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Consider the function $$f(x,y) := \frac{1}{(1 - xy)^\alpha}$$ where $\alpha \in \mathbb{R}$ and $(x,y) \in (0,1)\times (0,1)$. For which $\alpha$ is the function integrable?

First of all, if $(x,y) \in (0,1)\times (0,1)$ we see that $0 < xy < 1$. Therefore, $(1 - xy)^\alpha > 0$ and we can apply Tonelli's theorem since $f$ is clearly continuous and hence measurable. This yields $$\int_{(0,1)\times (0,1)} f(x,y) dx \otimes dy = \int_0^1 \int_0^1 \frac{1}{(1 - xy)^\alpha} dxdy$$ Now I think that we could substitute $z = 1 - xy$ and get $$\int_0^1 \int_0^1 \frac{1}{(1 - xy)^\alpha} dxdy = \int_0^1 \int_{1 - y}^1 \frac{1}{y}\frac{1}{z^\alpha} dxdy$$ Then I think one should split the cases $\alpha = 1$ and $\alpha \neq 0$, but the integrals get nasty. Is there an easier way or how would you solve this?

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    The integral converges when $\alpha=1$. This can be proved by writing $(1-xy)^{-1} = \sum_n x^ny^n$ and using monotone convergence to express the integral as an infinite sum. By comparison it converges for $\alpha<1$. The same method shows it diverges when $\alpha\geq 2$. But I haven't found a proof yet for $\alpha \in (1,2)$, though interpolation would suggest it does converge there. – 2017-02-09
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    @Shalop Nice ideas! But I am not sure if this is the way to go, since this exercise is taken from an old exam and usually, if the things are harder to prove, there is a hint (in this case there is no). – 2017-02-09
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    I used the power rule from undergraduate calculus and I am indeed getting convergence iff $\alpha <2$. I suppose that would be the boring approach haha. Though, I could be wrong. – 2017-02-09
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    A third way to show this would be to show that the function $f(x,y)=(1-xy)^{-1}$ is in weak $L^2$ but not in $L^2$. For this all you would need to show is $\sup_{x \geq 0} x^2 \lambda (|f|>x) < \infty$ which is easy. So again we get $\alpha<2$. – 2017-02-09

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