Convergence of $\iint_{\Bbb R^2}\frac{dxdy}{(1+x^2+xy+y^2)^\alpha}$ I need to determine for what values of $\alpha$ the following improper integral converges:  $$\iint_{\Bbb R^2}\frac{dxdy}{(1+x^2+xy+y^2)^\alpha}$$
Any hints on how to change variable? 
 A: $x^2+xy+y^2$ is a quadratic form associated to $\left(\begin{smallmatrix}1 & \frac{1}{2}\\\frac{1}{2}& 1\end{smallmatrix}\right)$, having eigenvalues $\frac{1}{2}$ and $\frac{3}{2}$. By enforcing the substitution $x=\frac{A+B}{\sqrt{2}}, y=\frac{A-B}{\sqrt{2}}$ the original integral is converted into
$$ I(\alpha)=\iint_{\mathbb{R}^2}\frac{dA\,dB}{\left(1+\frac{3}{2}A^2+\frac{1}{2}B^2\right)^\alpha}=\frac{2}{\sqrt{3}}\iint_{\mathbb{R}^2}\frac{du\,dv}{\left(1+u^2+v^2\right)^{\alpha}} $$
and by switching to polar coordinates:
$$ I(\alpha)=\frac{4\pi}{\sqrt{3}}\int_{0}^{+\infty}\frac{\rho}{(1+\rho^2)^{\alpha}}\,d\rho=\color{red}{\frac{2\pi}{\sqrt{3}(\alpha-1)}} $$
as soon as $\color{red}{\alpha>1}$.

Remark: an explicit diagonalization (in order to find the correct substitution) is not really needed. If $Ax^2+Bxy+Cy^2=q(x,y)$ is a positive definite quadratic form,
$$ \forall \alpha>1,\qquad \iint_{\mathbb{R}^2}\frac{dx\,dy}{(1+q(x,y))^\alpha}=\frac{2\pi}{(\alpha-1)\sqrt{4AC-B^2}} .$$
A: A hint:
From $2|xy|\leq x^2+y^2$ it follows that
$${1\over2}(x^2+y^2)\leq x^2+xy+y^2\leq{3\over2}(x^2+y^2)\ .$$
