show that $\int_{0}^{\infty} \frac {\sin^3(x)}{x^3}dx=\frac{3\pi}{8}$ show that
$$\int_{0}^{\infty} \frac {\sin^3(x)}{x^3}dx=\frac{3\pi}{8}$$
using different ways
thanks for all

Related technique. You can use the Laplace transform technique. Recalling the Laplace transform

$$F(s)= \int_{0}^{\infty} f(x) e^{-sx}dx. $$

Taking $ f(x) = \frac{\sin(x)^3}{x^3} $ gives 

$$ F(s)= \frac{\pi \,{s}^{2}}{8}+\frac{3\,\pi}{8}- \frac{3( {s}^{2}-1) }{8}\,\arctan \left( s \right) +\frac{( {s}^{2}-9)}{8}\,\arctan \left( \frac{s}{3} \right) $$ 
  $$+\frac{3s}{8}\, \left( -\ln  \left( {s}^{2}+9 \right) +\ln  \left( {s}
^{2}+1 \right)  \right). $$

Taking the limit as $s\to 0$ gives the desired result $\frac{3\pi}{8}$.  
Another Laplace transform approach: Referring to the problem, we can use the following relation

$$ \begin{align}
 \int_0^\infty F(u)g(u) \, du  & = \int_0^\infty f(u)G(u) \, du \\[6pt]
 L[f(t)] & = F(s) \\[6pt]
 L[g(t)] & = G(s)\end{align} $$

Let 

$$ G(u)=\frac{1}{u^3} \implies g(u)=\frac{u^2}{2!}, $$

and 

$$ f(u)= \sin(u)^3  \implies F(u) = {\frac {6}{ \left( {u}^{2}+1 \right)  \left( {u}^{2}+9 \right) }}. $$

Now, 

$$ \int_0^\infty \frac{\sin^3 x}{x^3} \, dx = \frac{6}{2}\int_0^\infty \frac{u^2}{\left( {u}^{2}+1 \right)  \left( {u}^{2}+9 \right)} \, du = \frac{3\pi}{8}$$.

