For which $p \in (0,\infty)$ is $\int_{y=1}^{\infty}\int_{x=1}^{x=\sqrt{y}}|\cos x\sin y |^p\frac{1}{y2}dxdy<\infty$

Question

For which $p \in (0,\infty)$ is

$$\int_{y=1}^{\infty}\int_{x=1}^{x=\sqrt{y}}\left|\cos x\sin y \right|^p\frac{1}{y^2}dxdy<\infty.$$

I was thinking for all $p \in (0,\infty)$ since $|\sin x \cos y |\leq1$.

But I'm not sure about this.

Answer 1

As you said, for every $p\in(0,+\infty)$ we get

$$ \begin{split} \int_{1}^{\infty}\int_{1}^{\sqrt{y}}\left|\cos x\sin y \right|^p\frac{1}{y^2}\mathrm dx\mathrm dy &\le \int_{1}^{\infty}\int_{1}^{\sqrt{y}} \frac{1}{y^2}\mathrm dx\mathrm dy \\ &=\int_{1}^{\infty}\frac{1}{y^2}\int_{1}^{\sqrt{y}} \mathrm dx\mathrm dy\\ &=\int_{1}^{\infty}\frac{\sqrt y -1}{y^2}\mathrm dy \\ &=\int_{1}^{\infty}\left(\frac{1}{y^{3/2}}-\frac{1}{y^2}\right)\mathrm dy \\ &= \lim_{R \to +\infty} \left( \frac{1}{R}-\frac{2}{\sqrt R}\right)+1 = 1 \end{split} $$

Actually with the same proof we could demonstrate a more general result:

$$\int_{1}^{\infty}\int_{1}^{\sqrt{y}}\frac{\left|\cos x\sin y \right|^p}{y^q}\mathrm dx\mathrm dy < +\infty$$ For every $p \in[0,+\infty)$ and $ q \in (2,+\infty) $