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 \[\operatorname{Li}_{-1}(x)=\sum_{n=1}^\infty nx^n=\dfrac{1}{(1-x)^2}.\] We can keep doing this to get closed form of \(Li_{-s} (x)\), but if we were given something like \(Li_{-25}(0.5)\), 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:

\[ \DeclareMathOperator{Li}{Li\,} \begin{array} { | c | c | l | } \hline s & \Li_s(z) & \text{ Algebraic expression } \\ \hline 1 & \Li_1(z) & -\ln(1-z) \\ \hline 0 & \Li_0(z) & -\dfrac z{1-z} \\ \hline -1 & \Li_{-1}(z) & \dfrac {z}{(1-z)^2} \\ \hline -2 & \Li_{-2}(z) & \dfrac {z(z+1)}{(1-z)^3} \\ \hline -3 & \Li_{-3}(z) & \dfrac {z(z^2+4z+1)}{(1-z)^4} \\ \hline -4 & \Li_{-4}(z) & \dfrac {z(1+z)(z^2+10z+1)}{(1-z)^5} \\ \hline -5 & \Li_{-5}(z) & \dfrac {z(z^4+26z^3+ 66z^2+26z+1)}{(1-z)^5} \\ \hline \end{array} \]

In general, we can express \( \DeclareMathOperator{Li}{Li\,} \Li_{-n}(z) \) as \[ \DeclareMathOperator{Li}{Li\,} \Li_{-n}(z) = \left( z \; \dfrac {\partial}{ \partial z} \right)^n \dfrac z{1-z} = \sum_{k=0}^n k! \; S(n+1,k+1) \left( \dfrac z{1-z} \right)^{k+1} \; , \] 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!