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In these notes, they use some reasoning to get 10.1, that is $\delta(ax)=\delta(x)/|a|$. I am able to follow that, but then they proceed from there to show (11), which I am having some trouble understanding. Could someone please clarify this?

Addendum: They then generalize it for a function with more than two roots in (12), but that doesn't even seem to work for the $\Theta(x^2-a^2)$ example they have, as using the first line I get $0$ for $x<-a$ and $x>a$, and $1$ for $-a<x<a$.
I am aware there are other demonstrations of identity (11), but I am specifically interested in this one.

xihiro
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1 Answers1

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They have written $\theta(x^2-a^2) = 1 - \theta(x+a) + \theta(x-a)$ (assuming $a>0$) and then taken the derivative of both sides using the chain rule on the left hand side: $$2x\,\delta(x^2-a^2) = -\delta(x+a) + \delta(x-a).$$ Then they have divided with $2x$: $$\delta(x^2-a^2) = -\frac{1}{2x}\delta(x+a) + \frac{1}{2x}\delta(x-a).$$ At this step they have used that $f(x)\,\delta(x-a) = f(a)\,\delta(x-a)$ to get $$\delta(x^2-a^2) = -\frac{1}{2(-a)}\delta(x+a) + \frac{1}{2a}\delta(x-a) = \frac{1}{2a}\delta(x+a) + \frac{1}{2a}\delta(x-a).$$

md2perpe
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  • They don't mention assuming $a>0$; why do consider so? I also thought they used the $f(x)\delta(x-a)$ relation (and it seems you're right, as it is one of the identities listed), but I always thought that was only true under the integral sign, so it seems like cheating as they don't integrate the LHS. In any case, they end with $|a|$. – xihiro Apr 17 '20 at 22:27
  • @xihiro. Thley are physicists, not mathematicians, and therefore a bit sloppy with assumptions and calculations. – md2perpe Apr 17 '20 at 22:30
  • Ahah, I too am a physicist, but I strive for rigour. Any way to prove it without the sloppiness but still using the derivative and getting $|a|$?As for 10.1, it's not useful here, after all, and 12 is still wrong, no (at least for $x^2-a^2$ it doesn't seem to work)? – xihiro Apr 17 '20 at 22:38