# An identity involving polylogarithms

Recall that $$\mathrm{Li}_2(x):=\sum_{n=1}^\infty\frac{x^n}{n^2}.$$ I have found the following identity: \begin{aligned}&\mathrm{Li}_2\left(\frac{-1-\sqrt{-7}}4\right)+\mathrm{Li}_2\left(\frac{-1+\sqrt{-7}}4\right) \\&+\mathrm{Li}_2\left(\frac{3-\sqrt{-7}}8\right)+\mathrm{Li}_2\left(\frac{3+\sqrt{-7}}8\right) \\&\qquad=\arctan^2\frac{\sqrt7}5-\frac{\log^22}4. \end{aligned}\label{1}\tag{1} This can be easily checked numerically.

Question. How to prove the identity \eqref{1}?

• Take $z_0=\frac{5-i\sqrt{7}}{8}$, then the given expression is $2 \mathcal{Re}[\mathrm{Li}_2(1-z_0)+\mathrm{Li}_2(1-\frac{1}{z_0})]$. The explicit formula for this is on wikipedia page en.m.wikipedia.org/wiki/Spence%27s_function. Using this we get the solution. Jul 6, 2023 at 9:09
• @AlapanDas Incidentally, we have the identities $\frac{5-\sqrt{-7}}8 = 2\left(\frac{-1+\sqrt{-7}}4\right)^3$, so it is almost a cube, while $\frac{3+\sqrt{-7}}8 = -\left(\frac{-1+\sqrt{-7}}4\right)^2$ is a square. Jul 6, 2023 at 18:35

I am elaborating my comment here: The expression given in the question can be written as $$2\Re [\mathrm{Li}_2(\alpha) + \mathrm{Li}_2(\beta)]$$ where $$\alpha=\frac{-1+i\sqrt{7}}{8}, \beta=\frac{3-i\sqrt{7}}{8}$$.
Now, take $$z_0=\frac{5+i\sqrt{7}}{8}$$. So, we have $$\alpha=1-\frac{1}{z_0}$$ and $$\beta=1-z_0$$. Having written this in a known form, we use the identity $$\mathrm{Li}_2(1-z) + \mathrm{Li}_2(1-\frac{1}{z})=-\frac{(\ln z)^2}{2}$$ which is mentioned in Wikipedia.
Hence we obtain the expression in question to be $$-\Re [(\ln (\frac{5-i\sqrt{7}}{8}))^2]=\theta^2-\ln(\sqrt{2})^2$$, where $$\theta=-\arctan(\frac{\sqrt{7}}{5})$$ is the argument of $$z_0$$.
A quick remark on the linked identity $$\mathrm{Li}_2(1-z) + \mathrm{Li}_2(1-\frac{1}{z})=-\frac12 \log^2 z$$. Since from the definition $$\displaystyle \mathrm{Li}_2(1-z)=\int_1^z\frac{\log t}{1-t}dt$$, it can be written $$\int_1^z\frac{\log t}{1-t}dt+\int_1^\frac1z \frac{\log t}{1-t}dt+\frac12 \log^2 z=0,$$ and can be checked immediately by derivation.
My way to go is to use the functional identities w.r.t. to the transformations $$z\to 1/z$$ and $$z\to 1-z$$, usually isolated in the world of the $$K_3$$-symbols. These identities are (i am citing Zagier, §2, page 8): \begin{aligned} \operatorname{Li}_2(z) + \operatorname{Li}_2\left(\frac 1z\right) &= -\frac{\pi^2}6-\frac 12\log^2(-z)\ , \\ \operatorname{Li}_2(z) + \operatorname{Li}_2(1-z) &= +\frac{\pi^2}6-\log (z)\log(1-z)\ . \end{aligned} Let $$a$$ be $$\sqrt{-7}$$ for short, and the numbers $$s,t;u,v$$ be the numbers that appear in the OP formula $$(1)$$, $$s=-\frac 14(1+a)\ ,\ t=-\frac 14(1-a)\ ;\ u=\frac 18(3-a)\ ,\ v=\frac 18(3+a)\ .$$ We use the equality $$1-\frac 1s=\frac 1v\ ,$$ and proceed from now on in a natural manner. Explicit full computation: \begin{aligned} \operatorname{Li}_2(s) + \operatorname{Li}_2(v) &= \operatorname{Li}_2(s) + \operatorname{Li}_2\left(\frac 1s\right) \color{blue}{ - \operatorname{Li}_2\left(\frac 1s\right) - \operatorname{Li}_2\left(\frac 1v\right)} + \operatorname{Li}_2\left(\frac 1v\right) + \operatorname{Li}_2(v) \\ &= -\frac{\pi^2}6-\frac 12\log^2(-s) -{\color{blue}{ \left(\frac{\pi^2}6 -\log\left(\frac 1s\right)\log\left(\frac 1v\right)\right) }} -\frac{\pi^2}6-\frac 12\log^2(-v) \\ &=-\frac{\pi^2}2 -\frac 12(\log (-s) -\underbrace{\log(-v)}_{\log v-i\pi})^2 +( \underbrace{\log s}_{\log(-s)-i\pi} \log(v) - \log(-s) \underbrace{\log(-v)}_{\log v-i\pi}) \\ &= -\frac 12\log^2\frac{-s}v \\ &= -\frac 12\log^2\left(\frac 14(5+a)\right) = -\frac 12\left(\log\sqrt2 +i\arctan\frac{\sqrt 7}5\right)^2 \\ &= \frac 12\arctan^2\frac{\sqrt 7}5-\frac 18\log^2 2 \qquad-\frac i2\log 2\arctan\frac{\sqrt 7}5\ , \\[3mm] \operatorname{Li}_2(t) + \operatorname{Li}_2(u) &= \operatorname{Li}_2(\bar s) + \operatorname{Li}_2(\bar v) \\ &= \frac 12\arctan^2\frac{\sqrt 7}5-\frac 18\log^2 2 \qquad+\frac i2\log 2\arctan\frac{\sqrt 7}5\ , \end{aligned} so the sum to be calculated is $$\operatorname{Li}_2(s) + \operatorname{Li}_2(t) + \operatorname{Li}_2(u) + \operatorname{Li}_2(v) = \bbox[lightyellow]{\arctan^2\frac{\sqrt 7}5-\frac 14\log^2 2} \ .$$