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 :).
 A: The convergence can be based on the alternating series test.
Let $0 < \alpha < \beta$. 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.
A: 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!
A: Hint: You can start from
$$\sin x=\frac{\mathrm{e}^{ix}-\mathrm{e}^{-ix}}{2i}$$
hence your integral can be written like:
$$\int_0^\infty x\sin(x^3) dx=\int_0^\infty\dfrac{ix\left(\mathrm{e}^{ix^3}-\mathrm{e}^{-ix^3}\right)}{2i^2}\,\mathrm{d}x=\int_0^\infty-\dfrac{ix\left(\mathrm{e}^{ix^3}-\mathrm{e}^{-ix^3}\right)}{2}\,\mathrm{d}x$$
$$=\frac i2\int_0^\infty x\left(\mathrm{e}^{ix^3}-\mathrm{e}^{-ix^3}\right)\,\mathrm{d}x$$
Last integral is simple to solve using the substitution $t=-x^2 \iff \mathrm{d}x=-\frac{1}{2x}\,\mathrm{d}t$ for $\int_0^\infty x\mathrm{e}^{ix^3}\,\mathrm{d}x$.
It is necessary to remember incomplete gamma function https://en.wikipedia.org/wiki/Incomplete_gamma_function
