POLYLOGARITHM IDENTITIES

We have already seen polylogarithms functions and it´s usefulness for computing sums and integrals. This post is a sort o Apendix that I want to go a little bit further and derive some formulas for $\operatorname{Li_{n}}(x)$ that will help us compute a nice integral in the next post.

Integral representation of Polylogarithms

The polylogarithm function is defined as

$\operatorname{Li_{n}}(x)=\sum_{k=1}^{\infty}\frac{x^k}{k^{n}}$

We can find a very conviniet integral representation for it as following: Start with

$\operatorname{Li_{n+1}}(x)=\sum_{k=1}^{\infty}\frac{x^k}{k^{n+1}}$

$=\sum_{k=1}^{\infty}\frac{1}{k^{n}}\int_{0}^{x}t^{k-1}dt=\sum_{k=1}^{\infty}\frac{1}{k^{n}}\int_{0}^{x}\frac{t}{t}t^{k-1}dt$

$=\int_{0}^{x}\frac{1}{t}\sum_{k=1}^{\infty}\frac{t^k}{k^{n}}dt$

$\boxed{\operatorname{Li_{n+1}}(x)=\int_{0}^{x}\frac{\operatorname{Li_{n}}(t)}{t}dt}$ (1)

For example

$\operatorname{Li_{2}}(x)=\int_{0}^{x}\frac{\operatorname{Li_{1}}(t)}{t}dt$

$\operatorname{Li_{2}}(x)=\int_{0}^{x}\frac{1}{t}\sum_{k=1}^{\infty}\frac{x^k}{k}dt$

$\boxed{\operatorname{Li_{2}}(x)=-\int_{0}^{x}\frac{\ln(1-x)}{t}dt}$ (2)

With (2) we can derive an important identitie.

$\operatorname{Li_{2}}\left(1-\frac{1}{x}\right)=-\int_{0}^{1-\frac{1}{x}}\frac{\ln(1-x)}{t}dt$

differentiating the above expression with respect to

$\frac{d}{dx}\operatorname{Li_{2}}\left(1-\frac{1}{x}\right)=-\frac{\ln\left(1-\left(1-\frac{1}{x} \right)\right)}{1-\frac{1}{x}}\frac{1}{x^2}$

$\operatorname{Li_{2}^{\prime}}\left(1-\frac{1}{x}\right)=-\frac{\ln\left(\frac{1}{x}\right)}{x^2-x}=\frac{\ln(x)}{x(x-1)}=-\frac{\ln(x)}{x(1-x)}$

By partial fractions we get

$\operatorname{Li_{2}^{\prime}}\left(1-\frac{1}{x}\right)=-\ln(x)\left(\frac{1}{x}+\frac{1}{1-x} \right)$

$\operatorname{Li_{2}}\left(1-\frac{1}{x}\right)=-\int\frac{\ln(x)}{x}dx-\int\frac{\ln(x)}{1-x}dx$

$\operatorname{Li_{2}}\left(1-\frac{1}{x}\right)=-\frac{\ln^2(x)}{2}-\operatorname{Li_{2}}(1-x)+C$

The first integral is evaluated through integration by parts and the second by the integral definition above of $\operatorname{Li_{2}}$

Letting x=1 we get that C=0

$\boxed{\operatorname{Li_{2}}\left(1-\frac{1}{x}\right)=-\frac{\ln^2(x)}{2}-\operatorname{Li_{2}}(1-x)}$ (3)

Another important formula for $Li_{2}$ is the Euler´s relflection formula given by

$Li_{2}(x)+Li_{2}(1-x)=\zeta(2)-\log(x)\log(1-x)$

To proof it, lets start again from the integral representation

$Li_{2}(x)+Li_{2}(1-x)=-\int_{0}^{x}\frac{\log(1-t)}{t}dt-\int_{0}^{1-x}\frac{\log(1-t)}{t}dt$

Make the change of variable 1-t=w in the second integral

$Li_{2}(x)+Li_{2}(1-x)=-\int_{0}^{x}\frac{\log(1-t)}{t}dt+\int_{1}^{x}\frac{\log(w)}{(1-w)}dw$

$Li_{2}(x)+Li_{2}(1-x)=-\int_{0}^{x}\frac{\log(1-t)}{t}dt+\int_{1}^{x}\frac{\log(t)}{(1-t)}dt$

Integrate by parts the second integral

$Li_{2}(x)+Li_{2}(1-x)=-\int_{0}^{x}\frac{\log(1-t)}{t}dt+ \Big[-\log(t)\log(1-t) \Big|_{1}^{x}+\int_{1}^{x}\frac{\log(1-t)}{t}dt \Big]$

$Li_{2}(x)+Li_{2}(1-x)=-\int_{0}^{x}\frac{\log(1-t)}{t}dt+\int_{1}^{x}\frac{\log(1-t)}{t}dt-\log(x)\log(1-x)$

$Li_{2}(x)+Li_{2}(1-x)=\underbrace{-\int_{0}^{1}\frac{\log(1-t)}{t}dt}_{=Li_{2}(1)}-\log(x)\log(1-x)$

$\boxed{Li_{2}(x)+Li_{2}(1-x)=\zeta(2)-\log(x)\log(1-x)}$ (4)

Once (3) and (4) are estabilished, we can now derive a very important formula for $Li_{3}$ , namely:

$Li_{3}(x)+Li_{3}(1-x)+Li_{3}\left(1-\frac{1}{x}\right)=\zeta(3)+\frac{\ln ^{3}(x)}{6}+\frac{\pi^{2} \ln (x)}{6}-\frac{\ln ^{2}(x) \ln (1-x)}{2}$

We proceed in the same fashion as we did above. Start with the integral representation

$\operatorname{Li_{3}}\left(1-\frac{1}{x}\right)=\int_{0}^{1-\frac{1}{x}}\frac{\operatorname{Li_{2}}(t)}{t}dt$

differentiating the above expression with respect to

$\operatorname{Li_{3}^{\prime}}\left(1-\frac{1}{x}\right)=\frac{\operatorname{Li_{2}}\left(1-\frac{1}{x}\right)}{1-\frac{1}{x}}\frac{1}{x^2}$

$\operatorname{Li_{3}^{\prime}}\left(1-\frac{1}{x}\right)=\frac{\operatorname{Li_{2}}\left(1-\frac{1}{x}\right)}{x^2-x}$

$\operatorname{Li_{3}}\left(1-\frac{1}{x}\right)=-\int\frac{\operatorname{Li_{2}}\left(1-\frac{1}{x}\right)}{x(1-x)}dx$

Performing a partial fraction decomposition on the right hand side

$\operatorname{Li_{3}}\left(1-\frac{1}{x}\right)=-\int\frac{\operatorname{Li_{2}}\left(1-\frac{1}{x}\right)}{x}dx-\int\frac{\operatorname{Li_{2}}\left(1-\frac{1}{x}\right)}{1-x}dx$

Using (3) we can rewrite $\operatorname{Li_{3}}\left(1-\frac{1}{x}\right)$ as a sum if four integrals, i.e.

$\operatorname{Li_{3}}\left(1-\frac{1}{x}\right)=\frac{1}{2}\int\frac{\ln^2(x)}{x}dx+\int\frac{\operatorname{Li_{2}}\left(1-x\right)}{x}dx+\frac{1}{2}\int\frac{\ln^2(x)}{1-x}dx+\int\frac{\operatorname{Li_{2}}\left(1-x\right)}{1-x}dx$

$\operatorname{Li_{3}}\left(1-\frac{1}{x}\right)=\frac{\ln^3(x)}{6}+\int\frac{\operatorname{Li_{2}}\left(1-x\right)}{x}dx+\frac{1}{2}\int\frac{\ln^2(x)}{1-x}dx-\operatorname{Li_{3}}\left(1-x\right)+ C$

The constant C here will take care of all constants steming from the indefinite integrals

$\operatorname{Li_{3}}\left(1-\frac{1}{x}\right)=\ln(x)\operatorname{Li_{2}}\left(1-x\right)+\frac{\ln^3(x)}{6}-\operatorname{Li_{3}}\left(1-x\right)-\frac{1}{2}\int\frac{\ln^2(x)}{1-x}dx + C$

Using

$\operatorname{Li_{3}}\left(1-\frac{1}{x}\right)=\ln(x)\operatorname{Li_{2}}\left(1-x\right)+\frac{\ln^3(x)}{6}-\operatorname{Li_{3}}\left(1-x\right)-\frac{1}{2}\left( -\ln^2(x)\ln(1-x)-2$\ln(x)Li_{2}(x)+2 Li_{3}(x) \right)+C$

$\operatorname{Li_{3}}\left(1-\frac{1}{x}\right)=\ln(x)\operatorname{Li_{2}}\left(1-x\right)+\frac{\ln^3(x)}{6}-\operatorname{Li_{3}}\left(1-x\right)+\frac{\ln^2(x)\ln(1-x)}{2}+$\ln(x)Li_{2}(x)- Li_{3}(x) + C$

$Li_{3}(x)+\operatorname{Li_{3}}\left(1-x\right)+\operatorname{Li_{3}}\left(1-\frac{1}{x}\right)=\frac{\ln^3(x)}{6}+\frac{\ln^2(x)\ln(1-x)}{2}+\ln(x)\operatorname{Li_{2}}\left(1-x\right)+\ln(x)Li_{2}(x) + C$

$Li_{3}(x)+\operatorname{Li_{3}}\left(1-x\right)+\operatorname{Li_{3}}\left(1-\frac{1}{x}\right)=\frac{\ln^3(x)}{6}+\frac{\ln^2(x)\ln(1-x)}{2}+\ln(x) \left\{\operatorname{Li_{2}}\left(1-x\right)+Li_{2}(x) \right\} + C$

and now using (4)

$Li_{3}(x)+\operatorname{Li_{3}}\left(1-x\right)+\operatorname{Li_{3}}\left(1-\frac{1}{x}\right)=\frac{\ln^3(x)}{6}+\frac{\ln^2(x)\ln(1-x)}{2}+\ln(x) \left\{\zeta(2)-\log(x)\log(1-x) \right\} + C$

$Li_{3}(x)+\operatorname{Li_{3}}\left(1-x\right)+\operatorname{Li_{3}}\left(1-\frac{1}{x}\right)=\frac{\ln^3(x)}{6}-\frac{\ln^2(x)\ln(1-x)}{2}+\ln(x)\zeta(2) + C$

setting z=1 in the above equation we easily find that $C=\zeta(3)$ . Recall that $\zeta(2)=\frac{\pi^2}{6}$ to finally get

$\boxed{Li_{3}(x)+Li_{3}(1-x)+Li_{3}\left(1-\frac{1}{x}\right)=\zeta(3)+\frac{\ln ^{3}(x)}{6}+\frac{\pi^{2} \ln (x)}{6}-\frac{\ln ^{2}(x) \ln (1-x)}{2}} \qquad \qquad (5)$