Integrability Question A question that has popped up while studying for qualifying exams is the following:
Prove that $\int_0^1 \int_0^1 \frac{1}{x^p + y^q} dx dy$ is integrable iff $p^{-1} + q^{-1} > 1$
I can handle a few special cases (e.g. $p=q=1$) by changing variables, but the general case seems to be quite messy. Any ideas?
 A: For any positive numbers $a$ and $b$, note that $\max\{a,b\} < a+b < 2\max\{a,b\}$; so
$$
\int_0^1 \int_0^1 \frac1{2\max\{x^p,y^q\}} \,dx \,dy < \int_0^1 \int_0^1 \frac1{x^p+y^q} \,dx \,dy < \int_0^1 \int_0^1 \frac1{\max\{x^p,y^q\}} \,dx \,dy. 
$$
Therefore your integral converges if and only if the integral $\int_0^1 \int_0^1 \frac1{\max\{x^p,y^q\}} \,dx \,dy$ converges. This integral is
$$
\int_0^1 \int_0^1 \frac1{\max\{x^p,y^q\}} \,dx \,dy = \int_0^1 \int_0^{x^{p/q}} \frac1{x^p} \,dy \,dx + \int_0^1 \int_0^{y^{q/p}} \frac1{y^q} \,dx \,dy;
$$
since the integrand is positive, the left-hand side converges if and only if both integrals on the right-hand side converge. But they can be evaluated explicitly:
\begin{align*}
\int_0^1 \int_0^{x^{p/q}} \frac1{x^p} \,dy \,dx = \int_0^1 x^{p/q}\frac1{x^p} \,dx &= \lim_{t\to0+} \frac{x^{p/q-p+1}}{p/q-p+1}\bigg|_t^1 \\
&= \frac1{p/q-p+1} \big( 1 - \lim_{t\to0+} (t^p)^{1/q-1+1/p} \big); \\
\int_0^1 \int_0^{y^{q/p}} \frac1{y^q} \,dx \,dy = \int_0^1 y^{q/p}\frac1{y^q} \,dy &= \lim_{t\to0+} \frac{y^{q/p-q+1}}{q/p-q+1}\bigg|_t^1 \\
&= \frac1{q/p-q+1} \big( 1 - \lim_{t\to0+} (t^q)^{1/p-1+1/q} \big).
\end{align*}
The two limits converge when $1/p+1/q-1>0$ and diverge when $1/p+1/q-1<0$, as desired. (The computation needs to be modified when $1/p+1/q-1=0$; the result is the desired divergence in that case.)
