# Existence of a complex integral

Let $U \subset \subset \mathbb{C}$ (that is, $\bar U \subset \mathbb{C}$ is compact) and let $f$ be a complex-valued function, which is continuously differentiable in a neighbourhood of $\bar U$.

Consider the integral \begin{align*} u(z) := \int\limits_U \frac{f(\xi)}{\xi - z} d\xi \wedge d\bar \xi, ~~~~~ \text{where} ~ z \in U. \end{align*}

My question is: Why does this integral always exist? -- as the integrand is not bounded.

To give some context: $u$ (divided by $2\pi i$) is the solution of the inhomogeneous Cauchy-Riemann equation $\frac{\partial u}{\partial \bar z} = f$.

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The integrand is not bounded, but its size is $O(1/r)$ as $r$, the distance from the singularity, approaches zero. Indeed, the area element is equal to $i r dr d\theta$ (in polar coordinates centered at $z$), and the denominator is $r \exp(i\theta)$ in the same coordinates, letting you rewrite $u(z) = i \int_{U} f(r,\theta)\exp(-i\theta) dr d\theta$, which is clearly well-behaved. – mjqxxxx Dec 14 '11 at 21:11

Since $d\xi\wedge d\bar\xi=-2idx\wedge dy$ and since $|f|$ is bounded on $\bar U$, it is enough to prove that $$\int\limits_\bar U \frac{1}{|\xi - z|} dx dy \lt \infty$$

Since $\bar U$ is compact and since $\frac{1}{|\xi - z|}$ is bounded outside any disk centered at $z$, it is enough to show that the above integral is finite on the compact disk $B=\lbrace \xi \in \mathbb C:|\xi-z| \leq a\rbrace$ .
But this is clear by using polar coordinates:

$$\int\limits_B \frac{1}{|\xi - z|} dx dy =\int\limits_B \frac{1}{r} dx dy=\int _0^{2\pi} d\theta \int _0^a \frac{1}{r}rdrd\theta=2\pi a \lt\infty$$

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You may assume $z=0$. Then write $\zeta:=r e^{i\phi}$ and obtain $d\zeta\wedge d\bar\zeta=- 2 i\ d\xi\wedge d\eta=-2 i\ r\ {\rm d}(r,\phi)$, where ${\rm d}(r,\phi)$ denotes Lebesgue measure in the $(r,\phi)$-plane. It follows that

$$u(0)=-2i \int\nolimits_V \ f\bigl(re^{i\phi}\bigr)e^{-i\phi}\ {\rm d}(r,\phi)\ ,$$

where $V$ denotes the pullback of $U$ in the $(r,\phi)$-plane. Now everything is bounded.

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