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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.

57Jimmy
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  • How do you define an integral with a complex bound ? (z is a complex number in your "definition") – FDP Feb 23 '18 at 08:53
  • @FDP It is just an integral along any path that leads to $z$. Of course, your path cannot cross the cut. But $\mathbb{C}\backslash [1,\infty)$ is simply connected, so all that matters are the start and end points. And it is not my "definition", it is really the "official" one. – 57Jimmy Feb 23 '18 at 09:11
  • Related: https://math.stackexchange.com/a/2885109. – Maxim Jan 11 '19 at 00:44

2 Answers2

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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!

57Jimmy
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Mark Viola
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}

Felix Marin
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