# 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?

• Note: this is called the dilog reflection formula. The duplication formula is something else. – David H Dec 7 '14 at 18:31

## 2 Answers


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

• Excellent, simple, and very useful. – Ron Gordon Dec 7 '14 at 18:50