Polylogarithm

The polylogarithm function is an important function for integration, and finding seemingly complicated sum.

The polylogarithm function is defined as $\operatorname{Li}_s(x) = \sum_{n=1}^\infty \dfrac{x^n}{n^s}$ for all complex $s$ and $|x|\leq 1$ and can be computed for $|x|>1$ by analytical continuation. $_\square$

$\operatorname{Li}_s(x)=\int_0^x \dfrac{\operatorname{Li}_{s-1}(u)}{u} du.\ _\square$

Consider $I=\int_0^x \dfrac{\operatorname{Li}_{s-1}(u)}{u} du=\int_0^x \sum_{n=1}^\infty \dfrac{u^{n-1}}{n^{s-1}} du.$ Interchanging the summation and integral sign, we end up with $I=\sum_{n=1}^\infty \int_0^x \dfrac{u^{n-1}}{n^{s-1}}du=\sum_{n=1}^\infty \dfrac{x^n}{n^s} = \operatorname{Li}_s(x).\ _\square$

Polylogarithm is connected to the infinite geometric progression sum $\operatorname{Li}_0(x)=\sum_{n=1}^\infty x^n=\dfrac{x}{1-x}.$ We can divide by $x$ and differentiate with respect to $x$ to get $\operatorname{Li}_{-1}(x)=\sum_{n=1}^\infty nx^n=\dfrac{x}{(1-x)^2}.$ We can keep doing this to get closed form of $\operatorname{Li}_{-s} (x)$, but if we were given something like $\operatorname{Li}_{-25}(0.5)$, it would become extremely tedious to divide by $x$ and differentiate with respect to $x$ 25 times. Similar to the binomial theorem which was a shortcut to the tedious Pascal's triangle, there is an easy way to calculate these:

$\begin{array} { | c | c | l | } \hline s & \operatorname{Li}_s(z) & \text{ Algebraic expression } \\ \hline 1 & \operatorname{Li}_1(z) & -\ln(1-z) \\ \hline 0 & \operatorname{Li}_0(z) & \dfrac z{1-z} \\ \hline -1 & \operatorname{Li}_{-1}(z) & \dfrac {z}{(1-z)^2} \\ \hline -2 & \operatorname{Li}_{-2}(z) & \dfrac {z(z+1)}{(1-z)^3} \\ \hline -3 & \operatorname{Li}_{-3}(z) & \dfrac {z(z^2+4z+1)}{(1-z)^4} \\ \hline -4 & \operatorname{Li}_{-4}(z) & \dfrac {z(1+z)(z^2+10z+1)}{(1-z)^5} \\ \hline -5 & \operatorname{Li}_{-5}(z) & \dfrac {z(z^4+26z^3+ 66z^2+26z+1)}{(1-z)^6} \\ \hline \end{array}$

In general, we can express $\def\Li{Li\,} \Li_{-n}(z)$ as $\def\Li{Li\,} \Li_{-n}(z) = \left( z \; \dfrac {\partial}{ \partial z} \right)^n \dfrac z{1-z} = \sum_{k=0}^n k! \; S(n+1,k+1) \left( \dfrac z{1-z} \right)^{k+1} \; ,$ where $n$ is a non-negative integer, and $S(n,k)$ denote the Stirling's number of the second kind. $_\square$

Try to solve the following problem using polylogarithms!