Proof of Dilogarithm Identity

Question

Wikipedia states that

$$\operatorname{Li}_2\left(z\right)+\operatorname{Li}_2\left(\frac{1}{z}\right)=-\frac{\pi ^2}{6}-\frac{1}{2}\left[\ln\left(-z\right)\right]^2 \\z\notin \left[0;1\right[$$

I tried to verify this identity by plotting it on matlab. Source code: syms x; m=polylog(2,x)+polylog(2,1/x)+pi^2/6+1/2*(log(-x)^2); ezplot(m, [0,10]); Main problem, and you can test it yourself, the plot is not as i expected it to be. Which, to me, can lead to three conclusions:

1. I'm wrong. There's something i'm missing.
2. Matlab's polylog is not precise enough (i've actually experienced the opposite in previous works on polylog so this should not be the case)
3. The identity is not true.

Ignore the fact that matlab's plot includes z for which the identity has no meaning because the error is actually not localized in such interval. So if anyone could actually give me further information about this or just prove the identity it would be great. Thank you in advance.

Answer 1

The question is to prove $$\operatorname{Li}_2\left(x\right)+\operatorname{Li}_2\left(\frac{1}{x}\right)=-\frac{\pi ^2}{6}-\frac{1}{2}\left[\ln\left(-x\right)\right]^2 $$ and since $\ln(-x)=\ln x+i\pi$

then the question can be written

$$\operatorname{Li}_2\left(x\right)+\operatorname{Li}_2\left(\frac{1}{x}\right)=i\pi-\frac12\ln^2x+2\zeta(2)$$

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Solution:

Since $$\frac{d}{dy}\operatorname{Li}_2(1/y)=\frac{\ln(1-1/y)}{y}=\frac{\ln(y-1)}{y}-\frac{\ln y}{y}$$

Then $$\operatorname{Li}_2(1/y)|_1^x=\underbrace{\int_1^x\frac{\ln(y-1)}{y}\ dy}_{IBP}-\int_1^x\frac{\ln y}{y}\ dy$$ \begin{align}\operatorname{Li}_2(1/x)-\zeta(2)&=\ln y\ln(y-1)|_1^x+\int_1^x\frac{ln y}{1-y}\ dy-\frac12\ln^2y|_1^x\\ &=\ln x\ln(x-1)+\operatorname{Li}_2(1-y)|_1^x-\frac12\ln^2x\\ &=\ln x\ln(x-1)+\operatorname{Li}_2(1-x)-\frac12\ln^2x\\ &=\ln x(\ln(1-x)+i\pi)+\zeta(2)-\ln x\ln(1-x)-\operatorname{Li}_2(x)-\frac12\ln^2x\\ &=i\pi\ln x-\operatorname{Li}_2(x)-\frac12\ln^2x+\zeta(2) \end{align}

Reagganging the terms gives

$$\operatorname{Li}_2(1/x)+\operatorname{Li}_2(x)=i\pi\ln x-\frac12\ln^2x+2\zeta(2)$$

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Note that we used the reflection formula $\ \operatorname{Li}_2(1-x)=\zeta(2)-\ln x\ln(1-x)+\operatorname{Li}_2(x)$ ( see here).

Answer 2

Differentiation Method

Let $$ f(x)=\operatorname{Li_2}(z)+ \operatorname{Li_2}\left(\frac{1}{z}\right) $$ Differentiating both sides w.r.t $z$ yields $$ \begin{aligned} f^{\prime}(z)= & \frac{-\ln (1-z)}{z}+\frac{\ln \left(\frac{z-1}{z}\right)}{z} \\ & =\frac{\ln \left(\frac{z-1}{z} \cdot \frac{1}{1-z}\right)}{z} \\ & =\frac{-\ln (-z)}{z} \end{aligned} $$ Integrating back gives $$ \begin{aligned} f(x) & =-\int \frac{\ln (-z)}{z} d z \\ & =-\frac{1}{2} \ln ^2(-z)+C \end{aligned} $$ Putting $z=1$ gives $$ \begin{aligned} 2 L_{i_2}(1) & =-\frac{1}{2} \ln ^2(-1)+C \\ \Rightarrow \quad \frac{\pi^2}{3} & =\frac{\pi^2}{2}+C\\ \Rightarrow \qquad C &=-\frac{\pi^2}{6} \end{aligned} $$ Now we can conclude that $$ \boxed{\operatorname{Li_2}(z)+ \operatorname{Li_2}\left(\frac{1}{z}\right)=-\frac{\pi^2}{6}-\frac{1}{2}[\ln (-z)]^2} $$