Several conjectured identities for polylogarithms

Question

I asked a question at M.SE a couple of years ago about polylogarithms $\!^{[1]}$$\!^{[2]}$$\!^{[3]}$$\!^{[4]}$ where I conjectured $$720\,\operatorname{Li}_4\!\left(\tfrac12\right)-2160\,\operatorname{Li}_4\!\left(\tfrac13\right)+2160\,\operatorname{Li}_4\!\left(\tfrac23\right)+270\,\operatorname{Li}_4\!\left(\tfrac14\right)+540\,\operatorname{Li}_4\!\left(\tfrac34\right)+135\,\operatorname{Li}_4\!\left(\tfrac19\right)\\ =19\pi^4+30\left(\pi^2-\beta^2\right)\left(10\alpha^2-12\alpha\beta+3\beta^2\right)-30\,\alpha^2\left(19\alpha^2-24\alpha\beta+8\beta^2\right)\!,\\ \text{where $\alpha=\ln2,\,\,\beta=\ln3$.}$$ I put a bounty on it recently, but it still remains unanswered. So, I decided to re-post it here. Also, I recently discovered several similar conjectures for $\operatorname{Li}_5(z)$ involving powers of the golden ratio $\phi=\frac{1+\sqrt5}2$: $$4455\,\operatorname{Li}_5\left(\phi^{-2}\right)-1215\,\operatorname{Li}_5\left(\phi^{-4}\right)-360\,\operatorname{Li}_5\left(\phi^{-6}\right)+15\,\operatorname{Li}_5\left(\phi^{-12}\right)\\ =3015\,\zeta(5)-702\ln^5\phi+180\,\pi^2\ln^3\phi-38\,\pi^4\ln\phi$$

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$$45000\,\operatorname{Li}_5\left(\phi^{-2}\right)-16875\,\operatorname{Li}_5\left(\phi^{-4}\right)-144\,\operatorname{Li}_5\left(\phi^{-10}\right)+9\,\operatorname{Li}_5\left(\phi^{-20}\right)\\ =28944\,\zeta(5)-6000\ln^5\phi+1600\,\pi^2\ln^3\phi-356\,\pi^4\ln\phi$$

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$$15660\,\operatorname{Li}_5\left(\phi^{-2}\right)-19440\,\operatorname{Li}_5\left(\phi^{-4}\right)+7680\,\operatorname{Li}_5\left(\phi^{-6}\right)+2430\,\operatorname{Li}_5\left(\phi^{-8}\right)-15\,\operatorname{Li}_5\left(\phi^{-24}\right)=5025\,\zeta(5)+2088\ln^5\phi-240\,\pi^2\ln^3\phi-28\,\pi^4\ln\phi$$

How can we prove these conjectures? Are there other similar identities of this kind?

Answer 1

(Updated answer):

Upon further research, it turns out your three equations involving $\phi$ are special cases of three polylogarithm ladders of index $12,\,20,\,24$ that can be found in "The Polylogarithm in Algebraic Number Fields" (1985) by M. Abouzahra and L. Levin. Let $\color{red}{\rho = \frac1{\phi}}$.

I. Ladder of index 12

$$A = \frac{\text{Li}_n(\rho^{12})}{12^{n-1}}- \frac32\frac{\text{Li}_n(\rho^6)}{6^{n-1}}- \frac{\text{Li}_n(\rho^4)}{4^{n-1}}+ \frac{11}{48}\frac{\text{Li}_n(\rho^2)}{2^{n-1}}$$

$$B = -\frac{13}{48}\frac{\ln^n\rho}{n!}+\frac{\zeta(2)}{48}\frac{\ln^{n-2}\rho}{(n-2)!}-\frac{19\,\zeta(4)}{1728}\frac{\ln^{n-4}\rho}{(n-4)!} $$

then for $n=1,2,3,4,5$,

$$A+B= \frac{67\,\zeta(5)}{6912}\frac{\ln^{n-5}\rho}{(n-5)!} $$

with $A+B = 0$ for $n<5$. If $n=5$, then we recover your experimentally found equation. Incidentally, a consequence (or is it the other way round?) of this ladder is the high-degree cyclotomic equation satisfied by the golden ratio,

$$(1 - \phi^{-12}) = (1 - \phi^{-6})^{3/2}(1 - \phi^{-4})(1 - \phi^{-2})^{-11/48}\,\phi^{13/48}$$

Notice that the powers of $(1 - \phi^{k})$ are coefficients in $A,B$.

II. Ladder of index 20

Similar to the ladder above. For $n=5$ yields the OP's second formula but, with modifications, can be extended to $n=7$.

III. Ladder of index 24

Likewise. If modified, can be extended even to $n=9$. This also explains the index $24$ formulas below for $n=6,7$ that I found empirically.

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(Old answer): As to your question if there are similar identities, it seems there is a whole family,

$$5\,\text{Li}_2(\phi^{-1}) =-5\ln^2\phi+\,3\,\color{blue}{\zeta(2)}$$

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$$15\,\text{Li}_3(\phi^{-2})=10\ln^3\phi\,-2\pi^2\ln\phi+12\,\color{blue}{\zeta(3)}$$

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$$144\text{Li}_4(\phi^{-1})-9\text{Li}_4(\phi^{-2})-64\text{Li}_4(\phi^{-3})+4\text{Li}_4(\phi^{-6})=27\ln^4\phi-6\pi^2\ln^2\phi+80\,\color{blue}{\zeta(4)}$$

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$$4455\,\text{Li}_5\left(\phi^{-2}\right)-1215\,\text{Li}_5\left(\phi^{-4}\right)-360\,\text{Li}_5\left(\phi^{-6}\right)+15\,\text{Li}_5\left(\phi^{-12}\right)\\ =-702\ln^5\phi+180\,\pi^2\ln^3\phi-38\,\pi^4\ln\phi+3015\,\color{blue}{\zeta(5)}$$

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$$m_1\text{Li}_6(\phi^{-2})+m_2\text{Li}_6(\phi^{-4})+m_3\text{Li}_6(\phi^{-6})+m_4\text{Li}_6(\phi^{-8})+m_5\text{Li}_6(\phi^{-12})+m_6\text{Li}_6(\phi^{-24})\\ =m_7\ln^6\phi+m_8\pi^2\ln^4\phi+m_9\pi^4\ln^2\phi +32155\,\color{blue}{\zeta(6)}$$

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$$n_1\text{Li}_7(\phi^{-2})+n_2\text{Li}_7(\phi^{-4})+n_3\text{Li}_7(\phi^{-6})+n_4\text{Li}_7(\phi^{-8})+n_5\text{Li}_7(\phi^{-12})+n_6\text{Li}_7(\phi^{-24})\\ =n_7\ln^7\phi+n_8\pi^2\ln^5\phi+n_9\pi^4\ln^3\phi+n_{10}\pi^6\ln\phi +44689995\,\color{blue}{\zeta(7)}$$

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and where the integers $m_i$ and $n_i$ were found using Integer Relations but are too tedious to write here.

Does this go on forever? Note the exponents of $\phi$ in $\text{Li}_k(\phi^{-n})$ match up for $k=4$ as $c=1\times6=2\times3$ then for $k=5$ as $c=2\times12=6\times4$ then for $k=6,7$ as $c=2\times24=4\times12 = 6\times8$.

P.S. Based on Abouzahra and Levin's paper cited in the update, I guess this can't be brought to arbitrarily high $k$.