I am trying to perform the following limit

\begin{equation} \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}} \end{equation} 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

\begin{equation} \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 \end{equation} 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.

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    $\begingroup$ 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. $\endgroup$ Sep 24, 2021 at 13:51
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    $\begingroup$ 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. $\endgroup$ Sep 24, 2021 at 13:55
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    $\begingroup$ 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 $\endgroup$ Sep 24, 2021 at 14:16
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    $\begingroup$ \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*} $\endgroup$
    – sined
    Sep 24, 2021 at 14:41
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    $\begingroup$ @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. $\endgroup$ Sep 24, 2021 at 16:04


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