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 $f(z) = u(x, y) + iv(x, y)$

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 $v(a,b)$.
