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I would like to show that $$\delta(\sin{x})=\sum_{n=-\infty}^{\infty}\delta(x-n\pi),$$ where $\delta(x)$ satisfies the relation $$\int_{-\infty}^{\infty} \delta(x-a)\phi(x) \, dx =\phi(a),$$ for some test function $\phi$.

The $n\pi$ term makes it clear that at some point I need to be considering the roots of $\sin{x}$, but aside from that, I'm not sure what to do.

MHW
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  • Depends on what you are allowed to assume here. You might want to take a look at https://math.stackexchange.com/questions/276583/dirac-delta-function-of-a-function and substitute in $\sin x $ for $f(x)$ there. – bob.sacamento May 03 '18 at 21:57
  • What's the definition of $\delta(\sin(x))$ Saying $\int_{-\infty}^{\infty} \delta(x-a)\phi(x) , dx =\phi(a)$ doesn't tell me what $\delta(\sin(x))$ is... – David C. Ullrich May 04 '18 at 13:50

1 Answers1

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HINT

You should read this, but the idea is basically

$$ \int _{-\infty }^{\infty }f(x)\,\delta (g(x))\,dx=\sum _{i}{\frac {f(x_{i})}{|g'(x_{i})|}}. $$

where $x_i$ are roots of $f$, that is $f(x_i) = 0$. Call $g(x) = \sin(x)$, so that $x_i = i \pi$

caverac
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