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

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!

## 2 Answers

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.

No need to justify switching integrals. Just obtain a closed form of $f(z)$, which allows to prove continuity and that $\int_{\gamma} f =0$:

$$f(z) := \int_{[0,1]} \frac{dw}{1-wz} = -\frac1z\operatorname{Ln}(1-z)1_{z\ne0} + 1_{z=0},$$

1. $f(z)$ is continuous on $D[0,1]$ because $$\lim_{z \to 0}-\frac1z\operatorname{Ln}(1-z) = 1 = f(0)$$

2. $$\int_{\gamma} f(z) dz = \int_{\gamma} -\frac1z\operatorname{Ln}(1-z)1_{z\ne0} + 1_{z=0} dz$$

$$= \int_{\gamma} -\frac1z\operatorname{Ln}(1-z)1_{z\ne0} dz + \int_{\gamma} 1_{z=0} dz = 0$$

Now all conditions of Morera's Thm 5.6 are satisfied without (explicit) reference to antiderivatives or Fubini's theorem, measure spaces, uniform continuity, etc