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We consider a following quantity: \begin{eqnarray} {\mathcal I}^{(p,q)}_r(\xi,t) := \int\limits_\xi^t \frac{[\log(\frac{t}{\eta})]^p}{p!} \cdot \frac{[\log(\frac{\eta}{\xi})]^q}{q!} \cdot \frac{Li_r(\eta)}{\eta} d\eta \end{eqnarray} where $p$,$q$ and $r$ are non-negative integers and $0\le \xi \le t \le 1$ and $L_r(\eta) := \sum\limits_{m=1}^\infty \eta^m/m^r$ is the poly-logarithm.

By using integration by parts we found that the quantities above satisfy following recurrence relations: \begin{eqnarray} {\mathcal I}^{(p,q-p)}_{W-q} &=& 1_{2 p \ge q} \cdot \sum\limits_{j=0}^{q-p-1} \binom{q-p}{j} (-1)^j {\mathcal I}^{(2p-q+j,q-p-j)}_{W-p} + 1_{2p < q}\sum\limits_{j=1}^p \binom{p}{j} (-1)^j {\mathcal I}^{(j,q-p-j)}_{W-q+p} + \\ &&1_{2 p \ge q} \cdot (-1)^{q-p}\left(Li_{W+1}(t) - \sum\limits_{l=1}^{p+1} Li_{W-p+l}(\xi) \cdot \frac{[\log(\frac{t}{\xi})]^{p+1-l}}{(p+1-l)!}\right)-\\ &&1_{2p < q} \cdot (-1)^{q-p}\left(Li_{W+1}(\xi)-\sum\limits_{l=1}^{q-p+1} Li_{W-q+p+l}(t) \frac{\log(\frac{\xi}{t})]^{q-p+1-l}}{(q-p+1-l)!}\right) \end{eqnarray} for $0\le p \le q \le W$. This is a system of $\binom{W+2}{2}$ linear equations for all the unknown quantities $\left\{ {\mathcal I}^{(p,q-p)}_{W-q} \right\}_{0\le p \le q \le W}$ which for a given $W \ge1$ is straightforward to solve on any CAS. For example we have the following: \begin{eqnarray} {\mathcal I}^{(1,2)}_1&=&\frac{1}{2} \text{Li}_3(t) \log ^2\left(\frac{\xi }{t}\right)+2 \text{Li}_4(t) \log \left(\frac{\xi }{t}\right)-\text{Li}_4(\xi ) \log \left(\frac{t}{\xi }\right)+3 \text{Li}_5(t)-3 \text{Li}_5(\xi )\\ {\mathcal I}^{(1,3)}_2&=& -\frac{1}{6} \text{Li}_4(t) \log ^3\left(\frac{\xi }{t}\right)-\text{Li}_5(t) \log ^2\left(\frac{\xi }{t}\right)-3 \text{Li}_6(t) \log \left(\frac{\xi }{t}\right)+\text{Li}_6(\xi ) \log \left(\frac{t}{\xi }\right)-4 \text{Li}_7(t)+4 \text{Li}_7(\xi )\\ {\mathcal I}^{(1,4)}_3 &=& \frac{1}{24} \text{Li}_5(t) \log ^4\left(\frac{\xi }{t}\right)+\frac{1}{3} \text{Li}_6(t) \log ^3\left(\frac{\xi }{t}\right)+\frac{3}{2} \text{Li}_7(t) \log ^2\left(\frac{\xi }{t}\right)+4 \text{Li}_8(t) \log \left(\frac{\xi }{t}\right)-\text{Li}_8(\xi ) \log \left(\frac{t}{\xi }\right)+5 \text{Li}_9(t)-5 \text{Li}_9(\xi ) \end{eqnarray} Now, my question is can we actually find a closed form expression for the quantities in question or otherwise do we always have to resort to CAS to solve the equations in question?

Przemo
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  • What is the definition of ${Li_r}(\eta)$? – unseen_rider Jun 19 '17 at 17:02
  • It is the polylogarithm. $Li_r(\eta) := \sum\limits_{m=1}^\infty \eta^m/m^r$. – Przemo Jun 19 '17 at 17:07
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    Ok please update your question with this – unseen_rider Jun 19 '17 at 17:09
  • I state the word poly-logarithm in the title explicitly and I use the standard notation for it as well. – Przemo Jun 19 '17 at 17:15
  • Ok,however there are some people that do not know what a polylogarithm is. Bear this in mind when asking future questions in general. More likely to get better answers when you explain what may be obvious to yourself but not all. – unseen_rider Jun 19 '17 at 17:19
  • did you use any kind of tech to write such a wonderful post without typos? – Aditya Jun 19 '17 at 17:36
  • @unseen_rider That's decent advice most of the time, but I'd wager an entire year's salary that anyone capable of answering this question doesn't need to be reminded what a polylogarithm is. ;) – David H Jun 20 '17 at 02:15
  • Just out of curiosity : in which context did you find this monster ? – Claude Leibovici Jun 20 '17 at 05:25
  • @ClaudeLeibovici I simply need this result to answer this question https://math.stackexchange.com/questions/2323993/a-list-of-multiple-zeta-values-of-depth-three .You just deal with such integrals in the context of zeta values of depth three. – Przemo Jun 20 '17 at 09:35

1 Answers1

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The closed form in question reads: \begin{eqnarray} &&{\mathcal I}^{(p,q)}_r = \frac{1}{p! q!} \sum\limits_{l_1=0}^p \sum\limits_{l_2=0}^q \sum\limits_{l_3=1}^{l_1+l_2+1} \binom{p}{l_1} \binom{q}{l_2} (l_1+l_2)_{(l_3-1)} (-1)^{q-1+l_1+l_2+l_3} \cdot \\ &&\left( Li_{r+l_3}(t) [\log(\xi)]^{q-l_2} [\log(t)]^{p+l_2+1-l_3} - Li_{r+l_3}(\xi) [\log(t)]^{p-l_1} [\log(\xi)]^{q+l_1+1-l_3}\right) \end{eqnarray} and it is obtained by expanding the two terms with logarithms into binomial series then by collecting all powers of logs together and then integrating term by term and by using integration by parts in order to evaluate the integrals.

Przemo
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