Proof of dilogarithm reflection formula $\zeta(2)-\log(x)\log(1-x)=\operatorname{Li}_2(x)+\operatorname{Li}_2(1-x)$ How to prove
$$\zeta(2)-\log(x)\log(1-x)=\operatorname{Li}_2(x)+\operatorname{Li}_2(1-x)$$
I havent started, any hints?
 A: Consider:
$$ f(x)=\log(x)\log(1-x)+\operatorname{Li}_2(x)+\operatorname{Li}_2(1-x).$$
We want to show that $f$ is constant, hence we compute $f'$:
$$ f'(x) =\left(\frac{\log(1-x)}{x}-\frac{\log(x)}{1-x}\right)-\frac{\log(1-x)}{x}+\frac{\log x}{1-x}=0. $$
To finish the proof, we just need to compute $f(x)$ in a point, or to compute the limit:
$$ \lim_{x\to 1^-} f(x) = \zeta(2)+\lim_{x\to 1^-}\log(x)\log(1-x)=\zeta(2).$$
Notice that we have a nice corollary:
$$\sum_{n\geq 1}\frac{1}{2^n n^2}=\operatorname{Li}_2\left(\frac{1}{2}\right)=\frac{1}{2}\left(f\left(\frac{1}{2}\right)-\log^2 2\right)=\frac{\pi^2}{12}-\frac{\log^2 2}{2}.$$
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\begin{align}&\color{#66f}{\large\Li{2}\pars{x} + \Li{2}\pars{1 - x}}
\\[5mm] & =
-\int_{0}^{x}{\ln\pars{1 - t} \over t}\,\dd t\ -\
\overbrace{\int_{0}^{1 - x}{\ln\pars{1 - t} \over t}\,\dd t}
^{\dsc{t\ \mapsto 1 - t}}
\\[5mm]&=-\int_{0}^{x}{\ln\pars{1 - t} \over t}\,\dd t
+\int_{1}^{x}{\ln\pars{t} \over 1 - t}\,\dd t
\\[8mm]&=-\int_{0}^{x}{\ln\pars{1 - t} \over t}\,\dd t
-\left.\vphantom{\Large A}\ln\pars{1 - t}\ln\pars{t}
\right\vert_{\, t\ \to\ 1^{-}}^{\, t\ =\ x}
\\[2mm] & +\int_{1}^{x}\ln\pars{1 - t}\,{1 \over t}\,\dd t
\\[8mm]&=-\ln\pars{1 - x}\ln\pars{x}
-\int_{0}^{1}{\ln\pars{1 - t} \over t}\,\dd t
\\[5mm] & =
-\ln\pars{1 - x}\ln\pars{x} + \int_{0}^{1}\Li{2}'\pars{t}\,\dd t
\\[5mm]&=-\ln\pars{1 - x}\ln\pars{x} +\ \overbrace{\Li{2}\pars{1}}^{\dsc{\zeta\pars{2}}}\ =\
\color{#66f}{\large\zeta\pars{2} - \ln\pars{x}\ln\pars{1 - x}}
\end{align}
A: Since $$\frac{d}{dy}\operatorname{Li}_2(1-y)=\frac{\ln y}{1-y}$$
Then 
\begin{align}
\operatorname{Li}_2(1-y)|_0^x&=\int_0^x\frac{\ln y}{1-y}\ dy \quad \text{apply integration by parts}\\
\operatorname{Li}_2(1-x)-\zeta(2)&=-\ln(1-y)\ln y|_0^x+\int_0^x\frac{\ln(1-y)}{y}\ dy\\
&=-\ln(1-x)\ln x-\operatorname{Li}_2(y)|_0^x\\
&=-\ln(1-x)\ln x-\operatorname{Li}_2(x)
\end{align}
