[TUT] Special values on the polylogarithm functions
Posted: Wed Sep 28, 2016 8:41 pm
1. $\displaystyle \Im\left [ {\rm Li}_2 (i) \right ] =\mathcal{G}$ since in general
\begin{align*}
{\rm Li}_2 (iz) &= - \int_{0}^{z} \frac{\log (1-it)}{t} \, {\rm d}t \\
&= -\int_{0}^{z} \frac{\log \left [ \left ( 1+t^2 \right )^{1/2} e^{-i \arctan t} \right ]}{t} \, {\rm d}t\\
&= -\frac{1}{2} \int_{0}^{z}\frac{\log \left ( 1+t^2 \right )}{t} \, {\rm d}t + i \int_{0}^{z} \frac{\arctan t}{t} \, {\rm d}t\\
&= \frac{1}{4}{\rm Li}_2 \left ( -z^2 \right ) + i {\rm Ti}_2 (z)
\end{align*}
Now for $z=1$ we have that ${\rm Ti}_2(1)=\mathcal{G}$ and thus the first equation follows.
2. $\displaystyle \Im \left [ {\rm Li}_2 \left ( 1+i \right ) \right ] = \mathcal{G} + \frac{\pi \log 2}{4}$. This pretty much follows from the fundamental equation the dilogarithm function satisfies, i.e
\begin{equation} {\rm Li}_2(z)+ {\rm Li}_2(1-z) = \zeta(2) - \log z \log (1-z) \end{equation}
For $z=-i$ we get the following
\begin{equation} {\rm Li}_2 \left ( -i \right ) + {\rm Li}_2 \left ( 1+i \right ) = \zeta(2) - \log (-i) \log (1+i) \end{equation}
Taking imaginary parts as well as into account the following fact
\begin{align*}
{\rm Li}_2 (-i) + {\rm Li}_2 \left ( 1+i \right ) &= \zeta(2) - \log (-i) \log (1+i) \\
&=\zeta(2) -\frac{\pi^2}{8} +\frac{i \pi \log 2 }{4}
\end{align*}
yields the result , since $\Im \left({\rm Li}_2(-i) \right)=-\mathcal{G}$.
3. $\displaystyle \Im \left [ {\rm Li}_2 \left ( \frac{1+i}{2} \right ) \right ] = \mathcal{G} - \frac{\pi \log 2}{8}$.
Well, we are using another fundamental relation of the dilog, namely:
\begin{equation} {\rm Li}_2 (z) +{\rm Li}_2 \left ( -\frac{z}{1-z} \right ) = - \frac{1}{2} \log^2 \left ( 1-z \right ) \; , \quad z \notin [1, +\infty) \end{equation}
as well as the trivial results
\begin{align}
{\rm Li}_2 \left ( e^{i \theta} \right ) &= {\rm Sl}_2 (\theta) + i {\rm Cl}_2 (\theta) \\
{\rm Sl}_2(\theta)&= \sum_{n=1}^{\infty} \frac{\sin n\theta}{n^2} \\
{\rm Cl}_2(\theta)&= \sum_{n=1}^{\infty} \frac{\cos n \theta}{n^2} = \zeta(2) - \frac{\pi \theta }{2} + \frac{\theta^2}{4}
\end{align}
(Note: Equation $(6)$ is just a Fourier series. Well, $\displaystyle \sum_{n=1}^{\infty} \frac{\sin n\theta}{n} = \frac{\pi-\theta}{2} , \; \theta \in (0, 2\pi)$. Since the convergence of that series is uniform we can integrate and get the result. )
Now for $z=\frac{1}{2} + \frac{i}{2}$ we get that
\begin{align*}
{\rm Li}_2 \left ( \frac{1+i}{2} \right ) + {\rm Li}_2 (-i) &= -\frac{1}{2} \log^2 \left ( 1-\frac{1+i}{2} \right )\\
&=\frac{\pi^2}{32} - \frac{\log^2 2}{8} - \frac{i \pi \log 2}{8}
\end{align*}
Taking imaginary part yields the result. As a side note $\displaystyle \Im \left [ {\rm Li}_2 \left ( \frac{1-i}{2} \right ) \right ] = -\mathcal{G} + \frac{\pi \log 2}{8}$.
Note: Did you think that these special values had no real part? Well, you're mistaken. Here there are:
\begin{align}
\Re \left ( {\rm Li}_2 (i) \right ) &= - \frac{\pi^2}{48}\\
\Re \left ( {\rm Li}_2(-i) \right )&= -\frac{\pi^2}{48}\\
\Re \left [ {\rm Li}_2 \left ( \frac{1+i}{2} \right ) \right ]&= \frac{5\pi^2}{96} - \frac{\log^2 2}{8} \\
\Re \left [ {\rm Li}_2 \left ( \frac{1-i}{2} \right ) \right ] &=\frac{5 \pi^2}{96} - \frac{\log^2 2}{8}
\end{align}
This post was migrated from here.
\begin{align*}
{\rm Li}_2 (iz) &= - \int_{0}^{z} \frac{\log (1-it)}{t} \, {\rm d}t \\
&= -\int_{0}^{z} \frac{\log \left [ \left ( 1+t^2 \right )^{1/2} e^{-i \arctan t} \right ]}{t} \, {\rm d}t\\
&= -\frac{1}{2} \int_{0}^{z}\frac{\log \left ( 1+t^2 \right )}{t} \, {\rm d}t + i \int_{0}^{z} \frac{\arctan t}{t} \, {\rm d}t\\
&= \frac{1}{4}{\rm Li}_2 \left ( -z^2 \right ) + i {\rm Ti}_2 (z)
\end{align*}
Now for $z=1$ we have that ${\rm Ti}_2(1)=\mathcal{G}$ and thus the first equation follows.
2. $\displaystyle \Im \left [ {\rm Li}_2 \left ( 1+i \right ) \right ] = \mathcal{G} + \frac{\pi \log 2}{4}$. This pretty much follows from the fundamental equation the dilogarithm function satisfies, i.e
\begin{equation} {\rm Li}_2(z)+ {\rm Li}_2(1-z) = \zeta(2) - \log z \log (1-z) \end{equation}
For $z=-i$ we get the following
\begin{equation} {\rm Li}_2 \left ( -i \right ) + {\rm Li}_2 \left ( 1+i \right ) = \zeta(2) - \log (-i) \log (1+i) \end{equation}
Taking imaginary parts as well as into account the following fact
\begin{align*}
{\rm Li}_2 (-i) + {\rm Li}_2 \left ( 1+i \right ) &= \zeta(2) - \log (-i) \log (1+i) \\
&=\zeta(2) -\frac{\pi^2}{8} +\frac{i \pi \log 2 }{4}
\end{align*}
yields the result , since $\Im \left({\rm Li}_2(-i) \right)=-\mathcal{G}$.
3. $\displaystyle \Im \left [ {\rm Li}_2 \left ( \frac{1+i}{2} \right ) \right ] = \mathcal{G} - \frac{\pi \log 2}{8}$.
Well, we are using another fundamental relation of the dilog, namely:
\begin{equation} {\rm Li}_2 (z) +{\rm Li}_2 \left ( -\frac{z}{1-z} \right ) = - \frac{1}{2} \log^2 \left ( 1-z \right ) \; , \quad z \notin [1, +\infty) \end{equation}
as well as the trivial results
\begin{align}
{\rm Li}_2 \left ( e^{i \theta} \right ) &= {\rm Sl}_2 (\theta) + i {\rm Cl}_2 (\theta) \\
{\rm Sl}_2(\theta)&= \sum_{n=1}^{\infty} \frac{\sin n\theta}{n^2} \\
{\rm Cl}_2(\theta)&= \sum_{n=1}^{\infty} \frac{\cos n \theta}{n^2} = \zeta(2) - \frac{\pi \theta }{2} + \frac{\theta^2}{4}
\end{align}
(Note: Equation $(6)$ is just a Fourier series. Well, $\displaystyle \sum_{n=1}^{\infty} \frac{\sin n\theta}{n} = \frac{\pi-\theta}{2} , \; \theta \in (0, 2\pi)$. Since the convergence of that series is uniform we can integrate and get the result. )
Now for $z=\frac{1}{2} + \frac{i}{2}$ we get that
\begin{align*}
{\rm Li}_2 \left ( \frac{1+i}{2} \right ) + {\rm Li}_2 (-i) &= -\frac{1}{2} \log^2 \left ( 1-\frac{1+i}{2} \right )\\
&=\frac{\pi^2}{32} - \frac{\log^2 2}{8} - \frac{i \pi \log 2}{8}
\end{align*}
Taking imaginary part yields the result. As a side note $\displaystyle \Im \left [ {\rm Li}_2 \left ( \frac{1-i}{2} \right ) \right ] = -\mathcal{G} + \frac{\pi \log 2}{8}$.
Note: Did you think that these special values had no real part? Well, you're mistaken. Here there are:
\begin{align}
\Re \left ( {\rm Li}_2 (i) \right ) &= - \frac{\pi^2}{48}\\
\Re \left ( {\rm Li}_2(-i) \right )&= -\frac{\pi^2}{48}\\
\Re \left [ {\rm Li}_2 \left ( \frac{1+i}{2} \right ) \right ]&= \frac{5\pi^2}{96} - \frac{\log^2 2}{8} \\
\Re \left [ {\rm Li}_2 \left ( \frac{1-i}{2} \right ) \right ] &=\frac{5 \pi^2}{96} - \frac{\log^2 2}{8}
\end{align}
This post was migrated from here.