We have
\begin{equation}
\int_{0}^{1/2} \log(1 - t z) dt
\end{equation}
Integrate over $\gamma$ with respect to $z$, consider
\begin{equation}
f(z)=\log(1-tz)
\end{equation}
Function $f(z)$ is analytic on $\mid z \mid <2$, by Cauchy's theorem, for any closed curve $\gamma$
\begin{equation}
\int_\gamma f(z)dz = 0 \\
\int_{0}^{1/2} (\int_\gamma \log(1-zt) \,dz)\,dt = \int_{0}^{1/2} 0\,dt
\end{equation}
So it is analytic on domain $\mid z\mid < 2$.
Since $\log(1 - t z) = \sum_{n=1}^\infty \frac{-( z t)^n}{n}$ is valid when $\mid zt \mid<1$, and since $\mid z \mid<2$, for all $t \in [0,\frac{1}{2}]$. We have
\begin{equation}
\int_{0}^{1/2} \log(1 - t z) dt = - \int_{0}^{1/2} \sum_{n=1}^\infty \frac{( z t)^n}{n} dt \\
= - \sum_{n=1}^\infty \int_{0}^{1/2} \frac{( z t)^n}{n} dt \\
= -\sum_{n=1}^\infty\frac{1}{2^{n+1} n (n+1)} z^n
\end{equation}
