# Dirac function squared?

I am trying to perform the following limit

$$$$\lim_{t\rightarrow\infty}\int\frac{d\omega}{2\pi}S\left(\omega\right)\frac{\sin^{2}\left[\left(\omega+\Omega\right)t/2\right]}{\left[\left(\omega+\Omega\right)/2\right]^{2}}$$$$ Ideally, I was thinking about using the relation $$\lim_{t\rightarrow\infty}\frac{\sin\left(\pi xt\right)}{\pi x}=\delta\left(x\right)$$ to simplify the equation above, and actually this kind of works, but leads to the following integral of a Dirac delta function squared, \begin{align*} \int\frac{d\omega}{2\pi}S\left(\omega\right)\lim_{t\rightarrow\infty}\frac{\sin^{2}\left[\left(\omega+\Omega\right)t/2\right]}{\left[\left(\omega+\Omega\right)/2\right]^{2}} & =\int\frac{d\omega}{2\pi}S\left(\omega\right)\lim_{t\rightarrow\infty}\frac{\sin\left[\pi\left(\frac{\omega+\Omega}{2\pi}\right)t\right]}{\pi\left(\frac{\omega+\Omega}{2\pi}\right)}\frac{\sin\left[\pi\left(\frac{\omega+\Omega}{2\pi}\right)t\right]}{\pi\left(\frac{\omega+\Omega}{2\pi}\right)}\\ & =\int\frac{d\omega}{2\pi}\delta\left[\left(\frac{\omega+\Omega}{2\pi}\right)\right]\delta\left[\left(\frac{\omega+\Omega}{2\pi}\right)\right]\\ & =\int\frac{d\omega}{2\pi}2\pi\delta\left(\omega+\Omega\right)\times2\pi\delta\left(\omega+\Omega\right)\rightarrow\infty \end{align*} I was reading about the square of the Dirac delta function, it turns out this is not even a distribution. [See: https://math.stackexchange.com/questions/2221429/why-is-the-square-of-dirac-delta-function-not-a-distribution];

or even if treated as would diverge after the integration [See: https://physics.stackexchange.com/questions/47934/dont-understand-the-integral-over-the-square-of-the-dirac-delta-function] In a related old post, one of our peers mentioned that is possible to prove

$$$$\lim_{t\rightarrow\infty}\frac{\sin^{2}\left[\left(\omega+\Omega\right)t/2\right]}{\left[\left(\omega+\Omega\right)/2\right]^{2}}=2\pi\delta\left(\omega+\Omega\right)t$$$$ Limit of $\frac{1}{2}\frac{\sin^2(\omega t/2)}{(\omega/2)^2}$which resolves easly my problem. However, I do not know how to prove that. It seems one only applies the relation above once.

• You're correct that the square of the delta distribution is bad news. In general, products of distributions are not well defined. There is some work on this but it's a very small field. Most people try to avoid products of distributions instead. Sep 24, 2021 at 13:51
• So here's how I would approach the problem at hand: Split up the square of your sinc function into two pieces. One gets paired with $S$, the other stays by itself. Then make use of Parseval/Plancherel. Then use the convolution theorem. I would write out a solution but I'm about to head to work so I don't have the time to type it out. Hopefully you can get the answer from this. If not and no one else answers, I'll answer after my first class. Sep 24, 2021 at 13:55
• I would not say this is a very small field given that it is the subject of quite a bit of ongoing work like Martin Hairer's theory of regularity structures which earned him the 2021 Breakthrough Prize in Mathematics. See also math.stackexchange.com/a/2301706/244562 Sep 24, 2021 at 14:16
• \begin{align*} \lim_{t\rightarrow\infty}\int\frac{d\omega}{2\pi}S\left(\omega\right)\frac{\sin\left[\pi\left(\frac{\omega+\Omega}{2\pi}\right)t\right]}{\pi\left(\frac{\omega+\Omega}{2\pi}\right)}\lim_{t\rightarrow\infty}\frac{\sin\left[\pi\left(\frac{\omega+\Omega}{2\pi}\right)t\right]}{\pi\left(\frac{\omega+\Omega}{2\pi}\right)} & =\lim_{t\rightarrow\infty}\int\frac{d\omega}{2\pi}S\left(\omega\right)\frac{\sin\left[\pi\left(\frac{\omega+\Omega}{2\pi}\right)t\right]}{\pi\left(\frac{\omega+\Omega}{2\pi}\right)}2\pi\delta\left(\omega+\Omega\right)\\ & =tS\left(-\Omega\right) \end{align*} Sep 24, 2021 at 14:41
• @AbdelmalekAbdesselam I don't mean that no one is working on it or that there isn't a lot of math to be done, but rather that not very many people are working on it (relative to other subfields). It is exceedingly difficult as a subject area. Sep 24, 2021 at 16:04