Integration

Hi Craig,

Thanks for writing to Dr. Math.

The function f(x) = exp(-ax^2) doesn't have an antiderivative that can 
be expressed in terms of other commonly used functions. What this 
means is you can't analytically integrate it.

Yet, the function pops up a lot in many places where it would be nice 
to be able to integrate it. So mathematicians have come up with 
several workarounds. Your options are:

  (i) Hope your limits of integration are zero and infinity.
      For these limits of integration, the integral does have an 
      anlytical solution. The integral of exp(-ax^2) from 0 to 
      infinity is (1/2)sqrt(pi/a). If you'd like to see a proof of 
      this, feel free to write back.

 (ii) Solve it numerically.
      So your limits of integration aren't zero and infinity. Use 
      Simpson's Rule or the numerical integration method of your 
      choice to approximate the integral.

(iii) Use the error function.
      Mathematicians like this integral a lot. They like it so much, 
      in fact, that they defined a function just so this integral 
      would have an "analytical" solution. The error function is 
      defined 

                   2      /x
      erf(x) = ---------  |  exp[-t^2] dt
                sqrt(pi)  /0

      and

      /x                 sqrt(pi/a)
      |  exp[-at^2] dt = ---------- erf(x)
      /0                     2

      The error function has been evaluated numerically at many, many 
      points, and the results tabulated in mathematical handbooks.  
      Essentially this is just doing (ii), but taking advantage of the 
      fact that someone else has probably already done the work for 
      you.

Does this help?  Write back if you'd like to talk about this some
more, or if you have any other questions.

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


Date: 11/23/2001 at 09:36:37
From: Craig Robinson
Subject: Re: Integration

Hi,

Could I see a proof of the integral of f(x) = exp(-ax^2) between 0 
and infinity, please? 

Thanks.


Date: 11/23/2001 at 12:12:37
From: Doctor Jubal
Subject: Re: Integration

Hi Craig, thanks for writing back.

We want to evaluate 

  /inf
  |    exp(-ax^2) dx
  /0

However, using any of our usual arsenal of integration tricks to 
produce an antiderivative of this will fail. When I was an 
undergraduate, I spent five hours one evening attempting various ways 
to integrate this by parts and trying various substitutions. Little 
did I know there was no antiderivative. The solution came to me over 
breakfast, within half an hour of my waking up, thus illustrating that 
in mathematics, there is no substitute for a good night's sleep. But I 
digress.

Let's call the value of the definite integral value I.

For reasons that will become apparent shortly, it's a lot easier to 
evaluate I^2.

        /inf                /inf 
  I^2 = |    exp(-ax^2) dx  |    exp(-ax^2) dx
        /0                  /0

It's rather irrelevant what symbols we use for the variables of 
integration, so I'm going to use x in the first integral and y in the 
second integral. This is okay because the two integrals are completely 
independent of each other.

        /inf                /inf 
  I^2 = |    exp(-ax^2) dx  |    exp(-ay^2) dy
        /0                  /0

And because the two integrals are independent of each other, we can 
combine them into a single double integral.

        /inf /inf
  I^2 = |    |    exp(-ax^2) exp(-ay^2) dx dy
        /0   /0

        /inf /inf
      = |    |    exp[-a(x^2 + y^2)] dx dy
        /0   /0

Now so far, it looks as if all we've managed to do is make a mess of 
things. This integral doesn't look any easier to deal with than what 
we started with, and we have an extra variable to boot. But we have 
managed to create an x^2 + y^2 term, which means this double integral 
may be more manageable in polar coordinates.

The region of integration is the first quadrant, so we're integrating 
over r from 0 to infinity and theta from 0 to pi/2. I'll use Q for 
theta. Remember that in polar coordinates, r^2 = x^2 + y^2, and 
dx dy = r dr dQ.

        /(pi/2) /inf
  I^2 = |       |     exp(-ar^2) r dr dQ
        /Q=0    /r=0

        /(pi/2)     /inf
      = |       dQ  |    exp(-ar^2) r dr
        /Q=0        /r=0

This integral can be solved analytically, but I'll leave this as 
something for you to work out on your own. (Hint: try the substitution 
u^2 = ar^2).

If there's any part of this that doesn't make sense to you, feel free 
to write back.

- Doctor Jubal, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
College Calculus
High School Calculus

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