I experimentally discovered (using PSLQ) the following conjectured tetralogarithm identity: $$720 \,\text{Li}_4\!\left(\tfrac{1}{2}\right)-2160 \,\text{Li}_4\!\left(\tfrac{1}{3}\right)+2160 \,\text{Li}_4\!\left(\tfrac{2}{3}\right)+270 \,\text{Li}_4\!\left(\tfrac{1}{4}\right)+540 \,\text{Li}_4\!\left(\tfrac{3}{4}\right)+135 \,\text{Li}_4\!\left(\tfrac{1}{9}\right)\\ \stackrel?=19 \pi ^4+30 \left(\pi ^2-\beta ^2\right) \left(10 \alpha ^2-12 \alpha \beta +3 \beta ^2\right)-30 \,\alpha ^2 \left(19 \alpha ^2-24 \alpha \beta +8 \beta ^2\right)\!,\\ \text{where $\alpha=\ln 2, \,\, \beta=\ln 3$.}$$ Numeric computations show that the absolute value of the difference between the left and right sides is smaller than $10^{-10^5}$, so I believe the identity must hold exactly. Moreover, this is the simplest vanishing linear combination with integer coefficients of elementary terms and tetralogarithm values at rational points in the interval $(0,1)$ that I found.

How can we prove it?

  • $\begingroup$ "the simplest identity of this form for tetralogarithms" What does it mean? $\endgroup$ – user153012 Jul 29 '15 at 18:29
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    $\begingroup$ @user153012 Well-order rationals in the open interval $(0,1)$ using lexicographic order $\mathcal L$ on $\langle denominator, numerator\rangle$ pairs. Assign to each identity the maximal (according to $\mathcal L$) argument among its tetralogarithm terms. Select identities that are assigned the least (according to $\mathcal L$) rational number, and among them select identities with the least number of tetralogarithm terms. In case of a tie, select identities whose second largest rational argument is the least, then third largest, etc. The remaining identity is the simplest. $\endgroup$ – Vladimir Reshetnikov Jul 29 '15 at 18:51
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    $\begingroup$ A known identitiy of your type of tetralogarithms, which lead us to two alternate forms of your stunning conjecture:$$\operatorname{Li}_4(1/64)-8 \operatorname{Li}_4(1/8)-54 \operatorname{Li}_4(1/4)+96 \operatorname{Li}_4(1/2)\\=5\ln^4 2-2\pi^2\ln^2 2 + 4\pi^4/9.$$ $\endgroup$ – user153012 Jul 29 '15 at 19:09
  • $\begingroup$ Just in case, here are corresponding Mathematica and Maple expressions: goo.gl/zR8axn goo.gl/h2Noxk $\endgroup$ – Vladimir Reshetnikov Jan 28 '17 at 0:36
  • $\begingroup$ Lol, this looks like one of Ramanujan's unbelievably random identities on $\pi$ or modular forms. $\endgroup$ – user335907 Feb 2 '17 at 18:19

How can we prove it?

This might be a viable method:

From "Structural Properties of Polylogarithms, Leonard Lewin, p.30 sec.3, Polylogarithmic Ladders" we have


However, due to restrictions on viewing I can't see the conditions required for $x,y,\eta,\xi$, checking in Mathematica it's not convinced its true for most combinations.

If we can understand that identity, we can probably pick a combination of $\{x,y,\eta,\xi\}\in\{\pm 1,\pm 2,\pm 3\}$ such that all of the relevant terms are generated from your example and the identity remains true and then use either $$ \frac{1}{8}\mathrm{Li}_4(z^2)=\mathrm{Li}_4(z)+\mathrm{Li}_4(-z) $$ or $$ \mathrm{Li}_4(z)=-\mathrm{Li}_4\left(\frac{1}{z}\right) - \frac{16\pi^4}{4!}B_4\left(\frac{\log(z)}{2\pi i}\right) $$ where $B_n(x)$ are Bernoulli Polynomials, to reduce the identity to your expression. We should then find that the resulting polynomial factors generated from this expression, when cancelled down, recreate your identity.

It does look like the right hand side of your conjecture will be generated by the Bernoulli polynomials, for example the first and last terms of your L.H.S $$ - 720\frac{16\pi^4}{4!}B_4\left(\frac{\log(\frac{1}{2})}{2\pi i}\right) - 135\frac{16\pi^4}{4!}B_4\left(\frac{\log(\frac{1}{9})}{2\pi i}\right) = 19\pi^4 + 30\pi^2(4\alpha^2+3\beta^2)-60i\pi(2\alpha^3+3\beta^3)-30(\alpha^4+3\beta^4) $$ may well be the origin of the $19\pi^4$!

There are also more identities on page 19 of this link


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