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Improper double integral over R2

  1. Jul 20, 2009 #1
    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. jcsd
  3. Jul 20, 2009 #2

    Dick

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    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
  4. Jul 20, 2009 #3
    If I change variables, I still integrate the new variables over the entire r2 right? A linear transformation of R2 covers R2 always?
     
  5. Jul 20, 2009 #4

    Dick

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    If it's linear and nonsingular, sure. And I think you can choose one that is.
     
    Last edited: Jul 20, 2009
  6. Jul 21, 2009 #5
    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
     
  7. Jul 21, 2009 #6

    Dick

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    That's what I get.
     
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