I need to find the values of $p, q \in \mathbb{R}$ for which the following integral converges: $$\int_ \limits{B}\frac{dxdy}{x^p+y^q}$$ where $B= \left \{ x^2+y^2\leq 1, x, y \geq0 \right \}$.
So far, I've shown convergence for $p=q=1$, which implies convergence for every $p, q \leq 1$.
In general, a useful fact is that $\frac{1}{x^p+y^q}\leq \frac{1}{x^P+y^Q}$ whenever $p\leq P, q\leq Q$.
Any ideas?
On
If either of $p,q<0,$ then the denominator of the integrand is $>1$ for $(x,y)\in B,$ and the integral converges.
Suppose $p,q>0.$ In the integrals that follow, we have convergence of the $B$ integral iff the $[0,1]\times [0,1]$ integral converges. So consider
$$\tag 1 \int_0^1\int_0^1 \frac{1}{x^p+y^q}\,dx\,dy.$$
Here make the change of variables $x=u^{1/p},y = v^{1/q}$ to see $(1)$ converges iff
$$\int_0^1\int_0^1 \frac{u^{1/p-1}v^{1/q-1}}{u+v}\,du\,dv <\infty.$$
Now go back to $B$ and use polar coordinates to get
$$\int_0^{\pi/2}\int_0^1 r^{1/p+1/q - 2}\frac{(\cos t)^{1/p-1}(\sin t)^{1/q-1}}{\cos t+\sin t}\,dr\,dt.$$
This splits nicely into the product of integrals. The $t$ integral converges for all $p,q>0.$ The $r$ integral converges iff $1/p+1/q >1.$
Conclusion: Your integral converges iff one of the following holds i) $p<0; $ii) $q<0;$ iii) both $p,q>0$ and $1/p+1/q >1.$
$$\int_B \frac{dx\,dy}{x^p+y^q} = \int_B \int_{0}^{+\infty}e^{-t x^p}e^{-t y^q}\,dt\,dx\,dy $$ can be approximated pretty well through $\left(0,\frac{1}{\sqrt{2}}\right)^2\subset B \subset (0,1)^2$. Assuming $p,q>0$, it follows that the LHS is convergent iff $\sqrt{t^2+1}^{-\frac{1}{p}-\frac{1}{q}}$ is an integrable function over $\mathbb{R}^+$, i.e. iff $$ \frac{1}{p}+\frac{1}{q}>1.$$