Im looking for which $\alpha\in\mathbb{R}$ the integral $\int_{\mathbb{R}^{2}}\frac{dxdy}{\left(1+x^{2}+xy+y^{2}\right)^{\alpha}}$ converges/diverges.
What I was looking for is an appropriate change of variables. I tried polar coordinates $\left(x,y\right)=T\left(r,\theta\right)=\left(r\cos\theta,r\sin\theta\right)$. So $\left|J_{T}\left(r,\theta\right)\right|=r$ and $T^{-1}\left(\mathbb{R}^{2}\right)=\left\{ \left(r,\theta\right)\mid r>0\:,\:0<\theta<2\pi\right\} $ up to a null set. So $$ \int_{\mathbb{R}^{2}}\frac{dxdy}{\left(1+x^{2}+xy+y^{2}\right)^{\alpha}}=\int_{T^{-1}\left(\mathbb{R}^{2}\right)}\frac{rdrd\theta}{\left(1+r^{2}+r^{2}\sin\theta\cos\theta\right)^{\alpha}}=\int_{T^{-1}\left(\mathbb{R}^{2}\right)}\frac{2rdrd\theta}{\left(2+r^{2}\left(2+\sin2\theta\right)\right)^{\alpha}} $$ Then with another change of variables $t=r^{2}$ we get that $dt=2rdr$ and so I got $$ \int\frac{dtd\theta}{\left(2+t\left(2+\sin2\theta\right)\right)^{\alpha}} $$ Questions:
- Does it help somehow?
- Is there a better change of variables here?
- How can I find the range of $\alpha$ that im looking for?
- Is it possible maybe to block the function with and easier one to integral and then use sandwich?
First note that if the integral over one of the four quadrants converge then the integral over the whole plane converges. [Use the transformations $x \to -x$ and $y \to -y$ for this]. So integrate over $x,y>0$ Note that $x^{2}+xy+y^{2} \geq x^{2}+y^{2}$ and $x^{2}+xy+y^{2} \leq 2(x^{2}+y^{2})$ since $2xy \leq x^{2}+y^{2}$. Now using polar coordinates you see that the integral converges iff $\frac {r} {r^{2\alpha}}$ is integrable near $\infty$ which is true iff $\alpha >1$. Note: I have used the inequalities above just to get rid of $\sin (\theta)$ and $\cos (\theta)$. When you have a function of $r$ alone it is easy to see when the integral converges.