# Decomposition of the Dirac delta + function composition $\delta(f(x))$ at extremum nullpoints

One of the properites of the Dirac delta distribution which is easily proved and very useful in practical calculations is the decomposition of the composite of Dirac delta and a well behaved function $f$, given by

$$\delta(f(x))=\sum_{x_i\in \mathcal{I}}\frac{\delta(x-x_i)}{|f'(x_i)|},$$ where the $x_i$ are nullpoints of the original function, i.e. $$\mathcal{I}=\{x_i|f(x_i)=0\}$$

This formula works in most cases, however it fails in a narrow class of problems where $f(x)=f'(x)=0$, and I couldn't find the procedure for continuing anywhere on the internet.

For example, we want to decompose $\delta(x\sin(x))$ on the interval $x\in[-\frac{3\pi}{2},\frac{3\pi}{2}]$. The function $f(x)=x\sin(x)$ looks something like this: We can see that there are three null-points in the given interval, namely $\mathcal{I}=\{0,\pm\pi\}$. Taking the derivative of $f(x)$ gives $f'(x)=\sin(x)+x\cos(x)$. It's immediatly obvious that $f'(\pm\pi)\neq 0$, so these don't pose a problem, but $f'(0)=0$, so the decomposition breaks down at that point.

My question is: What is the procedure in dealing with these kind of functions where the nullpoint is also an extremal (or inflexion) point? If a standard procedure doesn't exist, is there a way to circumnavigate the problem in the general (or this particular) case?

• I was wondering the same, here's my take on this – Tobias Kienzler Dec 12 '17 at 0:25

If my derivation is correct, then, since $|\sin(n\pi)+n\pi\cos(n\pi)| = n\pi$ and $f''(x) = 2\cos x - x\sin x\ \Big|_{x=0} = 2$ doesn't vanish at $x=0$,
$$\delta(x\sin x) = \frac{\delta(x)}{2\sqrt{x}} + \sum_{n=1}^\infty \frac{\delta(x-n\pi)+\delta(x+n \pi)}{|n\pi|},$$