# Integrating exp(-x^2)

Discussion in 'Undergraduate Math' started by Stan Brown, Dec 24, 2010.

1. ### Stan BrownGuest

This came up at work yesterday, and neither of us could remember how
to do it. I pulled out my copy of Thomas and couldn't find it,
though I did find a statement at the beginning of the "Methods of
Integration" chapter that it would be dealt with later and would
involve infinite series.

Can someone remind me, or point me toward an online reference?
Thanks!

P.S. We actually want the definite integral from x=0 to infinity.

And again, it's not the answer but the method that we're looking for.

Stan Brown, Dec 24, 2010

2. ### Paul SperryGuest

but not much of a method.

<http://mathforum.org/library/drmath/view/69832.html> gives a

The two answers do not agree - I think (but am not sure) that Wolfram
is correct.

Paul Sperry, Dec 24, 2010

3. ### PubkeybreakerGuest

Turn it into a double integral and convert to polar coordinates.

Pubkeybreaker, Dec 24, 2010
4. ### Brian M. ScottGuest

On Fri, 24 Dec 2010 01:47:26 -0500, Stan Brown
What you want is half of the integral over R; call its value
I. Then I^2 = int[exp(-x^2) dx] * int[exp(-y^2) dy] =
int int[exp(-(x^2 + y^2)) dy dx], where all integrals are
from -inf to inf, so that you're integrating over the whole
plane. Now convert everything to polar coordinates.

Brian

Brian M. Scott, Dec 24, 2010
5. ### VirgilGuest

If I = Integral_0^oo e^(-x^2) dx

Then I^2 = Integral_0^oo Integral_0^oo e^(-x^2-y^2 ) dx dy

= Integral_0^(pi/2) Integral_0^oo e^(-r^2) r dr dtheta

Which is fairly simple to evaluate.

To show that the two infinite double integrals actually converge to the same value is
also not difficult.

Consider a quarter circle region centered at origin contained in a minimal square
contained in a larger but also minimal quarter circle with center at origin.

Clearly the integral over the square region has a value between the integrals over the
two quarter circle regions.

Then show that as the radii of the quarter circle goes towards oo, the integral over the
region between the quarter circles goes to zero, thus the difference in the 3 integrals
over the finite regions goes to zero as the regions of integration become infinite.

Virgil, Dec 24, 2010
6. ### Stan BrownGuest

Wow! Thanks, Brian!

Stan Brown, Dec 24, 2010
7. ### Stan BrownGuest

match the method that Brian Scott posted, Dr. Math integrates from -
infinity to +infinity and gets sqrt(pi) = 1.772... Wolfram is
integrating from 0 to +infinity and gets 0.886... which is half of
sqrt(pi), so I think the two actually do agree.

Stan Brown, Dec 24, 2010
8. ### Stan BrownGuest

Thanks for responding.

Stan Brown, Dec 24, 2010