Improper integral of $\sin^2(x)/x^2$ evaluated via residues I have come across another improper integral I wish to evaluate via residues.
The integral is:
$$\int_{-\infty}^\infty{\frac{\sin(x)^2}{x^2}}dx$$ 
$\sin(z)$ behaves in an uneasy way so I tried using the function $\frac{{e^{iz}}^2}{z^2}$ with a half circle on the upper complex plane with radius R and a half-circle of radius 1/R which arcs below $0$.  
The problem is the small semi-circles integral does not go to $0$ and in fact doesn't exist.
What other types of contours or function substitutions should be used here?
 A: Note that $ \cos(2x)=1-2\sin(x)^2 $, this suggest to consider the integral
$$ \int_{C} \frac{ {\rm e}^{2 i z} - 1 }{ z^2} dz \,.$$ 
A: Since there are no singularities of $\frac{\sin^2(z)}{z^2}$ in the rectangle $[-R,R]\times[0,-i]$ and the function vanishes along the ends near $-R$ and $R$,
$$
\begin{align}
\int_{-\infty}^\infty\frac{\sin^2(x)}{x^2}\,\mathrm{d}x
&=\int_{-\infty}^\infty\frac{2-e^{-2iz}-e^{2iz}}{4z^2}\,\mathrm{d}z\\
&=\int_{-\infty-i}^{\infty-i}\frac{2-e^{-2iz}-e^{2iz}}{4z^2}\,\mathrm{d}z\\
\end{align}
$$
Next we can use two $D$-shaped contours
$$
\gamma_+=[-R-i,R-i]\cup Re^{[0,\pi]i}-i
$$
and
$$
\gamma_-=[-R-i,R-i]\cup Re^{-[0,\pi]i}-i
$$
Since the integral along the large semicircles vanishes and the singularity is only circled once counterclockwise by $\gamma_+$ and not by $\gamma_-$, we get
$$
\begin{align}
\int_{-\infty-i}^{\infty-i}\frac{2-e^{-2iz}-e^{2iz}}{4z^2}\,\mathrm{d}z
&=\int_{\gamma_-}\frac{2-e^{-2iz}}{4z^2}\,\mathrm{d}z
-\int_{\gamma_+}\frac{e^{2iz}}{4z^2}\,\mathrm{d}z\\
&=0-2\pi i\cdot\frac{i}{2}\\[9pt]
&=\pi
\end{align}
$$
since the residue of
$$
\frac{e^{2iz}}{4z^2}=\frac1{4z^2}+\color{#C00000}{\frac{2i\color{#000000}{z}}{4\color{#000000}{z^2}}}+\frac{-2z^2}{4z^2}+\dots
$$
at $z=0$ is $\frac i2$.
A: $\newcommand{\+}{^{\dagger}}%
 \newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
 \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
 \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
 \newcommand{\dd}{{\rm d}}%
 \newcommand{\isdiv}{\,\left.\right\vert\,}%
 \newcommand{\ds}[1]{\displaystyle{#1}}%
 \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
 \newcommand{\ic}{{\rm i}}%
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}%
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
 \newcommand{\sech}{\,{\rm sech}}%
 \newcommand{\sgn}{\,{\rm sgn}}%
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}%
 \newcommand{\verts}[1]{\left\vert #1 \right\vert}%
 \newcommand{\yy}{\Longleftrightarrow}$
Since
$\ds{%
\sin\pars{x} \over x}
=
{1 \over 2}\int_{-1^{-}}^{1^{+}}\expo{\ic kx}\,\dd k$, we have:
\begin{align}
{\large\int_{-\infty}^{\infty}{\sin^{2}\pars{x} \over x^{2}}\,\dd x}
&=
\int_{-\infty}^{\infty}\dd x\,{1 \over 2}\int_{-1^{-}}^{1^{+}}\expo{\ic kx}\,\dd k\,
{1 \over 2}\int_{-1^{-}}^{1^{+}}\expo{\ic qx}\,\dd q
\\[3mm]&=
{\pi \over 2}\int_{-1_{-}}^{1^{+}}\dd k\int_{-1_{-}}^{1^{+}}\dd q
\int_{-\infty}^{\infty}\expo{\ic\pars{k + q}x}\,{\dd x \over 2\pi}
=
{\pi \over 2}\int_{-1^{-}}^{1^{+}}\dd k
\int_{-1_{-}}^{1^{+}}\dd q\,\delta\pars{k + q}
\\[3mm]&=
{\pi \over 2}\int_{-1^{-}}^{1^{+}}\Theta\pars{1 - \verts{k}}\,\dd k
=
{\pi \over 2}\int_{-1^{-}}^{1^{+}}\dd k = {\large \pi}
\end{align}
A: The function $\mathrm{sinc}: x\mapsto \frac{\sin x}x$ is the Fourier transform of the gate function $\Pi:x\mapsto \mathbf{1}_{[-1,1]}(x)$
$$\mathcal{F}(\Pi)=2\mathrm{sinc}.$$
Parseval's identity for this Fourier pair yields
$$\int\Pi^2=\frac{1}{2\pi}\int (2\mathrm{sinc})^2.$$
Since $\int\Pi^2=2$, we get the result
$$ \int_{-\infty}^\infty \frac{\sin^2x}{x^2}\mathrm dx=\pi.$$
