# Improper double integral over R2

1. Jul 20, 2009

### theneedtoknow

1. The problem statement, all variables and given/known data

Recall that the integral from -∞ to +∞ of e^(-x^2) is equal to the square root of Pi. Use this fact to calculate the double integral of e^-(x^2 + (x-y)^2 + y^2) dx over the entire region R2.

2. Relevant equations

3. The attempt at a solution

I am not sure if it's even the right thing to do, but whats in the brackets simplifies to e^-(2x^2-2xy+2y^2) = e^(2xy-2x^2-2y^2)

Now to do the inner integral first, I will integrate with respect to x

I can use the propery of exponents to formt he integral from -∞ to +∞ of e^2xy * e^(-2x^2) * e^(-2y^2)

since y is a constant I can pull e^(-2y^2) and integrate just e^2xy * e^(-x^2)

However, this is where I get stuck...
I try to integrate by parts but that only gives me a more complicated integral to do, so I don't know if it's the right way to go. Any suggestions?

2. Jul 20, 2009

### Dick

Why don't you concentrate on trying to find a change of variables so that (2x^2-2xy+2y^2)=u^2+v^2 for u and v some linear functions of x and y? I.e. complete the square and use a jacobian to change the variables?

Last edited: Jul 20, 2009
3. Jul 20, 2009

### theneedtoknow

If I change variables, I still integrate the new variables over the entire r2 right? A linear transformation of R2 covers R2 always?

4. Jul 20, 2009

### Dick

If it's linear and nonsingular, sure. And I think you can choose one that is.

Last edited: Jul 20, 2009
5. Jul 21, 2009

### theneedtoknow

Hey, Thanks a lot for the help!
I have never done this "completing the square" procedure below, so I had no idea what to do with it, but I did look up how to do it online and I played around with it for a while and in the end I think I found a u and v that work so that i get e^(-u^2 - v^2)

And in the end my integral came out to be Pi/root(3), which I hope is right :D

6. Jul 21, 2009

### Dick

That's what I get.