Does the integral $$\int_0^\infty \sin(2x^4) \, dx$$ converge absolutely/conditionally?
I tried to evaluate $\int_0^b \sin(2x^4) dx$ by integrating by parts twice and got something relatively complicated, so I guess another way is preferable.
I also thought about using the Abel/Dirichlet test. Could it be applied here?
Via the substitution $2 x^4=u$, it can be shown that your integral is equivalent to $$I=C\int_0^\infty \!du \,\frac{\sin u}{u^{3/4}} ,$$ with $C= 512^{-1/4}$.
Potential problems might arise for $u\to 0$ or for $u\to\infty$. So it is useful to split the integral in $I= C( I_1 + I_2)$ with $$ I_1 = \int_0^\pi \!du \frac{\sin u}{u^{3/4}}$$ and $$ I_2= \int_\pi^\infty \!du \frac{\sin u}{u^{3/4}}.$$
For $u\to0$, we have that $$ \frac{\sin u}{u^{3/4}} = O(u^{1/4})$$ so $I_1$ converges absolutely.
The integral $I_2$ is a bit more tricky. To test the absolute convergence, we need to estimate $$ \int_\pi^\infty\!du\,\frac{|\sin u|}{u^{3/4}} = \sum_{n=1}^\infty \int_{n\pi}^{(n+1)\pi}\!du\,\frac{|\sin u|}{u^{3/4}}\geq \sum_{n=1}^\infty (n\pi)^{-3/4} \int_{n\pi}^{(n+1)\pi} |\sin u| = \frac{2}{\pi^{3/4}} \sum_{n=1}^\infty n^{-3/4}$$ where we used the fact that $u^{-3/4}$ is monotonously decreasing. As $\sum n^{-3/4}$ diverges, the integral is not absolutely convergent.
On the other hand, we have that $$ I_2 = \sum_{n=1}^\infty \int_{n\pi}^{(n+1)\pi}\!du\,\frac{\sin u}{u^{3/4}}\leq \sum_{n=1}^\infty ( (n+1)\pi)^{-3/4} \int_{n\pi}^{(n+1)\pi} \sin u = \frac{2}{\pi^{3/4}} \sum_{n=1}^\infty (-1)^{n+1}(n+1)^{-3/4}$$ which converges because of the Dirichlet test. So the integral is in fact conditionally convergent.