Prove that $$\lim_{\omega\rightarrow+\infty}\frac{\sin(\omega t)}{\pi t}=\delta(t)$$ where $\delta(t)$ is the impulse function.
This is a property that our teacher mentioned without strict proof. I've already tried to proof $$\int_{-\infty}^{+\infty}\frac{\sin(\omega t)}{t}=\pi(\omega>0)$$ But I have some trouble connecting this equation to the upper property.
We have to show that $\forall \varphi \in \mathcal{S}(\mathbb{R})$: $\lim\limits_{\omega \rightarrow \infty} \frac{\sin(\omega t)}{\pi t} [\varphi]=\delta[\varphi]$, ( $\mathcal{S}(\mathbb{R})$ is a space of test functions, f.e. the Schwartz-Space). More explicitly: $$ \forall \varphi \in \mathcal{S}(\mathbb{R}): \quad \lim\limits_{\omega \rightarrow \infty}\int_{-\infty}^{\infty} \frac{\sin(\omega t)}{\pi t} \varphi(t)\ dt=\varphi(0) $$ This follows by the substitution $u:=\omega t$, then above expression turns into: $$ \lim\limits_{\omega \rightarrow \infty}\int_{-\infty}^{\infty} \frac{\sin(u)}{\pi u} \varphi(u/\omega)\ du= $$ Using dominated convergence we find: $$ =\int_{-\infty}^{\infty} \lim\limits_{\omega \rightarrow \infty} \frac{\sin(u)}{\pi u} \varphi(u/\omega)\ du= \varphi(0)\int_{-\infty}^{\infty} \frac{\sin(u)}{\pi u}\ du=\varphi(0) $$
Edit: Maybe to further clarify: What you‘re trying to prove is not an equality of functions in the usual sense. Normally two functions $f(x)$ and $g(x)$ are equal if $\forall x : f(x)=g(x)$, but your „impulse function“ $\delta(x)$ doesn’t have defined values, particularly not for $x=0$. To deal with this problem in mathematics two functions of this kind (distributions) are said to be equal if $\int f(x)\varphi(x)dx= \int g(x)\varphi(x)dx$, for all $\varphi$ that „behave well enough“. (It actually gets even trickier in your case because you’re considering a limit of distributions, and to talk about limits your distribution space needs to be equipped with a topology and so on but these are just details that probably aren’t relevant here)