Does the integral $\int_0^\infty \sin(2x^4) \, dx$ converge absolutely/conditionally? 
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?
 A: 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.
