$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ 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}