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Given $g\in\mathcal{C}^1(\bar\Delta)$, and $z\in\Delta$, how can i prove that the 2-form $$ d\omega=\frac{\partial_{\bar{\zeta}}g(\zeta)}{\zeta-z}d\zeta\wedge d\bar{\zeta} $$ is integrable in $z$?

At MSE no one could help me, I hope to be more lucky here.

If you need more detail look at https://math.stackexchange.com/questions/1240685/detail-about-integration-in-cauchy-integral-formula many thanks

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    $\begingroup$ "At MSE no one could help me, I hope to be more lucky here." You posted here only an hour after posting on MSE. You should be a little bit more patient. $\endgroup$ Apr 19, 2015 at 2:14

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Griffiths & Harris (pp. 2-3) put it thus:

"Setting $\zeta - z=re^{i\theta}$, $$ d\zeta\wedge d\bar\zeta = -2i dx\wedge dy = -2i rdr\wedge d\theta $$ so $$ \left|\frac{\partial g(\zeta)}{\partial\bar\zeta}\frac{d\zeta\wedge d\bar\zeta}{\zeta - z}\right| = 2\left|\frac{\partial g}{\partial\bar\zeta}dr\wedge d\theta\right| \leqslant c\left|dr\wedge d\theta\right|. $$ So $(\partial g/\partial\bar\zeta)(d\zeta\wedge d\bar\zeta)/(\zeta - z)$ is absolutely integrable over $\Delta$, and..."

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In terms of the absolute value, this is a singularity no worse than 1/r, but you are integrating on two dimensions, so it's still absolutely intergrable.

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