The integral is:
$$ I =\int_{0}^{\infty} \frac{\sin^{5}(\alpha x)}{x^{3}}dx; $$
I think that the Dirichlet Integral can helps me: $$\int\limits_0^{+\infty} \frac{\sin \alpha x}{x}\,dx = \frac{\pi}{2}\mathrm{sgn}\,\alpha $$ But I have no idea how to get rid of the $x^{3}$ in the denominator.
Any hint or advice would be much appreciated!
Outline: You can write $$ \sin^5{\theta} = \frac{1}{16}\sin{5\theta} -\frac{5}{16}\sin{3\theta}+\frac{5}{8}\sin{\theta} $$ (this is basically the reverse of de Moivre's formula) to deal with the power of sine in the numerator.
For the power of $x$ in the denominator, you can with sufficient care integrate by parts twice to get down to things of the form $\frac{\sin{k\alpha x}}{x}$.