Convergence of $\int _0^{\infty }\:\left(-1\right)^{\left[x^2\right]}$ It's clear to me it doesn't converge absolutely as $\int_0^\infty\:1$ doesn't

Question

I have to find out if $\int _0^{\infty }\:\left(-1\right)^{\left[x^2\right]}$ converges / diverges / converges absolutely.

It's clear to me it doesn't converge absolutely because $\int _0^{\infty}\:1$ doesn't converge, but I don't know how to go on from here.

Answer 1

The integral from $\sqrt n$ to $\sqrt{n+1}$ is $$(-1)^n\left(\sqrt{n+1}-\sqrt n\right).$$ The improper integral is therefore the sum of an alternating series. The absolute values of the terms are monotonically decreasing to zero and therefore the series is convergent.

Therefore the integral converges.