Show the function is analytic

Question

Set $f(t)$ as a complex valued function on [0,1] s.t. $\int_0^1|f(t)|dt<\infty$. Show that $\int_0^1\frac{f(t)}{t-z}dt$ defines an analytic function on $\mathbb{C}\setminus[0,1]$.

Answer 1

Hint:\begin{align}\int_0^1\frac{f(t)}{t-z}\,\mathrm dt&=\int_0^1\frac{f(t)}{t-z_0-(z-z_0)}\,\mathrm dt\\&=\int_0^1\frac{f(t)}{t-z_0}\cdot\frac1{1-\frac{z-z_0}{t-z_0}}\,\mathrm dt\\&=\int_0^1\frac{f(t)}{t-z_0}\sum_{n=0}^\infty\left(\frac{z-z_0}{t-z_0}\right)^n\,\mathrm dt\\&=\sum_{n=0}^\infty\left(\int_0^1\frac{f(t)}{(t-z_0)^{n+1}}\mathrm dt\right)(z-z_0)^n.\end{align}