Jump of dilogarithm I am reading about the dilogarithm function
$$ \mathrm{Li}_2(z):= - \int_0^z \frac{\log(1-u)}{u}du, \quad z \in \mathbb{C} \backslash [1, \infty).$$
I found it stated that the "jump" of the dilogarithm across the axis where it is not defined is $2\pi i \log(r)$ for crossing at $r>1$. Why is that so? I can see that $\log(1-u)$ jumps by $2\pi i$ when $u$ crosses the axis, but I cannot see how to procede from there.
 A: Note that for $r=\text{Re}(z)>1$ and $\text{Im}(z)\to 0^\pm$, we have 
$$\begin{align}
-\int_0^z \frac{\log(1-u)}{u}\,du&=-\int_0^1 \frac{\log(1-u)}{u}\,du-\int_1^z \frac{\log(1-u)}{u}\,du\\\\
&=\frac{\pi^2}{6}-\int_1^r \frac{\log(|1-u|)\pm i\pi}{u}\,du\\\\
&=\frac{\pi^2}{6}-\int_1^r \frac{\log(|1-u|)}{u}\,du\mp i\pi \log(r)\\\\
\end{align}$$
Hence, the discontinuity is $2\pi i \log(r)$ as was to be shown! 
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
\mrm{Li}_{2}\pars{z} & = -\int_{0}^{z}{\ln\pars{1 - u} \over u}\,\dd u
\,\,\,\stackrel{u/z\ \mapsto\ u}{=}\,\,\,
-\int_{0}^{1}{\ln\pars{1 - zu} \over u}\,\dd u
\\[5mm] & \stackrel{\mrm{IBP}}{=}\,\,\,
\int_{0}^{1}\ln\pars{u}\,{-z \over 1 - zu}\,\dd u  \,\,\,\stackrel{u\ \mapsto\ 1/u}{=}\,\,\,
\int_{\infty}^{1}\ln\pars{1/u}\,{-z \over 1 - z/u}
\pars{-\,{\dd u \over u^{2}}}
\\[5mm] & =
-\int_{1}^{\infty}{\ln\pars{u} \over u}\,{z \over z - u}
\,\dd u =
-\int_{1}^{\infty}{\ln\pars{u} \over u}
\,\pars{1 + {u \over z - u}}\,\dd u
\end{align}


Then, with $\ds{r \in \mathbb{R}}$:

\begin{align}
&\bbox[10px,#ffd]{\mrm{Li}_{2}\pars{r + \ic 0^{+}} - \mrm{Li}_{2}\pars{r - \ic 0^{+}}}
\\[5mm] = &\
-\int_{1}^{\infty}\ln\pars{u}\
\overbrace{\pars{{1 \over r + \ic 0^{+} - u} - {1 \over r - \ic 0^{+} - u}}}
^{\ds{-2\pi\ic\,\delta\pars{r - u}}}\
\,\dd u
\\[5mm] = &\ \bbx{2\pi\ic\bracks{r > 1}\ln\pars{r}}
\end{align}
