# How do you handle a functional input in a Dirac delta function and prove these types of relations?

I have a quadratic relation inside of a Dirac delta function with the following relation \begin{align} \delta((x-x_1)(x-x_2)) = \dfrac{ \delta(x-x_1) + \delta(x-x_2) }{|x_1-x_2|}. \end{align}

How do you handle a functional input in a Dirac delta function and prove these types of relations?

$$\delta(xy)~=~\frac{\delta(y)}{|x|}+\frac{\delta(x)}{|y|}.$$
• Sorry I got stuck with your hint. I know that $\delta(ax)=\delta(x)/a$, but I am not sure how to extend that identity to the identity that you provided. $x$ and $a$ are assumed to be independent variables? – linuxfreebird Sep 3 '14 at 10:41
• Consider what the Dirac delta does to a function $f(x,y)$ when you integrate them together, i.e., the integral $\int dxdy \, \delta(xy)f(x,y) = \int dx \big( \int dy \delta(xy)f(x,y) \big)$. In the $y$ integral, $x$ is just a number. – Robin Ekman Sep 3 '14 at 11:00
• Here is a heuristic argument that OP might like and that reduces the problem to 1D: Since the support of the delta functions is the whole $x$- and $y$-axis, we can leave out a small $\epsilon$-disc around the origin. In polar coordinates this means that the radius $r\geq\epsilon>0$ is never zero. – Qmechanic Sep 4 '14 at 17:33
• (cont.) If we next substitute $x=r\cos\theta$ and $y=r\sin\theta$, then the eq. becomes $$\delta(r^2\cos\theta\sin\theta) ~=~ \frac{\delta(r\sin\theta)}{r|\cos\theta|} + \frac{\delta(r\cos\theta)}{r|\sin\theta|},$$ which follows from (among other things) the 1D product formula $$\delta(f(\theta)g(\theta)) ~=~ \frac{\delta(f(\theta))}{|g(\theta)|}+\frac{\delta(g(\theta))}{|f(\theta)|}.$$ – Qmechanic Sep 4 '14 at 17:33