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 ${\bf c}$.
 A: To simplify our task, note that WLOG we can choose the coordinates to set $c_{11},c_{22}$ to unity. (The overall dimensionful factor lost by this redefinition will be dealt with at the end of the calculation.) This renders the first two $\delta$-functions as $$\delta(1-\mathbf{x}_1^2)\delta(1-\mathbf{x}_2^2)=\frac{1}{4}\delta(1-\Vert\mathbf{x}_1\Vert)\delta(1-\Vert\mathbf{x}_2\Vert).$$ We furthermore may take the third constraint to be $\cos\Theta=\mathbf{x}_1\cdot\mathbf{x}_2=\cos\theta_{12}$ where $\theta_{12}$ is the angle between the two vectors in $\mathbb{R}^T$. This conveniently expresses the determinant as $1-\cos^2\Theta = \sin^2\Theta$. Integrating out the first two $\delta$-functions, we may write our integral of interest as
$$\Omega^{(T)}(\Theta):=\frac{\omega^{(T)}(\Theta))}{V(S_{T-1}}=
\frac{1}{4V(S_{T-1})}\int\!\!\!\int_{R} \delta\left(\cos\Theta-\cos\theta_{12}\right)\,d\Omega_1 \,d\Omega_2$$ where $R=S_{T-1}\times S_{T-1}$.
The switch to $\Omega$ may seem superfluous. But we may take the coordinates of $\mathbf{x_2}$ to have $\mathbf{x_1}$ as its azimuthal axis  so that the constraint is simply $\cos\Theta=\cos\theta_2$. Then the integration over $x_1$ simply gives as an overall factor the volume of $S_{n-1}$ and we may write
$$\Omega^{(T)}(\Theta)=\frac{1}{4}\int_{S_{T-1}} \delta(\cos\Theta-\cos\theta)\,d\Omega$$ where we have dropped the now-superfluous subscript.
To evaluate this integral, we reintroduce the $r$-integration to return to an integral over $\mathbb{R}^n$ i.e.
\begin{align}
\frac{1}{4}\int_{\mathbb{R}^n}d^n \! x \, \delta(\cos\Theta-\cos\theta)\delta(1-\Vert \mathbb{x}\Vert)
&=\frac{1}{2}\int_{\mathbb{R}^n}d^n\!x \, \delta(\cos\Theta-\cos\theta)\delta(1-\Vert \mathbb{x}\Vert^2)\\
&=\frac{1}{2}\int_{\mathbb{R}^n}d^n\!x \, \delta(\cos\Theta-x_n)\delta(1-\Vert \mathbb{x}\Vert^2)
\end{align}
where we have taken $\cos\theta$ to denote the $n$-th coordinate restricted to the unit $(n-1)$-sphere. Writing $\mathbf{y}=\mathbf{x}-x_n \hat{e}_n$, we can use the first $\delta$-function to express the second as $$\delta(1-\Vert \mathbb{x}\Vert^2)=\delta(1-x_n^2- \Vert \mathbb{y}\Vert^2)=\delta(1-\cos^2\theta-\Vert \mathbb{y}\Vert^2)=\delta(\sin^2\Theta-\Vert \mathbb{y}\Vert^2)$$ We can thus integrate out the $n$-th coordinate entirely to obtain
$$\frac{1}{2}\int_{\mathbb{R}^{n-1}}d^{n-1}\!x \, \delta(\sin^2\Theta-\Vert \mathbb{x}\Vert^2)
 =\frac{1}{4|\sin\Theta|}\int_{\mathbb{R}^{n-1}}d^{n-1}\!x \, \delta(|\sin\Theta|-\Vert \mathbb{x}\Vert)=\frac{1}{4}V(S_{n-2})|\sin\Theta|^{n-2}$$
from which we conclude that $\omega^{(T)}(\Theta)=\frac{1}{4}|\sin\Theta|^{n-2}V(S_{n-1})V(S_{n-2})$. All that remains is to express this in terms of $\det(\mathbf{c})$ and restore the overall constants lost in setting $c_{11}=c_{22}=1$...
A: 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.
A: For comparison with Przemo's additional answer, here is my calculation of the $T=2$ case. I'll compute in polar coordinates from the start:
$$\omega^{(2)}=\frac{1}{4}\int\limits_0^\infty d(r_1^2)\int\limits_{-\pi}^{\pi} d\phi_1\int\limits_0^\infty d(r_2^2)\int\limits_{-\pi}^{\pi} d\phi_2
\;\delta(c_{11}-r_1^2)\;\delta(c_{22}-r_2^2)\;\delta(c_{12}-r_1 r_2 \cos(\phi_2-\phi_1))$$
I've written the area element in a non-standard form to make the next step obvious: Since we're integrating over $r^2$'s, the delta functions simplify the integral to
$$\frac{1}{4}\int_{-\pi}^{\pi}\int_{-\pi}^{\pi} d\phi_1 d\phi_2\,\delta(c_{12}-\sqrt{c_{11}c_{22}} \cos(\phi_2-\phi_1))$$
Note that the integrand is periodic, and so we may substitute $\phi=\phi_2+\phi_1$ in place of $\phi_2$ without altering the bounds of integration. But then the delta function doesn't involve $\phi_1$ at all, and we may integrate immediately over $\phi_1$ to obtain 
$$\frac{\pi}{2}\int_{-\pi}^{\pi} d\phi\,\delta(c_{12}-\sqrt{c_{11}c_{22}} \cos\phi)=\pi \int_{0}^\pi d\phi \,\delta(c_{12}-\sqrt{c_{11}c_{22}}\cos\phi)$$
where we have used the symmetry of the integrand on $[-\pi,\pi]$.The substitution $\mu=c_{12}-\sqrt{c_{11}c_{22}} \cos\phi$ then yields
\begin{align}
\pi \int_{c_{12}-\sqrt{c_{11}c_{22}}}^{c_{12}+\sqrt{c_{11}c_{22}}}
 \frac{\delta(\mu)\,d\mu}{\sqrt{c_{11}c_{22}}\sin\phi(\mu)}
&=\frac{\pi}{\sqrt{c_{11}c_{22}}}\cdot\left(1-\frac{c_{12}^2}{c_{11}c_{22}}\right)^{-1/2}\\
&=\frac{\pi}{\sqrt{c_{12}^2-c_{11}c_{22}}}\\&=\pi \det{(\mathbf{c})}^{-1/2}
\end{align}
This differs from Przemo's answer---but by a factor of $2$, not $8$. So I'm a bit perplexed.
