eiraszero11cu

2021-12-31

I am looking for a short proof that
${\int }_{0}^{\mathrm{\infty }}{\left(\frac{\mathrm{sin}x}{x}\right)}^{2}dx=\frac{\pi }{2}$
What do you think?

twineg4

Expert

Well, it's not hard to reduce this integral to $f\left(x\right)=max\left\{0,1-|x|\right\}$. It is easy to calculate the Fourier transform
$\stackrel{^}{f}\left(\xi \right)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f\left(x\right){e}^{-ix\xi }={\left(\frac{\mathrm{sin}\left(\frac{\xi }{2}\right)}{\frac{\xi }{2}}\right)}^{2}$
Taking the inverse Fourier transform, we get
${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\left(\frac{\mathrm{sin}\left(\frac{\xi }{2}\right)}{\frac{\xi }{2}}\right)}^{2}{e}^{ix\xi }d\xi =2\pi f\left(x\right)$,
and the result follows.
The second integral can be computed in a similar way. Just take $f\left(x\right)={\chi }_{\begin{array}{cc}-1& 1\end{array}}\left(x\right)$ (the indicator function of the interval $\left[-1,1\right]\right)$

Karen Robbins

Expert

${\int }_{0}^{\mathrm{\infty }}\frac{{\mathrm{sin}}^{2}x}{{x}^{2}}dx={\int }_{0}^{\mathrm{\infty }}\frac{\frac{12}{1-\mathrm{cos}2x}}{{x}^{2}}dx$
$={\int }_{0}^{\mathrm{\infty }}\frac{1-\mathrm{cos}x}{{x}^{2}}dx$
$=\frac{1}{2}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1-\mathrm{cos}x}{{x}^{2}}dx$
$={\frac{12}{\int }}_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1-\mathrm{cos}x}{{x}^{2}}dx$
$=\frac{1}{2}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\mathfrak{R}\frac{1-e\left\{ix\right\}}{{x}^{2}}dx$
$=\frac{1}{2}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\mathfrak{R}\frac{1-{e}^{ix}+i\frac{x}{1+{x}^{2}}}{{x}^{2}}dx$
$=\frac{1}{2}\mathfrak{R}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1-{e}^{ix}+i\frac{x}{1+{x}^{2}}}{{x}^{2}}dx$
Now close the contour in the upper half plane, enclosing the pole at $x=i$ with residue $\frac{1}{2i}$, yielding
${\int }_{0}^{\mathrm{\infty }}\frac{{\mathrm{sin}}^{2}x}{{x}^{2}}dx=\frac{1}{2}\cdot 2\pi \cdot \frac{1}{2i}=\frac{\pi }{2}$

Vasquez

Expert

From squaring the identity
$\frac{\mathrm{sin}nx}{\mathrm{sin}x}=\frac{{e}^{inx}-{e}^{-inx}}{{e}^{ix}-e-ix}=\sum _{k=0}^{n-1}{e}^{\left(2k-n+1\right)ix}$
and integrating we get
$n\pi ={\int }_{-\pi /2}^{\pi /2}\frac{{\mathrm{sin}}^{2}nx}{{\mathrm{sin}}^{2}x}dx$
Let
${I}_{n}={\int }_{-\pi /2}^{\pi /2}\frac{{\mathrm{sin}}^{2}nx}{n{x}^{2}}dx={\int }_{-n\pi /2}^{n\pi /2}\frac{{\mathrm{sin}}^{2}y}{{y}^{2}}dy$
Then
$\pi -{I}_{n}=\frac{1}{n}{\int }_{-\pi /2}^{\pi /2}{\mathrm{sin}}^{2}nx\left({\mathrm{csc}}^{2}x-{x}^{-2x}\right)dx$
and so
$|\pi -{I}_{n}|\le \frac{1}{n}{\int }_{-\pi /2}^{\pi /2}|{\mathrm{csc}}^{2}x-{x}^{-2}|dx=O\left(1/n\right)$
as $x↦{\mathrm{csc}}^{2}x-{x}^{-2}$ extends to a continuous function on $\left[-\pi /2,\pi /2\right]$. Hence ${I}_{n}\to \pi$ as $n\to \mathrm{\infty }$ and
$\pi ={\int }_{\mathrm{\infty }}^{\mathrm{\infty }}\frac{{\mathrm{sin}}^{2}y}{{y}^{2}}dy$

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