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?
 A: I will use the following definition of the Fourier Transform, for $g\in L^1(\mathbb{R})$, 
$$
{\cal F}(g)(\omega)=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}g(t)e^{it\omega}dt
$$
If $g ={\bf 1}_{[-1,1]}$ the characteristic function of the interval $[-1,1]$ then
$$
{\cal F}(g)(\omega)=\sqrt{\frac{2}{\pi}}\frac{\sin \omega}{\omega}
$$
So, if $h=g*g$, then
$$
{\cal F}(h)(\omega)=\sqrt{2\pi}({\cal F}(g)(\omega))^2=2\sqrt{\frac{2}{\pi}}\left(
\frac{\sin \omega}{\omega}\right)^2
$$
Since ${\cal F}(h)\in L^1$, it follows that $$h(t)={\cal F}^{-1}({\cal F} (h))(t)=\frac{2}{\pi}\int_{\mathbb{R}}\left(
\frac{\sin \omega}{\omega}\right)^2e^{-it\omega}d\omega$$
or equivalently
$$\int_{\mathbb{R}}\left(
\frac{\sin \omega}{\omega}\right)^2e^{it\omega}d\omega=\frac{\pi}{2}g*g(t).$$
So, we only have to calculate the convolution product $g*g$ which is easy:
$$
g*g(t)=\int_{-1}^1{\bf 1}_{[-1,1]}(t-x)dx=\left\{
\matrix{0&\hbox{if}&2\leq t \cr
2-t&\hbox{if}&0\leq t<2\cr
2+t&\hbox{if}&-2\leq t<0\cr
0&\hbox{if}& t\leq-1 }\right.=\max(2-|t|,0)
$$ 
Hence,
$$\int_{\mathbb{R}}\left(
\frac{\sin \omega}{\omega}\right)^2e^{it\omega}d\omega=\frac{\pi}{2}\max(2-|t|,0).
$$
which is the desired limit since the integrand is integrable.
A: Hint:
Try convolution. $\sin x/x$ has a very nice Fourier transform (a box), and squaring the function convolves the fourier transform with itself (which in this case can be done with a pencil and a ruler).

$$\mathcal{F}\left(\frac{\sin x}{x}\right)=\int_{-\infty}^\infty\left(\frac{\sin x}{x}\right) e^{itx}dx=\pi \operatorname{rect}(t) $$
where
$$\operatorname{rect}(t)=\begin{cases} 1 & |t|<1 \\ 0 &\text{else}\end{cases}$$
Fourier transform of a product is a convolution of fourier transforms:
$$\mathcal{F}\left(\frac{\sin x}{x}\right)^2=\int_{-\infty}^\infty\left(\frac{\sin x}{x}\right)^2 e^{itx}dx=\frac{1}{2\pi}\pi^2 \operatorname{rect}(t)\ast \operatorname{rect}(t) $$
Convolution of a box with itself gives the maximum of $2$ and falls to $0$ linearly on both sides, reaching $0$ at $t=2$. You can therefore write
$$=\frac{\pi}{2}\begin{cases}2-|t| & |t|<2 \\0 & \text{else}\end{cases}$$
Why the $\frac{1}{2\pi}$ in the convolution theorem? Well, it all depends which definition of Fourier transform you have (the scaling factor). It's very easy to verify if you don't remember which one it is. Just plug in $t=0$ and evaluate the integral.
A: 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 $I(t)=I(-t)$.
A: $\newcommand{\+}{^{\dagger}}
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\begin{align}&\color{#66f}{\large%
\int_{-\infty}^{\infty}{\sin^{2}\pars{x} \over x^{2}}\,\expo{\ic tx}\,\dd x}
=\int_{-\infty}^{\infty}\pars{\half\int_{-1}^{1}\expo{-\ic k x}\,\dd k}
\pars{\half\int_{-1}^{1}\expo{-\ic qx}\,\dd q}\expo{\ic tx}\,\dd x
\\[3mm]&={1 \over 4}\int_{-1}^{1}\int_{-1}^{1}\
\overbrace{\int_{-\infty}^{\infty}\expo{\ic\pars{t - k - q}x}\,\dd x}
^{\ds{2\pi\,\delta\pars{t - k - q}}}\ \dd q\,\dd k
={\pi \over 2}\int_{-1}^{1}\Theta\pars{1 - \verts{k - t}}\,\dd k
\\[3mm]&=\left.{\pi \over 2}
\int_{-1}^{1}\dd k\,\right\vert_{t\ -\ 1\ <\ k\ <\ t\  +\ 1}
={\pi \over 2} \times  \left\lbrace\begin{array}{lcrcccl}
0 & \mbox{if} & t + 1 & < & -1
\\[1mm]
t + 2 & \mbox{if} & -1 & < & t + 1 & < & 1
\\[1mm]
2 - t & \mbox{if} & -1 & < & t - 1 & < & 1
\\[1mm]
0 & \mbox{if} & t - 1 & > & 1
\end{array}\right.
\end{align}

$\ds{\delta\pars{x}}$ is the
  Dirac Delta Function and $\ds{\Theta\pars{x}}$ is the
  Heaviside Step Function.

The limiting case $\ds{t \to 0^{\pm}}$ yields the correct result $\ds{\pi \over 2}$ when $\ds{t = 0}$ such that:
$$\color{#66f}{\large%
\int_{-\infty}^{\infty}{\sin^{2}\pars{x} \over x^{2}}\,\expo{\ic tx}\,\dd x}
={\pi \over 2} \times \left\lbrace\begin{array}{rcrccl}
t + 2 & \mbox{if} & -2 & < & t & \leq & 0
\\[1mm]
2 - t & \mbox{if} & 0 & \leq & t & < & 2
\\[1mm]
0 & \mbox{otherwise}&&&&&
\end{array}\right.
$$

This is equivalent to
  $$
\color{#66f}{\large%
\int_{-\infty}^{\infty}{\sin^{2}\pars{x} \over x^{2}}\,\expo{\ic tx}\,\dd x
={\pi \over 2}\,\Theta\pars{2 - \verts{t}}\verts{\vphantom{\huge A}2 - \verts{t}}}
$$
  

