Conditional convergence of $\int_{0}^{\infty}(-1)^{[ x^2]}dx$ I'm trying to prove that the following integral $\int_{0}^{\infty}(-1)^{[ x^2]}dx$ is conditional converges, ( The brackets stand for the floor function). I can't use the integral test since this is clearly not a positive function, but I'd like somehow to do an analog to  Leibniz series and use it, I'm not sure how. 
Any suggestions?
Thank you!

Note that
$$
\int\limits_{[0,+\infty)}(-1)^{[x^2]}dx=
\sum\limits_{k=0}^\infty\int\limits_{[\sqrt{k},\sqrt{k+1})}(-1)^{[x^2]}dx=
\sum\limits_{k=0}^\infty\int\limits_{[\sqrt{k},\sqrt{k+1})}(-1)^{k}dx=
$$
$$
\sum\limits_{k=0}^\infty(-1)^{k}(\sqrt{k+1}-\sqrt{k})=
\sum\limits_{k=0}^\infty\frac{(-1)^{k}}{\sqrt{k+1}+\sqrt{k}}
$$
Using Leibniz test we see that this sum converges, hence converges the integral
