Double integral requiring orthogonal transformations and quadric forms I have $\int_{-\infty}^\infty \int_{-\infty}^\infty exp(-x^T Ax) \;\mathrm{d}x_1 \; \mathrm{d}x_2$
$A = \left[ \begin{align} 3 && 2 \\ 2 && 3 \end{align} \right]$
Where $x^T = (x_1,x_2)$
$x_1 = \frac1{\sqrt{2}}(u_1 + u_2)$
$x_2 = \frac1{\sqrt{2}}(u_1 - u_2)$
Now I would think this constructs a $2 \times 1 * 2 \times 2 * 1 \times 2$ but these matrices are incompatible.
 A: You can just expand the matrix product to get the exponential of a scalar:
$$I=\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-(3x^2+4xy+3y^2)}\,dx\,dy\ .$$
In order to actually evaluate this you will need to write the quadric form as a sum of squares using an orthogonal transformation.  I assume you have done this in your course and can provide the working to get
$$3x^2+4xy+3y^2=X^2+5Y^2$$
where
$$X=\frac{x-y}{\sqrt2}\ ,\quad Y=\frac{x+y}{\sqrt2}\ .$$
Now making a substitution in the integral, you will find that the Jacobian is $1$ and the limits of integration do not change, so
$$I=\left(\int_{-\infty}^\infty e^{-X^2}dX\right)
  \left(\int_{-\infty}^\infty e^{-5Y^2}dY\right)\ .$$
You can evaluate these integrals by using the result
$$\int_{-\infty}^\infty e^{-at^2}\,dt=\sqrt{\frac{\pi}{a}}$$
provided $a>0$.
See if you can fill in the working I have omitted.
A: Hint:  A is a symmetric matrix. It can be diagonalized by an orthogonal matrix.
       let $A = O^{T}DO$ where $O$ is a suitable orthogonal matrix and $D$ i a diagonal matrix.
       now $X^{T}AX=X^{T}(O^{T}DO)X=(OX)^{T}D(OX)$. Now defining $Y=O^{T}X$ and explicitly 
   evaluating $Y^{T}DY$ the problem can be solved without any problem.
