Parseval identity to compute $ \int_{-\infty}^{\infty} (\frac{\sin x}{x})^2dx$

Question

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

$$ \int_{-\infty}^{\infty} \left(\frac{\sin x}{x}\right)^2dx$$

Answer 1

Note that$$\int_{-1}^1\frac12e^{ikx}dk=\frac{e^{ix}-e^{-ix}}{2ix}=\frac{\sin x}{x}.$$So$$\int_{\Bbb R}\left(\frac{\sin x}{x}\right)^2dx=2\pi\int_{-1}^1\left(\frac12\right)^2dk=\pi.$$

Answer 2

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.

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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