<s>[INST] How do I evaluate $\lim_{A\to\infty}\int_{-A}^A \left(\frac{\sin x}{x}\right)^2e^{itx}\,dx$? How do I compute
$$
\lim_{A\to\infty}\int_{-A}^A \left(\frac{\sin x}{x}\right)^2e^{itx}\,dx
$$
 for $t\in\mathbb{R}$ and $i$ the imaginary unit?
 [/INST] Let $ \displaystyle I(t) = \int_{-\infty}^{\infty} \frac{\sin^{2}x}{x^{2}} e^{itx} \ dx$.
Then
$$\begin{align} I(t) &= \int_{-\infty}^{\infty} \frac{\sin^{2} x}{x^{2}} \cos tx \ dx \\ &= 2 \int_{0}^{\infty} \frac{\sin^{2} x}{x^{2}} \cos tx\ dx  \\ &= \int_{0}^{\infty} \frac{\left(1- \cos 2x \right) \cos tx}{x^{2}} \\ &= \int_{0}^{\infty}\left(\frac{\cos tx}{x^{2}} - \frac{\cos(t+2)x}{2x^{2}} - \frac{\cos(t-2)x}{2x^{2}}  \right) \ dx \ . \end{align} $$
Integrating by parts,
$$ \begin{align} \int \frac{\cos ax}{x^{2}} \ dx &= - \frac{\cos ax}{x} - a \int \frac{\sin ax}{x} \ dx \  \\ &= - \frac{\cos ax}{x} - a \ \text{Si}(ax) \end{align}$$
where $\text{Si}(x)$ is the sine integral.
http://mathworld.wolfram.com/SineIntegral.html
Therefore,
$$I(t)  = - \frac{\cos tx}{x} - t \ \text{Si}(tx) + \frac{\cos (t+2)x}{2x} + \frac{(t+2) \ \text{Si}\big((2+t)x\big)}{2} + \frac{\cos (t-2)x}{2x}$$
$$  + \frac{(t-2) \ \text{Si}\big((2-t)x\big)}{2} \Bigg|^{\infty}_{0}$$
$$ = -t \ \text{sgn} (t) \frac{\pi}{2} + \frac{t+2}{2} \text{sgn}(t+2) \frac{\pi}{2} + \frac{t-2}{2} \text{sgn}(t-2) \frac{\pi}{2}$$
where I used the fact that 
$$ \lim_{x \to \infty} \text{Si} (ax) = \int_{0}^{\infty} \frac{\sin ax}{x} \ dx = \text{sgn}(a) \frac{\pi}{2} \ .$$
Then you can check to see what happens for different values of $t$.
If $t > 2$,
$$I(t) = - t \frac{\pi}{2} + \frac{t+2}{2} \frac{\pi}{2} + \frac{t-2}{2} \frac{\pi}{2} = 0$$
If $t=2$,
$$ I(2) = -2 \frac{\pi}{2} +2 \frac{\pi}{2} + 0 = 0$$
If $0 < t < 2$,
$$ I(t) = -t \frac{\pi}{2} + \frac{t+2}{2} \frac{\pi}{2} - \frac{t-2}{2} \frac{\pi}{2} = \frac{\pi}{2} (2-t)$$
If $t=0$,
$$ I(0) = 0 + \frac{\pi}{2} + \frac{\pi}{2} = \pi$$
And $I(t)=I(-t)$.
</s>