$$\frac{2}{\pi}\int_{-\infty}^{\infty}\left( \frac{\sin (\alpha t)}{t} \right)^2e^{i \nu t} \,\mathrm{d}t=\max(0,\alpha-|\nu|)$$
Apparently I found this in Montgomery's paper (1977). I need to know how exactly this happened. It looks strange when I am trying to do it by simple trigonometric expansion and integration by parts.
Note that this is simply the Fourier transform (inverse Fourier transform to be more precise) of \begin{align} f(t)^2 = \left(\frac{\sin \alpha t}{t} \right)^2 \end{align} which is a product. Hence we see that \begin{align} \mathcal{F}(f^2) = \mathcal{F}(f)\ast \mathcal{F}(f). \end{align} where $\mathcal{F}$ is the Fourier transform and $\ast$ denotes convolution of two functions. Moreover, the Fourier transform of $f(t)$ is a characteristic function. I will let you finish this on your own.