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PostPosted: Wed Sep 28, 2016 8:41 pm 

Joined: Sat Nov 14, 2015 6:32 am
Posts: 155
Location: Melbourne, Australia
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.

_________________
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$


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PostPosted: Wed Sep 28, 2016 9:04 pm 

Joined: Sat Nov 14, 2015 6:32 am
Posts: 155
Location: Melbourne, Australia
Based on the above fact here are some exercises that are left to the reader.

1. Prove that:

$$\Im\operatorname{Li}_2 \left[\left(i\left(2\pm\sqrt3\right)\right) \right] =\frac{2 \mathcal{G}}{3}-\frac{\pi\,(2\pm3)}{12}\ln\left(2-\sqrt3\right)$$

(Hint: Use another great equation the dilogarithm satisfies:

$$\operatorname{Li}_2(z)+\operatorname{Li}_2\left(\frac{1}{z} \right)=-\frac{\pi^2}{6}-\frac{\ln^2(-z)}{2}$$

as well as the very known fact $\displaystyle \Re (z)=\frac{z+\bar{z}}{2}$. )

2. Prove that:

$$\Re\left [ {\rm Li}_2 \left ( 1 \pm i \sqrt{3} \right ) \right ] = \frac{\pi^2}{24} - \frac{\log^2 2}{4} - \frac{1}{4} {\rm Li}_2 \left ( \frac{1}{4} \right )$$

(Hint: Apply equation $(1)$ along with the equation of ${\rm Li}_2(iz)$. )

3. Prove that:

$$\Re\left [ {\rm Li}_2 \left ( \frac{1}{2}+ \frac{i}{6} \right ) \right ] = \frac{7 \pi^2}{48} - \frac{\arctan^2 2}{3} - \frac{\arctan^2 3}{6} - \frac{1}{8} \log^2 \left ( \frac{18}{5} \right )$$

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PostPosted: Wed Sep 28, 2016 10:13 pm 

Joined: Sat Nov 14, 2015 6:32 am
Posts: 155
Location: Melbourne, Australia
Now let us move on to the trilogarithm function. It is defined as $\displaystyle {\rm Li}_3 (z) = \sum_{n=1}^{\infty} \frac{z^n}{n^3}= \int_{0}^{z}\frac{{\rm Li}_2 (t)}{t} \, {\rm d}t$.

1. $\displaystyle {\rm Li}_3(i) = -\frac{3 \zeta(3)}{32} + i \; \frac{\pi^3}{32}$ since:

1st way:

\begin{align*}
{\rm Li}_3(i) &=\sum_{n=1}^{\infty} \frac{i^n}{n^3} \\
&= \sum_{ 0 \mod n} \frac{i^n}{n^3} + \sum_{1 \mod n} \frac{i^n}{n^3} + \sum_{2 \mod n} \frac{i^n}{n^3} +\sum_{3 \mod n} \frac{i^n}{n^3}\\
&= -\frac{3 \zeta(3)}{32} + i \; \frac{\pi^3}{32}
\end{align*}

2nd way: One of the most fundamental equations all polylogarithms obey to is:

\begin{equation} {\rm Li}_{n+1} (z) = \frac{z}{(-1)^n n!} \int_{0}^{1}\frac{\log^n t}{1-zt} \, {\rm d}t \end{equation}

For $z=i$ and $n =2$ we have that:

\begin{align*}
{\rm Li}_3 (i) &=\frac{i}{2} \int_{0}^{1} \frac{\log^2 t}{1-it} \, {\rm d}t \\
&= \frac{i}{2}\int_{0}^{1} \log^2 t \sum_{n=0}^{\infty} i^n t^n \, {\rm d}t\\
&= \frac{i}{2}\sum_{n=0}^{\infty} i^n \int_{0}^{1} t^n \log^2 t \, {\rm d}t\\
&= i \sum_{n=0}^{\infty} \frac{i^n}{\left ( n+1 \right )^3}\\
&= i \left ( 1 + \frac{i}{2^3} - \frac{1}{3^3} -\frac{i}{4^3} + \cdots \right ) \\
&= i\sum_{n=0}^{\infty} \frac{(-1)^n)}{\left ( 2n+1 \right )^3} -i^2 \sum_{n=1}^{\infty} \frac{(-1)^n}{8n^3} \\
&= -\frac{3 \zeta(3)}{32} + i \; \frac{\pi^3}{32}
\end{align*}

Now using the equation

\begin{equation} {\rm Li}_3(z) + {\rm Li}_3(-z) = \frac{1}{4} {\rm Li}_3\left ( z^2 \right ) \end{equation}

we get that $\displaystyle {\rm Li}_3 (-i) = -\frac{3 \zeta(3)}{32} - i \; \frac{\pi^3}{32}$.

Note: Let $s \in \mathbb{N}$. Following the exact same procedure as before (or even easier) we get that:

$${\rm Li}_s \left ( \pm i \right ) = -2^{-s} \eta(s) \pm i \beta (s)$$

where $\eta, \; \beta$ are eta and beta Dirichlet functions respectively.

2. $\displaystyle \Re\left [ {\rm Li}_3 \left ( 1 \pm i \right ) \right ]= \frac{\pi^3 \log 2}{32} + \frac{35 \zeta(3)}{64}$.

Proof is left to the reader.

3. $\displaystyle \Re\left [ {\rm Li}_3 \left ( \frac{1+i}{2} \right ) \right ] = \frac{\log^3 2}{48} - \frac{5\pi^2 \log 2}{192} + \frac{35 \zeta(3)}{64}$.

Proof is left to the reader.

4. $\displaystyle \Im\left [ {\rm Li}_3 \left ( \frac{1\pm i}{\sqrt{2}} \right ) \right ] = \pm \frac{7 \pi^3}{256}$.

Proof is left to the reader.

_________________
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$


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PostPosted: Wed Sep 28, 2016 10:25 pm 

Joined: Sat Nov 14, 2015 6:32 am
Posts: 155
Location: Melbourne, Australia
General notes on the polylogs.

Let $n \in \mathbb{N}$. Then ${\rm Li}_n$ is analytic and single valued on $\mathbb{D}= \mathbb{C} \setminus [1,+\infty)$ and note that this domain is symmetric with respect to the real axis. $g(z)=\overline{{\rm Li}_n(\overline{z})}$ is also analytic on the same domain and it coincides with ${\rm Li}_n$ on $(-1, 1)$. Thus:

$$\forall\,z\in \mathbb{D},\qquad {\rm Li}_n(z)=\overline{{\rm Li}_n(\overline{z})}$$

We now conclude that

\begin{align}
\Re\left [ {\rm Li}_n (z) \right ] &=\frac{1}{2}\left ( {\rm Li}_n(z) + \overline{{\rm Li}_n (z)} \right )= \frac{1}{2}\left ( {\rm Li}_n(z) + {\rm Li}_n (\bar{z}) \right )\\
\Im \left [ {\rm Li}_n (z) \right ]&=\frac{1}{2i} \left ( {\rm Li}_n(z) - \overline{{\rm Li}_n (z)} \right ) = \frac{1}{2i} \left ( {\rm Li}_n(z) - {\rm Li}_n (\bar{z}) \right )
\end{align}

This last identity $(13)$ can be used to prove what T. stated at the linked post:

$$\Re{ \left[{\rm Li}_{2}\left(\frac{1}{2}+iq\right) \right]}=\frac{{\pi}^{2}}{12}-\frac{1}{8}{\ln^2{\left(\frac{1+4q^2}{4}\right)}}-\frac{{\arctan^2{(2q)}}}{2}$$

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