Convergence of $\int_0^1 \int_0^1 (1-xy)^{-a}dxdy$

Question

How can I investigate the convergence of $\int_0^1 \int_0^1 (1-xy)^{-a}dxdy$ where $a \in (0, \infty)$? I know that one approach is to explicitly find the antiderivative but that seemed too laborious. Also, I believe that the integral converges for $a \in (0,2)$ by numerical solutions provided by WolframAlpha. Any help is greatly appreciated :)

Answer 1

I'll use the notational convention $A\approx B$ if $B/C \le A \le CB$ for an absolute constant $C$. Also, if $E$ is a set, we use $|E|$ to denote the measure (or area) of $E$.

The integrand has a singularity along $1-xy = 0$. Furthermore, note that $1-xy\ge 0$ in $[0,1]^2$. Therefore, the convergence of the integral is equivalent to the convergence of the sum \begin{align*} \sum_{k=0}^\infty\iint_{[0,1]^2\cap\{1-xy\approx 2^{-k}\}}(1-xy)^{-a}\,dx\,dy &\approx \sum_{k=0}^\infty2^{ka}|\{(x,y)\in[0,1]^2\mid 1-xy\approx 2^{-k}\}|. \end{align*} Note that for each $k \ge 0$, the set of $(x,y)\in[0,1]^2$ satisfying $1-xy\approx 2^{-k}$ is approximately a square of diameter $2^{-k}$ centered at the point $(1,1)$. (More formally, the set contains and is contained in a square of side a constant times $2^{-k}$.) Therefore, the sum above is about $$ \iint_{[0,1]^2}(1-xy)^{-a}\,dx\,dy\approx \sum_{k=0}^\infty 2^{ka}(2^{-k})^2 = \sum_{k=0}^\infty 2^{(a-2)k}, $$ which converges if and only if $a\in (-\infty,2)$. In particular, the integral $\iint_{[0,1]^2}(1-xy)^{-a}\,dx\,dy$ converges if and only if $a\in(-\infty,2)$. It diverges otherwise.

This method of investigating integrals is a common tactic of Lebesgue integration if one only cares about the value of the integral up to a constant multiplicative factor (e.g., if one only cares whether the integral is finite).