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### 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 negative infinity 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/
```
Associated Topics:
College Calculus
High School Calculus

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