# 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?

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:

1. Does it help somehow?
2. Is there a better change of variables here?
3. How can I find the range of $$\alpha$$ that im looking for?
4. Is it possible maybe to block the function with and easier one to integral and then use sandwich?
• Since the integrand is asymptotic to $r^{-2 \alpha}$, you have convergence whenever $- 2 \alpha < -1$, i.e. for $\alpha >1$. – Crostul Jan 10 at 8:53
• Is there a way to actually compute the integral? – Jon Jan 10 at 8:55
• @Crostul There is a factor of $r^{2}$ coming in when you switch to polar coordinates. – Kabo Murphy Jan 10 at 8:57
• @Jon Explicit computation of integral may not be possible. – Kabo Murphy Jan 10 at 8:59

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.
• The Jacobian of the polar transformation is $r$, not $r^2$. Then we integrate $d\theta$, a fixed total angle - it should be $\frac{r}{r^{2\alpha}}$ we're testing for integrability at $\infty$. – jmerry Jan 10 at 9:07