# Definite integral using diagonalizatio

1. Feb 16, 2016

### Steve Turchin

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. Feb 16, 2016

### Samy_A

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.

3. Feb 16, 2016

### BvU

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...

4. Feb 16, 2016

### Ray Vickson

Last edited: Feb 16, 2016
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