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The polylogarithm is defined by the series $$\operatorname{Li}_n(x)=\sum_{k=1}^\infty\frac{x^k}{k^n}.$$ There are relations connecting values of the polylogarithm at certain rational points in the interval $(0,1)$ using combinations of logarithms and values of $\zeta$-function of integer arguments. Here are some examples for small integer orders:

$$\pi^2-12\ln^22+12\ln2\cdot\ln3-6\ln^23-12\operatorname{Li}_2\!\left(\tfrac13\right)-6\operatorname{Li}_2\!\left(\tfrac14\right)=0$$

$$2\!\;\pi^2\ln2-4\!\;\pi^2\ln3+8\ln^32-12\ln2\cdot\ln^23+8\ln^33\\+45\,\zeta(3)-24\operatorname{Li}_3\!\left(\tfrac13\right)-24\operatorname{Li}_3\!\left(\tfrac23\right)-6\operatorname{Li}_3\!\left(\tfrac14\right)=0$$

$$8\,\pi^4\ln2-12\,\pi^2\ln^32+18\ln^52-1209\,\zeta(5)\\+1728\operatorname{Li}_5\!\left(\tfrac12\right)-486\operatorname{Li}_5\!\left(\tfrac14\right)-48\operatorname{Li}_5\!\left(\tfrac18\right)+3\operatorname{Li}_5\!\left(\tfrac1{64}\right)=0$$

Are there any known similar relations for $\operatorname{Li}_6$ and higher orders?

I know there are some relations ("ladders") for non-rational algebraic arguments, but now I'm interested in rational arguments only.

MathWorld has formula $(19)$ attributed to Bailey et al. that is apparently supposed to hold for any positive integer order, but it does not check numerically for $m>5$. Perhaps, there is a missing term or condition. (Update: Indeed, the original paper has this identity as formula $(2.16)$ saying it only holds for $1\le m\le5$, and attributes it to Lewin)

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  • $\begingroup$ Do you know which denominators are likely to appear together in an identity? It's not completely obvious why $1/4$ would be related to $\{1,2\}/3$. $\endgroup$ – Kirill Sep 26 '14 at 2:33
  • $\begingroup$ @Kirill No, I don't know. $\endgroup$ – Vladimir Reshetnikov Sep 26 '14 at 18:00
  • $\begingroup$ Not quite what you're looking for, but the largest real root $\alpha$ of the polynomial with coefficients $[x^{0\ldots10}]=\{1,1,0,-1,-1,-1,-1,-1,0,1,1\}$ is not rational, but there is a conjectured integer relation between one hundred and fifteen values of the form $\mathrm{Li}_{17}(\alpha^{-k})$, and values of the form $\zeta(17)$, $\pi^{2j}\log^{17-2j}\alpha$, according to Bailey and Broadhurst. $\endgroup$ – Kirill Oct 2 '14 at 6:37
  • $\begingroup$ @Kirill Yes, this is a ladder. I think this question is an open problem. $\endgroup$ – user153012 Oct 2 '14 at 18:22
  • $\begingroup$ @Kirill: Zagier's result for $\text{Li}_7$ below. $\endgroup$ – Tito Piezas III Dec 22 '17 at 4:29
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Yes, there are known relations using rational arguments for higher orders such as $\color{blue}{\text{Li}_7}$. In Don Zagier's Special Values And Functional Equations Of Polylogarithms, we get an amazing identity. In page 4, first define the function,

$$P_7(x_n) = \sum_{j=0}^6\frac{2^j\,B_j}{j!}\,\ln^j|x_n|\,\text{Li}_{7-j}(x_n)$$

with Bernoulli number $B_j$ and absolute value $|x|$. Then the $29$ rationals,

$$x_n = \small 1/3, -1/3, -1/9, -1/27, 1/2, -1/2, 1/6, -1/6, -1/4374, -1/4, 1/8, -1/8, -1/24, -1/48, 2/3, -2/3,\\ \small 3/4, -3/4, 3/8, -3/32, 3/128, 2/9, 8/9, -9/16, 2/27, -8/27, 27/32, 32/81, -2/243$$

and $29$ coefficients,

$$a_n = \small -25111753072, -27461584367, -171330250, 57577037, -151540388696, -136446322032, -2209899405, -2199243270, 43524, -7089743800, -284585110, 470985412/3, 38987641, 17015061/2, -11528187258, -6563312469, 2802854628, -751304106, -785318380, 11883921, 1765911, 478706760, -66158750, 15912813, 23786119, \color{blue}{-2179429}, 2585366, 111363, -2372265$$

then,

$$\sum_{n=1}^{29} a_n\,P_7(x_n) = \frac{-1020149599795}{96}\,\zeta(7)$$

Note: The OP has verified the result above using Mathematica. But Zagier made a small typo and wrote $\small-2\color{red}879429$ instead of the correct $\small\color{blue}{-2179429}$, so I had to use Mathematica's Integer Relations subroutine to find that error.

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  • $\begingroup$ Remarkable! I think that this equation likely could be separated into several ones that do not mix polylogarithms of different orders, but might need to include $\zeta(3), \zeta(5), \pi^4, \pi^6$ and, perhaps, some additional logarithm values. $\endgroup$ – Vladimir Reshetnikov Dec 21 '17 at 20:17
  • $\begingroup$ @VladimirReshetnikov: The presence of lower-order polylogs does deflate the identity a bit. $\endgroup$ – Tito Piezas III Dec 22 '17 at 3:00

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