TITLE: How to integrate the dilogarithms?
QUESTION [10 upvotes]: $\def\Li{\operatorname{Li}}$
How can you integrate $\Li_2$? I tried from $0 \to 1$
$\displaystyle \int_{0}^{1} \Li_2(z) \,dz = \sum_{n=1}^{\infty} \frac{1}{n^2(n+1)}$
$$\frac{An + B}{n^2} + \frac{D}{n+1} = \frac{1}{n^2(n+1)}$$
$$(An + B)(n+1) + D(n^2) = 1$$
Let $n = -1, \implies D = 1$
Let $n = 0, \implies B = 1$
Let $n = 1, \implies A = -1$
$$\frac{-n + 1}{n^2} + \frac{1}{n+1} = \frac{1}{n^2(n+1)}$$
$$= \sum_{n=1}^{\infty} \frac{-n + 1}{n^2} + \frac{1}{n+1} = \sum_{n=1}^{\infty} \frac{1}{n^2(n+1)} = \sum_{n=1}^{\infty} \frac{1}{n^2} - \frac{1}{n} + \frac{1}{n+1} $$ 
The $1/n$ is the problem, it is the harmonic series, which diverges.

REPLY [3 votes]: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}&\color{#66f}{\large\int_{0}^{1}{\rm Li}_{2}\pars{x}\,\dd x}
=\left.\vphantom{\Large A}x\,{\rm Li}_{2}\pars{x}\right\vert_{0}^{1}
-\int_{0}^{1}x\,{\rm Li}_{2}'\pars{x}\,\dd x
\\[5mm]&=\overbrace{{\rm Li}_{2}\pars{1}}
^{\ds{\color{#c00000}{\sum_{n\ =\ 1}^{\infty}{1^{n} \over n^{2}}\ =\
{\pi^{2} \over 6}}}}\ -\ 
\int_{0}^{1}x\ \overbrace{\bracks{-\,{{\ln\pars{1 - x} \over x}}}}
^{\ds{=\ \color{#c00000}{{\rm Li}_{2}'\pars{x}}}}\ \,\dd x
={\pi^{2} \over 6} +\
\underbrace{\int_{0}^{1}\ln\pars{x}\,\dd x}_{\ds{=\ \color{#c00000}{-1}}}
\\[5mm]&=\color{#66f}{\large{\pi^{2} \over 6} - 1} \approx {\tt 0.6449} 
\end{align}