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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

(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.

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

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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