Definite integral using diagonalizatio

[Steve Turchin]

Homework Statement

$\int_{-\infty} ^{\infty} dx \int_{-\infty} ^{\infty} dy \ \ e^{-3x^2+2xy-3y^2}$

Homework Equations

$\int_{-\infty} ^{\infty} dx \int_{-\infty} ^{\infty} dy \ \ e^{-x^2-y^2} = \pi$

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 :

$\frac{1}{2\sqrt{2}} \int_{-\infty} ^{\infty} dx \int_{-\infty} ^{\infty} dy \ \ e^{-x^2-y^2} \\ =\frac{\pi}{2\sqrt{2}}$

 

[Samy_A]

Homework Statement

$\int_{-\infty} ^{\infty} dx \int_{-\infty} ^{\infty} dy \ \ e^{-3x^2+2xy-3y^2}$

Homework Equations

$\int_{-\infty} ^{\infty} dx \int_{-\infty} ^{\infty} dy \ \ e^{-x^2-y^2} = \pi$

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?) :

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

This should still be a definite integral.

And why can this be changed to :

$\frac{1}{2\sqrt{2}} \int_{-\infty} ^{\infty} dx \int_{-\infty} ^{\infty} dy \ \ e^{-x^2-y^2} \\ =\frac{\pi}{2\sqrt{2}}$

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.

 

[BvU]

I got this far, I know the solution but I don't understand what is done here

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...

 

[Ray Vickson]

Homework Statement

$\int_{-\infty} ^{\infty} dx \int_{-\infty} ^{\infty} dy \ \ e^{-3x^2+2xy-3y^2}$

Homework Equations

$\int_{-\infty} ^{\infty} dx \int_{-\infty} ^{\infty} dy \ \ e^{-x^2-y^2} = \pi$

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?) :

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

*************************************************

This should be $I = \int \int e^{-2 x'^2 - 4 y'^2} \, |J(x',y')| \, dx' \, dy'$ where $J(x',y')$ is the Jacobian determinant of the transformation from $(x,y)$ to $(x',y')$; that is,
J(x&#039;,y&#039;) = \left| \begin{array}{cc} \partial x / \partial x&#039; &amp; \partial x / \partial y&#039; \\<br /> \partial y / \partial x&#039; &amp; \partial y / \partial y&#039;<br /> \end{array} \right|<br />
****************************************************And why can this be changed to :

$\frac{1}{2\sqrt{2}} \int_{-\infty} ^{\infty} dx \int_{-\infty} ^{\infty} dy \ \ e^{-x^2-y^2} \\ =\frac{\pi}{2\sqrt{2}}$