Parseval identity to compute $ \int_{-\infty}^{\infty} (\frac{\sin x}{x})^2dx$ The Parseval identity in Hilbert space $H$ is 

If $h \in H$, $\lVert h\rVert = \sum \{|<h,e>|^2:e \in S\}  $  then when $S$ is a basis for $H$

How to use this to compute 
$$ \int_{-\infty}^{\infty} \left(\frac{\sin x}{x}\right)^2dx$$
 A: Note that$$\int_{-1}^1\frac12e^{ikx}dk=\frac{e^{ix}-e^{-ix}}{2ix}=\frac{\sin x}{x}.$$So$$\int_{\Bbb R}\left(\frac{\sin x}{x}\right)^2dx=2\pi\int_{-1}^1\left(\frac12\right)^2dk=\pi.$$
A: Plancherel and the Fourier Transform on the Line
Compute the Fourier transform of $\pi\left[-\frac1{2\pi}\le x\le\frac1{2\pi}\right]$:
$$
\begin{align}
\int_{-\infty}^\infty\pi\left[-\frac1{2\pi}\le x\le\frac1{2\pi}\right]\,e^{-2\pi ix\xi}\,\mathrm{d}x
&=\frac{e^{-i\xi}-e^{i\xi}}{-2i\xi}\\
&=\frac{\sin\left(\xi\right)}{\xi}
\end{align}
$$
Apply Plancherel:
$$
\begin{align}
\int_{-\infty}^\infty\left(\frac{\sin(\xi)}{\xi}\right)^2\mathrm{d}\xi
&=\int_{-\infty}^\infty\pi^2\left[-\frac1{2\pi}\le x\le\frac1{2\pi}\right]\mathrm{d}x\\
&=\pi^2\cdot\frac1\pi\\[6pt]
&=\pi
\end{align}
$$
Note that $[\cdots]$ are Iverson brackets.

Fourier Series on the Circle and Riemann Sums
$$
\begin{align}
\sum_{k=1}^\infty\frac{\sin(2kx)}{k}
&=\frac{\log\left(1-e^{-2ix}\right)-\log\left(1-e^{2ix}\right)}{2i}\tag1\\
&=\frac{i\pi-2ix}{2i}\tag2\\[6pt]
&=\frac{\pi-2x}2\tag3\\[3pt]
\sum_{k=1}^\infty\frac{\sin^2(kx)}{k^2}
&=\frac{\pi x-x^2}2\tag4\\
\sum_{k=1}^\infty\frac{\sin^2(kx)}{(kx)^2}x
&=\frac{\pi-x}2\tag5\\
\int_0^\infty\left(\frac{\sin(t)}{t}\right)^2\mathrm{d}t
&=\frac\pi2\tag6
\end{align}
$$
Explanation:
$(1)$: power series for $\log(1-x)$
$(2)$: $\log(-e^{-2ix})=i\pi-2ix$
$(3)$: simplify
$(4)$: integrate with respect to $x$
$(5)$: divide by $x$
$(6)$: Riemann Sum as $x\to0$
$\phantom{\text{(6):}}$ the total variation of $\left(\frac{\sin(t)}{t}\right)^2$ is less than $\frac43$ on $[0,\infty)$
$\phantom{\text{(6):}}$ so the Riemann Sum of the improper integral converges properly
