For what values of $a\in \mathbb{R}$ is the integral $\int\int_{\mathbb{R}^2} \frac{1}{(1+x^4+y^4)^a}dxdy$ convergent? Well, first notice that the above integral equals to $2\int_{0}^\infty \int_0^\infty \frac{1}{(1+x^4+y^4)^a}dxdy$ since the integrand is an even function of both $x$ and $y$.
There are two approaches which I thought to do here.


*

*Change of variables: $x^2 = r\cos \theta , y^2 = r\sin \theta$; which I am not sure it's valid since there are value of theta which give the rhs to be negative, and we are in real calculus.

*The legitimate approach is to break the integral into:
$$\int_0^1 \int_0^1+\int_0^1\int_1^\infty + \int_1^\infty\int_0^1+\int_1^\infty \int_1^\infty$$
I thought to compare the integrand with $x^2+y^2$, i.e when $x,y \in [0,1]$ we know that $x^2 \ge x^4$ and when $x>1$ then the opposite follows.
Seems a bit long calculation.
Can anyone help me with this?
Thanks!
 A: Assume $a > 0$, since divergence is obvious for $a \leqslant 0$. Changing to polar coordinates, $x = r \cos \theta, \,y = r \sin \theta$, we have
$$\int_0^\infty\int_0^\infty \frac{1}{(1+ x^4 + y^4)^a}\, dx\, dy =\int_0^\infty\int_0^{\pi/2} \frac{r\,dr\, d\theta }{(1+ r^4\cos^4 \theta + r^4 \sin^4 \theta)^a}\, $$
Since $\cos^4 \theta + \sin^4 \theta$ attains a minimum value of $1/2$ at $\theta = \pi/4$ and a maximum value of $1$ at $\theta = 0, \pi/2$, it follows that
$$\int_0^\infty\int_0^\infty \frac{1}{(1+ x^4 + y^4)^a}\, dx\, dy \leqslant\int_0^\infty\int_0^{\pi/2} \frac{1}{(1+ r^4/2 )^a}\, r\,dr\, d\theta \\= \frac{\pi}{2}\int_0^\infty \frac{r}{(1+ r^4/2 )^a}\,dr, \\ \int_0^\infty\int_0^\infty \frac{1}{(1+ x^4 + y^4)^a}\, dx\, dy \geqslant\int_0^\infty\int_0^{\pi/2} \frac{1}{(1+ r^4 )^a}\, r\,dr\, d\theta \\= \frac{\pi}{2}\int_0^\infty \frac{r}{(1+ r^4 )^a}\,dr$$
Note that $\dfrac{r}{(1+cr^4)^a} = \mathcal{O}(r^{1-4a})$ as $r \to \infty$. You should easily determine the values of $a$ for which the integral converges / diverges using the above comparisons.
