$\int_{0}^{\infty}(-1)^{[x^2]}$ converge? I need to tell if $\int_{0}^{\infty}(-1)^{[x^2]}dx$ converge, diverge or absolutely converges.
I managed to say it does not absolutely converges.
for it's converges, diverges i tried substituting $t=x^2$ so i get $\int_{0}^{\infty}\frac{(-1)^{[t]}}{2\sqrt{t}}dt$ but I still can't get rid of the $[]$ (which stands for round)
How can i continue from here? or is it a wrong way to go?

Hint
$$\int_{\sqrt{0.5}}^{\sqrt{0.5+M}}(-1)^{[x^2]}dx=\sum_{n=1}^M\int_{\sqrt{n-0.5}}^{\sqrt{n+0.5}}(-1)^{[x^2]}dx$$
