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 $f(z) = u(x, y) + iv(x, y)$
 A: Let $\gamma$ be a closed curve in $\mathbb{C} \setminus [-1,1]$, not enclosing $[-1,1]$. Then
\begin{align}
\int_\gamma f(z)\,dz 
&= \int_\gamma \left( \int_{-1}^1 \frac{\phi(t)}{z-t}\,dt \right) \, dz \\
&= \int_{-1}^1 \left( \int_{\gamma} \frac{\phi(t)}{z-t}\,dz \right) \, dt \\
&= \int_{-1}^1 \phi(t) \left( \int_{\gamma} \frac{1}{z-t}\,dz \right) \, dt \\  
&= \int_{-1}^1 \phi(t) \cdot 0 \, dt  = 0
\end{align}
since the inner integral vanishes for every $t$ (by Cauchy's integral theorem). Morera's theorem now shows that $f$ is holomorphic on $\mathbb{C} \setminus [-1,1]$. (I'll leave it to you to verify that $f$ is continuous.)
A: If you wanted to use CR:
(This answer only works for $\phi$ real, but if $\phi=\phi_1+i\phi_2$ with $\phi_1,\phi_2$ real, you can see that it follows from the real case.)
Write $z=a+bi$ then $$\frac{\phi(t)}{(t-a)-bi} = \frac{\phi(t)(t-a +bi)}{(t-a)^2+b^2}$$
So $$\int_{-1}^{1} \frac{\phi(t)}{t-z}\,dt = \int_{-1}^1 \frac{\phi(t)(t-a)}{(t-a)^2+b^2}\,dt + i\int_{-1}^1 \frac{b\phi(t)}{(t-a)^2+b^2}\,dt$$
So there are your $u(a,b)$ and $v(a,b)$.
A: 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
$f'(z) = \int_{-1,1]} {\partial \eta(z,t) \over \partial z} dt = \int_{-1,1]} {\phi(t) \over (t-z)^2} dt$.
A: Hint:
$$u(x,y)=\Re\int_{-1}^1\frac{\phi(t)\,dt}{t-z},\\v(x,y)=\Im\int_{-1}^1\frac{\phi(t)\,dt}{t-z}.$$
