How to correctly convolve two delta functions?

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The discussion centers on the convolution of two delta functions, specifically ##\delta(\omega)\otimes\delta(\omega)##. The initial computation suggests that the integral evaluates to ##2\delta(\omega)##, which raises concerns about its correctness. A key clarification is that the integral is non-zero only when both conditions, ##\omega - \Omega = 0## and ##\omega = 0##, are satisfied. This leads to the conclusion that the correct result of the convolution is indeed ##\delta(\omega)##, not ##2\delta(\omega)##. The resolution emphasizes the importance of correctly applying the properties of delta functions in convolution operations.
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How do I correctly compute the convolution of two delta functions? For example, if I want to compute ##\delta(\omega)\otimes\delta(\omega)##, I should integrate $$\int_{-\infty}^\infty \delta(\omega-\Omega)\delta(\Omega) d\Omega$$
This integrand "fires" at two places: ##\Omega = 0## and ##\Omega = \omega##, and evaluates to ##\delta(\omega)## in either case. That leads me to say that ##\delta(\omega)\otimes\delta(\omega) = 2\delta(\omega)##, which somehow feels wrong... I want it to be just ##\delta(\omega)##.

Anyone know the correct treatment here?
 
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In your example, ##\int_{-\infty}^\infty \delta(\omega-\Omega) \delta(\Omega) \, d\Omega## is only non-zero if both ## \omega - \Omega = 0 ## and ## \omega = 0 ##.
So for this, you end up with ##\delta(\omega)##.
 

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