A list of Multiple Zeta values of depth three

Solution 1:

This is not going to be a full answer but a starting point that will eventually lead to establishing recurrence relations for the quantities in question. Let us define: \begin{eqnarray} {\mathfrak F}^{(p,q,r)}_\xi &:=& (-1)^q \sum\limits_{l=p+1}^{q+p+1} \binom{q}{l-p-1} (l-1)! Li_{r+l}(1) [\log(\xi)]^{q+p+1-l}+\\ &&(-1)^p \sum\limits_{l=q+1}^{q+p+1} \binom{p}{l-q-1} (-1)^l (l-1)! Li_{r+l}(\xi) [\log(\xi)]^{q+p+1-l} \end{eqnarray} and $\theta_j := (j+r) \% 2$.

Now, by using An integral involving product of poly-logarithms and a power of a logarithm. we expressed the Multiple Zeta Values (MZVs) as follows. \begin{eqnarray} \zeta(p,q,r)&=& \sum\limits_{j=q}^{\lfloor \frac{q-1+r}{2} \rfloor} (-1)^{j-q} \binom{j-1}{q-1} \int\limits_0^1 \frac{[\log(1/\xi)]^{p-1}}{(p-1)!} \cdot \frac{Li_j(\xi) Li_{r+q-j}(\xi) Li_0(\xi)}{\xi} d\xi+\\ && \sum\limits_{j=0}^{q-1}(-1)^{\lfloor\frac{j+r}{2}\rfloor-j} \binom{\lfloor\frac{j+r-1}{2}\rfloor}{j}\frac{1}{2^{\theta_j}} \frac{1}{(p-1)!(q-1-j-\theta_j)!} \int\limits_0^1 {\mathfrak F}^{(p-1,q-1-j-\theta_j,0)}_\xi \cdot \frac{[Li_{\theta_j+\lfloor \frac{j+r}{2} \rfloor}(\xi)]^2}{\xi} d\xi \end{eqnarray}

Note 1: Let $q=r=1$ and $p\ge 2$. Then the only term that survives in the right hand side above is the $(j=0)$ term in the second sum. Then we have: \begin{eqnarray} &&\zeta(p,1,1) = \frac{(-1)^{p-1}}{2(p-1)!} \int\limits_0^1 \frac{Li_0(\xi)}{\xi} [\log(\xi)]^{p-1} [Li_1(\xi)]^2 d\xi\\ &&= \frac{(-1)^{p-1}}{2 (p-1)!} \int\limits_0^1 \frac{[\log(1-\xi)]^{p-1} [\log(\xi)]^2}{\xi} d\xi\\ &&=\frac{(-1)^{p-1}}{2 (p-1)!}\left[ -\frac{1}{3} \Psi^{(p+1)}(1) + \frac{1}{2} \sum\limits_{j=1}^{p-2} \binom{p-1}{j} \left\{ \Psi^{(j+1)}(1) \Psi^{(p-1-j)}(1) + \Psi^{(j)}(1) \Psi^{(p-j)}(1)\right\} -\frac{1}{3} \sum\limits_{1 \le j < j_1 \le p-2} \binom{p-1}{j,j_1-j,p-1-j_1} \Psi^{(j)}(1) \Psi^{(j_1-j)}(1) \Psi^{(p-1-j_1)}(1)\right]\\ &&=\left\{\zeta(4),2 \zeta (5)-\zeta (3) \zeta(2),\frac{23 \zeta(6)}{16}-\zeta (3)^2,-2 \zeta (5) \zeta(2)-\frac{5}{4} \zeta (3) \zeta(4)+5 \zeta (7),\cdots\right\} \end{eqnarray} Now let us take $p \leftarrow p-1$, $q=2$ and $r=1$. The again the only term which survives in the right hand side is the $(j=0)$ term in the second sum. Then we have: \begin{eqnarray} &&\zeta(p-1,2,1) =\\ && \frac{1}{2(p-1)!} \int\limits_0^1 {\mathfrak F}^{(p-2,0,0)}_\xi \cdot \frac{[Li_1(\xi)]^2}{\xi} d\xi\\ &&= \zeta(3) \zeta(p-1) - \frac{1}{2} \sum\limits_{l=1}^{p-1} \int\limits_0^1 \frac{[Li_1(\xi)]^2}{\xi} Li_l(\xi) \cdot \frac{[\log(1/\xi)]^{p-1-l}}{(p-1-l)!}d\xi\\ &&= \zeta(3) \zeta(p-1) - \frac{1}{2} \sum\limits_{l=1}^{p-1} \\ && \left( \sum\limits_{l_1=2}^l \binom{p-1-l_1}{p-1-l} (-1)^{l-l_1} \zeta(p-l_1+1,1) \zeta(l_1) + \sum\limits_{l_1=2}^{p-l} \binom{p-1-l_1}{l-1} (-1)^l \zeta(p-l_1+1,1) \zeta(l_1) + \sum\limits_{l_1=1}^{p-l} \left\{ \zeta(p-l_1+1,l_1,1)+\zeta(p-l_1+1,1,l_1)+\zeta(p-l_1+1,l_1+1)+\zeta(p+1,1)\right\} \right) \end{eqnarray} for $p\ge 3$. By simplifying we get the following identity: \begin{eqnarray} \zeta(2,1,p) + \zeta(2,p,1)+\zeta(p,2,1) = \zeta(3) \zeta(p) - \zeta(2,p+1)-\zeta(p+2,1) \end{eqnarray} for $p \ge 2$.

Now let us take $p\leftarrow p-1$, $q=1$ and $r=2$. We encounter an integral with three poly-logarithms in it. That integrals has been evaluated in my answer to Generating function for cubes of Harmonic numbers I . Taking this into account and simplifying the result we get the following identity: \begin{eqnarray} 2 \zeta(2,1,p) + \zeta(2,p,1) - \sum\limits_{j=2}^p \zeta(j,p-j+1,2) \cdot(1_{j<p} + 2 \cdot 1_{j=p})=\\ \zeta(p+3) + 2 \zeta(3) \zeta(p)-\zeta(2) \zeta(p+1) - \zeta(2) \zeta(p,1)-\zeta(2,p+1) - \zeta(p+2,1) \end{eqnarray} valid for $p\ge 2$.

Finally let us take $p \leftarrow p-2$, $q=3$ and $r=1$. Then only the $(j=0)$ term in the second sum survives. The only non-trivial integral we come across is the one that contains a square of a polylog of order one, another polylog and a power of a log. That integral has been calculated in my answer to Generating function for cubes of Harmonic numbers I .By simplifying everything we end up with the following identity: \begin{eqnarray} &&2 \zeta(p-2,3,1)+\sum\limits_{l=2}^{p-1} \sum\limits_{j=1}^l \sum\limits_{i=1}^{p-1-l}(l-1) A_{l,j} \zeta(p+1-l-i,i+j,l+1-j)=\\ &&\sum\limits_{l=1}^{p-1} \sum\limits_{j=1}^l(l-1)A_{l,j} \cdot \\ &&\left( \zeta(p-l+j+1,l+1-j)+\zeta(p-l+j,l+2-j)+\zeta(p-l+j,l+1-j,1)+\zeta(p-l+j,1,l+1-j)\right) +\\ &&-2(p-2) \zeta(3) \zeta(p-1)-(\zeta(2)^2-3 \zeta(4))\zeta(p-2) \end{eqnarray} where $A_{l,j} :=(1_{l<j}+2 \cdot 1_{l=j})$. Here $p\ge 4$.

Note 2: In the generic case of arbitrary $p$, $q$ and $r$ there are only two types of integrals that we are dealing with firstly an integral that involves a square of a polylogarithm and a power of a log and secondly an integral that involves products of three polylogarithms and a power of a log. From An integral involving product of poly-logarithms and a power of a logarithm. we know that the first type of integral always reduces to single zeta values and to bi-variate zeta values (where the later in many cases reduce to single zeta values as well). The second type of integral is more complicated however we conjecture that it also always reduces to single, bi-variate and to triple-variate zeta values all of which have the same weight. As such we will obtain a system of linear equations for all triple-zeta values of a given weight. We will derive those equations and solve them asap.