<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] Well, there's a lot of creative answers but it seems that no one bothered to put here the "follow your nose" one, so I'll add it here for completeness.
Recall the identity: $$\sin^3x = \frac{3 \sin x - \sin(3x)}{4}.$$
So we have: $$\int_{-\infty}^{+\infty} \left(\frac{\sin x}{x}\right)^3\,{\rm d}x = \int_{-\infty}^{\infty} \frac{3 \sin x - \sin (3x)}{4x^3}\,{\rm d}x. $$Let $0 < r < R$. Consider $\gamma_1$ the line segment joining $r$ to $R$, $\gamma_2$ the circular arc oriented counterclockwise joining $R$ to $-R$, $\gamma_3$ the line segment joining $-R$ to $-r$ and $\gamma_4$ the circular arc oriented clockwise joining $-r$ to $r$. Consider $\gamma = \gamma_1 \ast \gamma_2 \ast \gamma_3 \ast \gamma_4$ the concatenation. Sketch:

Consider the function: $$ f(z) = \frac{3e^{iz}-e^{3iz}}{4z^3}. $$The only singularity is $z = 0$ (triple pole). Expanding in  Laurent: $$ f(z) = \frac{1}{4z^3}\left(3\sum_{n \geq 0}\frac{i^nz^n}{n!} - \sum_{n \geq 0}\frac{3^ni^nz^n}{n!}\right)  = \sum_{n \geq 0} \left(\frac{(3-3^n)i^n}{4\cdot n!}\right)z^{n-3}. $$ Since $f$ is holomorphic in $\gamma$ and inside it, by Cauchy-Goursat we get: $$ \oint_{\gamma} f(z)\,{\rm d}z = \int_{\gamma_1} f(z)\,{\rm d}z + \int_{\gamma_2} f(z)\,{\rm d}z + \int_{\gamma_3}f(z)\,{\rm d}z + \int_{\gamma_4}f(z)\,{\rm d}z = 0.$$ Let's analyze everything sistematically.
Parametrizing $\gamma_1(x) = x$, with $r \leq x \leq R$, we have: $$ \int_{\gamma_1}f(z)\,{\rm d}z = \int_r^R \frac{3e^{ix}-e^{3ix}}{4x^3}\,{\rm d}x. $$For $\gamma_2$, we have that its lenght is $\pi R$, $|e^{iz}| = e^{{\rm Re}(iz)} = e^{-{\rm Im}(z)} < 1$, and similarly $|e^{3iz}| < 1$, since for all $z$ in $\gamma_2$ we have ${\rm Im}(z) > 0$. Hence: $$ \left|\int_{\gamma_2} f(z)\,{\rm d}z\right| \leq \frac{\pi R(3+4)}{4 R^3} = \frac{\pi}{R^2} \stackrel{R\, \to \,+\infty}{\longrightarrow} 0. $$
Parametrizing $\gamma_3^-(x) = -x$, with $r \leq x \leq R$, we have: $$ \int_{\gamma_3}f(z)\,{\rm d}z =- \int_r^R \frac{3e^{-ix}-e^{-3ix}}{4x^3}\,{\rm d}x, $$once the signs in $(-x^3) = -x^3$ and ${\rm d}z = -{\rm d}x$ cancel each other. Notice here that: $$ \int_{\gamma_1}f(z)\,{\rm d}z + \int_{\gamma_3}f(z)\,{\rm d}z = 2i\int_r^R \left(\frac{\sin x}{x}\right)^3\,{\rm d}x.$$
For $\gamma_4$, we have: $$\begin{align} \int_{\gamma_4}f(z)\,{\rm d}z &= \int_{\gamma_4} \sum_{n \geq  0}\left(\frac{(3-3^n)i^n}{4 \cdot n!}\right)z^{n-3}\,{\rm d}z \\ &= \int_{\gamma_4}\frac{1}{2z^3}\,{\rm d}z + \int_{\gamma_4}\frac{3}{4z}\,{\rm d}z + \int_{\gamma_4} \sum_{n \geq 3} \left(\frac{(3-3^n)i^n}{4 \cdot n!}\right)z^{n-3}\,{\rm d}z \end{align}$$
We have: $$ \int_{\gamma_4} \frac{1}{2z^3}\,{\rm d}z = -\frac{1}{4z^2}\Bigg|_{-r}^{r} = 0, \quad \int_{\gamma_4} \frac{3}{4z}\,{\rm d}z = \frac{3}{4}\int_{\gamma_4} -i \frac{{\rm d}z}{-iz} = \frac{3}{4}\int_0^\pi -i\,{\rm d}t = -\frac{3\pi i}{4}, $$ and: $$  \left|\int_{\gamma_4} \sum_{n \geq 3}\frac{(3-3^n)i^n}{4\cdot n!}z^{n-3}\right| \leq \pi r \sum_{n \geq 3}\frac{3^n-3}{4\cdot n!}r^{n-3} = \sum_{n \geq 3}\frac{\pi(3^n-3)}{4 \cdot n!}r^{n-2} \stackrel{r \to 0}{\longrightarrow} 0. $$
Making first $r \to 0$, and then $R \to +\infty$, in $\oint_\gamma f = 0$, we get: $$  2i\int_0^{+\infty}\left(\frac{\sin x}{x}\right)^3\,{\rm d}x - \frac{3\pi i}{4} = 0 \implies \int_0^{+\infty}\left(\frac{\sin x}{x}\right)^3\,{\rm d}x = \frac{3\pi}{8}, $$as desired.
</s>