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I experimentally discovered the following conjectures: $$\Re\Big[1800\operatorname{Li}_3(i\,\phi)-24\operatorname{Li}_3\left(i\,\phi^5\right)\Big]\stackrel{\color{gray}?}=100\ln^3\phi-47\,\pi^2\ln\phi-150\,\zeta(3),\tag1$$ $$\Im\Big[720\operatorname{Li}_3(i\,\phi)-320\operatorname{Li}_3\left(i\,\phi^3\right)-48\operatorname{Li}_3\left(i\,\phi^5\right)\Big]\stackrel{\color{gray}?}=9\,\pi^3-780\,\pi\ln^2\phi,\tag2$$ where $\phi=\frac{1+\sqrt5}2$ is the golden ratio and $\operatorname{Li}_3(z)$ is the trilogarithm. They check numerically with at least $20000$ decimal digits. It appears that Maple and Mathematica know nothing about these identities.

Are these known identities? How can we prove them?


Update: It also appears that $$\begin{align}&\Re\operatorname{Li}_3(i\,\phi)\stackrel{\color{gray}?}=\frac1{32}\operatorname{Li}_3\left(\phi^{-4}\right)+\frac3{16}\operatorname{Li}_3\left(\phi^{-2}\right)-\frac38\ln^3\phi-\frac14\zeta(3)\\ \,\\ &\Re\operatorname{Li}_3\left(i\,\phi^3\right)\stackrel{\color{gray}?}=\frac9{32}\operatorname{Li}_3\left(\phi^{-4}\right)+\frac12\operatorname{Li}_3\left(\phi^{-3}\right)+\frac38\operatorname{Li}_3\left(\phi^{-2}\right)-\frac38\operatorname{Li}_3\left(\phi^{-1}\right)-\frac{43}8\ln^3\phi-\frac{15}{32}\zeta(3)\\ \,\\ &\Re\operatorname{Li}_3\left(i\,\phi^5\right)\stackrel{\color{gray}?}=\frac{75}{32}\operatorname{Li}_3\left(\phi^{-4}\right)-\frac58\operatorname{Li}_3\left(\phi^{-2}\right)-\frac{45}2\ln^3\phi-\frac34\zeta(3)\end{align}$$ These together with a known value for $\operatorname{Li}_3\left(\phi^{-2}\right)$ imply $(1)$.

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I'll try to prove Tito Piezas III's (+1) neat conjecture : $$\tag{1}32\;\Re\operatorname{Li}_3\left(i\,\phi^k\right)\stackrel{\color{blue}?}=\operatorname{Li}_3\left(\phi^{-4k}\right)-4\operatorname{Li}_3\left(\phi^{-2k}\right)+10\,k\operatorname{Li}_3\left(\phi^{-2}\right)-\frac{4k^3+5k}{6}\,\ln^3(\phi^2)-8k\,\zeta(3)$$ The classical identity $\;\displaystyle\frac {\operatorname{Li}_3(x^2)}4=\operatorname{Li}_3(x)+\operatorname{Li}_3(-x)\;$ (c.f. Lewin $1981$ "Polylogaritms and associated functions" p.$154$) allows to rewrite the two first terms at the right as : $$\tag{2}32\;\Re\operatorname{Li}_3\left(i\,\phi^k\right)\stackrel{\color{blue}?}=4\operatorname{Li}_3\left(-\phi^{-2k}\right)+10\,k\operatorname{Li}_3\left(\phi^{-2}\right)-\frac{4k^3+5k}{6}\,\ln^3(\phi^2)-8k\,\zeta(3)$$ and gives further : $\;\dfrac {\operatorname{Li}_3\left(-\phi^{2k}\right)}4=\operatorname{Li}_3\left(i\,\phi^k\right)+\operatorname{Li}_3\left(-i\,\phi^k\right)=2\Re\operatorname{Li}_3\left(i\,\phi^k\right)\;$ that is : $$\tag{3}4\operatorname{Li}_3\left(-\phi^{2k}\right)\stackrel{\color{blue}?}=4\operatorname{Li}_3\left(-\phi^{-2k}\right)+10\,k\operatorname{Li}_3\left(\phi^{-2}\right)-\frac{4k^3+5k}{6}\,\ln^3(\phi^2)-8k\,\zeta(3)$$

Another usual identity is $\;\displaystyle \operatorname{Li}_3(-x)-\operatorname{Li}_3(-1/x)=-\zeta(2)\log(x)-\frac{\ln^3(x)}6\;$ (same Lewin page)
applied to $\,x:=\phi^{2k}\,$ $(3)$ becomes : \begin{align} -8k\zeta(2)\ln(\phi)-\frac{4k^3\ln^3(\phi^{2})}6\stackrel{\color{blue}?}=&10\,k\operatorname{Li}_3\left(\phi^{-2}\right)-\frac{4k^3+5k}{6}\,\ln^3(\phi^2)-8k\,\zeta(3)\\ -8k\zeta(2)\ln(\phi)+\frac{5k\ln^3(\phi^{2})}6+8k\,\zeta(3)\stackrel{\color{blue}?}=&10\,k\operatorname{Li}_3\left(\phi^{-2}\right)\\ \tag{4}\operatorname{Li}_3\left(\phi^{-2}\right)=&\frac 45(\zeta(3)-\zeta(2)\ln(\phi))+\frac 23{\ln^3(\phi)}\\ \end{align} This last identity is proved in Lewin's $1991$ book (page $2$ with $\,\rho=\dfrac 1{\phi}\,$) and may be compared with alpha's evaluation knowing that $\,\operatorname{csch}^{-1}(2)=\ln(\phi)$.

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  • $\begingroup$ Ah, great derivation. Bravo! $\endgroup$ – Tito Piezas III Oct 6 '16 at 2:07
  • $\begingroup$ By the way, note that the case $k=3,6$ turns out to be expressible as polylogarithmic sums in two ways. I've tried $k=9$ but couldn't find a second relation. $\endgroup$ – Tito Piezas III Oct 6 '16 at 4:45
  • $\begingroup$ Thanks @Tito of course your general formula helped much! It was also an indication that Vladimir's results came from general polylogarithmic identities rather than sparse polylogarithmic ladders (valid for some consecutive integers $k$ only). Excellent continuation (I have no idea for $k=9$ sorry) anyway, $\endgroup$ – Raymond Manzoni Oct 6 '16 at 7:14
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(Too long for a comment.)

You may have missed the small even powers $k=2,4,6$ $$\begin{align} &32\,\Re\operatorname{Li}_3\left(i\,\phi^2\right)\stackrel{\color{blue}?}=\operatorname{Li}_3\left(\phi^{-8}\right)-4\operatorname{Li}_3\left(\phi^{-4}\right)+20\operatorname{Li}_3\left(\phi^{-2}\right)-7\ln^3(\phi^2)-16\,\zeta(3)\\ \,\\ &32\,\Re\operatorname{Li}_3\left(i\,\phi^4\right)\stackrel{\color{blue}?}=\operatorname{Li}_3\left(\phi^{-16}\right)-4\operatorname{Li}_3\left(\phi^{-8}\right)+40\operatorname{Li}_3\left(\phi^{-2}\right)-46\ln^3(\phi^2)-32\,\zeta(3)\\ \,\\ &32\,\Re\operatorname{Li}_3\left(i\,\phi^6\right)\stackrel{\color{blue}?}=18\operatorname{Li}_3\left(\phi^{-8}\right)-72\operatorname{Li}_3\left(\phi^{-4}\right)+64\operatorname{Li}_3\left(\phi^{-3}\right)+90\operatorname{Li}_3\left(\phi^{-2}\right)-48\operatorname{Li}_3\left(\phi^{-1}\right)-1196\ln^3\phi-35\,\zeta(3) \end{align}$$ More generally, for any real number $k$ it seems, $$32\,\Re\operatorname{Li}_3\left(i\,\phi^k\right)\stackrel{\color{blue}?}=\operatorname{Li}_3\left(\phi^{-4k}\right)-4\operatorname{Li}_3\left(\phi^{-2k}\right)+10\,k\operatorname{Li}_3\left(\phi^{-2}\right)-\frac{4k^3+5k}{6}\,\ln^3(\phi^2)-8k\,\zeta(3)$$ though I have no proof yet.

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