Prove $f(z)=\int_{[0,1]}\frac{1}{1-wz}dw$ is holomorphic in the open unit disk.

Question

Define $f:D[0,1] \rightarrow \mathbb C$ through $$f(z)=\int_{[0,1]}\frac{1}{1-wz}dw$$ The integration path is from 0 to 1 along the real line.

Prove that $f$ is holomorphic in the open unit disk $D[0,1]$.

I was trying to use the Cauchy's Integral Formula $$\begin{aligned} f'(w) & =\frac{1}{2\pi i}\int_{\gamma}\frac{f(z)}{(z-w)^2}dz\\\ & =\frac{1}{2\pi i}\int_{\gamma}\frac{1}{(z-w)^2}\left[\int^1_0\frac{1}{1-tz}dt\right]dz\\ & =\frac{1}{2\pi i}\int^1_0\left[\int_{\gamma} \frac{\frac{1}{1-tz}}{(z-w)^2}dz\right]dt\\ & =\frac{1}{2\pi i}\int^1_0 \left[\frac{\partial}{\partial z}\frac{1}{1-tz}\left.\right|_{z=w}\right]dt\\ & =\frac{1}{2\pi i}\int^1_0 \frac{t}{(1-tw)^2}dt\\ & =\cdots \end{aligned}$$ But the Cauchy's Integral formula need $f$ to be holomorphic at the first place...Any help with this? Many Thanks!

Answer 1

Use Morera's theorem. First prove that $f$ is continuous.Then, if $\gamma$ is any closed $C^1$ curve in the unit disk, we have $$ \int_\gamma f(z)\,dz=\int_\gamma\Bigl(\int_{[0,1]}\frac{dw}{1-w\,z}\Bigr)\,dz=\int_{[0,1]}\Bigl(\int_\gamma\frac{dz}{1-w\,z}\Bigr)\,dw=0, $$ since $1/(1-w\,z)$ is holomorphic on the unit disk as a function of $z$.

Note: you have to justify the change of order in the integration.