Determine for which $t\in \mathbb{R}$ the integral $\int_0^\infty \sin(x^t)\ dx$ converges!

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Obviously for $t=0$ the integral diverges.

For $t>1$ I have tried to subsitute $u=x^t$ and then split the integral up into segments of pi length such that I can write it as an alternating sum. Then (I think) I managed to prove that I can use Leibniz test since the alternating integrals get ever so smaller as $x \to \infty$.

Also for $x\in(0,1]$ I can bound the sine from below by inscribing ever increasing triangles thus proving that the sequence doesn't go to 0, which implies that the series is divergent.

Now for $t<0$ I am completely clueless, partly because the graph is vastly different in that case. Substituting $1/x = u$ gets ugly very quickly, so I didn't follow through.

I'd highly appreciate any ideas and solutions, particularly if they are more elegant than my tries so far have been :)

Thanks in advance!