Show that the product of two delta functions $\delta{(x)}$$\delta{(y)}$ is invariant under rotation around the origin. This is a problem from Zee's textbook on Gravity on page 51.
The book was speaking about Generators $J_x, J_y, J_z$ of a rotation group SO(3), and writing A as a linear combination of these Generators ($A = \theta_xJ_x + \theta_yJ_y+\theta_zJ_z$) and then making the claim that rotations about an angle $\theta$, $R(\theta)$ can be written as $e^A$. Probably this has something to do with $e^X = \sum_{n=0}^\infty \frac{X^n}{n!}$. I also only know that $$\delta(x) =\begin{cases} \infty & \text{ if } x=0\\ 0 & \text{ otherwise}\end{cases}$$ s.t. $\int_{-\infty}^{\infty}f(t)\delta(t)dt=f(0)$
But it also brought up $R(\theta)$ as the plain rotation matrix from elementary linear algebra.
Presumably you want to show it is invariant under 2D rotations (because it isn't rotationally invariant in 3D). So for any smooth compactly supported test function $\varphi$, we have $$ \int_{-\infty}^\infty \! \int_{-\infty}^\infty \varphi(x,y) \delta(x) \delta(y) \, dy \, dx = \varphi(0,0) .$$ Now suppose that $R = \begin{bmatrix} c & -s \\ s & c \end{bmatrix}$ is any 2D-rotation, with $c = \cos\theta$ and $s = \sin\theta$. You want to show that $$ \delta(cx - sy) \delta(sx + cy) = \delta(x)\delta(y) .$$ So consider $$ I = \int_{-\infty}^\infty \! \int_{-\infty}^\infty \varphi(x,y) \delta(cx - sy) \delta(sx + cy) \, dy \, dx .$$ Make the substitution $\xi = cx - sy$, $\eta = sx + cy$. Note that $x = c\xi+s\eta$, $y = -s\xi + c \eta$, and that $dx\,dy = d\xi \, d\eta$, the last equality coming from the fact that the determinant of $R$ is 1. Therefore $$ I = \int_{-\infty}^\infty \! \int_{-\infty}^\infty \varphi(c\xi+s\eta,-s\xi + c \eta) \delta(\xi) \delta(\eta) \, d\eta \, d\xi = \varphi(c0+s0,-s0 + c 0) = \varphi(0,0) .$$