Sum of Dilogarithm and its Complex Conjugate

Question

$$\newcommand{\dl}[1]{\operatorname{Li}_2\left( #1\right)}$$

I have the following expression involving Dilogarithims, where $z\in\mathbb{R}$ and $b\in\mathbb{C}\setminus\lbrace 0\rbrace$:

$$f(z,b) = \dl{\frac{z}{b}}-\dl{-\frac{z}{b}}+\dl{\frac{z}{b^*}}-\dl{-\frac{z}{b^*}}$$

In the above, I mean that $b^*$ is the complex conjugate. I have this result from a definite integral, and would like to evaluate $h(b)=\lim_{z\to\infty} f(z,b)$, giving me a result that only depends on $b$. In the above definition of $f$, each term individually diverges when I take the limit. I'm certain that $\lim_{z\to\infty}f(z,b)$ is convergent though, so I'd like to find some way of simplifying $f$ to allow me to take the limit. I'm hoping someone can point me to a relation involving $\dl{z}$, such as $$\dl{z}-\dl{-z}$$ or $$\dl{\frac{z}{b}}+\dl{\frac{z}{b^*}}$$ I am certain one of those relations will help me, but I couldn't find anything like that.

Does someone maybe have an idea, how to evaluate this?

Thanks in advance!

Regards

Zyrax

Answer 1

$$\newcommand{\dl}[1]{\operatorname{Li}_2\left( #1\right)}$$

From the definition $\;\displaystyle\dl z:=\sum_{n=1}^\infty \frac {z^n}{n^2}\;$ we have of course :

\begin{align} \tag{1}\dl{\overline{z}}&=\overline{\dl{z}},\quad z\in\mathbb{C}\\ &\text{and}\\ \tag{2}\dl{-z}&=-\dl{z}+\frac 12 \dl{z^2}\\ &\text{since}\\ \dl{-z}&=\sum_{n=1}^\infty \frac {(-z)^n}{n^2}=-\sum_{n=1}^\infty \frac {z^n}{n^2}+2\sum_{k=1}^\infty \frac {z^{2k}}{(2k)^2}\\ &\text{but also}\\ \tag{3}\dl{z}-\dl{-z}&=\sum_{n=1}^\infty \frac {z^n-(-z)^n}{n^2}=2\sum_{k=0}^\infty \frac {z^{2k+1}}{(2k+1)^2},\quad |z|<1\\ \end{align}

This allows to rewrite (for $\,x\in\mathbb{R}\,$ and $\,z:=\dfrac xb$) : \begin{align} f(x,b) &= \dl{\frac{x}{b}}-\dl{-\frac{x}{b}}+\dl{\frac{x}{\overline{b}}}-\dl{-\frac{x}{\overline{b}}}\\ &= \dl{\frac{x}{b}}-\dl{-\frac{x}{b}}+\overline{\dl{\frac{x}{b}}}-\overline{\dl{-\frac{x}{b}}}\\ \tag{4}&= 2\;\Re\left(\dl{\frac{x}{b}}-\dl{-\frac{x}{b}}\right)\\ &= 4\;\Re\left(\sum_{k=0}^\infty \frac {\left(\frac{x}{b}\right)^{2k+1}}{(2k+1)^2}\right),\quad \text{for $\ \left|\frac{x}{b}\right|<1$ using $(3)$}\\ \tag{5}&=\Re\left(\frac{x}{b}\;\Phi\left(\left(\frac{x}{b}\right)^2, 2, \frac 12\right)\right),\quad\text{(without bounds on $|z|$)}\\ \tag{6}&=4\;\Re\left(\chi_2\left(\frac xb\right)\right) \end{align} i.e. as an analytic continuation of the Lerch transcendent $\;\displaystyle\Phi(z, s, \alpha) := \sum_{n=0}^\infty \frac { z^n} {(n+\alpha)^s}$
or of the Legendre chi function $\;\displaystyle\chi_s(z) := \sum_{k=0}^\infty \frac{z^{2k+1}}{(2k+1)^s}$.

Concerning the asymptotic expression as $x\to +\infty$ the Wikipedia link contains the interesting $$\tag{7}\chi_2(x) + \chi_2(1/x)= \frac{\pi}2\left(\frac{\pi}{2}-i|\ln x|\right),\quad x>0$$

The $x>0$ restriction appears not really required and we have in fact for $z\neq 0$ : $$\tag{8}\chi_2(z) + \chi_2(1/z)= \frac{\pi}2\left(\frac{\pi}{2}+\epsilon\;(\arg(z)-i\ln|z|)\right)\,\quad\text{with}\ \ \epsilon:=\begin{cases} -1 & \arg(z)>0\\ +1 & \arg(z)\le 0 \end{cases}$$

Since $\;\chi_2(1/z) \to 0\;$ as $\;|z|\to +\infty\;$ with $\,z=\dfrac xb\;$ we obtain simply using $(6)$ and $\arg \dfrac xb=-\arg b$ ("arg" is the complex argument i.e. $\;\arg b=\arctan\dfrac {\Im(b)}{\Re(b)}$ if $\Re(b)>0$) : $$\tag{9}f(x,b)\sim 2\,\pi\,\left(\frac {\pi}2-|\arg(b)|\right),\quad x\to \infty$$

Answer 2

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \mrm{f}\pars{z,b} & \equiv \Li{2}\pars{z \over b} - \Li{2}\pars{-\,{z \over b}} + \Li{2}\pars{z \over b^{*}} - \Li{2}\pars{-\,{z \over b^{*}}} \end{align}

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The following approach is valid whenever $\ds{\pm\,{z \over b}, \pm\,{z \over b^{*}} \not\in \,\left[\, 0,\infty\, \right)}$. In such a case we can use the identity: $$ \Li{2}\pars{\xi} = -\Li{2}\pars{1 \over \xi} - \,{\pi^{2} \over 6} - \half\,\ln^{2}\pars{-\xi} \,,\qquad\xi \not\in \,\left[\, 0,\infty\, \right) $$ and $\ds{\ln}$-function branch-cut is the principal one ( along the 'negative $\ds{x}$-axis' ). $\ds{\lim_{\verts{z} \to \infty}\,\mrm{f}\pars{z,b}}$ is reduced to \begin{align} \lim_{\verts{z} \to \infty}\,\,\mrm{f}\pars{z,b} & = \half\,\lim_{\verts{z} \to \infty}\bracks{\ln^{2}\pars{z \over b} - \ln^{2}\pars{-\,{z \over b}} + \ln^{2}\pars{z \over b^{*}} - \ln^{2}\pars{-\,{z \over b^{*}}}} \end{align} Other situations $\ds{\pars{~\mbox{besides}\ \xi \not\in \,\left[\, 0,\infty\, \right)~}}$ can be explored by means of identities in this link.