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?
  • $\begingroup$ Since the integrand is asymptotic to $r^{-2 \alpha}$, you have convergence whenever $- 2 \alpha < -1$, i.e. for $\alpha >1$. $\endgroup$ – Crostul Jan 10 at 8:53
  • $\begingroup$ Is there a way to actually compute the integral? $\endgroup$ – Jon Jan 10 at 8:55
  • $\begingroup$ @Crostul There is a factor of $r^{2}$ coming in when you switch to polar coordinates. $\endgroup$ – Kabo Murphy Jan 10 at 8:57
  • $\begingroup$ @Jon Explicit computation of integral may not be possible. $\endgroup$ – 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.

  • $\begingroup$ 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 $\infty$. $\endgroup$ – jmerry Jan 10 at 9:07
  • $\begingroup$ @jmerry Thanks for the correction. I have edited my answer. $\endgroup$ – Kabo Murphy Jan 10 at 9:11

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