For what values of $a > 0$ does $f(x,y)=(x^{2}+y^{a})^{-1} $ belong to $ L^{1}([0,1]^{2})?$

Question

I am trying to understand for what values of $a>0$ the function $$f(x,y) = \frac{1}{x^2+y^a}$$ belongs to $L^1([0,1]^2)$.

I think $a \geq 2$ should work. But how to show that it is not the case for other values of $a$?

Answer 1

When we consider integrability, estimates within a constant factor are enough. Then the sum of two positive numbers is as good as their maximum, because $$\max(a,b)\le a+b\le 2\max(a,b)$$ So, replace $x^2+y^a$ by $\max(x^2,y^a)$. Then we only need to integrate $$ \iint \chi_{\{x^2<y^a\}} y^{-a} \,dx\,dy + \iint \chi_{\{x^2>y^a\}} x^{-2} \,dy\,dx $$ The inner integral is easily found, since it's the integral of $1$ over an interval. The whole thing boils down to $$ \int_0^1 y^{a/2} y^{-a} \,dy + \int_0^1 x^{2/a} x^{-2} \,dx $$ which I'm sure you can handle.