I am trying to compute the improper integarl $$\int_0^\infty \frac{|\sin x|^3}{x^2} \mathrm dx.$$ First of all, since $$ \lim_{x\to 0} \frac{\sin x}{x} = 1,$$ the integrand $$ \frac{|\sin x|^3}{x^2} = \frac{\sin ^2 x}{x^2} \sin x$$ is bounded around $x=0$. Also, the bound $$ \frac{|\sin x|^3}{x^2} \le \frac{1}{x^2}$$ and the integral test implies that the improper integral is bounded. But I don't see how to calculate the exact value.
I come to this problem when I try to calculate the limit: \begin{gather*} \lim_{n \to \infty} \frac{1}{n} \int_0^{\frac{\pi}{2}} x \left|\frac{\sin nx}{\sin x} \right|^3 \, \mathrm{d}x = \int_0^{+\infty} \frac{\left| \sin x \right|^3}{x^2} \, \mathrm{d}x \end{gather*} One may see the answer below to see how the above limit is derived.
We indeed have a very simple answer:
$$ \int_{0}^{\infty} \frac{\left| \sin u \right|^3}{u^2} \, \mathrm{d}u = 1 $$
However, all of my solutions rely on some additional knowledge. For instance, one of my solutions relies on the identity
$$ \sum_{n=-\infty}^{\infty} \frac{1}{(x+n\pi)^2} = \frac{1}{\sin^2 x}, $$
whose proof usually involves on some degree of complex analysis. Then, assuming this, we have
\begin{align*} \int_{0}^{\infty} \frac{\left| \sin u \right|^3}{u^2} \, \mathrm{d}u &= \frac{1}{2} \int_{-\infty}^{\infty} \frac{\left| \sin u \right|^3}{u^2} \, \mathrm{d}u \\ &= \frac{1}{2} \sum_{n=-\infty}^{\infty} \int_{n\pi}^{(n+1)\pi} \frac{\left| \sin u \right|^3}{u^2} \, \mathrm{d}u \\ &= \frac{1}{2} \sum_{n=-\infty}^{\infty} \int_{0}^{\pi} \frac{\sin^3 u}{(u + n\pi)^2} \, \mathrm{d}u \tag{$u \mapsto u + n\pi$} \\ &= \frac{1}{2} \int_{0}^{\pi} \sum_{n=-\infty}^{\infty} \frac{\sin^3 u}{(u + n\pi)^2} \, \mathrm{d}u \tag{$\because$ Tonelli} \\ &= \frac{1}{2} \int_{0}^{\pi} \sin u \, \mathrm{d}u \\ &= 1. \end{align*}
In the third to last step, Fubini-Tonelli Theorem is applied to interchange the order of summation and integration.
Currently I am seeking for a more self-contained solution for this.
Answer to the initial question (or so I misunderstood). Substitute $x=u/n$ to obtain
$$ \frac{1}{n} \int_{0}^{\frac{\pi}{2}} x \left| \frac{\sin nx}{\sin x} \right|^3 \, \mathrm{d}x = \int_{0}^{\frac{n\pi}{2}} u \left| \frac{\sin u}{n \sin (u/n)} \right|^3 \, \mathrm{d}u. $$
Now use the fact that $ \sin x \geq \frac{2}{\pi}x$ for $0 \leq x \leq \frac{\pi}{2}$ to produce the bound
$$ u \left| \frac{\sin u}{n \sin (u/n)} \right|^3 \leq u \left| \frac{\sin u}{(2/\pi)u} \right|^3 = (\pi/2)^3 \frac{\left| \sin u \right|^3}{u^2}. $$
But since $ \int_{0}^{\infty} \frac{\left| \sin u \right|^3}{u^2} \, \mathrm{d}u < \infty$, the dominated convergence allows to interchange the order of integration and limit to yield
\begin{align*} \lim_{n\to\infty} \int_{0}^{\frac{n\pi}{2}} u \left| \frac{\sin u}{n \sin (u/n)} \right|^3 \, \mathrm{d}u &= \int_{0}^{\infty} \lim_{n\to\infty} u \left| \frac{\sin u}{n \sin (u/n)} \right|^3 \mathbf{1}_{[0, n\pi/2]}(u) \, \mathrm{d}u \\ &= \int_{0}^{\infty} \frac{\left| \sin u \right|^3}{u^2} \, \mathrm{d}u \end{align*}