Average value of $x\sin(x)$ from $0$ to $\infty$

Question

Are there any arguments I could use to tell that the average value of $x\sin(x)$ is zero? The average value oscillates around zero with the increasing amplitude as $x$ goes to infinite.

Answer 1

$$A_N=\int_0^N x\sin(x)dx$$ $$ = -x\cos(x)|_{0}^N + \int_0^N\cos(x)dx$$ $$=-N\cos(N)-\sin(N)$$ Thus, $\langle x\sin(x) \rangle= \lim_{N\to\infty}\frac{A_N}{N} = \lim_{N\to\infty} -\cos N$

Which is not defined

Answer 2

Hint. Note that for $t>0$ the (integral) mean value of $x\sin(x)$ over $[0,t]$ is, by definition, equal to $$\frac{1}{t}\int_0^t x\sin(x)\,dx=-\cos(t)+\frac{1}{t}\int_0^t \cos(x)\,dx=-\cos(t)+\frac{\sin(t)}{t}$$ which has no limit as $t$ goes to $+\infty$. If you take the average for interval of length $t_n=\pi n+\frac{\pi}{2}$ then
$$\lim_{n\to +\infty}\frac{1}{t_n}\int_0^{t_n} x\sin(x)\,dx=\lim_{n\to +\infty}\left(-\cos(t_n)+\frac{\sin(t_n)}{t_n}\right)=\lim_{n\to +\infty}\left(0+\frac{(-1)^n}{t_n}\right)=0.$$