Proof of dilogarithm reflection formula $\zeta(2)-\log(x)\log(1-x)=\operatorname{Li}_2(x)+\operatorname{Li}_2(1-x)$

Question

How to prove

$$\zeta(2)-\log(x)\log(1-x)=\operatorname{Li}_2(x)+\operatorname{Li}_2(1-x)$$ I havent started, any hints?

Answer 1

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: $$\sum_{n\geq 1}\frac{1}{2^n n^2}=\operatorname{Li}_2\left(\frac{1}{2}\right)=\frac{1}{2}\left(f\left(\frac{1}{2}\right)-\log^2 2\right)=\frac{\pi^2}{12}-\frac{\log^2 2}{2}.$$

Answer 2

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \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}

Answer 3

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}