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Recall that $$\mathrm{Li}_2(x):=\sum_{n=1}^\infty\frac{x^n}{n^2}.$$ I have found the following identity: \begin{equation}\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}\end{equation} This can be easily checked numerically.

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

Your comments are welcome!

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    $\begingroup$ 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. $\endgroup$
    – Alapan Das
    Jul 6, 2023 at 9:09
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    $\begingroup$ @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. $\endgroup$ Jul 6, 2023 at 18:35

3 Answers 3

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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 $z_0$.

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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.

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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 $$ \operatorname{Li}_2(s) + \operatorname{Li}_2(t) + \operatorname{Li}_2(u) + \operatorname{Li}_2(v) = \bbox[lightyellow]{\arctan^2\frac{\sqrt 7}5-\frac 14\log^2 2} \ . $$

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