Does $\int_{0}^{\infty} \sin(x^3) \, dx$ converge?

Question

Does

$$ \int_{0}^{\infty} \sin (x^3) \, dx $$

converge? And If it does, why?

You may substitute $u=x^3$ (or equivalently, $x=u^{1/3}$). A more general discussion can be found here. — Sangchul Lee, Jan 27 '22 at 08:49
It does. I can't provide a real analysis driven proof, but I can provide a closed form for $I(n), \ n\gt 1$ which will be enough to decide whether the integral converges, or not. $$I(n) = \int_0^\infty \sin(x^n), \mathrm dx$$ — Laxmi Narayan Bhandari, Jan 27 '22 at 08:50
An integration by part should probably help to show the convergence, like with the integral of sin(x^2) ? — Lelouch, Jan 27 '22 at 08:58
Yes, Lelouch is right. An integration by parts does this. Personally, I would first substitute $y=x^3$ and then try to find an integration by parts to solve this. — GEdgar, Jan 27 '22 at 09:03
I did it, but still didn't find a solution. Can you explain more please? @GEdgar — Dolores_Haze, Jan 27 '22 at 10:23
https://math.stackexchange.com/questions/1155150/integrate-using-cauchy-integral-theorem — cineel, Jan 27 '22 at 10:56
Apparent duplicate: https://math.stackexchange.com/questions/4367333/how-to-prove-that-the-improper-integral-int-0-infty-sinx3-dx-converges?noredirect=1&lq=1 — Travis Willse, Jan 27 '22 at 17:08

GEdgar · Answer 1 · 2022-01-27T08:55:15.663

3

not a solution

Graphical evidence of convergence: $$ \int_0^X \sin(x^3)\;dx $$

edited Jan 27 '22 at 08:55

answered Jan 27 '22 at 08:44

GEdgar

1 Answers1