When does $\int \frac{1}{(1+x^2+xy+y^2)^\alpha}$ converge?

Question

I wish to find for which $\alpha \in \mathbb R$ the integral $\int_{\mathbb R^2}\frac{1}{(1+x^2+xy+y^2)^\alpha}dxdy$ converges.

What I tried:

Firstly notice that the integrand is always positive since $xy \leq x^2+y^2$.

If $\alpha \leq 0$ the integral trivially diverges. Assume $\alpha >0$. Then:

$\int_{\mathbb R^2}\frac{1}{(1+x^2+xy+y^2)^\alpha}dxdy \geq \int_{\mathbb R^2}\frac{1}{(1+2x^2+2y^2)^\alpha}dxdy= 2\pi\int_{0}^{\infty}\frac{r}{(1+2r^2)^\alpha}dr = \pi\int_{0}^{\infty}\frac{1}{(1+2t)^\alpha}dt = \frac{\pi}{2}\int_{1}^{\infty}\frac{1}{s^\alpha}ds$

This integral diverges when $\alpha \leq 1$ so it follows that our original integral diverges when $\alpha \leq 1$.

Now what? I'm stuck.

Answer 1

$$ \frac{1}{2} \left( x^2 + y^2 \right) \leq \; x^2 + xy + y^2 \; \leq \frac{3}{2} \left( x^2 + y^2 \right) $$

Answer 2

Hint:

For $\alpha > 0$, changing to polar coordinates,

$$\frac{1}{(1+x^2+xy+y^2)^\alpha} = \frac{1}{(1+r^2+r^2\cos \theta \sin \theta)^\alpha} = \frac{1}{(1+r^2(1+\sin (2\theta)/2))^\alpha} \\ \begin{cases}\leqslant \frac{1}{(1+r^2/2)^\alpha}\\ \geqslant \frac{1}{(1+3r^2/2)^\alpha}\end{cases}$$

and as $r \to \infty$

$$\frac{r}{(1+r^2/2)^\alpha} =\mathcal{O}( r^{1-2\alpha})$$

Answer 3

Let $q(x,y)$ be a positive definite binary quadratic form with associated matrix $Q$, i.e. $$ q(x,y) = (x,y) Q (x,y)^T. $$ By the spectral theorem $Q$ can be written as $J^{-1} D J$, where $D$ is a diagonal matrix with real entries $\lambda_1,\lambda_2>0$ and $\left|\det J\right| = 1$. It follows that $$ \iint_{\mathbb{R}^2}\frac{1}{(1+q(x,y))^{\alpha}}\,dx\,dy=\iint_{\mathbb{R}^2}\frac{1}{(1+(x,y)J^{-1} D J(x,y)^T)^{\alpha}}\,dx\,dy $$ equals $$\iint_{\mathbb{R}^2}\frac{1}{(1+(x,y)D (x,y)^T)^{\alpha}}\,dx\,dy=\iint_{\mathbb{R}^2}\frac{1}{(1+\lambda_1 x^2+\lambda_2 y^2)^{\alpha}}\,dx\,dy $$ or $$ \frac{1}{\sqrt{\lambda_1 \lambda_2}}\iint_{\mathbb{R}^2}\frac{1}{(1+x^2+y^2)^{\alpha}}\,dx\,dy=\frac{2\pi}{\sqrt{\det Q}}\int_{0}^{+\infty}\frac{\rho}{(1+\rho^2)^{\alpha}}\,d\rho$$ or $$ \iint_{\mathbb{R}^2}\frac{1}{(1+q(x,y))^{\alpha}}\,dx\,dy=\frac{\pi}{(\alpha-1)\sqrt{\det Q}} \tag{Q}$$ as soon as $\text{Re}(\alpha)>1$. In your case $Q=\left(\begin{smallmatrix}1 & \frac{1}{2}\\ \frac{1}{2} & 1\end{smallmatrix}\right)$, hence $\det Q=\frac{3}{4}$ and $$\boxed{ \iint_{\mathbb{R}^2}\frac{dx\,dy}{(1+x^2+xy+y^2)^{\alpha}}=\frac{2\pi}{\sqrt{3}(\alpha-1)}\qquad \text{for }\text{Re}(\alpha)>1.} $$ In arbitrary dimension,

$$\int_{\mathbb{R}^n}\frac{dx_1\cdots dx_n}{(1+q(x_1,\ldots,x_n))^{\alpha}}=\frac{\pi^{n/2}}{(\alpha-1)\Gamma(n/2)\sqrt{\det Q}}\qquad \text{for }\text{Re}(\alpha)>1.\tag{Qn}$$