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Definite integral using diagonalizatio

  1. Feb 16, 2016 #1
    1. The problem statement, all variables and given/known data
    ##\int_{-\infty} ^{\infty} dx \int_{-\infty} ^{\infty} dy \ \ e^{-3x^2+2xy-3y^2} ##

    2. Relevant equations
    ## \int_{-\infty} ^{\infty} dx \int_{-\infty} ^{\infty} dy \ \ e^{-x^2-y^2} = \pi ##

    3. The attempt at a solution
    ##3x^2-2xy+3y^2 = (x,y)
    \left( \begin{array}{ccc}
    3 & -1 \\
    -1 & 3 \end{array} \right)

    \left( \begin{array}{ccc}
    x \\
    y \end{array} \right) = \\

    =(x',y')
    \left( \begin{array}{ccc}
    2 & 0 \\
    0 & 4 \end{array} \right)
    \left( \begin{array}{ccc}
    x' \\
    y' \end{array} \right) = 2x'^2+4y'^2 \\
    \\ \\
    ## I got this far, I know the solution but I don't understand what is done here:
    (copied answer from my tutor):

    Why is this suddenly an indefinite integral? (or maybe it's just a ## \LaTeX ## mistake?) :

    ## I = \int dx \int dy \ \ e^{-2x^2-4y^2} ##

    And why can this be changed to :

    ##\frac{1}{2\sqrt{2}} \int_{-\infty} ^{\infty} dx \int_{-\infty} ^{\infty} dy \ \ e^{-x^2-y^2} \\
    =\frac{\pi}{2\sqrt{2}}
    ##
     
  2. jcsd
  3. Feb 16, 2016 #2

    Samy_A

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    This should still be a definite integral.
    He did a substitution to change ##2x²+4y²## into ##u²+v²## (I deliberately used other variable names, the tutor kept the variable names ##x,\ y##).
    You should be able to see what substitution he did.
     
  4. Feb 16, 2016 #3

    BvU

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    Gold Member

    A change of integration variables gets rid of the ##xy## in the exponent. Now the integrations can be done 'one after the other'.

    It wasn't a ##\LaTeX## mistake, just a bit of laziness.

    The next substitution is just to get rid of the 2 and the 4

    And I assume you know where the ##\sqrt \pi## cames from...
     
  5. Feb 16, 2016 #4

    Ray Vickson

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    Last edited: Feb 16, 2016
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