Multidimensional integral involving delta functions The question is to compute the following multidimensional integral:
\begin{equation}
\omega^{(T)}({\bf c}) := \int\limits_{{\mathbb R}^{2 T}} 
\delta\left( c_{1,1} - \sum\limits_{j=1}^T x_{1,j}^2 \right)
\delta\left( c_{2,2} - \sum\limits_{j=1}^T x_{2,j}^2 \right)
\delta\left( c_{1,2} - \sum\limits_{j=1}^T x_{1,j} x_{2,j} \right)
\prod\limits_{j=1}^T d x_{1,j} d x_{2,j}
\end{equation}
This is the ``density of states'' of the estimator of covariances in a random matrix ensemble.
Using the definition of delta function and elementary integration I have checked that :
\begin{eqnarray}
\omega^{(1)}({\bf c}) &=&    \delta\left(\det(c)\right) \\
\omega^{(2)}({\bf c})  &=& \pi (\det(c))^{-1/2} \\
\omega^{(3)}({\bf c})  &=& \pi^2 1_{\det(c) >0} \\
\omega^{(4)}({\bf c})  &=& \pi^3 \left(\det(c)\right)^{1/2}
\end{eqnarray}
where 
\begin{equation}
{\bf c} := 
\left(\begin{array}{cc} c_{1,1} & c_{1,2} \\ c_{1,2} & c_{2,2}
\end{array}
\right)
\end{equation}
The question is what is the result for generic values of $T$. I suspect that the result depends only on the determinant of the matrix ${\bf c}$.

Below is the result for $T=2$. We have:
\begin{eqnarray}
\int\limits_{{\mathcal R}^4}
\delta\left(c_{1,1} - x_{1,1}^2 - x_{1,2}^2\right)
\delta\left(c_{2,2} - x_{2,1}^2 - x_{2,2}^2\right)
\delta\left(c_{1,2} - x_{1,1}x_{2,1} - x_{1,2}x_{2,2}\right) d x_{1,1} d x_{1,2} d x_{2,1} d x_{2,2} = \\
\int\limits_{{\mathcal R}^3}
\delta\left(c_{1,1} x_{2,1}^2 - \left({c_{1,2} - x_{1,2} x_{2 2}}\right)^2 - x_{1,2}^2 x_{2,1}^2\right)
\delta\left(c_{2,2} - x_{2,1}^2 - x_{2,2}^2\right) {x_{2,1}}d x_{1,2} d x_{2,1} d x_{2,2} = \\
\frac{1}{2}  \int\limits_{{\mathcal R}^2} 
\delta\left((c_{1,1}-x_{1,2}^2)(c_{2,2}-x_{2,2}^2) - (c_{1,2}-x_{1,2} x_{2,})^2\right) d x_{1,2} d x_{2,2} = \\
\frac{1}{2} \frac{1}{\det(c)} \int\limits_{{\mathcal R}^2}
\delta\left(1 - \vec{x}^T c^{-1} \vec{x}\right) d^2 x = \\
\frac{\pi}{2} \frac{1}{\sqrt{\det(c)}}
\end{eqnarray}
In the second line we have integrated over $x_{1,1}$ using the definition of the delta function.In the third line we integrated over $x_{2,1}$.In the fourth line we took out the determinant from the inside of the delta function and we wrote the remaining terms as a quadratic form.Finally, we diagonalized the quadratic form and computed the integral by going to radial coordinates.
