When does $\int \frac{1}{(1+x^2+xy+y^2)^\alpha}$ converge? I wish to find for which $\alpha \in \mathbb R$ the integral $\int_{\mathbb R^2}\frac{1}{(1+x^2+xy+y^2)^\alpha}dxdy$ converges.
What I tried:
Firstly notice that the integrand is always positive since $xy \leq x^2+y^2$.
If $\alpha \leq 0$ the integral trivially diverges. Assume $\alpha >0$. Then:
$\int_{\mathbb R^2}\frac{1}{(1+x^2+xy+y^2)^\alpha}dxdy \geq \int_{\mathbb R^2}\frac{1}{(1+2x^2+2y^2)^\alpha}dxdy= 2\pi\int_{0}^{\infty}\frac{r}{(1+2r^2)^\alpha}dr = \pi\int_{0}^{\infty}\frac{1}{(1+2t)^\alpha}dt = \frac{\pi}{2}\int_{1}^{\infty}\frac{1}{s^\alpha}ds$
This integral diverges when $\alpha \leq 1$ so it follows that our original integral diverges when $\alpha \leq 1$.
Now what? I'm stuck.

Hint:
For $\alpha > 0$, changing to polar coordinates,
$$\frac{1}{(1+x^2+xy+y^2)^\alpha} = \frac{1}{(1+r^2+r^2\cos \theta \sin \theta)^\alpha} = \frac{1}{(1+r^2(1+\sin (2\theta)/2))^\alpha} \\ \begin{cases}\leqslant  \frac{1}{(1+r^2/2)^\alpha}\\ \geqslant \frac{1}{(1+3r^2/2)^\alpha}\end{cases}$$
and as $r \to \infty$
$$\frac{r}{(1+r^2/2)^\alpha} =\mathcal{O}( r^{1-2\alpha})$$
