# Thread: Double Integration

1. ## Double Integration

Any help with the following problem is appreciated (I have no idea how to approach it):

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

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

Consider the integral $I= \int^{\infty}_{-\infty} e^{-x^2} dx$. Express $I^2$ as a double integral involving $x$ and $y$.
It's well known that

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

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

3. Originally Posted by Prove It
It's well known that

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

So if $I = \sqrt{\pi}$, surely $I^2 = \pi$...
But the problem says "express $I^2$ as a double integral". How do we need to express it like that?

4. $I= \int_{-\infty}^\infty e^{-x^2}dx$

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

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

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

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

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

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

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

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

$I^2= \int_{x= -\infty}^\infty \int_{y= -\infty}^\infty e^{-(x^2+y^2)} dy dx$
It should also be pointed out that this double integral

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

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

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

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

can be evaluated to the value of $\pi$ by converting to polars...
How did you evaluate it to the value of $\pi$?

Here's my attempt:

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

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

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

But how does this equal to $\pi$?

7. You know that

$= I^2$.

You can also say

Therefore

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

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

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

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

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

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

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

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

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

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

So that means $\Gamma\left(\frac{1}{2}\right) = \sqrt{\pi}$.

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

Here's my attempt:

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

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

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

But how does this equal to $\pi$?
It was a double integral, remember?

$I^2= \int_{\theta= 0}^{ 2\pi} \int_{r= 0}^\infty r e^{-r^2}drd\theta$ $= 2\pi \int_{r= 0}^\infty r e^{-r^2}dr$

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

9. Originally Posted by HallsofIvy
It was a double integral, remember?

$I^2= \int_{\theta= 0}^{ 2\pi} \int_{r= 0}^\infty r e^{-r^2}drd\theta$ $= 2\pi \int_{r= 0}^\infty r e^{-r^2}dr$

Let $u= r^2$ so that $du= \frac{1}{2}r dr$ and $2du= rdr$. The integral becomes
$4\pi\int_0^\infty e^{-u}du$.
Actually, I believe you'll find that

$du = 2r\,dr$.

So the integral becomes

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

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

10. Prove It,

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

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

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

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

11. Originally Posted by demode
Prove It,

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

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

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

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.

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

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

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

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

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

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

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

$= 2\pi$.

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

We have shown that

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

So that means

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

Therefore

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

which is what we require for any probability density function.