Improper integral $\int_0^\infty x\sin(x^3) dx$ I need to check if the following integral converges / diverges conditionally.  $$\int_0^\infty x\sin(x^3) dx$$
I have tried integrating by parts and it didn't work.
Will appreciate any help :).
The convergence can be based on the alternating series test.
Let $0 &lt; \alpha &lt; \beta$. Then
$$
\int_{\alpha}^{\beta} x \sin(x^3) \, dx 
= \int_{\alpha^3}^{\beta^3} t^{1/3} \sin(t) \cdot \frac13 t^{-2/3} \, dt
= \frac13 \int_{\alpha^3}^{\beta^3} t^{-1/3} \sin(t)\, dt
$$ by change of variables, so your integral converges if and only if
$$
\int_{2\pi}^{\infty} t^{-1/3} \sin(t)\, dt 
$$
converges; it does since the subsequent integrals
$$
\int_{2\pi}^{3\pi} t^{-1/3} \sin(t)\, dt, \int_{3\pi}^{4\pi} t^{-1/3} \sin(t)\, dt, \cdots
$$
form an alternating sequence which decreases monotonically in absolute value and converges to zero.
Ok so applying Yves' suggestion I get:
$\int_0^\infty x\sin(x^3) dx = \int_0^\infty\sin(t)/t^{-3}dt$ which converges. Thank you!