Show a function is holomorphic Let $\phi: [0,1] \rightarrow \mathbb{C}$ be a continuous function. For all $z \in \mathbb{C} \setminus [0,1] $ define $f(z) = \int_0^1\frac{\phi(t)}{t-z} \ dt$. Prove that f is holomorphic on $\mathbb{C} \setminus [0,1]$.
I can't express $f$ as a composition of holomorphic functions, is there another way to prove $f$ is holomorphic? Thanks in advance!
 A: By standard theorems on differentiation of parameter dependent integrals, for $z\notin [0,1]$,
(Edit: to make it a bit clearer: $f$ is differentiable in $z$ and)$$\frac{\partial}{\partial \bar{z}}f(z)=\int_0^1\left(\frac{\partial}{\partial \bar{z} }\frac{\phi(t)}{t-z}\right)dt = 0$$
A: By Morera's theorem:
Let $\triangle$ be any triangle not containing values in $[0,1]$.  Then $\int_{\triangle} f(z) = \int_{\triangle}\int_0^1 \frac{\phi(t)}{t-z}dtdz = \int_0^1\phi(t)\left(\int_\triangle\frac{dz}{t-z}\right)dt = 0$.  As the integral is $0$ over  all triangles in the domain, $f$ is analytic.
A: Here is a more analytical proof.
First we prove that the function is continuous on $\mathbb{C}\setminus[0,1]$. Since $\phi$ is contiuous and $[0,1]$ is compact, $|\phi|$ attains some maximum, say $M$. Also $\mathbb{C}\setminus[0,1]$ is open. Fix some $z_0$ such that $D(z_0, 2r) \subseteq \mathbb{C}\setminus[0,1]$. Then
$$|f(z)-f(z_0)|=\left|\int_0^1\frac{\phi(t)}{t-z} - \frac{\phi(t)}{t-z_0}\ dt\right|$$
$$=\left|\int_0^1\phi(t)\cdot\frac{z-z_0}{(t-z)(t-z_0)}\ dt\right|$$
from the $LM$ estimate, we then have as $\epsilon \rightarrow 0$
$$\le |z-z_0|\cdot \max_{t\in[0,1]}\left|\frac{\phi(t)}{(t-z)(t-z_0)}\right|$$
Taking $|z-z_0|<\epsilon<r$ we have
$$\le\frac{M}{2r^2}\epsilon \rightarrow 0$$ With continuity at hand, consider then the differential quotient
$$\frac{f(z)-f(z_0)}{z-z_0}=\int_0^1\frac{\phi(t)}{(t-z)(t-z_0)}\ dt$$
now the function $\varphi(t) \equiv \frac{\phi(t)}{t-z_0}$ is also continuous. So that as $z\rightarrow z_0$
$$\int_0^1\frac{\varphi(t)}{(t-z)}\ dt\rightarrow \int_0^1\frac{\varphi(t)}{(t-z_0)}\ dt$$
which shows that
$$f'(z) = \int_0^1\frac{\phi(t)}{(t-z_0)^2}\ dt$$
