<s>[INST] 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
 [/INST] $$
\begin{aligned}
\int_{0}^{\infty} \frac{\sin ^{3} x}{x^{3}} d x 
=& \frac{1}{4} \int_{0}^{\infty} \frac{3 \sin x-\sin 3 x}{x^{3}} d x \\
\stackrel{IBP }{=} & \frac{1}{8} \int_{0}^{\infty} \frac{(3 \sin x-\sin 3 x)^{(2)}}{x} d x\\
=&\frac{1}{8} \int_{0}^{\infty} \frac{-3 \sin x+9 \sin 3 x}{x} d x \\
=&\frac{1}{8}\left(-\frac{3 \pi}{2}+\frac{9 \pi}{2}\right) \\
=&\frac{3 \pi}{8}
\end{aligned}
$$
</s>