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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:

enter image description here

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?

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  • $\begingroup$ I was wondering the same, here's my take on this $\endgroup$ Dec 12, 2017 at 0:25

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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 $x=0$,

$$\delta(x\sin x) = \frac{\delta(x)}{2\sqrt{x}} + \sum_{n=1}^\infty \frac{\delta(x-n\pi)+\delta(x+n \pi)}{|n\pi|},$$

i.e. the extremum simply gets a square root singularity.

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  • $\begingroup$ Thank you, the link is most illuminating! $\endgroup$ Dec 17, 2017 at 12:54
  • $\begingroup$ @Sobanoodles Glad to hear :) I just hope it's actually correct $\endgroup$ Dec 17, 2017 at 19:49

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