(This summarizes my posts on Nielsen polylogs.)

I. Question 1: How to complete the table below? Consider the special cases $z=-1$ and $z=\frac12$. Given the Nielsen generalized polylogarithm,

$$S_{n,p}(z) = C_1\int_0^1\frac{(\ln t)^{n-1}\big(\ln(1-z\,t)\big)^p}{t}dt$$


$$C_1 = \frac{(-1)^{n+p-1}}{(n-1)!\,p!}$$

then for what $n,p$ are there closed-forms in terms of ordinary polylogarithms $\rm{Li}_m(x)$?

Note: For $p=1$, then the Nielsen polylog just reduces to $\rm{Li}_m(x)$.

MSE has a lot of posts asking for close-forms (see this, this, etc), most of which do not name the integral as a Nielsen polylog. The table below summarizes known results (so far),

$$\begin{array}{|c|c|c|c|c|} \hline n+p&n&p&z=-1&z=\tfrac12\\ \hline 3&1 &2 &Y&Y\\ 3&2 &1 &Y&Y\\ \hline 4&1 &3 &Y&Y\\ 4&2 &2 &Y&Y\\ 4&3 &1 &Y&Y\\ \hline 5&1 &4 &Y&Y\\ 5&2 &3 &Y&Y\\ 5&3 &2 &Y&Y\\ 5&4 &1 &Y&Y\\ \hline 6&1 &5 &Y&Y\\ 6&2 &4 &-&-\\ 6&3 &3 &\color{red}N&\color{red}Y\\ 6&4 &2 &-&-\\ 6&5 &1 &Y&Y\\ \hline 7&1 &6 &Y&Y\\ 7&2 &5 &-&-\\ 7&3 &4 &-&-\\ 7&4 &3 &-&-\\ 7&5 &2 &\color{red}Y&\color{red}N\\ 7&6 &1 &Y&Y\\ \hline \end{array}$$

Surprisingly, $S_{3,3}\big(\tfrac12\big)$ is expressible, but $S_{3,3}(-1)$ is not. Let $a=\ln 2$, then,

$$2\,S_{3,3}\big(\tfrac12\big) =\tfrac{23}{16}\zeta(6)-2a\zeta(5)+\tfrac18a^2\zeta(4)-\tfrac1{16}a^3\zeta(3)+\tfrac1{72}a^6-\zeta^2(3)+a\big(S_{3,2}\big(\tfrac12\big)-S_{2,3}\big(\tfrac12\big)+\zeta(2)\zeta(3)\big)$$

Since the two Nielsen polylogs in RHS are expressible as ordinary polylogs (see here), then so is the LHS. Conversely, $S_{5,2}(-1)$ is expressible,

$$128S_{5,2}(-1) = 64\zeta(2)\zeta(5)+112\zeta(3)\zeta(4)-251\zeta(7)$$ but $S_{5,2}\big(\tfrac12\big)$ apparently is not.

II. Question 2: What simple relations are there between the inexpressible(?) Nielsen polylogs in the table? Two are,



which is mentioned in this and this post. (The two seem to belong to a general form.) Are there others?

  • $\begingroup$ I just remembered that $S_{n,2}(-1)$ for odd $n$ belongs to the expressible class. $\endgroup$ Jun 3, 2019 at 15:22
  • $\begingroup$ If I didn't make a mistake, this question leads to an identity for sums of values when $n+p=6$ for $z=-1$. $\endgroup$ Jul 6, 2019 at 21:24
  • $\begingroup$ @SimplyBeautifulArt: I gave a comment to your answer. $\endgroup$ Jul 7, 2019 at 4:04

2 Answers 2


$\newcommand{\Li}{\mathrm{Li}}$This is not intended to be a complete or totally rigorous answer, but I want to highlight some of the known and conjectured results in this area. To skip to the Nielsen identities, scroll down. This is the result of some recenty finished joint work with Herbert Gangl and Danylo Radchenko (see the arXiv preprint On functional equations for Nielsen polylogarithms), which first started last year when we realised that $ S_{3,2} $ satisfies the 5-term dilogarithm relation modulo $ \Li_5 $.


To understand what sort of reductions we can expect and not expect, we can bring in results from the theory of motives. More specifically there are motivic version of the iterated integrals $$ I(a_0; a_1, \ldots, a_n; a_{n+1}) = \int_{a_0 < t_1 < \cdots < t_n < a_{n+1}} \frac{\mathrm{d}t_1}{t_1 - a_1} \wedge \cdots \wedge \frac{\mathrm{d}t_n}{t_n - a_n} \,. $$ One can think of the motivic version $ I^\mathfrak{m}(a_0; a_1, \ldots, a_n; a_{n+1}) $ as some formal algebraic object, which reflects the good (i.e. expected) behaviours of the analytic version.

(There is a precise definition, a version of which involves triples $ [ H^n(X,Y), \nu, \phi ] $. Here $$ H^n(X,Y) = ( H^n_{dR}(X,Y) , H^n_{B}(X,Y), \mathrm{comp} ) $$ is a 'motive' arising from the relative cohomology of the variety $ X $ with respect to a subvariety $ Y $ and $ \mathrm{comp} $ is the comparison isormophism between the complexified versions $ H_dR \otimes_\mathbb{Q} \mathbb{C} \cong H_B \otimes_\mathbb{Q} \mathbb{C} $. Then $ \nu \in H_{dR}^n(X,Y) $ specifies the class of differential form to integral, and $ \phi \in H_B^n(X,Y)^\vee $ repesents integration cycle. The various rules of integration (bilinearity, Stokes, chance of variables/pullback) are imposed between the appropriate triples. By keeping track of the integration form and integration cycle, without actually integrating, one has a richer and more rigid structure, that can't be see on the level of numbers.)

Hopf algebras:

An important consequence of this is that the motivic iterated integrals form a graded Hopf algebra, meaning there is a coproduct which allows us to decompose a complicated object into simpler pieces, in the hopes of better understanding it. I'll use a variant of this which defines a cobracket $ \delta $on the Lie coalgebra of irreducibles, this is what remains after neglecting non-trivial products everywhere. (See Brown Notes on Motivic Periods and Goncharov Galois symmetries of fundamental groupoids for rigorous definitions, and perhaps Duhr Hopf algebras, coproducts and symbols for a more down to earth approach in some applications.)

To compute $ \delta $ one can use the formula $$ \delta I(a_0; a_1, \ldots, a_n; a_{n+1}) = \sum_{i < j} I(a_0; a_1, \ldots, a_i, a_j, \ldots, a_n; a_{n+1}) \wedge I(a_i; a_{i+1}, \ldots, a_{j-1}; a_j) \,, $$ which comes from killing the explicit product terms in Goncharov's semicircular coproduct. (There is some technical point that killing $ i \pi $ is necessary, to deal with branch cuts.)

Under the cobraket $ \delta $, one has $$ \delta \Li_n(x) = 0 \,, $$ moreover, one expects that $ \Li_n $ is the kernel of $ \delta $. I.e. any combination which vanishes under $ \delta $ can be written as a sum of the form $ \sum_i \alpha_i \Li_n(x_i) $ for some $ \alpha_i, x_i \in \mathbb{C} $. (This is from a conjecture of Goncharov, about the structure of the motivic Lie coalgebra.)

Nielsen polylogs:

Now one can write the Nielsen polylogarithm $ S_{n,p}(z) $ as $$ S_{n,p}(x) = (-1)^p I(0; \underbrace{1, \ldots, 1}_{\text{$p$ times}}, \overbrace{0, \ldots, 0}^{\text{$n$ times}}; x) \,. $$

Nielsen $S_{3,2}$: From this, one can compute that $$ \delta S_{3,2}(x) = \Li_2(x) \wedge \zeta(3) \,. $$ At $ x = 1, \tfrac{1}{2} $ this vanishes because $ \Li_2(1) = \frac{\pi^2}{6} $, and $ \Li_2(\tfrac{1}{2}) = \frac{\pi^2}{12} + \text{products} $. So we expect $ S_{3,2}(1) $ and $ S_{3,2}(\frac{1}{2}) $ to be expressible in terms of $\Li_5$, and indeed these are given in various links in the question above.

More surprisingly, this vanishes at $ x = \phi^{-2} $, where $ \phi $ the golen ratio, because of the identity $$ \Li_2(\phi^{-2}) = \frac{\pi^2}{15} - \log^2(\phi) \,. $$ So one expects that $ S_{3,2}(\phi^{-2}) $ can be expressed in terms of $ \Li_5 $. This turns out to be the case and we have (using the shorthand $ f(\sum_i \alpha_i [x_i]) = \sum_i \alpha_i f(x_i) $) \begin{align*} S_{3,2}(\phi ^{-2}) ={} & \frac{1}{33} \Li_5\big({}-8 [\phi ^{-3}]{}+{}780 [\phi ^{-1}]{}+{}804 [-\phi ]{}+{}8 [-\phi ^3]\big) + \Li_4(\phi ^{-2})\log (\phi ) \\ & +\frac{1}{2}\zeta (5) +\frac{481}{11} \zeta (4) \log (\phi ) -\zeta(3)\Li_2(\phi ^{-2}) +\frac{50}{11} \zeta (2) \log ^3(\phi ) +\frac{14}{15}\log ^5(\phi ) \,. \end{align*} The cobracket suggests more generally that $ S_{3,2}(x) $ should behave like $ \Li_2 $ modulo $ \Li_5 $ terms. The main result (Theorem 14) in On functional equations for Nielsen polylogarithms is to give explicit $ \Li_5 $ terms for this, from which one gets the above analytic identity, by following the same steps as to obtain the corresponding $ \Li_2 $ identity. We also found similar identities for $ -\phi^{\pm1} $, $ \phi^{-1} $.

Nielsen $S_{4,2}$: One can compute that $$ \delta S_{4,2}(x) = \Li_3(x) \wedge \zeta(3) \,. $$ This vanishes by antisymmetry at $ x = 1 $ and $ x = 1/2 $ since $ \Li_3(1) = \zeta(3) $ and $ \Li_3(1/2) = \frac{7}{8} \zeta(3) + \text{products} $. So in fact we expect to be able to find expressions for both $ S_{4,2}(1) $ and $ S_{4,2}(\frac{1}{2}) $ in terms of classical polylogs (and products of lower weight things). This would then give an expression for $ S_{3,3}(-1) $ based on the identity $$ 16 S_{3,3}(-1) - 24 S_{4,2}(-1) = -4 \zeta(3)^2 + 5 \zeta(6) $$ from the second question. The reduction of $ S_{4,2}(-1) $ is sufficient to get $ S_{4,2}(\frac{1}{2}) $ using the identity \begin{align*} S_{4,2}\bigg(\big[-1\big] + 2\bigg[\frac{1}{2}\bigg]\bigg) = {} & 4\Li_6\Big(\frac{1}{2}\Big) +2\Li_5\Big(\frac{1}{2}\Big) \log (2) -\frac{51}{16} \zeta(6) -\frac{1}{2}\zeta (3)^2 \\& -\Big( \frac{1}{16} \zeta (5) -\zeta(2) \zeta (3) \Big) \log (2) +\frac{1}{4 \cdot 2!} \zeta(4) \log ^2(2) \\& -\frac{2}{3!} \zeta (3) \log^3(2) +\frac{2}{4!} \zeta(2) \log ^4(2) -\frac{6}{6!} \log ^6(2) \,. \end{align*} This identity follows from know identities for Nielsen polylogarithms, namely reflection relating $ S_{n,p}(x) $ and $ S_{p,n}(1-x) $ and inversion relating $ S_{n,p}(x) $ and $ S_{n,p}(x^{-1}) $. Applying these to all elements of the anharmonic group, one can solve the resulting system of equations to obtain in weight 6 $$ S_{4,2}(x) + S_{4,2}(1-x) + S_{4,2}(1-\tfrac{1}{x}) = \text{$\Li_6$'s and products.} $$ This combination is just the trilogarithm 3-term relation, and reduces at $ x = \frac{1}{2} $ to $$ S_{4,2}(-1) + 2 S_{4,2}(\tfrac{1}{2}) = \text{$\Li_6$'s and products.} $$

The cobracket also vanishes for $ x = \phi^{-2} $ since $ \Li_3(\phi^{-2}) = \frac{4}{3} \zeta(3) + \text{products} $.

To find the analytic version of these results (up to a conjectural coefficient of $ \pi^6 $), one can use the full coproduct to analyse $ S_{4,2}(-1) $, and $ S_{4,2}(\phi^{-2}) $. One eventually finds \begin{align*} S_{4,2}(-1) \overset{?}{=}{} & \frac{1}{13} \bigg(\frac{1}{3}\Li_6\Big(-\frac{1}{8}\Big)-162 \Li_6\Big(-\frac{1}{2}\Big)-126 \Li_6\Big(\frac{1}{2}\Big)\bigg) -\frac{1787 }{624}\zeta (6) +\frac{3}{8} \zeta (3)^2 \\ & {}+\frac{31}{16} \zeta (5) \log(2) -\frac{15}{26} \zeta (4) \log ^2(2) +\frac{3}{104} \zeta (2) \log^4(2) -\frac{1}{208} \log ^6(2) \,. \end{align*} \begin{align*} S_{4,2}(\phi ^{-2}) \overset{?}{=} {} & \frac{1}{396} \Li_6\Big( 2 \big[ \phi ^{-6} \big] -128 \big[ \phi ^{-3}\big]+801 \big[\phi ^{-2}\big]-576 \big[\phi^{-1} \big] \Big) +\frac{35 }{99}\zeta (6) +\frac{2}{5} \zeta (3)^2 \\ & + \Li_5 \left(\phi ^{-2}\right) \log (\phi) -\zeta (5)\log (\phi ) +\frac{2}{11} \zeta (4) \log ^2(\phi ) -\zeta (3) \Li_3\left(\phi ^{-2}\right) \\ & +\frac{10}{33} \zeta (2) \log ^4(\phi ) -\frac{79}{990} \log ^6(\phi ) \,. \end{align*} Both of these give some evidence that $ S_{4,2} $ satisfies the trilogarithm duplication relation $$ \tfrac{1}{2^2} S_{4,2}(x^2) - (S_{4,2}(x) + S_{4,2}(-x)) \overset{?}{=} \text{$\Li_6$'s, products} $$ modulo $ \Li_6 $ terms, since the corresponding $ \Li_3 $ evaluations are obtained from it. More generally (via Goncharov's conjecture on the structure of the motivic Lie coalgebra), $ S_{4,2} $ should satisfy all $ \Li_3 $ functional equations modulo $ \Li_6 $ terms.

The structure of the $S_{4,2}(-1)$ reduction is explained quite well with the coproduct. In particular, we need to find a combination $ L = \sum_i \beta_i [y_i] $ such that $ \sum_i \beta_i \Li_5(y_i) \nu_p(y_i) \in \mathbb{Q}^\times \zeta(5) $, and $ \sum_i \beta_i \Li_5(y_i) \nu_p(y_i) = 0 $, p $ \neq 2 $, where $ \nu_p(x) $ is the exponent of $p$ in the prime factorisation of $x$. This comes from pseudo-integrating equation (7.100) in Lewin's Polylogarithms and associated functions: $$ \Li_5(-\tfrac{1}{8}) - 126 \Li_5(\tfrac{1}{2}) - 162 \Li_5(-\tfrac{1}{2}) = \tfrac{403}{16} \zeta(5) - \tfrac{3}{8} \log(2)^5 + \tfrac{3}{2} \zeta(2) \log(2)^3 - 15 \zeta(4) \log(2) \,. $$ Then $ \Li_6(L) $ will be the leading polylog terms in the reduction of $ S_{4,2}(-1) $.

Nielsen $S_{6,2}$: One can compute that $$ \delta S_{6,2}(x) = \zeta(3) \wedge \Li_5(x) - \Li_3(x) \wedge \zeta(5) $$ Unfortunately now, at $ x = -1 $, we obtain (using the $\Li_3$ and $\Li_5$ duplication identities) that $$ \delta S_{6,2}(-1) = - \frac{3}{16} \zeta(3) \wedge \zeta(5) $$ This is non-zero, so we cannot reduce $S_{6,2}(-1)$ to purely polylogarithms. Nevertheless, if we introduce the multiple zeta value $$ \zeta(3,5) = \sum_{0< n_1 < n_2} \frac{1}{n_1^3 n_2^5} = I(0; 1, 0, 0, 1, 0, 0, 0, 0; 1) \,, $$ we stand a chance since $$ \delta \zeta(3,5) = -5 \zeta(3) \wedge \zeta(5) \,. $$ In particular $$ \delta( S_{6,2}(-1) - \frac{3}{80} \zeta(3,5) ) = 0 $$ After understanding how the structure of the coproduct gives rise to the reduction of $ S_{4,2}(-1) $, one can attempt to find this reduction. It turns out that we must find a combination $ L' = \sum_i \gamma_i [z_i] $ such that \begin{align*} \sum_i \gamma_i \Li_7(z_i) \nu_2(z_i) &\in \mathbb{Q}^\times \zeta(7) \\ \sum_i \gamma_i \Li_7(z_i) \nu_p(z_i) &= 0 \,,\ \quad p > 2 \end{align*} Then $ \Li_8(L) $ becomes the leading order polylog terms in the reduction.

Already to find a combination such that even conjecturally $ \sum_i \gamma_i \Li_7(z_i) \in \mathbb{Q} \zeta(7) $ is tricky; one must recursively build a good combination via the higher weight Bloch groups up to weight $7$, as in Zagier's Polylogarithms, Dedekind zeta functions, and the algebraic $K$ theory of fields . One can in fact adapt this procedure to get a combination satisfying the $ \nu_2 $ and $ \nu_p $ requirements, and hence a conjectural reduction of $ S_{6,2}(-1) $. Both of these results are long ($L'$ has ~60 terms), and involve very large coefficients (order $10^{20}$); they are given in Appendix C of On functional equations for Nielsen polylogarithms.

Higher: In principle, this could work for any higher $S_{2n,2}(-1)$ since the cobracket will only involve wedges $ \zeta(n) \wedge \zeta(m) $ which can be matched by the MZV $ \zeta(n,m) $. However the reductions are likely to become significantly more complicated to find and to write.


Consider the generalization of this integral as follows:

$$I(a,b,c)=\int_0^\infty\ln^a(x)\ln^b(1+x)\ln^c\left(1+\frac1x\right)~\frac{\mathrm dx}x$$

We can apply the substitution $x\mapsto1/x$ to get $I(a,b,c)=(-1)^aI(a,c,b)$. We can also apply the substitution $x\mapsto1/x$ on $(1,\infty)$ to get two integrals:

$$I_1(a,b,c)=\int_0^1\ln^a(x)\ln^b(1+x)\ln^c\left(1+\frac1x\right)~\frac{\mathrm dx}x$$

$$I_2(a,b,c)=(-1)^a\int_0^1\ln^a(x)\ln^c(1+x)\ln^b\left(1+\frac1x\right)~\frac{\mathrm dx}x$$

Since $\ln(1+1/x)=\ln(1+x)-\ln(x)$, it is then possible to rewrite our initial integral in terms of the Nielsen polylog at $z=-1$. Let $\displaystyle S_{n,p}=\int_0^1\ln^n(x)\ln^p(1+x)~\frac{\mathrm dx}x$. Then we have

$$I_1(a,b,c)=\sum_{j=0}^c\binom cj(-1)^jS_{a+j,b+c-j}$$

$$I_2(a,b,c)=(-1)^a\sum_{k=0}^b\binom bk(-1)^kS_{a+k,b+c-k}$$

$$I(a,b,c)=\sum_{j=0}^c\binom cj(-1)^jS_{a+j,b+c-j}+(-1)^a\sum_{k=0}^b\binom bk(-1)^kS_{a+k,b+c-k}$$

On the other hand, one can perform the substitution $x\mapsto\frac1x-1$ to get

\begin{align}I(a,b,c)&=\int_0^1\ln^a\left(\frac{1-x}x\right)\ln^b\left(\frac1x\right)\ln^c\left(\frac1{1-x}\right)~\frac{\mathrm dx}{x(1-x)}\\&=(-1)^{b+c}\int_0^1[\ln(1-x)-\ln(x)]^a\ln^b(x)\ln^c(1-x)\left[\frac1{1-x}+\frac1x\right]~\mathrm dx\end{align}

For brevity we define another integral:

$$J(a,b,c)=\int_0^1[\ln(1-x)-\ln(x)]^a\ln^b(x)\ln^c(1-x)~\frac{\mathrm dx}x$$

Splitting this into two integrals and applying $x\mapsto1-x$ for the part with $1/(1-x)$ gives us


Let $J(b,c)=J(0,b,c)$. It is possible to rewrite $J(a,b,c)$ in terms of the two argument version by binomially expanding it out:

$$J(a,b,c)=\sum_{k=0}^a\binom ak(-1)^kJ(b+k,c+a-k)$$

and it is easy to solve $J(b,c)$ as partial derivatives of the Beta function, which gives us solutions in terms of the Riemann zeta function at natural arguments:


Conclusion: we can generate identities for sums of the form $\displaystyle\sum_{n+p=a+b+c}a_{a,b,c,n,p}S_{n,p}$ in terms of the Riemann zeta function.

For example, with $(a,b,c)=(2,1,2)$ we get


and with $(a,b,c)=(0,2,3)$ we get


and with $(a,b,c)=(0,1,4)$ we get


which sadly doesn't give any values.

  • $\begingroup$ I'll look into it later, but by choosing different values such that $a+b+c=k$, one might actually be able to solve for some values of $S_{n,p}(-1)$. $\endgroup$ Jul 7, 2019 at 13:32
  • $\begingroup$ Given your $a_{n,p} S_{n,p}(-1)$, I understand the Nielsen polylog part $S_{n,p}(-1)$, but what is $a_{n,p}$? $\endgroup$ Jul 7, 2019 at 15:43
  • $\begingroup$ Hm I suppose it depends on $a,b,c$, and they result from expanding $I_1$ and $I_2$. $\endgroup$ Jul 7, 2019 at 16:51
  • $\begingroup$ Considering $(a,b,c)=(2n,1,1)$ also gives your claim that $S_{2n+1,2}(-1)$ is solvable. When considering $(a,b,c)=(2n-1,1,1)$ however, a cancellation occurs. $\endgroup$ Jul 7, 2019 at 21:15
  • $\begingroup$ Actually it may be possible to solve $S_{n,9-n}(-1)$... oh dear time to work out these values $\endgroup$ Jul 7, 2019 at 21:32

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