Determine for which $\alpha$ the following integral is finite: $$\int_Q \frac{1}{(x^2+2y^2)^{\alpha}}dxdy \qquad Q=[0,1]\times[0,1]$$
I don't actually know what is the most efficient way to proceed. It's clear that the only problem is in the origin so we just need to describe what happens in a neighbourhood of it. I have now different ideas in mind: I can maybe try to do it in polar coordinates, but I think it really makes the task more difficult. Should I use Taylor formulas? What is the most intuitive and efficient way to do this? Thanks in advance.
Define the region $P$ as $$P=\{(x,y)\ \ |\ \ x>0 \ \ ,\ \ y>0\quad,\quad x^2+y^2<4\}\supset Q$$
Hence the integral can be bounded above by $$\iint_Q {dxdy\over (x^2+2y^2)}{<\iint_P {dxdy\over (x^2+2y^2)^\alpha} \\=\int_0^2\int_0^{\pi\over 2} r^{1-2\alpha}{1\over (1+\sin^2\phi)^\alpha}drd\phi \\=\left[\int_0^2 r^{1-2\alpha}dr\right]\left[\int_0^{\pi\over 2}{1\over (1+\sin^2\phi)^\alpha}d\phi\right] }$$The integral $\left[\int_0^{\pi\over 2}{1\over (1+\sin^2\phi)^\alpha}d\phi\right]$ is positive and bounded and the integral $\left[\int_0^1 r^{1-2\alpha}dr\right]$ becomes bounded if $\alpha< 1$ and unbounded if $\alpha\ge1$ by choosing $$P=\{(x,y)\ \ |\ \ x>0 \ \ ,\ \ y>0\quad,\quad x^2+y^2<1\}\subset Q$$