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



Integrability at $z$ of the 2-form $ d\omega=\frac{\partial_{\bar{\zeta}}g(\zeta)}{\zeta-z}d\zeta\wedge d\bar{\zeta} $
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..."


