Dilogarithm inversion formula: $ \text{Li}_2(z) + \text{Li}_2(1/z) = -\zeta(2) - \log^2(-z)/2$

Question

I have been chugging through some proofs regarding the dilogarithm, also known as [Spencer's function][1].

\begin{alignat}{2} & \operatorname{Li}_2(z) + \operatorname{Li}_2(-z) = \frac{1}{2} \operatorname{Li}_2(z^2) && \text{(Double Identity)} \tag{1} \\ & \operatorname{Li}_2(z) + \operatorname{Li}_2(1-z) = \frac{\pi^2}{6} - \log z \log (1 - z) \ && \text{(Eulers reflection formula)} \tag{2} \\ & \operatorname{Li}_2(-z) + \operatorname{Li}_2\left( \frac{z}{1+z} \right) = -\frac{1}{2} \log^2(z+1) && \text{(Landen's Identity)} \tag{3} \\ & \operatorname{Li}_2(z) + \operatorname{Li}_2\left( \frac{1}{z} \right) = - \frac{\pi^2}{6} - \frac{1}{2}\log^2(-z) \ \ && \text{(Inversions formula)} \tag{4} \end{alignat} With the basis of $\text{Abel's Identity}$ (see Proving Abel's identity for the Dilogarithm.). I have been able to prove all the the identities except the last. (My proof for $(3)$, was somewhat convoluted, so hints there would be appreciated..). I have yet to prove $(4)$, although I have given it two attempts below

Attempt 1

In the Abel Identity let $x=y=1-z$ and divide by $2$ to obtain $$ \frac{1}{2}\log(-z)^2 = \operatorname{Li}_2\left( -\frac{1+z}{z}\right) - \frac{1}{2} \operatorname{Li}_2\left( \left[ \frac{1+z}{z} \right]^2 \right) - \operatorname{Li}_2(1+z) $$ By now using $(1)$ with $z = [1+1/z]^2$ and inserting it into the equation above I obtain $$ \frac{1}{2}\log(-z)^2 = - \left[ \operatorname{Li}_2\left( \frac{1}{z} + 1\right)- \operatorname{Li}_2(1 + z)\right] $$ and from here I am stuck. It is close to what I want but I can not find any way to transform the right hand side.

Attempt 2

From chat the suggestion was to instead look at the integral definition, this gives \begin{align*} \operatorname{Li}_2(z) + \operatorname{Li}_2\left( \frac{1}{z}\right) & = - \int_0^z \frac{\log(1-t)}{t}\,\mathrm{d}t - \int_0^{1/z} \frac{\log(1-t)}{t}\,\mathrm{d}t \\ & = - \int\limits_0^1 {\frac{{\log \left( {z - t} \right) + \log \left( {1 - zt} \right) - \log z}}{t}dt} \\ & = -\zeta(2) + \int_0^1 \frac{\log z - \log(1-zt)}{t}\,\mathrm{d}t \end{align*} The last step used that $$ \int_0^1 \frac{\log(1-t)}{t}\,\mathrm{d}t = \int_0^1 -\frac{1}{t} \sum_{n=1}^\infty \frac{t^n}{n} \,\mathrm{d}t = \sum_{n=1}^\infty \frac{1}{n^2} = \zeta(2) $$ And this is where I stopped. I think this argument can be finished by series expansion, but I got lost in the algebra. If possible I would very much like to prove this identity from Abel's Identity and the three equations stated above. Any hint or solutions is much appreciated as always =) [1]: http://en.wikipedia.org/wiki/Spence%27s_function

Answer 1

The classical proof (Lewin) of the 'inversion formula' $(4)$ is simply to use : $$\operatorname{Li}_2(z)=-\int\frac{\log(1-z)}z\,dz$$ and set $z:=-\frac 1x\,$ to get the derivative : $$\frac d{dx}\operatorname{Li}_2\left(-\frac 1x\right)=-\frac{\log\left(1+\frac 1x\right)}{-\frac 1x}\frac 1{x^2}=\frac{\log\left(1+\frac 1x\right)}x=\frac{\log(1+x)-\log(x)}x$$ Integrating this again gives (with the constant determined at $x=1$) : $$\operatorname{Li}_2\left(-\frac 1x\right)+\operatorname{Li}_2(-x)=2\,\operatorname{Li}_2(-1)-\frac12\log^2(x)$$

($\,x:=-z\,$ gives your relation of course)

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Concerning the 'Landen identity' $(3)$ the same method applies : \begin{align} \frac d{dx}\operatorname{Li}_2\left(\frac x{1+x}\right)&=-\frac{\log\left(1-\frac x{1+x}\right)}{\frac x{1+x}}\frac 1{(x+1)^2}=-\frac{\log\left(\frac 1{1+x}\right)}{x(x+1)}\\ &=\log(1+x)\left(\frac 1x-\frac 1{1+x}\right)\\ \end{align}

so that integrating again gives (with constant determined by $x=0$) : $$\operatorname{Li}_2\left(\frac x{1+x}\right)=-\operatorname{Li}_2(-x)-\frac 12\log^2(1+x)$$

Answer 2

For the proof of $\operatorname{Li}_2(x) +\operatorname{Li}_2(-x) =\frac{1}{2}\operatorname{Li}_2(x^2)$ is by rewriting the dilogs on the right-hand side as sums. We have $$\sum_{n=1}^{\infty}\left(\frac{x^n}{n^2} +\frac{(-x)^n}{n^2}\right).$$ Then notice that the odd terms keep cancelling out. Leaving only the even terms so rewrite sum $$\sum_{n=1}^{\infty}\left(\frac{x^{2n}}{4n^2} +\frac{(-x)^{2n}}{4n^2}\right)=\frac{2}{4}\sum_{n=1}^{\infty}\frac{x^{2n}}{n^2},$$ which is recognized as $\frac{1}{2}\operatorname{Li}_2(x^2)$.