I'm trying to simplify the following integral
\begin{equation}
N(\psi_1,\psi_2)=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}w(x_0,y_0)\delta(g)dx_0dy_0
\end{equation}
Where $\delta$ is the Dirac delta function and $g(x_0,y_0,\psi_1,\psi_2)$ is given by:
\begin{equation}
g=x_0\cos(\frac{\psi_1+\psi_2}{2})+y_0\sin(\frac{\psi_1+\psi_2}{2})-R\cos(\frac{\psi_1-\psi_2}{2})
\end{equation}
I'm looking for a general solution, but in case you need it: $R$ is a positive constant, and $w(x_0,y_0)$ is zero for all values that satisfy $x_0^2+y_0^2>R^2$
Best Answer
You'll have to use the coarea formula and integrate over the zero level set of $g$, i.e., $g^{-1}(0) = \{(x,y) \in \mathbb{R}^2: g(x, y) = 0\}$.
$$\int_{\mathbb{R}^2} w(x,y) \, \delta(g(x,y)) \,dx\,dy = \int_{g^{-1}(0)}\frac{f(\xi)}{|\mathbf{\nabla}g|}\,d\sigma(\xi),$$
where $\xi$ are coordinates on $g^{-1}(0)$ and $d\sigma(\xi)$ is the surface measure (i.e., the "area element") on $g^{-1}(0)$.
In your example, $|\nabla g(x,y)| = 1$. Also, if we assume that $\psi_1,\psi_2$ are independent variables, then $g^{-1}(0)$ is a straight line in $\mathbb{R}^2$. Parameterizing this line using $\xi = x$, a point $\phi(\xi)$ on the the level set $g^{-1}(0)$ can be written as
$$\phi(\xi) = \left[\xi, R\frac{\cos{\psi''}}{\sin{\psi'}} - \xi\dfrac{\cos{\psi'}}{\sin{\psi'}} \right],\quad \psi' = \frac{1}{2}(\psi_1+\psi_2), \enspace \psi'' = \frac{1}{2}(\psi_1-\psi_2),$$
so that the surface measure $d\sigma(\xi)$ is
$$d\sigma(\xi) = \sqrt{\det(\nabla \phi^T \phi)}\,d\xi = \left|\frac{d\phi}{d\xi}\right|\,d\xi = \csc\psi'\,d\xi.$$
Thus, the integral you're seeking is
$$N(\psi_1, \psi_2) = \int_{\xi_-}^{\xi_+}d\xi\,w(\phi(\xi))\csc\left(\frac{\psi_1+\psi_2}{2}\right).$$
Because of the extra condition on $w(x,y)$, the limits of the above integral $\xi_{\pm}$ are the two intersection points of the circle $x^2 + y^2 = R^2$ with the line $\phi(\xi)$ given by the two roots of $|\phi(\xi)|^2 - R^2 = 0.$