Discuss the convergence and divergence for the following integral: $$ \int_A\dfrac{\mathrm{d}x\mathrm{d}y}{|x|^p+|y|^q}, A=\left\{|x|+|y|\geq 1\right\} $$
I have just started working with the multi-generalized integral. In the question, some condition is easy to determine (e.g. $p<0,q<0$; $p=0$ or $q=0$), for the remaining condition, I tried comparitive method but got stuck. Thanks in advance if you can provide a complete solution.
Let $I_{p,q}$ denote the integral in question. It suffices to integrate over the first quadrant region and then quadruple it. The integral is then $$ I_{p,q}=4\left(\underbrace{\int _0^1 \int _{1-x}^{\infty} \frac{1}{x^p+y^q}\,dydx}_{I_1;p,q}+\underbrace{\int _{1}^{\infty} \int _{0}^{\infty} \frac{1}{x^p+y^q}\,dydx}_{I_2;p,q}\right) $$We can actually say quite a lot about both of these integrals, particularly $I_{2;p,q}$. A clear necessary condition for $I_{2;p,q}$ to converge is $q>1$ by the Power Rule (and hence $p>1$ by symmetry). In this case, a beautiful closed-form exists: $$ I_2 = \int _{1}^{\infty} \int _{0}^{\infty} \frac{1}{x^p+y^q}\,dydx = \frac{\pi \csc \left(\frac{\pi }{q}\right)}{q}\int_1^{\infty} x^{p \left(\frac{1}{q}-1\right)}\,dx $$By Power Rule again, we must have $p \left(\frac{1}{q}-1\right)<-1$, or the more symmetric $pq-p-q> 0$, in which case we have: $$ I_{2;p,q} = \frac{\pi \csc \left(\frac{\pi }{q}\right)}{pq-p-q} $$For $I_{1;p,q}$, the condition $q>1$ is sufficient since the range of integration and integrand are bounded in $x$. A closed-form is more elusive: the best I could get was integrating in $y$ and expressing it as a single integral in $x$: $$ I_{1;p,q} = \frac{1}{q-1}\int _0^1 {(1-x)^{1-q} \, _2F_1\left(1,\frac{q-1}{q};2-\frac{1}{q};-(1-x)^{-q} x^p\right)} \,dx $$So in summary, the integral converges for $1<p,1<q$ and $pq-p-q>0$. The case $q=2,p>2$ is worth mentioning: it is $$ I_{p,2}=4\left(\frac{\pi}{p-2}+\int _0^1 x^{-p/2}\arctan\left(\frac{x^{p/2}}{1-x}\right)\,dx\right), $$which might admit evaluation for certain values of $p$.