Divergence of double integral $\iint_{x^2 + y^2 \leq 1} \frac{\mathop{d}x\mathop{d}y}{(x^2 + xy + y^2)^p},\quad p \geq 1$

Question

How to prove the divergence of the integral

$$ \iint_{x^2 + y^2 \leq 1} \frac{\mathop{d}x\mathop{d}y}{(x^2 + xy + y^2)^p},\quad p \geq 1 $$

without transition to generic polar coordinate system?

Answer 1

Consider $$ I=\iint_{x^2 + y^2 \leqslant 1} \frac{\mathop{d}x\mathop{d}y}{(x^2 + xy + y^2)^p},\qquad J= \iint_{x^2 + y^2 \leqslant1/4} \frac{\mathop{d}x\mathop{d}y}{(x^2 + xy + y^2)^p},$$ then these are integrals of positive functions hence $J\leqslant I$. Furthermore, the change of variable $(x,y)\mapsto(2x,2y)$ shows that $$I=\frac{2^2}{2^{2p}}J,$$ thus, $$I=4^{1-p}J\leqslant4^{1-p}I.$$ If $p\gt1$, this shows that $I$ is infinite. If $p=1$, the finiteness of $I$ would imply that $J=I$, hence the integral on the annulus $1/4\leqslant x^2+y^2\leqslant1$ would be zero, which is absurd.