# Prove int_0^infty sin^4x/x^4 dx=pi/3

Prove ${\int }_{0}^{\mathrm{\infty }}\frac{{\mathrm{sin}}^{4}x}{{x}^{4}}dx=\frac{\pi }{3}$
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Zoe Andersen
You are likely expected to integrate by parts (twice)
$\begin{array}{rcl}\int \frac{{\mathrm{sin}}^{4}\left(x\right)}{{x}^{4}}\mathrm{d}x& =& -\frac{1}{3}\frac{{\mathrm{sin}}^{4}\left(x\right)}{{x}^{3}}+\frac{4}{3}\int \frac{\mathrm{cos}\left(x\right){\mathrm{sin}}^{3}\left(x\right)}{{x}^{3}}\mathrm{d}x\\ & =& -\frac{1}{3}\frac{{\mathrm{sin}}^{4}\left(x\right)}{{x}^{3}}-\frac{2\mathrm{cos}\left(x\right){\mathrm{sin}}^{3}\left(x\right)}{3{x}^{2}}+\frac{2}{3}\int \frac{3{\mathrm{cos}}^{2}\left(x\right){\mathrm{sin}}^{2}\left(x\right)-{\mathrm{sin}}^{4}\left(x\right)}{{x}^{2}}\mathrm{d}x\\ & =& -\frac{1}{3}\frac{{\mathrm{sin}}^{4}\left(x\right)}{{x}^{3}}-\frac{2\mathrm{cos}\left(x\right){\mathrm{sin}}^{3}\left(x\right)}{3{x}^{2}}+\frac{2}{3}\int \left(\frac{{\mathrm{sin}}^{2}\left(2x\right)}{{x}^{2}}-\frac{{\mathrm{sin}}^{2}\left(x\right)}{{x}^{2}}\right)\mathrm{d}x\end{array}$
where the last equality used
$\begin{array}{rcl}3{\mathrm{cos}}^{2}\left(x\right){\mathrm{sin}}^{2}\left(x\right)-{\mathrm{sin}}^{4}\left(x\right)& =& 3{\mathrm{cos}}^{2}\left(x\right){\mathrm{sin}}^{2}\left(x\right)-{\mathrm{sin}}^{2}\left(x\right)\left(1-{\mathrm{cos}}^{2}\left(x\right)\right)\\ & =& {\left(2\mathrm{sin}\left(x\right)\mathrm{cos}\left(x\right)\right)}^{2}-{\mathrm{sin}}^{2}\left(x\right)={\mathrm{sin}}^{2}\left(2x\right)-{\mathrm{sin}}^{2}\left(x\right)\end{array}$
Now
$\begin{array}{rcl}{\int }_{0}^{\mathrm{\infty }}\frac{{\mathrm{sin}}^{4}\left(x\right)}{{x}^{4}}\mathrm{d}x& =& \frac{2}{3}{\int }_{0}^{\mathrm{\infty }}\frac{{\mathrm{sin}}^{2}\left(2x\right)}{{x}^{2}}\mathrm{d}x-\frac{2}{3}{\int }_{0}^{\mathrm{\infty }}\frac{{\mathrm{sin}}^{2}\left(x\right)}{{x}^{2}}\mathrm{d}x\\ & =& \frac{4}{3}{\int }_{0}^{\mathrm{\infty }}\frac{{\mathrm{sin}}^{2}\left(y\right)}{{y}^{2}}\mathrm{d}y-\frac{2}{3}{\int }_{0}^{\mathrm{\infty }}\frac{{\mathrm{sin}}^{2}\left(x\right)}{{x}^{2}}\mathrm{d}x\\ & =& \frac{2}{3}{\int }_{0}^{\mathrm{\infty }}\frac{{\mathrm{sin}}^{2}\left(x\right)}{{x}^{2}}\mathrm{d}x=\frac{\pi }{3}\end{array}$
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evitagimm9h
That is, the FT of $\left(\mathrm{sin}x/x{\right)}^{2}$ is
${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}dx\phantom{\rule{mediummathspace}{0ex}}\frac{{\mathrm{sin}}^{2}x}{{x}^{2}}{e}^{ikx}=\left\{\begin{array}{l}\\ \pi \left(1-\frac{|k|}{2}\right)& |k|\le 2\\ 0& |k|>2\end{array}$
Plancherel/Parseval says that
${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}dx\phantom{\rule{mediummathspace}{0ex}}\frac{{\mathrm{sin}}^{4}x}{{x}^{4}}=\frac{1}{2\pi }{\int }_{-2}^{2}dk\phantom{\rule{mediummathspace}{0ex}}{\pi }^{2}{\left(1-\frac{|k|}{2}\right)}^{2}=\frac{\pi }{2}\frac{4}{3}=\frac{2\pi }{3}$