Finding values for integral $\iint_A \frac{dxdy}{|x|^p+|y|^q}$ converges Given the following integral
$$\iint_A \frac{dxdy}{|x|^p+|y|^q}$$
where $A=|x|+|y|>1$. How can one find for which $p$ and $q$ values the integral converges?
Since the function is  non-negative it is sufficient to show convergence/divergence on any Jordan exhaustion of $A$ in order to show convergence/divergence on $A$. 
I tried to use polar coordinates here, but I don't know how to fit it to match $p$ and $q$ so it'll actually be helpful. 
 A: It's enough to observe $x, y > 0$ (from symmetry) and instead of $x+y\geq 1$ take $\max(x,y)\geq 1$ - since we only got rid of a finite integral it doesn't change the convergence.
Substituting $\phi(x,y) = (u, v) = (x^p, y^q)$ gives
$$ J_\phi = \det \frac{d(x,y)}{d(u,v)} = \frac{1}{\det\frac{d(u,v)}{d(x,y)}} = \frac{1}{pq\cdot u^a \cdot s^b} $$
where $a = \frac{p-1}{p}= 1- \frac{1}{p}, b = \frac{q-1}{q}= 1-\frac{1}{q}$. 
Now:
$$\int_{min(x,y) \geq 1} \frac{dx \, dy}{x^p+y^q}= \frac{1}{pq}[\int_{u\geq s\geq 1} \frac{du \, ds}{(u+s)\cdot u^a \cdot s^b} + \int_{s > u \geq 1} \dots]$$
Observe the first integrand ($u\geq s$) - it converges and diverges the same as $$\int_{u\geq max(1,s)} \frac {du \, ds}{2u \cdot u^a\cdot s^b} = \int_{1}^{\infty} \int_{1}^{u}\frac{ds \, du}{2u^{a+1}\cdot s^b} = \int_{1}^{\infty} \frac{\frac{1}{u^{b-1}}-1}{2u^{a+1}(1-b)}\,du$$
So for them to converge we need $a+1+b-1 > 1 \implies \frac{1}{p} + \frac{1}{q} < 1$, and also $a+1 > 1\implies \frac{1}{p}<1$, and if we'll take the second integrand instead we get $b+1 > 1 \implies \frac{1}{q} < 1$ so overall we just need $\frac{1}{p} + \frac{1}{q} < 1 \, \cap \,max(\frac{1}{p}, \frac{1}{q}) < 1$. 
