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

Using the formula found in my answer,
$$
\begin{aligned}
\int_{0}^{\infty} \frac{\sin ^{3} x}{x^{3}} &=\frac{\pi}{2^{3} \cdot 2 !}\left[\left(\begin{array}{l}
3 \\
0
\end{array}\right) 3^{2}-\left(\begin{array}{l}
3 \\
1
\end{array}\right) 1^{2}\right] \\
&=\frac{3 \pi}{8}
\end{aligned}
$$
