I currently working on the following exercise:
Let $f$ be holomorphic in some neighbourhood $\Omega$ of the closed circular disc $\overline D$. Then \begin{align*} g(z) = \int_{\partial D} \frac{f(\xi)}{\xi - z} \, d \xi \end{align*} is holomorphic on $\mathbb C \setminus \overline D$. Which form has $g$?
I could easily prove that $g$ is holomorphic. I used dominated convergence to show that $g$ is continuous on $\mathbb C \setminus \overline D$ and then used Fubini's and Morera's theorem to show that $g$ is even holomorphic on $\mathbb C \setminus \overline D$. But now I struggle to get a more explicit form for $g$. My intuition tells me that it may be that $g(z) = 0$ (with a view to Cauchy's integral formular). I think it is like that since I got $g(z) = 0$ at least for $z \in \Omega$ "close enough" to $D$ (depending on the version of Cauchy's integral formular we are using) and then the identity theorem would yield $g(z) = 0$ for all $z \in \mathbb C \setminus \overline D$. But I doubt my arguments in this point. Am I right or is there something I miss out on?
The integrand is analytic in $\mathbb C \setminus \{z\}$. In ths region the circle $\partial D$ is homotopic to any smaller concentric circle. Hence we can take teh integral over any smaller circle and, by letting the radius $\to 0$ we get $g(z)=0$.