Francisca Rodden

2022-01-03

Show that:
${\int }_{0}^{\mathrm{\infty }}\frac{{\mathrm{sin}}^{3}\left(x\right)}{{x}^{3}}dx=\frac{3\pi }{8}$

Ana Robertson

Let
$f\left(y\right)={\int }_{0}^{\mathrm{\infty }}\frac{{\mathrm{sin}}^{3}yx}{{x}^{3}}dx$
Then,
${f}^{\prime }\left(y\right)=3{\int }_{0}^{\mathrm{\infty }}\frac{{\mathrm{sin}}^{2}yx\mathrm{cos}yx}{{x}^{2}}dx={\frac{34}{\int }}_{0}^{\mathrm{\infty }}\frac{\mathrm{cos}yx-\mathrm{cos}3yx}{{x}^{2}}dx$
$f{}^{″}\left(y\right)=\frac{3}{4}{\int }_{0}^{\mathrm{\infty }}\frac{-\mathrm{sin}yx+3\mathrm{sin}yx}{x}dx$
Therefore,
$f{}^{″}\left(y\right)=\frac{9}{4}{\int }_{0}^{\mathrm{\infty }}\frac{\mathrm{sin}3yx}{x}dx-\frac{3}{4}{\int }_{0}^{\mathrm{\infty }}\frac{sunyx}{x}dx$
Now, it is quite easy to prove that

Therefore,

Then,
${f}^{\prime }\left(y\right)=\frac{3\pi }{4}|y|+C$
Note that, ${f}^{\prime }\left(0\right)=0$, therefore, C=0

Again, $f\left(0\right)=0$, therefore, $D=0$
Hence,
$f\left(1\right)={\int }_{0}^{\mathrm{\infty }}\frac{{\mathrm{sin}}^{3}x}{{x}^{3}}=\frac{3\pi }{8}$

Janet Young

Now, it is quite easy to prove that
${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}dxf\left(x\right)g\cdot \left(x\right)=\frac{1}{2\pi }{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}dkF\left(k\right)G\cdot \left(K\right)$
where f,g and F,G are respective Fourier transform pairs, e.g.,
$F\left(k\right)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}dxf\left(x\right){e}^{ikx}$
etc. If $f\left(x\right)=\frac{\mathrm{sin}x}{x}$, then
$F\left(k\right)=\left\{\begin{array}{cc}\pi & |k|\le 1\\ 0& |k|>2\end{array}$
Further, if $g\left(x\right)=\frac{{\mathrm{sin}}^{2}x}{{x}^{2}}$, then
$G\left(k\right)=\left\{\begin{array}{cc}\pi \left(1-\frac{|k|}{2}\right)& |k|\le 2\\ 0& |k|>2\end{array}$
Then ${\int }_{0}^{\mathrm{\infty }}dx\frac{{\mathrm{sin}}^{3}x}{{x}^{3}}=\frac{1}{2\pi }{\int }_{-1}^{1}dk{\pi }^{2}\left(1-\frac{|k|}{2}\right)=\pi -\frac{\pi }{2}{\int }_{0}^{1}dkk=\pi -\frac{\pi }{4}$
therefore
${\int }_{0}^{\mathrm{\infty }}dx\frac{{\mathrm{sin}}^{3}x}{{x}^{3}}=\frac{3\pi }{8}$

karton

${\int }_{0}^{\mathrm{\infty }}\frac{{\mathrm{sin}}^{3}x}{{x}^{3}}dx\phantom{\rule{0ex}{0ex}}=\frac{1}{4}{\int }_{0}^{\mathrm{\infty }}\frac{3\mathrm{sin}x-3\mathrm{sin}x}{{x}^{3}}dx\phantom{\rule{0ex}{0ex}}=\frac{1}{8}{\int }_{0}^{\mathrm{\infty }}\frac{\left(3\mathrm{sin}x-\mathrm{sin}3x{\right)}^{\left(2\right)}}{x}dx\phantom{\rule{0ex}{0ex}}=\frac{1}{8}{\int }_{0}^{\mathrm{\infty }}\frac{-3\mathrm{sin}x+9\mathrm{sin}3x}{x}dx\phantom{\rule{0ex}{0ex}}=\frac{1}{8}\left(-\frac{3\pi }{2}+\frac{9\pi }{2}\right)\phantom{\rule{0ex}{0ex}}=\frac{3\pi }{8}$