For which $\alpha$ the integral converges $$\iint_{\mathbb{R}^2}\frac{dx\,dy}{(x^2-xy+y^2+1)^\alpha}\,?$$ I changed to polar coordinates but got stuck at after calculating the integral of $r$
$$\int_{0}^{2\pi}dt\int_{0}^{n}\frac{r\,dr}{(r^2(1-1/2(\sin(2t))+1)^\alpha}=\int_{0}^{2\pi}dt\frac{((r^2(1-1/2(\sin(2t))+1)^{-\alpha+1}}{2(-\alpha+1)(1-(1/2)\sin(2t)}\big|_{0}^{n}$$ and after i plug the parameters i don't know how to continue .
To make it independent of the angular variable use the following change of coordinates
$$u = \frac{x-y}{\sqrt{2}} \hspace{24 pt} v = \frac{x+y}{\sqrt{2}}$$
with the factor chosen such that the Jacobian is $1$. Then the integral becomes
$$I = 2^\alpha\iint_{\Bbb{R}^2} \frac{du\:dv}{\left(3u^2+v^2+2\right)^\alpha}$$
Tweaking $u$ by the interchange $u \leftrightarrow \sqrt{3}u$ we get the integral
$$I=\frac{2^\alpha}{\sqrt{3}}\iint_{\Bbb{R}^2}\frac{du\:dv}{(u^2+v^2+2)^\alpha}$$
Using polar coordinates at this stage then gives us
$$I = \frac{2^{\alpha+1}\pi}{\sqrt{3}} \int_0^\infty \frac{r}{(r^2+2)^\alpha}dr = \begin{cases} \frac{\pi}{\sqrt{3}(\alpha-1)} & \alpha > 1 \\ \mbox{Diverges} & \alpha \leq 1 \\ \end{cases}$$