# I Dirac Delta δ(sin(ωx))?

1. Oct 16, 2017

### Kyle Nemeth

If we were to replace δ(x), the orginal Dirac Delta, with δ(sin(ωx)), what would be the result?

Would we have an infinite spike everywhere on the graph of sinx where x is a multiple integer of π/ω? and 0 everywhere else?

I apologize in advance if I had posted in the wrong category.

2. Oct 16, 2017

### Staff: Mentor

Seems reasonable that $\delta(\sin(\omega x))$ would be infinite wherever $\omega x$ is an odd multiple of $\pi/2$, and zero everywhere else. In other words, where $x = \frac{(2k + 1)\pi}{2\omega}$, with k in the integers.

3. Oct 16, 2017

### Orodruin

Staff Emeritus
Yes. For a fuction $f(x)$ with a countable set of zeroes $x_i$, it holds that
$$\delta(f(x)) = \sum_i \frac{\delta(x-x_i)}{|f’(x_i)|}$$

4. Oct 16, 2017

### Orodruin

Staff Emeritus
The zeroes of the sine function are integer multiples of pi.

5. Oct 16, 2017

### Kyle Nemeth

Thank you guys for answering my question. Very much appreciated.

6. Oct 16, 2017

### andrewkirk

Speaking very loosely, the answer is yes.

Speaking more carefully, the dirac delta is a distribution, not a function, and only gives a value when included in an integral. That is
$\int_a^b\delta(x)f(x)dx$ is equal to $f(0)$ if $a\leq 0 \leq b$ and to zero otherwise. So your question could be expressed more precisely as:
what is the value of $\int_a^b \delta(\sin\omega x)f(x)dx$?

This can be answered by substitution. Set $u=\sin\omega x$ so that $du=\omega\cos x\,dx=\omega\sqrt{1-u^2}\,dx$. Then the Dirac delta part of the integrand becomes just $\delta(u)$.

We'll end up with an answer that is equal to the sum of the values of $f(x)$ for all values of $x$ in $[a,b]$ that are multiples of $\pi/\omega$.

7. Oct 17, 2017

Doh!