Does the double integral diverge? $\iint_{\Omega} \sqrt{\frac{x^2 + y^2}{xy(1 - x^2 - y^2)}} ~ dx dy$

Question

Does the double integral converge $$\iint_{\Omega} \sqrt{\frac{x^2 + y^2}{xy(1 - x^2 - y^2)}} ~ dx dy$$ I tried using polar coordinate but then half way I realised it was not a good idea so what do I do?

I got $$\iint_{\Omega} \frac{r}{\sqrt{\cos(\theta)\sin(\theta) (1-r^2)}}drd\theta = \int \frac1{\sqrt{\cos(\theta)\sin(\theta)}}d\theta \times \int \frac{r}{\sqrt{(1-r^2)}},dr$$

$$\Omega = \left\{ (x, y) \in \mathbb{R}^2 : x^2 + y^2 < 1, x > 0, y > 0 \right\}$$

But I don't know how to continue, I haven't written the bounds yet because I was trying to simplify it.

Answer 1

As noticed, by polar coordinates we obtain

$$\iint_{\Omega} \sqrt{\frac{x^2 + y^2}{xy(1 - x^2 - y^2)}} dx dy= \int_0^{\frac \pi 2} \frac{1}{\sqrt{\cos \theta\sin \theta}}\int_0^1 \frac{r}{\sqrt{1-r^2}} \;dr \;d\theta $$

and since $\int_0^1 \frac{r}{\sqrt{1-r^2}} \;dr=1$ all boils down with

$$\int_0^{\frac \pi 2} \frac{1}{\sqrt{\cos \theta\sin \theta}}\;d\theta=2\int_0^{\frac \pi 4} \frac{1}{\sqrt{\cos \theta\sin \theta}}\;d\theta$$

and since for $\theta \to 0^+$ we have that $\cos \theta\sin \theta=\theta+O(\theta^3)$ the integral converges by limit comparison test with

$$\int_0^{\frac \pi 4} \frac{1}{\sqrt{\theta}}\;d\theta=\sqrt \pi$$

Answer 2

Integral on the radial part $$\int_0^1 \frac{r}{\sqrt{1-r^2}}dr=1$$ Integral on the angle part $$\begin{align}\int_0^{\pi/2} \frac1{\sqrt{\cos(\theta)\sin(\theta)}} d\theta &=\frac12\cdot2\int_0^{\pi/2} \sin^{-\frac12}(\theta)\cos^{-\frac12}(\theta) d\theta\\ \\ &=\frac12\cdot B\left(\frac14,\frac14\right)=\frac{\Gamma^2\left(\frac14\right)}{2\sqrt\pi}\end{align}$$

Therefore,

$$\boxed{\iint_{\Omega} \sqrt{\frac{x^2 + y^2}{xy(1 - x^2 - y^2)}} ~ dx dy=\frac{\Gamma^2\left(\frac14\right)}{2\sqrt\pi}~}$$