16
$\begingroup$

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)

$\endgroup$
5
  • $\begingroup$ 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
5
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.