# How to prove $\int_{0}^{-1} \frac{\operatorname{Li}_2(x)}{(1-x)^2} dx=\frac{\pi^2}{24}-\frac{\ln^2(2)}{2}$

$\def\Li{\operatorname{Li}}$ I wonder how to prove: $$\int_{0}^{-1} \frac{\Li_2(x)}{(1-x)^2} dx=\frac{\pi^2}{24}-\frac{\ln^2(2)}{2}$$ I'm not used to polylogarithm, so I don't know how to tackle it. So any help is highly appreciated.

The basic idea behind most polylogarithmic integrals is integration by parts. Note that we have the indefinite integral,

$$\int\frac{1}{(1-x)^2}\,\mathrm{d}x=\frac{1}{1-x}+\color{grey}{constant}.$$

And the derivative of the dilogarithm is of course:

$$\frac{d}{dx}\operatorname{Li}_{2}{\left(x\right)}=-\frac{\ln{\left(1-x\right)}}{x}.$$

Thus, integration by parts can transform this polylog integral into an integral with just ordinary logarithms and rational functions, which can be evaluated by the usual methods such as partial fractions::

\begin{align} \int_{-1}^{0}\frac{\operatorname{Li}_{2}{\left(x\right)}}{(1-x)^2}\,\mathrm{d}x &=\left[\frac{\operatorname{Li}_{2}{\left(x\right)}}{1-x}\right]_{-1}^{0}+\int_{-1}^{0}\frac{\ln{\left(1-x\right)}}{x(1-x)}\,\mathrm{d}x\\ &=-\frac{\operatorname{Li}_{2}{\left(-1\right)}}{2}+\int_{-1}^{0}\frac{\ln{\left(1-x\right)}}{x(1-x)}\,\mathrm{d}x\\ &=-\frac{\operatorname{Li}_{2}{\left(-1\right)}}{2}+\int_{-1}^{0}\frac{\ln{\left(1-x\right)}}{x}\,\mathrm{d}x+\int_{-1}^{0}\frac{\ln{\left(1-x\right)}}{1-x}\,\mathrm{d}x\\ &=-\frac{\operatorname{Li}_{2}{\left(-1\right)}}{2}+\left[-\operatorname{Li}_{2}{\left(x\right)}\right]_{-1}^{0}+\left[-\frac12\ln^2{\left(1-x\right)}\right]_{-1}^{0}\\ &=-\frac{\operatorname{Li}_{2}{\left(-1\right)}}{2}+\operatorname{Li}_{2}{\left(-1\right)}+\frac{\ln^2{\left(2\right)}}{2}\\ &=\frac{\operatorname{Li}_{2}{\left(-1\right)}}{2}+\frac{\ln^2{\left(2\right)}}{2}\\ &=-\frac{\pi^2}{24}+\frac{\ln^2{\left(2\right)}}{2}.\\ \end{align}

$$\def\Li{{\rm{Li}}_2}$$Set $$x\mapsto -x$$ followed by integration by parts, we have \begin{align} \int_0^{-1}\frac{\Li(x)}{(1-x)^2}\,dx&=-\int_0^{1}\frac{\Li(-x)}{(1+x)^2}\,dx\qquad\Rightarrow\qquad u=\Li(-x)\,\,\mbox{and}\,\,dv=\frac{1}{(1+x)^2}\\ &=\left.\frac{\Li(-x)}{1+x}\right|_0^{1}+\int_0^{1}\frac{\ln(1+x)}{x(1+x)}\,dx\\ &=-\frac{\pi^2}{24}+\int_0^{1}\frac{\ln(1+x)}{x}\,dx-\int_0^{1}\frac{\ln(1+x)}{1+x}\,dx\\ &=-\frac{\pi^2}{24}-\Li(-x)\bigg|_0^{1}-\frac{1}{2}\ln^2(1+x)\bigg|_0^{1}\\ &=\frac{\pi^2}{24}-\frac{\ln^22}{2} \end{align}