Let $\log$ be the principal branch of the logarithm, and assume that $-\pi < \theta < \pi$.
If $$ F_{n} (\theta) = \int_{1}^{e^{-i \theta}} \frac{(- \log z)^{n-1}}{1-z} \, dz, \quad (n =2,3, \ldots ) \ ,$$
how do you argue that $$ \overline{F_{n} (\theta)} = {\color{red}{(-1)^{n-1}}}\int_{1}^{e^{i \theta}} \frac{(\log z)^{n-1}}{1-z} \, dz \ ?$$
(The integration is along a line segment from $1$ to $e^{- i \theta}$.)
It's tempting to say that $$ \overline{F_{n} (\theta)} = \int_{1}^{\overline{e^{-i \theta}}} \frac{(-\log z)^{n-1}}{1-z} \, dz, $$ but I don't know why that would be necessarily true.
I get, assuming small enough angles and the principal branch of the logarithm:
$$\begin{align} \overline{F_n(\theta)} &= \overline{\int_0^\theta \frac{(-\log e^{-i\varphi})^{n-1}}{1 - e^{-i\varphi}} (-ie^{-i\varphi})\, d\varphi}\\ &= \int_0^\theta \frac{(-\overline{\log e^{-i\varphi}})^{n-1}}{1-e^{i\varphi}} ie^{i\varphi}\, d\varphi\\ &= \int_0^\theta \frac{(-\log e^{i\varphi})^{n-1}}{1-e^{i\varphi}} ie^{i\varphi}\, d\varphi\\ &= \int_1^{e^{i\theta}} \frac{(-\log z)^{n-1}}{1-z}\, dz\\ &= F_n(-\theta), \end{align}$$
or equivalently
$$\int_1^{e^{i\theta}} \frac{(\log z)^{n-1}}{1-z}\, dz = (-1)^{n-1}\overline{F_n(\theta)}. $$
Somebody messed up their signs (could have been me).