Limit calculation

Question

How to show this limit? Is there any way to check this, say by integrating around $x=0$? I am out of ideas. Any help is appreciated. Thanks $$\lim_{\alpha\rightarrow\infty} \frac{sin^2\alpha x}{\alpha x^2} =\pi\delta(x)$$

Answer 1

If you look at $\int^\epsilon_{-\epsilon} \frac{\sin^2(\alpha x)}{\alpha x^2} dx$, you can rewrite as $\int^{\epsilon/\alpha}_{-\epsilon/\alpha} \frac{\sin^2y}{y^2} dy$. According to Wolfram the integral is given as $-\frac{1}{2y} + \frac{1}{2}\frac{cos(2y)}{2y} + Si(2y)$, where $Si(y)$ is the sine integral. This can be written in a variety of ways, for example as $Si(2y) - \frac{sin^2(y)}{y}$.

It follows the definite integral is $[Si(2y) - \frac{sin^2(y)}{y}]^{y = \epsilon \alpha}_{y = -\epsilon \alpha}$. Can you take it from here?