$\iint_{\mathbb{R}^2} \frac{1}{\sqrt{1+x^4+y^4}}$ converges or diverges?

Question

$$\iint_{\mathbb{R}^2} \frac{1}{\sqrt{1+x^4+y^4}}$$ converges or diverges?

I've tried to change to polar coordinates but i got stuck really quick
$$\int_{0}^{2\pi}\int_{0}^{\infty}\frac{r}{\sqrt{1+r^4(1-2\sin^2(t)\cos^2(t))}}drdt$$
any hint please?

Answer 1

If $r>0$ and $\theta\in[0,2\pi]$, then$$(r\cos\theta)^4+(r\sin\theta)^4\leqslant r^4(\cos^2\theta+\sin^2\theta)=r^4.$$Therefore\begin{align}\int_0^{2\pi}\int_0^\infty\frac r{\sqrt{1+(r\cos\theta)^4+(r\sin\theta)^4}}\,\mathrm dr\,\mathrm d\theta&\geqslant\int_0^{2\pi}\int_0^\infty\frac r{\sqrt{1+r^4}}\,\mathrm dr\,\mathrm d\theta\\&=2\pi\int_0^\infty\frac r{\sqrt{1+r^4}}\,\mathrm dr.\end{align}This integral diverges, since$$\lim_{r\to\infty}\frac{\frac r{\sqrt{1+r^4}}}{\frac1r}=1$$and the integral $\int_1^\infty\frac{\mathrm dr}r$ diverges.

Answer 2

Hint: $x^4+y^4 = r^4(\cos^4\theta + \sin^4 \theta) \le 2r^4$. The integral diverges !