Question about an extension of an analytic function Setting: Let $\Omega \subseteq \mathbb{C}$ and suppose that $f$ is analytic on $\Omega' = \Omega - \{a\}$.  Suppose that $a \in \Omega$ satisfies the crucial property that
$$
\lim_{z \to a} (z - a)f(z) = 0
$$
Now if we let $C$ be a circle in $\Omega$ about $a$, we have that Cauchy's Integral Formula is valid for all $z \in \Omega'$ so that we can say
$$
f(z) = {1 \over 2 \pi i} \int_C {f(\zeta)\ d\zeta \over \zeta - z}
$$
Since $f(z)$ is analytic by hypothesis, we can know immediately that the function
$$
\int_C {f(\zeta)\ d\zeta \over \zeta - z}
$$
is analytic on $\Omega'$ in its own right.
Question: Why is it that this integral function is analytic on all of $\Omega$?  That is, why do we know that
$$
\int_C {f(\zeta)\ d\zeta \over \zeta - a}
$$
has a derivative?
 A: 
Since $f(z)$ is analytic by hypothesis, we can know immediately that the function... 

No, this is not what really goes on. For any integrable function $f$ on the circle (or other contour $C$), the function 
$$F(z) = \int_C \frac{f(\zeta)\,d\zeta}{\zeta -z }$$
is analytic in $\mathbb C\setminus C$, i.e., in the complement of the set of integration. One way to prove this is to fix $z_0\notin C$ and expand 
$$\frac{1}{\zeta-z} = \frac{1}{\zeta-z_0}  \frac{1}{1-\frac{z-z_0}{\zeta-z_0}} = 
 \frac{1}{\zeta-z_0} \sum_{n=0}^\infty \left(\frac{z-z_0}{\zeta-z_0}\right)^n $$
The power series can be integrated term by term within the disk $|z-z_0|<r$ where $r$ is the distance from $z_0$ to $C$. Hence, $F$ is holomorphic. 

The main question is why $F$ agrees with $f$ in the disk bounded by $C$ (which is the only place where we need $F$ anyway). To see this, apply Cauchy integral formula to $f$ in the annulus whose outer boundary is $C$ and the inner boundary is a much smaller circle $C_\epsilon$ centered at $a$: 
$$f(z) = \int_C \frac{f(\zeta)\,d\zeta}{\zeta -z } - \int_{C_\epsilon} \frac{f(\zeta)\,d\zeta}{\zeta -z }$$
 The integral over $C_\epsilon$ goes to zero as  $\epsilon\to 0$; this is where the assumption $\lim_{z \to a} (z - a)f(z) = 0$ is used. We conclude that $f(z) = F(z)$ in the disk bounded by $C$, except at $a$. The function $F$ is the holomorphic extension of $f$ that we wanted.
