Verifying that $\lim_{\alpha\to \infty} \frac{1}{\pi} \frac{\sin^2\alpha x}{\alpha x^2}= \delta(x)$ I'd like to show that:
$$
\lim_{\alpha\to \infty} \frac{1}{\pi} \frac{\sin^2\alpha x}{\alpha x^2}= \delta(x).
$$
 A: We have a lemma stating the following:

Given a succession of $L^1(\mathbb{R}^n)$ functions $f_j$, such that
$$
f_j(x)=j^nf(jx),
$$
then the corresponding function-type distribution converge to $C\delta(x)$ as $j\to \infty$, where:
$$
C=\int_{\mathbb{R}^n}f(y) \, d^ny.
$$

A quic way to see this is that, for any test function $\varphi(x)$ (say, with compact support), $\int \varphi(x) f_j(x) dx = \int \varphi(x/j) f(x) dx \simeq C\varphi(0)$ for large $j$.
In our case we have:
$$
\frac{1}{\pi} \frac{\sin^2\alpha x}{\alpha x^2} = \frac{1}{\pi} \alpha \frac{\sin^2\alpha x}{(\alpha x)^2} \to_{\alpha\to\infty} C\delta(x),
$$
where
$$
C=\frac{1}{\pi}\int_{-\infty}^{+\infty}\frac{\sin^2y}{y^2} \, dy.
$$
Now
$$
\frac{\sin^2y}{y^2} = -\frac{e^{2iy}-1+e^{-2iy}-1}{4y^2}
$$
therefore we have:
$$
\int_{-\infty}^{+\infty}\frac{1-e^{2iy}+1-e^{-2iy}}{4y^2}dy = \int_{-\infty}^{+\infty}\frac{1-e^{2iy}}{2y^2} \, dy.
$$
Using contour integration on an indented semicircle in the upper half-plane,
this yields:$$
\int_{-\infty}^{+\infty}\frac{1-e^{2iy}}{2y^2}dy=
\int_0^{\pi}\frac{-2i\varepsilon e^{i\varphi}+O(\varepsilon^2)}{2\varepsilon^2e^{i2\varphi}}i\varepsilon e^{i\varphi} \, d\varphi = \pi.
$$
and $C=1$ as needed.
EDIT:
Another way to calculate the integral without contour integration is the following. The Plancherel theorem ensures that, if $f,g\in L^2(\mathbb R)$
$$
(\hat f, \hat g) = 2\pi (f,g),
$$
where
$$
\hat f(\xi ) =\int_{-\infty}^{+\infty}f(x)e^{-i\xi x}dx
$$
and $(\cdot,\cdot)$ is the $L^2$ inner product
$$
(f,g) = \int_{-\infty}^{+\infty}\bar f(x) g(x) dx.
$$
Now, our integrand is $(\sin x/x)^2$, which is summable, which in turn makes $\sin x/x\in L^2(\mathbb R)$.
But
$$
\int_{-\infty}^{+\infty}\chi_{[-1,1]}(x)e^{-i\xi x}dx = \int_{-1}^{+1}e^{-i\xi x}dx = \frac{2\sin \xi}{\xi}.
$$
Thus
$$
C=\frac{1}{\pi}\int_{-\infty}^{+\infty}\frac{\sin^2y}{y^2} \, dy=2\pi\frac{1}{4\pi}\int_{-\infty}^{+\infty}\chi^2_{[-1,1]}(x)dx = \frac{1}{2}\cdot 2 = 1.
$$
