Prove this integral is analytic Let $\phi$ be a continuous (complex valued) function on the real interval [−1, 1] inside C, and deﬁne
$$f(z)=\int_{-1}^1\frac{\phi(t)}{t-z}dt$$
Show that f is analytic on C less the interval [−1, 1].
I thought about CR equation to prove analytic but the function is not of the form $f(z) = u(x, y) + iv(x, y)$

The functions $\eta(z,t) = {\phi(t) \over t-z}$  and ${\partial \eta \over \partial z}$are continuous on $[-1,1]^c \times [-1,1]$, hence
by the Leibnitz rule we see that $f$ is differentiable and
$f'(z) = \int_{-1,1]} {\partial \eta(z,t) \over \partial z} dt = \int_{-1,1]} {\phi(t) \over (t-z)^2} dt$.
