Product of composition of Dirac Delta functions I am wrestling with an integral of the following form:
$$\int_{\mathbb{R}^4} d\textbf{x}\: f(\textbf{x})\delta(g(\textbf{x}))\delta(h(\textbf{x}))$$
Where $\textbf{x}=(x_1,x_2,x_3,x_4)$, $h(x)=x_1+x_2-x_3-x_4$ and $g(x)$ is smooth almost everywhere. I additionally have a parameterization of the system
$$\begin{cases}
g(\textbf{x})=0\\
h(\textbf{x})=0
\end{cases}$$
Via Wikipedia I am aware of the formula for a single dimensional Delta function with a scalar substitution
$$\int d\textbf{x}\: f(\textbf{x})\delta(g(\textbf{x}))=\int_{g^{-1}(0)}\frac{f(\textbf{x})}{|\nabla g(\textbf{x})|}d\sigma(\textbf{x})$$
However it has been challenging to find information on generalizations to higher order Delta functions.
Any resources or analysis would be appreciated!
 A: I'm not aware of a good reference for the formula I gave; here's an explanation instead, (albeit a somewhat hand-waving one).
If integrals like the one you gave are well defined, multiplying by $\delta(x_i)$ where $x_i$ is a coordinate ought be the same as setting those coordinates to zero:
$$
\int_{U\subseteq\mathbb{R}^3}f(x_1,x_2,x_3)\delta(x_2)\delta(x_3)dx_1dx_2dx_3=\int_{\{x_1\in\mathbb{R}:(x_1,0,0)\in U\}}f(x_1,0,0)dx_1
$$
and so on for other dimensions and other numbers of deltas. Additionally, we would expect such integrals to still obey the change of variables formula:
$$
\int_Vf(\mathbf{x})d\mathbf{x}=\int_Uf(\varphi(\mathbf{y}))|\det(D\varphi(\mathbf{y}))|d\mathbf{y}
$$
Where $\varphi:U\to V$ is a diffeomorphism between open subsets of $\mathbb{R}^n$.
If we assume that the zero sets of $g$ and $h$ are transverse (i.e. $\nabla g$ and $\nabla h$ are linearly independent on $g^{-1}(0)\cap h^{-1}(0)$), then this is already enough. Let $\mathbf{p}\in g^{-1}(0)\cap h^{-1}(0)$. By inverse function theorem there exist a set of local coordinates $\mathbf{y}:V\to U$ on a neighborhood $V$ of $\mathbf{p}$ such that $\mathbf{y}(p)=\mathbf{0}$, $y_3=g$, and $y_4=h$. Let $\varphi:U\to V$ be the inverse of these coordinates. Applying the change of variables formula gives
$$
\int_Vf(\mathbf{x})\delta(g(\mathbf{x}))\delta(h(\mathbf{x}))d\mathbf{x}=\int_Uf(\varphi(\mathbf{y}))\delta(g(\varphi(\mathbf{y}
))\delta(h(\varphi(\mathbf{y}))|\det(D\varphi(\mathbf{y})|d\mathbf{y}
$$
Since $\varphi$ is inverse to $\mathbf{y}$, we can rewrite this as
$$
=\int_U\frac{f(\varphi(\mathbf{y}))\delta(y_3)\delta(y_4)}{|\det(D\mathbf{y}(\varphi(\mathbf{y})))|}d\mathbf{y}
$$
Now that the deltas are composed with coordinates, we can restrict to a surface integral.
$$
=\int_{U\cap\mathbb{R}^2}\frac{f(\varphi(y_1,y_2,0,0))}{|\det(D\mathbf{y}(\varphi(y_1,y_2,0,0)))|}dy_1dy_2
$$
The columns of $D\mathbf{y}$ are just the gradients of the coordinates.
$$
=\int_{U\cap\mathbb{R}^2}\frac{f(\varphi(\mathbf{y}))\delta(y_3)\delta(y_4)}{|\det([\nabla y_1,\nabla y_2,\nabla g,\nabla h])|(\varphi(y_1,y_2,0,0))}dy_1dy_2
$$
Let $\pi$ be the orthogonal projection onto $T(g^{-1}(0)\cap h^{-1}(0))$. Since this amounts to subtracting multiples of $\nabla g$ and $\nabla h$, we can apply it to $\nabla y_1$ and $\nabla y_2$ without changing the determinant:
$$
=\int_{U\cap\mathbb{R}^2}\frac{f(\varphi(y_1,y_2,0,0))}{|\det([\pi(\nabla y_1),\pi(\nabla y_2),\nabla g,\nabla h])|(\varphi(y_1,y_2,0,0))}dy_1dy_2
$$
To split the determinant, note that $|\det([a,b,c,d])|$ is equal to the volume of the parallelepiped spanned by $a,b,c,d$. If $\operatorname{span}(a,b)\perp\operatorname{span}(c,d)$, then this is equal to the product of areas the parallelograms spanned by $(a,b)$ and $(c,d)$. In terms of wedge products, this gives
$$
=\int_{U\cap\mathbb{R}^2}\frac{f(\varphi(y_1,y_2,0,0))}{\|\nabla g\wedge\nabla h\|(\varphi(y_1,y_2,0,0))}\frac{dy_1dy_2}{\|\pi(\nabla y_1)\wedge\pi(\nabla y_2)\|(\varphi(y_1,y_2,0,0))}
$$
Where the last term is just the standard area element of $g^{-1}(0)\cap h^{-1}(0)$ in the coordinates $y_1,y_2$. Writing this in a more coordinate-independent way, we (almost) have the desired formula.
$$
=\int_{V\cap(g^{-1}(0)\cap h^{-1}(0))}\frac{f}{\|\nabla g\wedge\nabla h\|}dA
$$
This result is only local, but you can recover the global version using a partition of unity.
