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Squaring an Integral ?

  1. Jul 31, 2012 #1
    1. The problem statement, all variables and given/known data

    Well iv been looking around for a similar problem and found the use of the following equation. what I would like to know is if the following equation is actually correct? I believe it may be a extension of the Fubini's theorem but cant find a reference or derivation on the net. any help would be great thanks

    I=∫0 e-x2
    Show I=sqrt(pi/4) by first evaluating I^2

    2. Relevant equations

    [∫f(x) dx]^2 = ∫∫f(x)f(y) dxdy ?

    link- https://www.physicsforums.com/showthread.php?t=243670

    3. The attempt at a solution

    I2=∫∫e-2x2dxdy?
     
    Last edited: Jul 31, 2012
  2. jcsd
  3. Jul 31, 2012 #2
    I2=∫∫exp(-2x2)dxdy is incorrect, where is y except in its differential?
     
  4. Jul 31, 2012 #3
    When you multiply two integrals together, their arguments are different (even though initially you use same letter for both): use x for one and y for the other.
     
  5. Jul 31, 2012 #4
    ok so we would have I^2 = ∫∫e-x^2.y^2 dxdy

    i would love to know where this rule comes from, seems very weird that we can just introduce a function dependent only on x as f(x,y).
    and how do we know which order the dxdy goes?
     
  6. Aug 1, 2012 #5
    You have number I that happens to be an integral of some f(x). You might as well look at it as the integral of f(y). You can multiply those, and you can treat them as a double integral - in any order - because their variables are independent.
     
  7. Aug 1, 2012 #6
    You're misunderstanding something here...

    [tex] I^2 = \int e^{-x^2} dx \int e^{-y^2} dy = \int e^{-x^2 - y^2} dx dy[/tex]

    This is basically a special case of Fubini where f(x,y) = g(x) h(y), if that helps your thinking.
     
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