Evaluating $\int^1_0 \frac{\operatorname{Li}_3(x)}{1-x} \log(x)\, \mathrm dx$

Question

How would you solve the following

$$\int^1_0 \frac{\operatorname{Li}_3(x)}{1-x} \log(x)\, \mathrm dx$$

I might be able to relate the integral to Euler sums .

Answer 1

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\int_{0}^{1}{{\rm Li}_{3}\pars{x}\ln\pars{x} \over 1 - x}\,\dd x: \ {\large ?}}$.

In this post it's shown, in general grounds, that: $$ \int_{0}^{1}{{\rm Li}_{q}\pars{x}\ln^{r - 1}\pars{x} \over 1 - x}\,\dd x =\pars{-1}^{r - 1}\pars{r - 1}!\bracks{\zeta\pars{r}\zeta\pars{q}% -\sum_{n = 1}^{\infty}{H_{n}^{\rm\pars{r}} \over n^{q}}} $$ where $\ds{\zeta\pars{z}}$ is the Riemann Zeta Function and $\ds{H_{n}^{\rm \pars{r}} \equiv \sum_{k = 1}^{n}{1 \over k^{\rm r}}}$ is a Generalized Harmonic Number

such that $\ds{\pars{~\mbox{with}\ q = 3\ \mbox{and}\ r = 2~}}$: $$ \int_{0}^{1}{{\rm Li}_{3}\pars{x}\ln\pars{x} \over 1 - x}\,\dd x =\sum_{n = 1}^{\infty}{H_{n}^{\rm\pars{2}} \over n^{3}} - \zeta\pars{2}\zeta\pars{3}\,,\qquad H_{n}^{\rm\pars{2}} = \sum_{k = 1}^{n}{1 \over k^{2}} $$

In a comment of this answer, the OP $\pars{\tt @Zaid\ Alfayeai}$ pointed out that $\ds{\sum_{n = 1}^{\infty}{H_{n}^{\rm\pars{2}} \over n^{3}} = 3\zeta\pars{2}\zeta\pars{3} - {9 \over 2}\,\zeta\pars{5}}$ such that

$$\color{#66f}{\large% \int_{0}^{1}{{\rm Li}_{3}\pars{x}\ln\pars{x} \over 1 - x}\,\dd x =2\zeta\pars{2}\zeta\pars{3} - {9 \over 2}\,\zeta\pars{5}} \approx {\tt -0.7115} $$

Answer 2

A related problem. You can have the following closed form

$$\int^1_0 \frac{\operatorname{Li}_3(x)}{1-x} \log(x)\, dx= -\sum _{k=1}^{\infty }{\frac {\psi' \left( k+1 \right) }{{k}^{3}}}\sim -0.7115661976, $$

where $\psi(x)$ is the digamma function.

Another possible solution:

$$\int^1_0 \frac{\operatorname{Li}_3(x)}{1-x} \log(x)\, dx = \zeta(5) -\sum _{k=1}^{\infty }{\frac {\psi' \left( k \right) }{{k}^{3}}}\sim -0.7115661976. $$

Added: For the first one just use the power series expansions of $\operatorname{Li}_3(x)$ and $\frac{1}{1-x}$ and you end up with

$$\sum_{k=1}^{\infty}\frac{1}{k^3}\sum_{n=0}^{\infty} \int_{0}^{1} x^{k+n}\ln(x)dx=-\sum_{k=1}^{\infty}\frac{1}{k^3}\sum_{n=0}^{\infty}\frac{1}{(n+k+1)^2} $$

$$ = -\sum_{k=1}^{\infty}\frac{\psi'(k+1)}{k^3}. $$

If you manipulate the last sum, you will be able to relate it to the Euler sums as

$$ -\sum_{k=1}^{\infty}\frac{\psi'(k+1)}{k^3}= \zeta(5)-\sum_{n=1}^{\infty}\frac{H_n^{(3)}}{n^2}.$$

Note: Notice that, we are getting identities for $\zeta(5)$.

Answer 3

A nice way to calculate the main sum with a bonus

Using the well-known identity

$$\sum_{n=1}^\infty\frac{H_n^{(a)}}{n^b}+\sum_{n=1}^\infty\frac{H_n^{(b)}}{n^a}=\zeta(a)\zeta(b)+\zeta(a+b)$$

gives

$$\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}+\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^2}=\zeta(2)\zeta(3)+\zeta(5)\tag1$$

On the other hand, by Cauchy product we have

$$\operatorname{Li}_2(x)\operatorname{Li}_3(x)=\sum_{n=1}^\infty x^n\left(\frac{6H_n}{n^4}+\frac{3H_n^{(2)}}{n^3}+\frac{H_n^{(3)}}{n^2}-\frac{10}{n^5}\right)$$

set $x=1$ and use $\sum_{n=1}^\infty \frac{H_n}{n^4}=3\zeta(5)-\zeta(2)\zeta(3)$ we get

$$3\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}+\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^2}=7\zeta(2)\zeta(3)-8\zeta(5)\tag2$$

By $(1)$ and $(2)$ we get

$$\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}=3\zeta(2)\zeta(3)-\frac92\zeta(5)$$

and

$$\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^2}=\frac{11}2\zeta(5)-2\zeta(2)\zeta(3)$$

Answer 4

\begin{align} I&=\int_0^1\frac{\operatorname{Li}_3(x)\ln x}{1-x}\ dx=\sum_{n=1}^\infty\left(H_n^{(3)}-\frac1{n^3}\right)\int_0^1x^{n-1}\ln x\ dx\\ &=\sum_{n=1}^\infty \left(H_n^{(3)}-\frac1{n^3}\right)\left(-\frac{1}{n^2}\right)=\zeta(5)-\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^2}\tag{1} \end{align} lets start with the following sum \begin{align} S&=\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}=\sum_{n=1}^\infty\frac1{n^3}\left(\zeta(2)-\sum_{k=1}^\infty\frac{1}{(n+k)^2}\right)\\ &=\zeta(2)\zeta(3)-\sum_{k=1}^\infty\left(\sum_{n=1}^\infty\frac1{n^3(n+k)^2}\right)\\ &=\zeta(2)\zeta(3)-\sum_{k=1}^\infty\left(\sum_{n=1}^\infty\left(-\frac3{k^4}\left(\frac1n-\frac1{n+k}\right)+\frac2{k^3n^2}+\frac1{k^3(n+k)^2}-\frac1{k^2n^3}\right)\right)\\ &=\zeta(2)\zeta(3)-\sum_{k=1}^\infty\left(-\frac{3H_k}{k^4}+\frac{2\zeta(2)}{k^3}+\frac1{k^3}\left(\zeta(2)-H_k^{(2)}\right)-\frac{\zeta(3)}{k^2}\right)\\ &=\zeta(2)\zeta(3)-3\sum_{k=1}^\infty\frac{H_k}{k^4}+2\zeta(2)\zeta(3)+\zeta(2)\zeta(3)-S-\zeta(2)\zeta(3)\\ 2S&=3\zeta(2)\zeta(2)-3\left(3\zeta(5)-\zeta(2)\zeta(3)\right)\\ S&=3\zeta(2)\zeta(3)-\frac92\zeta(5) \end{align} using the well known formula $$\sum_{n=1}^\infty\frac{H_n^{(a)}}{n^b}+\sum_{n=1}^\infty\frac{H_n^{(b)}}{n^a}=\zeta(a)\zeta(b)+\zeta(a+b)$$ therefore $$\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^2}=\zeta(2)\zeta(3)+\zeta(5)-\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}=\frac{11}{2}\zeta(5)-2\zeta(2)\zeta(3)\tag{2}$$ plugging $(2)$ in $(1)$, we get $$I=2\zeta(2)\zeta(3)-\frac{9}{2}\zeta(5)$$

Answer 5

In Eq (12) of this article we have, for integers $q,p\ge1, q\ne1$ with $q+p$ is odd, the following identity:

\begin{equation*} \int_0^1\frac{\ln^{q-1}(x)\operatorname{Li}_{p}(x)}{1-x}\mathrm{d}x=\frac12(-1)^{q}(q-1)!\zeta(q+p)-\frac12(1+(-1)^q)(q-1)!\zeta(q)\zeta(p) \end{equation*} \begin{equation*} +(q-1)!\sum_{k=0}^{\lfloor{\frac{q}{2}}\rfloor}\binom{q+p-2k-1}{p-1}\zeta(2k)\zeta(q+p-2k) \end{equation*} \begin{equation} +(q-1)!\sum_{k=0}^{\lfloor{\frac{p}{2}}\rfloor}\binom{q+p-2k-1}{q-1}\zeta(2k)\zeta(q+p-2k). \end{equation}

Let $(q,p)=(2,3)$, we find

$$\int_0^1\frac{\ln(x)\operatorname{Li}_{3}(x)}{1-x}\mathrm{d}x=2\zeta(2)\zeta(3)-\frac{9}{2}\zeta(5).$$

Answer 6

\begin{align} \int^1_0\frac{\log{x} \ {\rm Li}_3(x)}{1-x}{\rm d}x &=\sum^\infty_{n=1}H_n^{(3)}\int^1_0x^n\log{x} \ {\rm d}x\\ &=-\sum^\infty_{n=1}\frac{H_n^{(3)}}{(n+1)^2}\\ &=\sum^\infty_{n=1}\frac{1}{(n+1)^5}-\sum^\infty_{n=1}\frac{H_{n+1}^{(3)}}{(n+1)^2}\\ &=\zeta(5)-\underbrace{\sum^\infty_{n=1}\frac{H_{n}^{(3)}}{n^2}}_{S} \end{align} Consider $\displaystyle f(z)=\frac{\pi\cot{\pi z} \ \Psi^{(2)}(-z)}{z^2}$. We know that \begin{align}\pi\cot{\pi z}&=\frac{1}{z-n}-2\sum^\infty_{k=1}\zeta(2k)(z-n)^{2k-1}\\&\approx\frac{1}{z-n}-2\zeta(2)(z-n)\end{align} (see here for a proof) and \begin{align}\Psi^{(2)}(-z)&=\frac{2}{(z-n)^3}+\sum^\infty_{k=2}(-1)^{k}k(k-1)\left(H_n^{(k+1)}+(-1)^{k+1}\zeta(k+1)\right)(z-n)^{k-2}\\&\approx\frac{2}{(z-n)^3}+2\left(H_n^{(3)}-\zeta(3)\right)\end{align} At the positive integers, \begin{align} {\rm Res}(f,n) &=\operatorname*{Res}_{z=n}\left[\frac{2}{z^2(z-n)^4}-\frac{4\zeta(2)}{z^2(z-n)^2}+\frac{2\left(H_n^{(3)}-\zeta(3)\right)}{z^2(z-n)}\right]\\ &=-\frac{8}{n^5}+\frac{8\zeta(2)}{n^3}+\frac{2H_n^{(3)}}{n^2}-\frac{2\zeta(3)}{n^2} \end{align} At the negative integers, \begin{align} {\rm Res}(f,-n) &=\frac{\Psi^{(2)}(n)}{n^2}\\ &=\frac{2H_{n}^{(3)}}{n^2}-\frac{2\zeta(3)}{n^2}-\frac{2}{n^5} \end{align} At $z=0$, \begin{align} {\rm Res}(f,0) &=[z^1]\left(\frac{1}{z}-2\zeta(2)z\right)\left(-2\zeta(3)-12\zeta(5)z^2\right)\\ &=-12\zeta(5)+4\zeta(2)\zeta(3) \end{align} Hence \begin{align}4S&=8\zeta(5)-8\zeta(2)\zeta(3)+2\zeta(2)\zeta(3)+2\zeta(2)\zeta(3)+2\zeta(5)+12\zeta(5)-4\zeta(2)\zeta(3)\\&=22\zeta(5)-8\zeta(2)\zeta(3)\end{align} which implies $$\color{blue}{\int^1_0\frac{\log{x} \ {\rm Li}_3(x)}{1-x}}=\zeta(5)-\frac{22\zeta(5)-8\zeta(2)\zeta(3)}{4}=\color{blue}{2\zeta(2)\zeta(3)-\frac{9}{2}\zeta(5)}$$

Answer 7

Solution $100\%$ Euler sums free. \begin{align}J&=\int^1_0 \frac{\operatorname{Li}_3(x)\ln x}{1-x} \, \mathrm dx\\ &=\frac12\int_0^1\frac{\ln x}{1-x}\left(\int_0^1\frac{x\ln^2t}{1-tx}dt\right)dx\\ &=\frac12\int_0^1\int_0^1\left(\frac{\ln^2t\ln x}{(1-t)(1-x)}-\frac{\ln^2t\ln x}{(1-t)(1-tx)}\right)dtdx\\ &=-\zeta(2)\zeta(3)-\frac12\int_0^1\int_0^1\frac{\ln^2t\left(\ln(tx)-\ln t\right)}{(1-t)(1-tx)}dtdx\\ &=-\zeta(2)\zeta(3)-\frac12\int_0^1\frac{\ln^2t}{t(1-t)}\left(\int_0^t\frac{\ln x}{1-x}dx\right)dt+\frac12\int_0^1\frac{\ln^3t}{t(1-t)}\left(\int_0^t\frac{1}{1-x}dx\right)dt\\ &=-\zeta(2)\zeta(3)-\frac12\underbrace{\int_0^1\frac{\ln^2t}{t}\left(\int_0^t\frac{\ln x}{1-x}dx\right)dt}_{\text{IBP}}-\frac12\underbrace{\int_0^1\frac{\ln^2t}{1-t}\left(\int_0^t\frac{\ln x}{1-x}dx\right)dt}_{=J_1}+\\ &\frac12\underbrace{\int_0^1\frac{\ln^3t}{t}\left(\int_0^t\frac{1}{1-x}dx\right)dt}_{\text{IBP}}+\frac12\underbrace{\int_0^1\frac{\ln^3t}{1-t}\left(\int_0^t\frac{1}{1-x}dx\right)dt}_{=J_2}\\ &=-\zeta(2)\zeta(3)+\frac{1}{24}\int_0^1\frac{\ln^4t}{1-t}dt-\frac12J_1+\frac12J_2=\boxed{-\zeta(2)\zeta(3)+\zeta(5)-\frac12J_1+\frac12J_2}\\ J_1&\overset{\text{IBP}}=-2\zeta(2)\zeta(3)-\int_0^1\int_0^1\frac{t\ln^2(tx)\ln t}{(1-t)(1-tx)}dtdx=-2\zeta(2)\zeta(3)-\\ &\underbrace{\int_0^1\int_0^1\frac{t\ln(tx)\ln^2 t}{(1-t)(1-tx)}dtdx}_{=J_1}-\underbrace{\int_0^1\int_0^1\frac{t\ln(tx)\ln t\ln x}{(1-t)(1-tx)}dtdx}_{=J_3}=\boxed{-\zeta(2)\zeta(3)-\frac12J_3}\\ J_3&=\int_0^1\int_0^1\left(\frac{\ln(tx)\ln t\ln x}{(1-t)(1-x)}-\frac{\ln(tx)\ln t\ln x}{(1-x)(1-tx)}\right)dtdx\\ &=-4\zeta(2)\zeta(3)-J_3-\int_0^1\int_0^1\frac{\ln(tx)(\ln(tx)-\ln x)\ln x}{1-tx}dtdx\\ &=-2\zeta(2)\zeta(3)-\frac12\int_0^1\frac{\ln x}x\left(\int_0^x\frac{\ln^2t}{1-t}dt\right)dx+\frac12\int_0^1\frac{\ln^2 x}x\left(\int_0^x\frac{\ln t}{1-t}dt\right)dx\\ &\overset{\text{IBP}}=-2\zeta(2)\zeta(3)+\frac1{12}\int_0^1\frac{\ln^4x}{1-x}dx\\ &\boxed{J_3=-2\zeta(2)\zeta(3)+2\zeta(5)}\\ &\boxed{J_1=-\zeta(5)} \end{align} \begin{align}J_2&=-\int_0^1\frac{\ln^3t\ln(1-t)}{1-t}dt\\ &\overset{\text{IBP}}=-\underbrace{\left[\left(\int_0^t\frac{\ln^3x}{1-x}dx-\int_0^1\frac{\ln^3x}{1-x}dx\right)\ln(1-t)\right]_0^1}_{=0}-\\&\int_0^1 \frac1{1-t}\left(\int_0^t\frac{\ln^3x}{1-x}dx-\int_0^1\frac{\ln^3x}{1-x}dx\right)dt\\ &=-\int_0^1\int_0^1\frac{1}{1-t}\left(\frac{t\ln^3(tx)}{1-tx}-\frac{\ln^3x}{1-x}\right)dtdx\\ &=-\int_0^1\int_0^1\left(\frac{\ln^3(tx)-\ln^3x}{(1-t)(1-x)}-\frac{\ln^3(tx)}{(1-tx)(1-x)}\right)dtdx\\ &\overset{\text{Fubini}}=12\zeta(2)\zeta(3)-\int_0^1\int_0^1\left(\frac{\ln^3t}{(1-t)(1-x)}-\frac{\ln^3(tx)}{(1-tx)(1-x)}\right)dtdx\\ &=12\zeta(2)\zeta(3)-\int_0^1\frac1{1-x}\left(\int_0^1\frac{\ln^3t}{1-t}dt-\frac1x\int_0^x\frac{\ln^3t}{1-t}dt\right)dx\\ &=12\zeta(2)\zeta(3)+\int_0^1\frac1x\left(\int_0^x\frac{\ln^3t}{1-t}dt\right)dx-\int_0^1\frac1{1-x}\left(\int_x^1\frac{\ln^3t}{1-t}dt\right)dx\\ &\overset{\text{IBP}}=12\zeta(2)\zeta(3)-\underbrace{\int_0^1\frac{\ln^4x}{1-x}dx}_{=24\zeta(5)}-J_2\\ &\boxed{J_2=6\zeta(2)\zeta(3)-12\zeta(5)}\\ &\boxed{J=2\zeta(2)\zeta(3)-\frac92\zeta(5)} \end{align} NB: I assume \begin{align}|a|\leq 1,\text{Li}_{r+1}(a)=\frac{(-1)^r}{r!}\int_0^1\frac{a\ln^rx}{1-ax}dx\end{align}

Answer 8

$$\int_0^1\ln(x)\frac{Li_1(x)}{1-x}dx=-\zeta(3)$$ $$\int_0^1\ln(x)\frac{Li_2(x)}{1-x}dx=-\frac{3}{10}\zeta^2(2)$$ The same method may work for the present case. (To be continued.)