Determine if $\int^\infty_0x\sin(x^3)dx$ is divergent, convergence or absolute convergence I'm trying to determine whether $\int^\infty_0x\sin(x^3)dx$ is convergence, absolute convergence or divergent. 
Let $f(x) = x \sin(x^3)$, It isn't true that $f(x)\ge 0$ for $[0,\infty)$.
Therefore the tools I'm familiar with are only showing that $|f|$ is convergence and this will prove that $f$ is also convergence.
Or use Dirichlet test to show convergence, which doesn't fit in here.
How should I apporoach this question?
 A: A $u$-substitution and integration by parts will help to determine convergence of the integral. The $u$-sub $u = x^3$ gives
$$\int_0^\infty x\sin(x^3)\, dx = \int_{0}^{\infty} u^{1/3}\sin(u)\cdot \frac{1}{3}u^{-2/3}\, du = \frac{1}{3}\int_{0}^{\infty} \frac{\sin u}{u^{1/3}}\, du$$
and by integration by parts, 
$$\int_{0}^{\infty} \frac{\sin u}{u^{1/3}}\, du = \frac{1-\cos u}{u^{1/3}}\bigg|_{u = 0}^{\infty} + \int_{0}^{\infty} \frac{1 - \cos u}{u^{4/3}}\, du = \int_0^\infty \frac{1 - \cos u}{u^{4/3}}\, du$$
Since $\lim\limits_{u\to 0}\, (1 - \cos u)/u^{4/3} = 0$, the integral $\int_0^1 (1 - \cos u)/u^{4/3}\, du$  is proper. As $0 \le 1 - \cos u \le 2$ for all $u \ge 0$ and $\int_1^\infty 2/u^{4/3}\, du$ converges, $\int_1^\infty (1 - \cos u)/u^{4/3}\, du$ converges. Therefore, $\int_0^\infty (1 - \cos u)/u^{4/3}\, du$ converges. Consequently, $\int_0^\infty x\sin(x^3)\, dx$ converges.
However, $\int_0^\infty x\sin(x^3)\, dx$ does not converge absolutely. Indeed, this is because $\int_0^\infty (\sin u)/u^{1/3}\, du$ does not converge absolutely. If $N$ is a positive integer, 
$$\int_0^{N\pi} \left\lvert \frac{\sin u}{u}\right\rvert\, du = \sum_{n = 1}^{N} \int_{(n-1)\pi}^{n\pi} \left\lvert \frac{\sin u}{u^{1/3}}\right\rvert\, du = \sum_{n = 1}^N \int_0^\pi \left\lvert\frac{\sin(v + (n-1)\pi)}{[v + (n-1)\pi]^{1/3}}\right\rvert\, dv$$ 
using the $v$-sub $v = u - (n-1)\pi$. Since $\sin(v + (n-1)\pi) = (-1)^{n-1}\sin v$ for all $n$ and $v$, the latter sum equals
$$\sum_{n = 1}^N \int_0^\pi \frac{\sin v}{[v + (n-1)\pi]^{1/3}}\, dv$$
which is greater than or equal to 
$$\sum_{n = 1}^N \int_0^\pi \frac{\sin v}{[\pi + (n-1)\pi]^{1/3}}\, dv = \frac{2}{\pi^{1/3}}\sum_{n = 1}^N \frac{1}{n^{1/3}}$$
Since the series $\sum\limits_{n = 1}^\infty n^{-1/3}$ diverges, it follows that $\int_0^\infty \lvert (\sin u)/u^{1/3}\rvert\, du$ diverges.
