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!
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.