= Solution
<Analytic continuation> extends a <holomorphic function> by holomorphic elements agreeing on overlaps; along a path, a chain of overlapping neighbourhoods transports the initial element. The <identity theorem> makes compatible continuation unique on connected overlaps, although different paths can lead to different branches.
Choose the principal <complex logarithm> and the slit domain $D=\mathbb C\setminus[1,\infty)$. On $D$, the function
$$
h(z)=-\frac{\operatorname{Log}(1-z)}z
$$
is holomorphic, with the removable value $h(0)=1$. The domain $D$ is simply connected, so $h$ has a primitive and its integral from $0$ to $z$ is independent of the path in $D$. For $|z|<1$, the uniformly convergent <power series> on compact subdiscs gives
$$
f(z)=\int_0^zh(u)\,du=\int_0^z\sum_{n\geq1}\frac{u^{n-1}}n\,du=\sum_{n\geq1}\frac{z^n}{n^2}.
$$
Therefore \b[$f$ is a single-valued <analytic continuation> of the <dilogarithm> to $\mathbb C\setminus[1,\infty)$]. The apparent problem at $0$ is removable; $1$ is a genuine branch point, even though the function has a finite one-sided limit there, because its derivative has a logarithmic singularity. Other cuts from $1$ to infinity give corresponding branches.
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