# Closed form of $\int_{[0,1]} \frac{dw}{1-wz}$ involving Ln?

I think if

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

then $f(z)$ is continuous on $D[0,1]$.

For $z=0$,

$$\int_{[0,1]} \frac{dw}{1-wz} = \int_{[0,1]} \frac{dw}{1} = \int_0^1 dw = 1\tag{*}$$

For $z \ne 0$, consider the real integral where $z \in \mathbb R \setminus \{0\}$

$$\int_{0}^1 \frac{dw}{1-wz} = -\frac{1}{z}\ln|1-z|$$

1. Can we extend this to $\mathbb C$ to conclude that (Wolfram Alpha)

$$\int_{[0,1]} \frac{dw}{1-wz} = -\frac{1}{z}\operatorname{Ln}|1-z|?$$

It seems pretty tough to integrate $$\int_{[0,1]} \frac{dw}{1-wz} = \int_{0}^1 \Re \frac{dw}{1-wz} + i\int_{0}^1 \Im \frac{dw}{1-wz}$$

1. What's the relevance of $\operatorname{Re}(z) \le 1$ (Wolfram Alpha) ? Luckily that's the assumption given in the problem, but I don't see how that's relevant in computing the integral.

2. (*) Btw, why don't we actually have $\int_{[0,1]} \frac{dw}{1} = 0$ by Cor 4.20 to Cauchy's Theorem? I guess we don't have that $[0,1] \sim_{D[0,1]} 0$.

For $w\in[0,1]$, $\log(1-wz)$ is well-defined on $\mathbb{C}\setminus[1,\infty)$, so we can integrate along the path $[0,1]$: \begin{align} \int_{[0,1]}\frac{\mathrm{d}w}{1-wz} &=\frac1z\int_{[0,1]}\frac{z\,\mathrm{d}w}{1-wz}\\ &=\left.-\frac1z\log(1-wz)\,\right|_0^1\\[3pt] &=-\frac{\log(1-z)}z \end{align} with a branch cut along $[1,\infty)$. Note that if $z\not\in[1,\infty)$, then $1-wz$ is bounded away from $0$ for $w\in[0,1]$.
Another approach using a series which converges for $|wz|\lt1$: \begin{align} \int_{[0,1]}\frac{\mathrm{d}w}{1-wz} &=\int_{[0,1]}\left(\sum_{k=0}^\infty w^kz^k\right)\mathrm{d}w\\ &=\sum_{k=0}^\infty\int_0^1w^kz^k\,\mathrm{d}w\\ &=\sum_{k=0}^\infty\frac{z^k}{k+1}\\ &=-\frac{\log(1-z)}z \end{align} which converges for $|z|\lt1$.
• As mentioned, there is a branch cut in the first approach along $[1,\infty)$. In the second approach, the series converges where $|z|\lt1$. – robjohn Jul 25 '18 at 13:18