# Double Integration

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• Apr 30th 2010, 11:55 PM
demode
Double Integration
Any help with the following problem is appreciated (I have no idea how to approach it):

Consider the integral $\displaystyle I= \int^{\infty}_{-\infty} e^{-x^2} dx$. Express $\displaystyle I^2$ as a double integral involving $\displaystyle x$ and $\displaystyle y$.
• Apr 30th 2010, 11:58 PM
Prove It
Quote:

Originally Posted by demode
Any help with the following problem is appreciated:

Consider the integral $\displaystyle I= \int^{\infty}_{-\infty} e^{-x^2} dx$. Express $\displaystyle I^2$ as a double integral involving $\displaystyle x$ and $\displaystyle y$.

It's well known that

$\displaystyle \int_{-\infty}^{\infty}e^{-x^2}\,dx = \sqrt{\pi}$.

So if $\displaystyle I = \sqrt{\pi}$, surely $\displaystyle I^2 = \pi$...
• May 1st 2010, 12:06 AM
demode
Quote:

Originally Posted by Prove It
It's well known that

$\displaystyle \int_{-\infty}^{\infty}e^{-x^2}\,dx = \sqrt{\pi}$.

So if $\displaystyle I = \sqrt{\pi}$, surely $\displaystyle I^2 = \pi$...

But the problem says "express $\displaystyle I^2$ as a double integral". How do we need to express it like that?
• May 1st 2010, 12:13 AM
Prove It
• May 1st 2010, 02:33 AM
HallsofIvy
$\displaystyle I= \int_{-\infty}^\infty e^{-x^2}dx$

$\displaystyle I= \int_{-\infty}^\infty e^{-y^2}dy$ since that is just the same integral with "dummy variable" y rather than x.

Then $\displaystyle I^2= \left(\int_{-\infty}^\infty e^{-x^2}dx\right)\left(\int_{-\infty}^\infty e^{-y^2}dy\right)$

by Fubini's theorem, that is the same as
$\displaystyle I^2= \int_{x= -\infty}^\infty \int_{y= -\infty}^\infty e^{-x^2}e^{-y^2} dy dx$

$\displaystyle I^2= \int_{x= -\infty}^\infty \int_{y= -\infty}^\infty e^{-(x^2+y^2)} dy dx$
• May 1st 2010, 03:27 AM
Prove It
Quote:

Originally Posted by HallsofIvy
$\displaystyle I= \int_{-\infty}^\infty e^{-x^2}dx$

$\displaystyle I= \int_{-\infty}^\infty e^{-y^2}dy$ since that is just the same integral with "dummy variable" y rather than x.

Then $\displaystyle I^2= \left(\int_{-\infty}^\infty e^{-x^2}dx\right)\left(\int_{-\infty}^\infty e^{-y^2}dy\right)$

by Fubini's theorem, that is the same as
$\displaystyle I^2= \int_{x= -\infty}^\infty \int_{y= -\infty}^\infty e^{-x^2}e^{-y^2} dy dx$

$\displaystyle I^2= \int_{x= -\infty}^\infty \int_{y= -\infty}^\infty e^{-(x^2+y^2)} dy dx$

It should also be pointed out that this double integral

$\displaystyle I^2= \int_{-\infty}^\infty \int_{-\infty}^\infty e^{-(x^2+y^2)} \,dy\, dx$

can be evaluated to the value of $\displaystyle \pi$ by converting to polars...
• May 1st 2010, 06:13 PM
demode
Quote:

Originally Posted by Prove It
It should also be pointed out that this double integral

$\displaystyle I^2= \int_{-\infty}^\infty \int_{-\infty}^\infty e^{-(x^2+y^2)} \,dy\, dx$

can be evaluated to the value of $\displaystyle \pi$ by converting to polars...

How did you evaluate it to the value of $\displaystyle \pi$?

Here's my attempt:

$\displaystyle \int_{0}^\infty [e^{-(r^2cos^2 \theta+r^2sin^2 \theta)}]r \,dr$

$\displaystyle = \lim_{k \to \infty}\int_{0}^k re^{-r^2} \,dr$

$\displaystyle = \frac{-1}{2} \lim_{k \to \infty} e^{-r^2} \,dr$

But how does this equal to $\displaystyle \pi$?
• May 1st 2010, 11:36 PM
Prove It
You know that
http://upload.wikimedia.org/math/d/6...1e85083865.png
$\displaystyle = I^2$.

You can also say

http://upload.wikimedia.org/math/f/1...7e77514b2f.png
Therefore

$\displaystyle I^2 = \pi$ which means $\displaystyle I = \sqrt{\pi}$.

Another handy application is using the Gaussian Integral to compute $\displaystyle \Gamma\left(\frac{1}{2}\right)$.

Since $\displaystyle e^{-x^2}$ is an even function, that means

$\displaystyle \int_{-\infty}^{\infty}{e^{-x^2}\,dx} = 2\int_0^{\infty}{e^{-x^2}\,dx}$.

Using the substitution $\displaystyle x =t^{\frac{1}{2}}$ yields

$\displaystyle -x^2 = -t$ and $\displaystyle dx = \frac{1}{2}t^{-\frac{1}{2}}$

So that the integral $\displaystyle 2\int_0^{\infty}{e^{-x^2}\,dx}$ becomes

$\displaystyle 2\int_0^{\infty}{\frac{1}{2}\,e^{-t}\,t^{-\frac{1}{2}}\,dt}$

$\displaystyle = \int_0^{\infty}{e^{-t}\,t^{-\frac{1}{2}}\,dt}$

$\displaystyle = \Gamma\left(\frac{1}{2}\right)$.

So that means $\displaystyle \Gamma\left(\frac{1}{2}\right) = \sqrt{\pi}$.
• May 2nd 2010, 02:47 AM
HallsofIvy
Quote:

Originally Posted by demode
How did you evaluate it to the value of $\displaystyle \pi$?

Here's my attempt:

$\displaystyle \int_{0}^\infty [e^{-(r^2cos^2 \theta+r^2sin^2 \theta)}]r \,dr$

$\displaystyle = \lim_{k \to \infty}\int_{0}^k re^{-r^2} \,dr$

$\displaystyle = \frac{-1}{2} \lim_{k \to \infty} e^{-r^2} \,dr$

But how does this equal to $\displaystyle \pi$?

It was a double integral, remember?

$\displaystyle I^2= \int_{\theta= 0}^{ 2\pi} \int_{r= 0}^\infty r e^{-r^2}drd\theta$$\displaystyle = 2\pi \int_{r= 0}^\infty r e^{-r^2}dr Let \displaystyle u= r^2 so that \displaystyle du= \frac{1}{2}r dr and \displaystyle 2du= rdr. The integral becomes \displaystyle 4\pi\int_0^\infty e^{-u}du. • May 2nd 2010, 05:30 AM Prove It Quote: Originally Posted by HallsofIvy It was a double integral, remember? \displaystyle I^2= \int_{\theta= 0}^{ 2\pi} \int_{r= 0}^\infty r e^{-r^2}drd\theta$$\displaystyle = 2\pi \int_{r= 0}^\infty r e^{-r^2}dr$

Let $\displaystyle u= r^2$ so that $\displaystyle du= \frac{1}{2}r dr$ and $\displaystyle 2du= rdr$. The integral becomes
$\displaystyle 4\pi\int_0^\infty e^{-u}du$.

Actually, I believe you'll find that

$\displaystyle du = 2r\,dr$.

So the integral becomes

$\displaystyle 2\pi \int_{r= 0}^\infty {r\,e^{-r^2}\,d} = \pi \int_{r = 0}^{\infty}{e^{-r^2}\,2r\,dr}$

$\displaystyle = \pi \int_{r = 0}^{\infty}{e^{-u}\,du}$
• May 2nd 2010, 07:36 PM
demode
Prove It,

We know that $\displaystyle I=\int^{\infty}_{-\infty}e^{-x^2}= \sqrt{\pi}$

If I want to integrate the density function of the normal distribution:

$\displaystyle \int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi}}e^{\frac{-x^2}{2}}dx$

using the value of I, what would be a suitable change of variable in this case?
• May 7th 2010, 05:27 PM
Prove It
Quote:

Originally Posted by demode
Prove It,

We know that $\displaystyle I=\int^{\infty}_{-\infty}e^{-x^2}= \sqrt{\pi}$

If I want to integrate the density function of the normal distribution:

$\displaystyle \int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi}}e^{\frac{-x^2}{2}}dx$

using the value of I, what would be a suitable change of variable in this case?

You can use the exact same change of variable as given to integrate the Gaussian function.

$\displaystyle I^2 = \left(\int_{-\infty}^{\infty}{e^{-\frac{1}{2}x^2}\,dx}\right)^2 = \left(\int_{-\infty}^{\infty}{e^{-\frac{1}{2}x^2}\,dx}\right)\left(\int_{-\infty}^{\infty}{e^{-\frac{1}{2}y^2}\,dy}\right)$

$\displaystyle = \int_{-\infty}^{\infty}{\int_{-\infty}^{\infty}{e^{-\frac{1}{2}(x^2 + y^2)}\,dy}\,dx}$

$\displaystyle = \int_0^{2\pi}{\int_0^{\infty}{e^{-\frac{1}{2}r^2}\,r\,dr}\,d\theta}$

$\displaystyle = 2\pi \int_0^{\infty}{r\,e^{-\frac{1}{2}r^2}\,dr}$

$\displaystyle = 2\pi \int_{-\infty}^0{e^u\,du}$

$\displaystyle = 2\pi \left[e^0 - e^{-\infty}\right]$

$\displaystyle = 2\pi \left[1 - 0\right]$

$\displaystyle = 2\pi$.

Since $\displaystyle I^2 = 2\pi$ that means $\displaystyle I = \sqrt{2\pi}$.

We have shown that

$\displaystyle \int_{-\infty}^{\infty}{e^{-\frac{1}{2}x^2}\,dx} = \sqrt{2\pi}$

So that means

$\displaystyle \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}{e^{-\frac{1}{2}x^2}\,dx} = \frac{1}{\sqrt{2\pi}}\sqrt{2\pi}$.

Therefore

$\displaystyle \int_{-\infty}^{\infty}{\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}\,dx} = 1$

which is what we require for any probability density function.