$\int_0^{\infty} (-1)^{\lfloor{x^2}\rfloor} dx$ converges or diverges?

Question

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.

Answer 1

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?

Answer 2

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.

Answer 3

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 $$ \sum_{n\geq 0}(-1)^n\left(\sqrt{n+1}-\sqrt{n}\right)=2\,\eta\left(\tfrac{1}{2}\right)-\frac{4}{\sqrt{\pi}}\int_{0}^{+\infty}\frac{ds}{(e^{s^2}+1)^2}\approx 0.76. $$