I have been unable to find a definitive answer to this question anywhere online. Some say the integral *does* converge and its value is $0$, while others say that it does not. I have seen similar questions on here where reputable users (in excess of 15k reputation) had different answers (the function was $\sin x$). So as for this one.

$|x|\sin x$ is clearly odd, meaning that any sequence
$$a\_n=\int\_{-n\pi}^{n\pi}{|x|\sin x\, dx},\: \text{or}\:\: b\_n=\int\_{-2\pi n}^{2\pi n}{|x|\sin x\, dx}$$
and so on, will converge to $0$ as $n\to\infty$. The question is: Does this imply the convergence of
$$\int\_{-\infty}^{\infty}{|x|\sin x\,dx}\: ?$$
Any help would be appreciated. Level of mathematics is anything up to "Infinite Sequences" in Spivak, or anything of a similar level.