Prove ${\int}_{0}^{\mathrm{\infty}}\frac{{\mathrm{sin}}^{4}x}{{x}^{4}}dx=\frac{\pi}{3}$

Demarion Ortega
2022-11-20
Answered

Prove ${\int}_{0}^{\mathrm{\infty}}\frac{{\mathrm{sin}}^{4}x}{{x}^{4}}dx=\frac{\pi}{3}$

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Zoe Andersen

Answered 2022-11-21
Author has **16** answers

You are likely expected to integrate by parts (twice)

$\begin{array}{rcl}\int \frac{{\mathrm{sin}}^{4}(x)}{{x}^{4}}\mathrm{d}x& =& -\frac{1}{3}\frac{{\mathrm{sin}}^{4}(x)}{{x}^{3}}+\frac{4}{3}\int \frac{\mathrm{cos}(x){\mathrm{sin}}^{3}(x)}{{x}^{3}}\mathrm{d}x\\ & =& -\frac{1}{3}\frac{{\mathrm{sin}}^{4}(x)}{{x}^{3}}-\frac{2\mathrm{cos}(x){\mathrm{sin}}^{3}(x)}{3{x}^{2}}+\frac{2}{3}\int \frac{3{\mathrm{cos}}^{2}(x){\mathrm{sin}}^{2}(x)-{\mathrm{sin}}^{4}(x)}{{x}^{2}}\mathrm{d}x\\ & =& -\frac{1}{3}\frac{{\mathrm{sin}}^{4}(x)}{{x}^{3}}-\frac{2\mathrm{cos}(x){\mathrm{sin}}^{3}(x)}{3{x}^{2}}+\frac{2}{3}\int (\frac{{\mathrm{sin}}^{2}(2x)}{{x}^{2}}-\frac{{\mathrm{sin}}^{2}(x)}{{x}^{2}})\mathrm{d}x\end{array}$

where the last equality used

$\begin{array}{rcl}3{\mathrm{cos}}^{2}(x){\mathrm{sin}}^{2}(x)-{\mathrm{sin}}^{4}(x)& =& 3{\mathrm{cos}}^{2}(x){\mathrm{sin}}^{2}(x)-{\mathrm{sin}}^{2}(x)(1-{\mathrm{cos}}^{2}(x))\\ & =& {(2\mathrm{sin}(x)\mathrm{cos}(x))}^{2}-{\mathrm{sin}}^{2}(x)={\mathrm{sin}}^{2}(2x)-{\mathrm{sin}}^{2}(x)\end{array}$

Now

$\begin{array}{rcl}{\int}_{0}^{\mathrm{\infty}}\frac{{\mathrm{sin}}^{4}(x)}{{x}^{4}}\mathrm{d}x& =& \frac{2}{3}{\int}_{0}^{\mathrm{\infty}}\frac{{\mathrm{sin}}^{2}(2x)}{{x}^{2}}\mathrm{d}x-\frac{2}{3}{\int}_{0}^{\mathrm{\infty}}\frac{{\mathrm{sin}}^{2}(x)}{{x}^{2}}\mathrm{d}x\\ & =& \frac{4}{3}{\int}_{0}^{\mathrm{\infty}}\frac{{\mathrm{sin}}^{2}(y)}{{y}^{2}}\mathrm{d}y-\frac{2}{3}{\int}_{0}^{\mathrm{\infty}}\frac{{\mathrm{sin}}^{2}(x)}{{x}^{2}}\mathrm{d}x\\ & =& \frac{2}{3}{\int}_{0}^{\mathrm{\infty}}\frac{{\mathrm{sin}}^{2}(x)}{{x}^{2}}\mathrm{d}x=\frac{\pi}{3}\end{array}$

$\begin{array}{rcl}\int \frac{{\mathrm{sin}}^{4}(x)}{{x}^{4}}\mathrm{d}x& =& -\frac{1}{3}\frac{{\mathrm{sin}}^{4}(x)}{{x}^{3}}+\frac{4}{3}\int \frac{\mathrm{cos}(x){\mathrm{sin}}^{3}(x)}{{x}^{3}}\mathrm{d}x\\ & =& -\frac{1}{3}\frac{{\mathrm{sin}}^{4}(x)}{{x}^{3}}-\frac{2\mathrm{cos}(x){\mathrm{sin}}^{3}(x)}{3{x}^{2}}+\frac{2}{3}\int \frac{3{\mathrm{cos}}^{2}(x){\mathrm{sin}}^{2}(x)-{\mathrm{sin}}^{4}(x)}{{x}^{2}}\mathrm{d}x\\ & =& -\frac{1}{3}\frac{{\mathrm{sin}}^{4}(x)}{{x}^{3}}-\frac{2\mathrm{cos}(x){\mathrm{sin}}^{3}(x)}{3{x}^{2}}+\frac{2}{3}\int (\frac{{\mathrm{sin}}^{2}(2x)}{{x}^{2}}-\frac{{\mathrm{sin}}^{2}(x)}{{x}^{2}})\mathrm{d}x\end{array}$

where the last equality used

$\begin{array}{rcl}3{\mathrm{cos}}^{2}(x){\mathrm{sin}}^{2}(x)-{\mathrm{sin}}^{4}(x)& =& 3{\mathrm{cos}}^{2}(x){\mathrm{sin}}^{2}(x)-{\mathrm{sin}}^{2}(x)(1-{\mathrm{cos}}^{2}(x))\\ & =& {(2\mathrm{sin}(x)\mathrm{cos}(x))}^{2}-{\mathrm{sin}}^{2}(x)={\mathrm{sin}}^{2}(2x)-{\mathrm{sin}}^{2}(x)\end{array}$

Now

$\begin{array}{rcl}{\int}_{0}^{\mathrm{\infty}}\frac{{\mathrm{sin}}^{4}(x)}{{x}^{4}}\mathrm{d}x& =& \frac{2}{3}{\int}_{0}^{\mathrm{\infty}}\frac{{\mathrm{sin}}^{2}(2x)}{{x}^{2}}\mathrm{d}x-\frac{2}{3}{\int}_{0}^{\mathrm{\infty}}\frac{{\mathrm{sin}}^{2}(x)}{{x}^{2}}\mathrm{d}x\\ & =& \frac{4}{3}{\int}_{0}^{\mathrm{\infty}}\frac{{\mathrm{sin}}^{2}(y)}{{y}^{2}}\mathrm{d}y-\frac{2}{3}{\int}_{0}^{\mathrm{\infty}}\frac{{\mathrm{sin}}^{2}(x)}{{x}^{2}}\mathrm{d}x\\ & =& \frac{2}{3}{\int}_{0}^{\mathrm{\infty}}\frac{{\mathrm{sin}}^{2}(x)}{{x}^{2}}\mathrm{d}x=\frac{\pi}{3}\end{array}$

evitagimm9h

Answered 2022-11-22
Author has **5** answers

That is, the FT of $(\mathrm{sin}x/x{)}^{2}$ is

${\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}dx\phantom{\rule{mediummathspace}{0ex}}\frac{{\mathrm{sin}}^{2}x}{{x}^{2}}{e}^{ikx}=\{\begin{array}{l}\\ \pi (1-\frac{|k|}{2})& |k|\le 2\\ 0& |k|>2\end{array}$

Plancherel/Parseval says that

${\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}dx\phantom{\rule{mediummathspace}{0ex}}\frac{{\mathrm{sin}}^{4}x}{{x}^{4}}=\frac{1}{2\pi}{\int}_{-2}^{2}dk\phantom{\rule{mediummathspace}{0ex}}{\pi}^{2}{(1-\frac{|k|}{2})}^{2}=\frac{\pi}{2}\frac{4}{3}=\frac{2\pi}{3}$

${\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}dx\phantom{\rule{mediummathspace}{0ex}}\frac{{\mathrm{sin}}^{2}x}{{x}^{2}}{e}^{ikx}=\{\begin{array}{l}\\ \pi (1-\frac{|k|}{2})& |k|\le 2\\ 0& |k|>2\end{array}$

Plancherel/Parseval says that

${\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}dx\phantom{\rule{mediummathspace}{0ex}}\frac{{\mathrm{sin}}^{4}x}{{x}^{4}}=\frac{1}{2\pi}{\int}_{-2}^{2}dk\phantom{\rule{mediummathspace}{0ex}}{\pi}^{2}{(1-\frac{|k|}{2})}^{2}=\frac{\pi}{2}\frac{4}{3}=\frac{2\pi}{3}$

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