The book, ‘Introduction To Random Matrices - Theory and Practice’, Reference 1, has been illuminating. See (1.6) and Section 6.2. I am having trouble with (6.11) though, see below
Quote \begin{equation*} Vol( \mathbb{V}_2)= \int_0^{2 \pi} d\theta \int_0^{2 \pi} d\phi \int_0^\infty~dr ~r~\delta(r-1) \int_0^\infty~dR ~R~\delta(R-1)~ \delta ( rR\cos(\theta-\phi)) \end{equation*}
\begin{equation*} = \int_0^{2 \pi} d\theta \int_0^{2 \pi} d\phi ~\delta ( \cos(\theta-\phi))=4\pi ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\tag{6.11} \end{equation*} End Quote
\begin{equation*} \end{equation*}
My question is : Ignoring the “$Vol( \mathbb{V}_2)”$ part, how do I show, that (6.11) of ‘Introduction To Random Matrices - Theory and Practice’, is correct?
Reference
1, Giacomo Livan, Marcel Novaes, Pierpaolo Vivo; Introduction to Random Matrices - Theory and Practice https://arxiv.org/abs/1712.07903
See also https://link.springer.com/book/10.1007/978-3-319-70885-0
If you have a look at (1.6) and Section 6.2, then I think you might get the general idea behind obtaining an eigenvalue pdf, from a random matrix entry, pdf.
Other Information
My attempt at proving most of (6.11) is shown below, there are three issues that I would like addressed, two involve the two assumptions made about how the delta function behaves, and the third is, how to finish off the proof.
The two assumptions follow,
$\mathbf{1)}$ In an integral
\begin{equation*} \delta(abx) \equiv \frac {\delta (bx) } { |a| } \end{equation*} c.f a definite property of the delta function \begin{equation*} \delta(ax) \equiv \frac {\delta (x) } { |a| } \end{equation*}
$\mathbf{2)}$ with the integral ‘$\int dR$’ contained in an integral ‘$\int d\theta~\int d\phi$’ \begin{equation*} \int_0^\infty~dR ~R~ \delta ( R\cos(\theta-\phi))~\delta(R-1) \end{equation*} can be thought of as \begin{equation*} \int_0^\infty~dR~ f(R)~\delta(R-1) \end{equation*} with \begin{equation*} f(R)= R~\delta ( R\cos(\theta-\phi)) \end{equation*}
Hence \begin{equation*} \int_0^\infty~dR ~R~ \delta ( R\cos(\theta-\phi))~\delta(R-1)=f(1)=1 \delta(~1\cos(\theta-\phi))=\delta(\cos(\theta-\phi))\end{equation*}
So, with these assumptions \begin{equation*} \int_0^{2 \pi} d\theta \int_0^{2 \pi} d\phi \int_0^\infty~dr ~r~\delta(r-1) \int_0^\infty~dR ~R~\delta(R-1)~ \delta (r R\cos(\theta-\phi)) \end{equation*}
\begin{equation*} =\int_0^{2 \pi} d\theta \int_0^{2 \pi} d\phi \int_0^\infty~dr ~r~\delta(r-1) \int_0^\infty~dR ~R~\delta(R-1)~ \frac{\delta ( R\cos(\theta-\phi))} { |r| } \end{equation*}
\begin{equation*} =\int_0^{2 \pi} d\theta \int_0^{2 \pi} d\phi \int_0^\infty~dr ~\delta(r-1)~ \delta (\cos(\theta-\phi))~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \end{equation*}
\begin{equation*} =\int_0^{2 \pi} d\theta \int_0^{2 \pi} d\phi ~\delta (\cos(\theta-\phi))~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \end{equation*}
NB: The '$\int_0^\infty~dr$' evaluates to one.
The above integral is supposed to evaluate to $4\pi$, but I don’t know how to justify this.
One way to continue the calculation is to argue that, since $\cos(\theta-\phi)$ is a periodic function of $\phi$, we can shift $\phi\mapsto \phi+\theta$ without changing the result. (This can likely be said more rigorously but it's a bit tedious.) Hence the integral becomes $$\int_0^{2\pi}d\theta \int_0^{2\pi}d\phi ~\delta(\cos \phi)=2\pi \int_0^{2\pi}d\phi~\delta(\cos \phi)=4\pi \int_0^{\pi}d\phi~\delta(\cos \phi)$$ where in the second step we have used the symmetry of $\cos\phi$ with respect to $\phi\mapsto 2\pi-\phi$. Since $u=\cos \phi$ is a bijection from $[0,\pi]$ to $[-1,1]$, we may write this integral as $$4\pi\int_{-1}^1 \frac{du}{\sqrt{1-u^2}}\delta(u)=4\pi.$$ For $\delta(f(x))$ more generally, note that the above worked because we could restrict the integration to an interval on which the substitution $x=f^{-1}(u)$ was well-defined. If $f(x)$ has only simple zeros $\{x_i\}_{i=1}^k$ in an interval $[a,b]$, we may split up $[a,b]$ into subintervals with well-defined inverses; this approach yields the formula $${\displaystyle \int _a^b dx~g(x)\,\delta (f(x))=\sum _{i=1}^k{\frac {g(x_{i})}{|f'(x_{i})|}}.}$$