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I am trying to calculate the dilogarithm of the golden ratio and its conjugate $\Phi = \varphi-1$. Eg the solutions of the equation $u^2 - u = 1$. From Wikipdia one has the following

\begin{align*} \operatorname{Li}_2\left( \frac{1 + \sqrt{5}}{2} \right) & = -\int_0^\varphi \frac{\log(1-t)}{t}\,\mathrm{d}t = \phantom{-}\frac{\pi^2}{10} - \log^2\left( \Phi\right) \\ \operatorname{Li}_2\left( \frac{1 - \sqrt{5}}{2} \right) & = -\int_0^\Phi \frac{\log(1-t)}{t}\,\mathrm{d}t = -\frac{\pi^2}{15} - \log^2\left( -\Phi\right) \\ \end{align*}

I am quite certain that these two special values can be shown by combining the identites for the dilogarithm, and forming a system of equations. But I am having some problems obtaining a set of equations only involving $\operatorname{Li}_2(\varphi)$ and $\operatorname{Li}_2(\Phi)$. Can anyone show me how to set up the system of equations from the identites, or perhaps a different path in showing these two values?

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2 Answers 2

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The most curious and important dilogarithm identity is the pentagonal one: \begin{align} \mathrm{Li}_2\left(\frac{x}{1-y}\right)+\mathrm{Li}_2\left(\frac{y}{1-x}\right)-\mathrm{Li}_2(x)-\mathrm{Li}_2(y)-\mathrm{Li}_2\left(\frac{xy}{(1-x)(1-y)}\right)=\\ =\ln(1-x)\ln(1-y).\tag{1} \end{align} Denote $\alpha=\frac{\sqrt{5}-1}{2}$. Then it is very easy to check that \begin{align} \left(\frac{x}{1-y},\frac{y}{1-x},x,y,\frac{xy}{(1-x)(1-y)}\right)_{x=y=1-\alpha}&=\left(\alpha,\alpha,1-\alpha,1-\alpha,1-\alpha\right).\tag{2} \end{align} On the other hand, one has $$\mathrm{Li}_2(x)+\mathrm{Li}_2(1-x)=\frac{\pi^2}{6}-\ln x\ln(1-x).\tag{3}$$ Combining (1), (2) and (3), we can express $\mathrm{Li}_2(\alpha)$ and $\mathrm{Li}_2(1-\alpha)$ in terms of elementary functions. These are basic special values from which one can deduce the others.

Now concerning the values you are interested in. The first of your formulas is meaningless as written, since the dilogarithm has branch cut $[1,\infty)$. The second is almost okay, if you correct the sign in front of $\Phi$ on the right. To obtain the corresponding evaluation, it suffices to use the identity $$\mathrm{Li}_2(x)+\mathrm{Li}_2(-x)-\frac12\mathrm{Li}_2(x^{2})=0\tag{4}$$ with $x=\alpha$ and the fact that $\alpha^2=1-\alpha$.

Addition (for completeness): Denoting $$A=\mathrm{Li}_2(\alpha),\qquad B=\mathrm{Li}_2(1-\alpha),\qquad C=\mathrm{Li}_2(-\alpha),$$ we have from the equations (1)-(4): \begin{align} &2A-3B=\ln^2\alpha,\\ &A+B=\frac{\pi^2}{6}-2\ln^2\alpha,\\ &A+C-\frac12 B=0. \end{align} The solution of this system is given by \begin{align} &A=\frac{\pi^2}{10}-\ln^2\alpha,\\ &B=\frac{\pi^2}{15}-\ln^2\alpha,\\ &C=-\frac{\pi^2}{15}+\frac12\ln^2\alpha. \end{align}

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  • $\begingroup$ Thanks a bunch =) Combining those equations gave me $2h(\alpha) - 3h(1-\alpha) = \log^2\alpha$ and $h(\alpha) + h(1-\alpha) = \frac{\pi^2}{6} - \log(\sqrt{5}-2)$ does this seem correct? Here h h denotes the dilogarithm =) $\endgroup$
    – N3buchadnezzar
    Commented Aug 9, 2013 at 18:59
  • $\begingroup$ @N3buchadnezzar Yes for the 1st, whereas the 2nd will be $h(\alpha)+h(1-\alpha) = \pi^2/6 -\ln\alpha \ln(1- \alpha) =\pi^2/6 -2\ln^2\alpha $. This gives $h(\alpha)$ and $h(1- \alpha)$, and allows to subsequently find $h(-\alpha)$ from $h(\alpha)+h(-\alpha)-\frac12 h(1-\alpha)=0$. $\endgroup$
    – Start wearing purple
    Commented Aug 9, 2013 at 19:11
  • $\begingroup$ So the wikipedia article en.wikipedia.org/wiki/Spence%27s_function is wrong in saying that $\operatorname{Li}_2(\varphi)$ converges? Or does they implicitly take the P.V of the integral? $\endgroup$
    – N3buchadnezzar
    Commented Aug 9, 2013 at 21:49
  • $\begingroup$ @N3buchadnezzar I have no idea what they meant. The dilogarithms on two sides of the branch cut $[1,\infty)$ have opposite imaginary parts but Wiki expression for $\mathrm{Li}_2(1+\alpha)$ does not coincide with the common real part of these dilogarithms. $\endgroup$
    – Start wearing purple
    Commented Aug 9, 2013 at 22:08
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This solution is very similar to @Start wearing purple's solution above.

Using the following identities (check Eq $3$,$4$ and $5$).

$$\text{Li}_2(1-x)+\text{Li}_2\left(\frac{x-1}{x}\right)=-\frac12\ln^2\left(\frac{1}{x}\right)\tag1$$

$$\text{Li}_2(x)+\text{Li}_2(-x)-\frac12\text{Li}_2(x^2)=0\tag2$$

$$\text{Li}_2(x)+\text{Li}_2(1-x)=\zeta(2)-\ln(x)\ln(1-x)\tag3$$

If we set $x=\frac{\sqrt{5}-1}{2}$ we notice that

$$1-x=\frac{3-\sqrt{5}}{2}$$

$$\frac{x-1}{x}=\frac{1-\sqrt{5}}{2}$$

$$\frac1x=\frac{1+\sqrt{5}}{2}$$

$$x^2=\frac{3-\sqrt{5}}{2}$$

$$\ln(x)\ln(1-x)=2\ln^2\left(\frac{1+\sqrt{5}}{2}\right)$$

Plugging these values in $(1)$, $(2)$ and $(3)$ we get

$$\color{blue}{\text{Li}_2\left(\frac{3-\sqrt{5}}{2}\right)}+\color{red}{\text{Li}_2\left(\frac{1-\sqrt{5}}{2}\right)}=-\frac12\ln^2\left(\frac{1+\sqrt{5}}{2}\right)$$

$$\text{Li}_2\left(\frac{\sqrt{5}-1}{2}\right)+\color{red}{\text{Li}_2\left(\frac{1-\sqrt{5}}{2}\right)}-\frac12\color{blue}{\text{Li}_2\left(\frac{3-\sqrt{5}}{2}\right)}=0$$

$$\text{Li}_2\left(\frac{\sqrt{5}-1}{2}\right)+\color{blue}{\text{Li}_2\left(\frac{3-\sqrt{5}}{2}\right)}=\zeta(2)-2\ln^2\left(\frac{1+\sqrt{5}}{2}\right)$$

Solving this system of equations, we have

$$\color{blue}{\text{Li}_2\left(\frac{3-\sqrt{5}}{2}\right)}=\frac{\pi^2}{15}-\ln^2\left(\frac{1+\sqrt{5}}{2}\right)$$

$$\color{red}{\text{Li}_2\left(\frac{1-\sqrt{5}}{2}\right)}=-\frac{\pi^2}{15}+\frac12\ln^2\left(\frac{1+\sqrt{5}}{2}\right)$$

$$\text{Li}_2\left(\frac{\sqrt{5}-1}{2}\right)=\frac{\pi^2}{10}-\ln^2\left(\frac{1+\sqrt{5}}{2}\right)$$

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