# Challenging integral: Evaluate $\int_0^1\frac{\ln^3(1-x)\operatorname{Li}_3(x)}{x}dx$

How to evaluate $$I=\int_0^1\frac{\ln^3(1-x)\operatorname{Li}_3(x)}{x}dx\ ?$$

I came across this integral $$I$$ while I was trying to compute two advanced sums of weight 7. The problem with my approach is that when I tried to evaluate $$I_5$$ (shown below), the main integral $$I$$ appeared there which cancels out from both sides, so any idea how to evaluate $$I_5$$ or $$I$$?

Thanks.

Here is my trial:

Using the two generalized integral expressions of the polylogrithmic function which can be found in the book (Almost) Impossible Integrals, Sums and series page 4.

$$\int_0^1\frac{x\ln^n(u)}{1-xu}du=(-1)^n n!\operatorname{Li}_{n+1} (x)\Longrightarrow \operatorname{Li}_{3}(x)=\frac12\int_0^1\frac{x\ln^2(u)}{1-xu}du\tag{1}$$

$$\small{u\int_0^1\frac{\ln^n(x)}{1-u+ux}dx=(-1)^{n-1}n!\operatorname{Li}_{n+1}\left(\frac{u}{u-1}\right)\Longrightarrow\int_0^1\frac{\ln^3x}{1-u+ux}dx=\frac6u\operatorname{Li}_{3}\left(\frac{u}{u-1}\right)}\tag{2}$$

We have

\begin{align} I&=\int_0^1\frac{\ln^3(1-x)\operatorname{Li}_3(x)}{x}dx\overset{\text{use} (1)}{=}\frac12\int_0^1\frac{\ln^3(1-x)}{x}\left(\int_0^1\frac{x\ln^2u}{1-xu}du\right)dx\\ &=\frac12\int_0^1\ln^2u\left(\frac{\ln^3(1-x)}{1-xu}dx\right)\ du\overset{1-x\ \mapsto\ x}{=}\frac12\int_0^1\ln^2u\left(\int_0^1\frac{\ln^3x}{1-u+ux}dx\right)\ du\\ &\overset{\text{use}\ (2)}{=}3\int_0^1\frac{\ln^2u}{u}\operatorname{Li}_4\left(\frac{u}{u-1}\right)du\overset{IBP}{=}-\int_0^1\frac{\ln^3u}{u(1-u)}\operatorname{Li}_3\left(\frac{u}{u-1}\right)du \end{align}

Now we need the trilogarithmic identity:

$$\operatorname{Li}_3\left(\frac{x-1}{x}\right)=\zeta(2)\ln x-\frac12\ln^2x\ln(1-x)+\frac16\ln^3x-\operatorname{Li}_3(1-x)-\operatorname{Li}_3(x)+\zeta(3)$$

set $$1-x=u$$ to get

$$\small{\operatorname{Li}_3\left(\frac{u}{u-1}\right)=\zeta(2)\ln(1-u)-\frac12\ln^2(1-u)\ln u+\frac16\ln^3(1-u)-\operatorname{Li}_3(u)-\operatorname{Li}_3(1-u)+\zeta(3)}$$

Going back to our integral \begin{align} I&=\small{-\int_0^1\frac{\ln^3u}{u(1-u)}\left(\zeta(2)\ln(1-u)-\frac12\ln^2(1-u)\ln x+\frac16\ln^3(1-u)-\operatorname{Li}_3(u)-\operatorname{Li}_3(1-u)+\zeta(3)\right)du}\\ &=-\zeta(2)\underbrace{\int_0^1\frac{\ln^3u\ln(1-u)}{u(1-u)}du}_{\Large I_1}+\frac12\underbrace{\int_0^1\frac{\ln^4u\ln^2(1-u)}{u(1-u)}du}_{\Large I_2}-\frac16\underbrace{\int_0^1\frac{\ln^3u\ln^3(1-u)}{u(1-u)}du}_{\Large I_3}\\ &\quad+\underbrace{\int_0^1\frac{\ln^3u\operatorname{Li}_3(u)}{u(1-u)}\ du}_{\Large I_4}+\underbrace{\int_0^1\frac{\ln^3u}{u(1-u)}\left(\operatorname{Li}_3(1-u)-\zeta(3)\right)du}_{\Large I_5} \end{align}

\begin{align} I_1=\int_0^1\frac{\ln^3u\ln(1-u)}{u(1-u)}du=-\sum_{n=1}^\infty H_n\int_0^1 u^{n-1}\ln^3udu=6\sum_{n=1}^\infty\frac{H_n}{n^4} \end{align} .

\begin{align} I_2&=\int_0^1\frac{\ln^4u\ln^2(1-u)}{u(1-u)}du=\sum_{n=1}^\infty\left(H_n^2-H_n^{(2)}\right)\int_0^1 u^{n-1}\ln^4udu\\ &=24\sum_{n=1}^\infty\frac{H_n^2-H_n^{(2)}}{n^5}=24\sum_{n=1}^\infty\frac{H_n^2}{n^5}-24\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^5} \end{align}

\begin{align} I_3&=\int_0^1\frac{\ln^3u\ln^3(1-u)}{u(1-u)}du=\int_0^1\frac{\ln^3u\ln^3(1-u)}{u}du+\underbrace{\int_0^1\frac{\ln^3u\ln^3(1-u)}{1-u}du}_{1-x\ \mapsto\ x}\\ &=2\int_0^1\frac{\ln^3u\ln^3(1-u)}{u}\ du\overset{IBP}{=}\frac32\int_0^1\frac{\ln^4u\ln^2(1-u)}{1-u}du\\ &=\frac32\sum_{n=1}^\infty\left(H_n^2-H_n^{(2)}\right)\int_0^1 u^n\ln^4udu, \quad \text{reindex}\\ &=\frac32\sum_{n=1}^\infty\left(H_n^2-H_n^{(2)}-\frac{2H_n}{n}+\frac2{n^2}\right)\int_0^1 u^{n-1}\ln^4u du\\ &=\frac32\sum_{n=1}^\infty\left(H_n^2-H_n^{(2)}-\frac{2H_n}{n}+\frac2{n^2}\right)\left(\frac{24}{n^5}\right)\\ &=36\sum_{n=1}^\infty\frac{H_n^2}{n^5}-36\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^5}-72\sum_{n=1}^\infty\frac{H_n}{n^6}+72\zeta(7) \end{align} .

\begin{align} I_4&=\int_0^1\frac{\ln^3u\operatorname{Li}_3(u)}{u(1-u)}du=\sum_{n=1}^\infty H_n^{(3)}\int_0^1 u^{n-1}\ln^3u du=-6\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^4} \end{align}

\begin{align} I_5&=\int_0^1\frac{\ln^3u}{u(1-u)}\left(\operatorname{Li}_3(1-u)-\zeta(3)\right)du\\ &=\underbrace{\int_0^1\frac{\ln^3u}{u}\left(\operatorname{Li}_3(1-u)-\zeta(3)\right)du}_{IBP}+\underbrace{\int_0^1\frac{\ln^3u}{1-u}\left(\operatorname{Li}_3(1-u)-\zeta(3)\right)\ du}_{1-u\ \mapsto\ u}\\ &=\frac14\int_0^1\frac{\ln^4u\operatorname{Li}_2(1-u)}{1-u}du+\underbrace{\int_0^1\frac{\ln^3(1-u)\operatorname{Li}_3(u)}{u}du}_{\large \text{our main integral}}-\zeta(3)\int_0^1\frac{\ln^3u}{1-u}du\\ &=\frac14\int_0^1\frac{\ln^4u\operatorname{Li}_2(1-u)}{1-u}du+I+6\zeta(3)\zeta(4) \end{align}

In my solution here I came across the remaining integral and here is the result:

$$\frac14\int_0^1\frac{\ln^4u\operatorname{Li}_2(1-u)}{1-u}du=6\zeta(2)\zeta(5)+36\zeta(7)-30\sum_{n=1}^\infty\frac{H_n}{n^6}-6\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^5}$$

Then

$$I_5=I+6\zeta(3)\zeta(4)+6\zeta(2)\zeta(5)+36\zeta(7)-30\sum_{n=1}^\infty\frac{H_n}{n^6}-6\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^5}$$ .

Note: We can not use the two sums $$\sum_{n=1}^\infty\frac{H_n^3}{n^4}$$ and $$\sum_{n=1}^\infty\frac{H_nH_n^{(2)}} {n^4}$$ in our solution because the integral $$I$$ is the key to evaluate these two sums.

• I assume $\mbox{Li}_3$ denotes a 3-atom Lithium molecule? Aug 7, 2019 at 15:08
• @Michael here is the definition $\operatorname{Li}_a(x)=\sum_{n=1}^\infty \frac{x^n}{n^a}$. its called polylogarithmic function. Aug 7, 2019 at 21:10
• So you want a solution in terms of Euler sums? Aug 8, 2019 at 15:35
• @SimplyBeautifulArt it's fine if you write solution in terms of easy sums not the sums I mentioned in the note. Aug 8, 2019 at 15:53

We can write $$\ln(1-x)=-\int_0^x \frac{dt}{1-t},\hspace{7mm}\operatorname{Li}_3(x)=\int_{0\leq t_1\leq t_2\leq t_3\leq x}\frac{dt_1\,dt_2\,dt_3}{(1-t_1)t_2t_3}.$$ We can multiply out $$\ln(1-x)^3\operatorname{Li}_3(x)$$ and break the result into a sum over the different possible orderings of the variables of integration. This will allow us to write $$I$$ as an integer linear combination of terms of the form $$\int_{0\leq t_1\leq t_2\leq t_3\leq t_4\leq t_5\leq t_6\leq t_7\leq 1}\frac{dt_1\,dt_2\,dt_3\,dt_4\,dt_5\,dt_6\,dt_7}{f_1(t_1)f_2(t_2)f_3(t_3)f_4(t_4)f_5(t_5)f_6(t_6)f_7(t_7)},$$ where three of the functions $$f_i$$ are $$f_i(x)=x$$ and the other four are $$f_i(x)=1-x$$. Each of these latter integrals is a multiple zeta value of depth $$4$$ and weight $$7$$. It's a bit messy, but I find $$I = -24\zeta(4, 1, 1, 1)-18\zeta(3, 2, 1, 1)-18\zeta(2, 3, 1, 1)-12\zeta(3, 1, 2, 1)-12\zeta(2, 2, 2, 1)-12\zeta(2, 1, 3, 1)-6\zeta(3, 1, 1, 2)-6\zeta(2, 2, 1, 2)-6\zeta(2, 1, 2, 2)-6\zeta(2, 1, 1, 3).$$ Every multiple zeta value of weight $$7$$ can be written in terms of the Riemann zeta function. There are a number of different ways to work out these expressions. One way is to use the generalized double-shuffle relations (see the paper ), and the expressions have been tabulated here. The result is $$I=-\frac{327}{8}\zeta(7)+\frac{21}{5}\zeta(2)^{2}\zeta(3)+12\zeta(2)\zeta(5).$$

 Ihara, Kentaro; Kaneko, Masanobu; Zagier, Don, Derivation and double shuffle relations for multiple zeta values, Compos. Math. 142, No. 2, 307-338 (2006). ZBL1186.11053.

• thanks for the efforts. Using MZV seems very helpful for high weight series. To be honest i didn't understand the solution as i am not familiar with it so i need to study this topic well. I prefer common methods though as they are more understandable for me and people. Aug 18, 2019 at 7:36
• @JulianRosen: Nice answer, interesting references (+1). Aug 21, 2019 at 8:55

This post is a supplement to the nice answer from @JulianRosen. We go into some details thereby confirming his results, the intermediate result as well as the final one. As this post is rather lengthy albeit simple, here is short overview:

• Step 1: Representation of $$I$$ by multiple zeta values (MZV) confirming Julian Rosen's intermediate result.

• Step 2: Simplification of $$I$$ by application of a sum relation and the duality theorem for MZVs.

• Step 3: We make a shortcut and use EZ-Face, an MZV calculator, to find linear dependencies of MZVs. This way we obtain and thereby confirm Julian Rosen's final result.

Step1: Representation of $$I$$ by multiple zeta values (MZV)

We start with \begin{align*} I&=\int_0^1\frac{1}{x}\ln^3(1-x)\mathrm{Li}_3(x)\,dx\\ &=\int_0^1\frac{1}{x}\left(-\int_0^x\frac{dt}{1-t}\right)^3\int_{0

In (1) we rewrite the integral to easily see how the region of integration \begin{align*} \{(t_1,t_2,t_3,t_4,t_5,t_6,t_7): 0 is to split in order to obtain an integral representations of the MZVs of the form \begin{align*} \int_{0 See e.g. chapter 7 of Lectures on Multiple Zeta Values by W. Zudilin for more information.

Each of the following integrals has a factor $$6$$ since according to the region of integration in (2) we can permute $$t_4,t_5,t_6$$.

We obtain from (1) \begin{align*} I&=(-6)\left(\int_{0<\color{blue}{t_1} in accordance with the intermediate result of Julian Rosen.

In (3) we marked the indices $$t_1,t_4,t_5,t_6$$ corresponding to $$\frac{1}{1-t_j}$$ blue, to better see the structure of the integral. In (4) we collect all terms with the same structure (appropriately substituing indices thereby).

Step 2: Simplification of $$I$$

Here we recall the Sum theorem (Theorem 2.5 in W. Zudilins paper): For any integer $$s>1$$ and $$l\geq 1$$ the following holds, \begin{align*} \sum_{{s_1>1,s_2\geq 1,\ldots,s_l\geq 1}\atop{s_1+s_2+\cdots+s_l=s}}\zeta(s_1,s_2,\ldots,s_l)=\zeta(s) \end{align*}

Setting $$l=4$$ and $$s=7$$ we obtain \begin{align*} \zeta(7)&=\zeta(4,1,1,1)+\zeta(3,2,1,1)+\zeta(2,3,1,1)\\ &+\zeta(3,1,2,1)+\zeta(2,2,2,1)+\zeta(2,1,3,1)\\ &+\zeta(3,1,1,2)+\zeta(2,2,1,2)+\zeta(2,1,2,2)\\ &+\zeta(2,1,1,3) \end{align*}

Putting this relation into (5) we obtain

\begin{align*} \color{blue}{I}&\color{blue}{=(-6)\left(3\zeta(4,1,1,1)+2\zeta(3,2,1,1)+2\zeta(2,3,1,1)\right.}\\ &\qquad\qquad\quad\color{blue}{+\zeta(3,1,2,1)+\zeta(2,2,2,1)+\zeta(2,1,3,1)}\\ &\qquad\qquad\quad\color{blue}{\left.+\zeta(7)\right)}\tag{6} \end{align*}

Next we apply the duality theorem (Theorem 3.7 in W. Zudilins paper) and replace this way the MZVs with length $$4$$ with MZVs of length $$3$$.

We obtain from (6) \begin{align*} \zeta(4,1,1,1)&=Z(x^3y\cdot y\cdot y\cdot y)=Z(x^3y^4)=Z(x^4y^3)=\zeta(5,1,1)\\ \zeta(3,2,1,1)&=Z(x^2y\cdot xy\cdot y\cdot y)=Z(x^2yxy^3)=Z(x^3yxy^2)=\zeta(4,2,1)\\ \zeta(2,3,1,1)&=Z(xy\cdot x^2y\cdot y\cdot y)=Z(xyx^2y^3)=Z(x^3y^2xy)=\zeta(4,1,2)\\ \zeta(3,1,2,1)&=Z(x^2y\cdot y\cdot xy\cdot y)=Z(x^2y^2xy^2)=Z(x^2yx^2y^2)=\zeta(3,3,1)\\ \zeta(2,2,2,1)&=Z(xy\cdot xy\cdot xy\cdot y)=Z(xyxyxy^2)=Z(x^2yxyxy)=\zeta(3,2,2)\\ \zeta(2,1,3,1)&=Z(xy\cdot y\cdot x^2y\cdot y)=Z(xy^2x^2y^2)=Z(x^2y^2x^2y)=\zeta(3,1,3)\tag{7} \end{align*}

From (6) and (7) we derive a simpler representation \begin{align*} \color{blue}{I}&\color{blue}{=(-6)\left(3\zeta(5,1,1)+2\zeta(4,2,1)+2\zeta(4,1,2)\right.}\\ &\qquad\color{blue}{+\zeta(3,3,1)+\zeta(3,2,2)+\zeta(3,1,3)}\\ &\qquad\color{blue}{\left.+\zeta(7)\right)}\tag{8} \end{align*}

Step 3: Representation of $$I$$ with single zeta values

We know that MZVs of weight $$7$$ are in the $$\mathbb{Q}$$-linear span of $$\zeta(7),\zeta(2)^2\zeta(3)$$ and $$\zeta(2)\zeta(5)$$. See e.g. exercise 3.3 (iv) in W. Zudilins paper. We use EZ-Face to find the linear relations.

E.g. typing the input

\begin{align*} \mathrm{lindep}([z(7),z(2)*z(2)*z(3), z(2)*z(5), z(5,1,1)]) \end{align*}

The output is:

\begin{align*} [10, -1., -4.,-2.] \end{align*}

which means \begin{align*} 10\zeta(7)-\zeta(2)^2\zeta(3)-4\zeta(2)\zeta(5)-2\zeta(5,1,1,1)=0 \end{align*}

We find this way the coefficients of the linear expressions for all MZVs of length $$4$$ in (8) as

$$\begin{array}{rrr|r|c} \zeta(7)&\zeta(2)^2\zeta(3)&\zeta(2)\zeta(5)&\zeta(\mathrm{arg})&\mathrm{arg}\\ \hline 10&-1&-4&-2&(5,1,1)\\ 1105&-112&-440&80&(4,2,1)\\ -5&12&-20&8&(4,1,2)\\ -61&0&36&8&(3,3,1)\\ -785&-72&600&80&(3,2,2)\\ -5&2&0&-20&(3,1,3) \end{array}$$

Substituting the MZVs in (8) with the linear representation of the table we finally obtain \begin{align*} \color{blue}{I}&\color{blue}{=(-6)\left(3\zeta(5,1,1)+2\zeta(4,2,1)+2\zeta(4,1,2)\right.}\\ &\qquad\color{blue}{+\zeta(3,3,1)+\zeta(3,2,2)+\zeta(3,1,3)}\\ &\qquad\color{blue}{\left.+\zeta(7)\right)}\\ &=(-6)\left(\zeta(7)\left(3\cdot\frac{10}{2}-2\cdot\frac{1105}{80}+2\cdot\frac{5}{8}+\frac{61}{8}+\frac{785}{80}-\frac{5}{20}+1\right)\right.\\ &\qquad\qquad\quad+\zeta(2)^2\zeta(3)\left(-3\cdot\frac{1}{2}+2\cdot\frac{112}{80}-2\cdot\frac{12}{8}+0+\frac{72}{80}+\frac{2}{20}\right)\\ &\qquad\qquad\left.\quad+\zeta(2)\zeta(5)\left(-3\cdot\frac{4}{2}+2\cdot\frac{440}{80}+2\cdot\frac{20}{8}-\frac{36}{8}-\frac{600}{80}+0\right)\right)\\ &\color{blue}{=-\frac{327}{8}\zeta(7)+\frac{21}{5}\zeta(2)^2\zeta(3)+12\zeta(2)\zeta(5)}\\ \end{align*}

in accordance with Julian Rosen's result.

• Fantastic answer. Aug 26, 2019 at 7:32
• @Klangen: Many thanks for your nice comment and the credit. Aug 26, 2019 at 8:28