What happens when we replace the Dirac Delta function with a sine function?

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SUMMARY

Replacing the Dirac Delta function, δ(x), with δ(sin(ωx)) results in an infinite spike at points where ωx is an odd multiple of π/2, specifically at x = (2k + 1)π/(2ω) for integer k, and zero elsewhere. The Dirac Delta function is a distribution that only yields a value when integrated. The integral of δ(sin(ωx))f(x) can be evaluated using substitution, leading to the conclusion that the result is the sum of f(x) at all multiples of π/ω within the integration limits.

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Kyle Nemeth
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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.
 
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Kyle Nemeth said:
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?
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.
 
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Yes. For a fuction ##f(x)## with a countable set of zeroes ##x_i##, it holds that
$$
\delta(f(x)) = \sum_i \frac{\delta(x-x_i)}{|f’(x_i)|}
$$
 
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Mark44 said:
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.
The zeroes of the sine function are integer multiples of pi.
 
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Thank you guys for answering my question. Very much appreciated.
 
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##.
 
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Orodruin said:
The zeroes of the sine function are integer multiples of pi.
Doh!
 

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