Integral and trilogarithm

Evaluate the integral:

$$\int_{0}^{1}\frac{\ln^2 x}{2-x}\, {\rm d}x$$

Solution

We are using the following formulae:

${\color{gray}\bullet} \;\;\;{\rm Li}_3(z)= \sum \limits_{n=1}^{\infty}\frac{z^n}{n^3}, \; z \in [-1, 1] $
${\color{gray}\bullet} \;\;\; {\rm Li}_3(z)+ {\rm Li}_3(1-z)+{\rm Li}_3 \left ( 1-\frac{1}{z} \right )= \zeta(3)+ \frac{1}{6}\ln^3 z +\zeta(2)\ln z - \frac{1}{2}\ln^2 z \ln (1-z)$

Setting at the second equation $z=1/2$ we have that:

\begin{equation} 2{\rm Li}_3 \left ( \frac{1}{2} \right )+ {\rm Li}_3(-1)= \zeta(3)- \frac{1}{6}\ln^3 2 -\zeta(2)\ln 2+\frac{1}{2}\ln^3 2\end{equation}

However:

$$\begin{align*}
{\rm Li}_3(-1) &=\sum_{n=1}^{\infty}\frac{(-1)^n}{n^3} \\
 &= \sum_{n=1}^{\infty}\frac{1}{8n^3} - \sum_{n=0}^{\infty}\frac{1}{\left ( 2n+1 \right )^3}\\
 &= \frac{\zeta(3)}{8} - \sum_{n=0}^{\infty}\frac{1}{\left ( 2n+1 \right )^3}
\end{align*}$$

However, if we take a look at $\zeta(3)$ and split it up to even and odd components we have that:

$$\zeta(3)= \sum_{n=1}^{\infty}\frac{1}{n^3}= \frac{1}{8}\zeta(3)+ \sum_{n=0}^{\infty} \frac{1}{(2n+1)^3}\Leftrightarrow \sum_{n=0}^{\infty}\frac{1}{(2n+1)^3}= \frac{7}{8}\zeta(3)$$

Going back we have that:

\begin{equation} {\rm Li}_3 (-1)= - \frac{3\zeta(3)}{4} \end{equation}

Returning back at $(1)$ and doing the calculation we have that:

${\color{gray} \bullet} \;\;\;  {\rm Li}_3\left ( \frac{1}{2} \right )= \frac{1}{24}\left [ 21\zeta(3) +4\ln^3 2 -2\pi^2 \ln 2 \right ]$

Back to our integral we apply integration by parts, hence:

$$\begin{align*}
\int_{0}^{1}\frac{\ln^2 x}{2-x}\, {\rm d}x &=\left [ \ln \left ( 1- \frac{x}{2} \right )\ln^2 x \right ]_0^1 -2 \int_{0}^{1}\frac{\ln x \ln \left ( 1- \frac{x}{2} \right )}{x}\, {\rm d}x \\
 &= \left [ 2 {\rm Li}_2\left ( \frac{x}{2} \right )\ln x \right ]_0^1 + 2 \int_{0}^{1}\frac{{\rm Li}_2 \left ( \frac{x}{2} \right )}{x}\, {\rm d}x\\
 &= 2{\rm Li}_3 \left ( \frac{1}{2} \right )\\
 &= \frac{1}{12}\left [ 21\zeta(3)+4\ln^3 2- 2\pi^2 \ln 2 \right ]
\end{align*}$$

ending the calculations.