Assume that $q>1$. Let $U$ be the unit disk. For which $a\in (2,4)$, integral $$I=\int_U\left[\int_U 1/|1-z \bar w|^a dA(z)\right]^q dA(w)$$ converges. Here $dA(z)$ is the area measure of the unit disk.
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1$\begingroup$ expansion near the pole at $z\bar{w}=1$ suggests the double integral diverges as $\epsilon^{(2-a)q}$ with $\epsilon\rightarrow 0$, so divergent for any $a>2$ and $q>0$ $\endgroup$– Carlo BeenakkerCommented Nov 23, 2019 at 11:49
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$\begingroup$ @Carlo Beenakker. No it converges for example if q=3/2 and a=2,1 $\endgroup$– HengyCommented Nov 23, 2019 at 14:08
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1 Answer
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For $(a-2)q>-1 $ the integral converges.