Why $\int_{ \mathbb{R}^2 } \frac{dx\,dy }{(1+x^4+y^4)} $ converges? Why $\int_{ \mathbb{R}^2 } \frac{dx\,dy }{(1+x^4+y^4)} $ converges?
Apparently this integral is quite similar to the integral
$\iint_{\mathbb R^2} \frac{dx \, dy}{1+x^{10}y^{10}}$ diverges or converges?
and it converges.
So this is quite remarkable that it does converge.
 A: Answer. Yes.
Note that
$$
1+x^4+y^4\ge 1+\frac{1}{2}(x^2+y^2)^2
$$
and hence, using polar coordinates ($x=r\cos\theta, \,y=r\sin\theta$), we have
$$
\int_{\mathbb R^2}\frac{dx\,dy}{1+x^4+y^4}\le \int_{\mathbb R^2}\frac{dx\,dy}{1+\frac{1}{2}(x^2+y^2)^2}=
\int_0^{2\pi}\int_0^\infty\frac{r\,dr\,d\theta}{1+\frac{1}{2}r^4}\le 
\int_0^{2\pi}\int_0^\infty\frac{2r\,dr\,d\theta}{1+r^4}\\ 
=2\pi \int_0^\infty\frac{ds}{1+s^2}=2\pi\cdot\frac{\pi}{2}=\pi^2.
$$
Note. If we set
$$
A=\{(x,y): |xy|\le 1\},
$$
then $A$ has infinite area, since $\int_0^\infty\frac{dx}{x}=\infty$. Meanwhile,
$$
\frac{1}{1+x^{10}y^{10}}\ge \frac{1}{2}, \quad \text{for all $(x,y)\in A$} 
$$
and hence
$$
\int_{\mathbb R^2}\frac{dx\,dy}{1+x^{10}y^{10}}\ge \int_{A}\frac{dx\,dy}{1+x^{10}y^{10}}\ge \int_{A}\frac{1}{2}\,dx\,dy=\infty.
$$
A: The two functions are not so similar, so it is not surprising. Apart from calculus-tricks, visual inspection is useful to understand what is going on.
I plot the function $f(x,y) = \frac{1}{1+x^4 y^4}$ in the $(x,y,z)$ space.. the integral is the volume between the yellow surface and the $(x,y, 0)$ plane (i.e. the $z=0$ plane). Clearly, since $x^4 y^4=0$ along the x-axis and the y-axis, here the function is constant ($f(x,0)=f(0,y)=1$). Similarly if you replace the $4$th power with the $10$th power, as in the question you linked.

Now plot the function $g(x,y) = \frac{1}{1+x^4+ y^4}$. The shape is really different. In fact, the function has a single maximum at the origin, where you have $g(0,0) =1$ and then decays quite fast to zero in every direction.

Clearly this is far from rigorous, but it gives you the flavor of why $f$ behaves differently from $g$.
