I am looking for a short proof that ∫0∞(sinxx)2dx=π2 What do you think?

eiraszero11cu

Answered

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?

Answer & Explanation

twineg4

Expert

2022-01-01Added 33 answers

Well, it's not hard to reduce this integral to $f\left(x\right)=max\{0,1-\left|x\right|\}$. It is easy to calculate the Fourier transform $\hat{f}\left(\xi \right)={\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}f\left(x\right){e}^{-ix\xi}={\left(\frac{\mathrm{sin}\left(\frac{\xi}{2}\right)}{\frac{\xi}{2}}\right)}^{2}$ 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 $[-1,1])$

Karen Robbins

Expert

2022-01-02Added 49 answers

${\int}_{0}^{\mathrm{\infty}}\frac{{\mathrm{sin}}^{2}x}{{x}^{2}}dx={\int}_{0}^{\mathrm{\infty}}\frac{\frac{12}{1-\mathrm{cos}2x}}{{x}^{2}}dx$ $={\int}_{0}^{\mathrm{\infty}}\frac{1-\mathrm{cos}x}{{x}^{2}}dx$ $=\frac{1}{2}{\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}\frac{1-\mathrm{cos}x}{{x}^{2}}dx$ $={\frac{12}{\int}}_{-\mathrm{\infty}}^{\mathrm{\infty}}\frac{1-\mathrm{cos}x}{{x}^{2}}dx$ $=\frac{1}{2}{\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}\mathfrak{R}\frac{1-e\left\{ix\right\}}{{x}^{2}}dx$ $=\frac{1}{2}{\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}\mathfrak{R}\frac{1-{e}^{ix}+i\frac{x}{1+{x}^{2}}}{{x}^{2}}dx$ $=\frac{1}{2}\mathfrak{R}{\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}\frac{1-{e}^{ix}+i\frac{x}{1+{x}^{2}}}{{x}^{2}}dx$
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

2022-01-09Added 457 answers

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