Calculate integral using Dirichlet integral

Question

I have to calculate the following integral: $$ \int\limits_0^{+\infty} \left(\frac{\sin \alpha x}{x}\right)^3\,dx, $$ using the Dirichlet integral: $$ \int\limits_0^{+\infty} \frac{\sin \alpha x}{x}\,dx = \frac{\pi}{2}\mathrm{sgn}\,\alpha. $$

It seems to me, that there exists some substitution.

Answer 1

Exploiting $3\sin z-\sin(3z)=4\sin^3 z$ and integrating by parts twice, $$\int_{0}^{+\infty}\frac{\sin^3(\alpha x)}{x^3}\,dx = \frac{1}{8}\int_{0}^{+\infty}\frac{3\alpha^2\sin(\alpha x)+9\alpha^2\sin(3\alpha x)}{x}\,dx=\color{red}{\frac{3\pi \alpha^2}{8}\text{Sign}(\alpha)}.$$

Answer 2

Hint

A clumsy way is to really calculate that:

$$\int u dv = u v -\int v du$$

So

$$\int (\frac{\sin \alpha x}{x})^3 dx = (\frac{\sin \alpha x}{x})^3 -\int x d((\frac{\sin \alpha x}{x})^3) dx =(\frac{\sin \alpha x}{x})^3-\int\frac{3x}{-x^2} (\alpha x \cos \alpha x - 1 \times sin \alpha x )(\frac{\sin \alpha x}{x})^2 dx$$

and so on ...