$\int_0^{\infty} (-1)^{\lfloor{x^2}\rfloor} dx$ converges or diverges? Determine whether the following integral converges: $$\int_{0}^{\infty} (-1)^{\lfloor{x^2}\rfloor}  dx$$
I have no idea how to approach this one. It seems like it diverges similarly to $\sin x$ but I don't know how to show it.
Thank you.
 A: You can write this integral as $\int_0^{\infty}(-1)^{\lfloor x^2\rfloor}=\sum_{n=0}^{\infty}(-1)^n(\sqrt{n+1}-\sqrt{n})$. Can you take it from there?
A: Take the partial integral:
$$
I(t)=\int_0^t (-1)^{\lfloor x^2 \rfloor} dx
$$
Then cut it into chunks:
$$
I(t) = \sum_{k=0}^{\lfloor t^2 \rfloor -1} \int_\sqrt{k}^\sqrt{k+1} (-1)^{\lfloor x^2 \rfloor} dx + \int_{\sqrt{\lfloor t^2 \rfloor}}^t (-1)^{\lfloor x^2 \rfloor} dx
$$
Since $\lfloor x^2 \rfloor$ is constant on each interval, you may know write:
$$
I(t) = \sum_{k=0}^{\lfloor t^2 \rfloor} (-1)^k(\sqrt{k+1}-\sqrt{k}) + (-1)^{\lfloor t^2 \rfloor+1} (\sqrt{\lfloor t^2 \rfloor+1}-t)
$$
Then, because $x\mapsto \sqrt{x}$ is increasing over $\mathbb{R}^+$, $\sqrt{k+1}-\sqrt{k}>0$. And you may know conclude by using the alternating series test.
A: You can do much more than proving it is convergent, you may find its explicit value. As already remarked,
$$ \int_{0}^{+\infty}(-1)^{\lfloor x^2\rfloor}\,dx = \sum_{n\geq 0}(-1)^n\left(\sqrt{n+1}-\sqrt{n}\right) $$
is convergent by Leibniz' rule, since $\sqrt{n+1}-\sqrt{n}=\frac{1}{\sqrt{n+1}+\sqrt{n}}$ is positive and decreasing towards zero. By the (inverse) Laplace transform and integration by parts
$$ \sum_{n\geq 0}(-1)^n\left(\sqrt{n+1}-\sqrt{n}\right)=\frac{1}{2\sqrt{\pi}}\int_{0}^{+\infty}\frac{e^s-1}{s^{3/2}(e^s+1)}\,ds=\frac{2}{\sqrt{\pi}}\int_{0}^{+\infty}\frac{e^s}{\sqrt{s}(e^s+1)^2}\,ds $$
hence
$$  \sum_{n\geq 0}(-1)^n\left(\sqrt{n+1}-\sqrt{n}\right)=2\,\eta\left(\tfrac{1}{2}\right)-\frac{4}{\sqrt{\pi}}\int_{0}^{+\infty}\frac{ds}{(e^{s^2}+1)^2}\approx 0.76.  $$
