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

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 .

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 $\zeta(5)$.

• You could complete it , it is not that difficult! – Zaid Alyafeai Aug 13 '13 at 17:51
• @ZaidAlyafeai: Ok. I'll be adding more material. – Mhenni Benghorbal Aug 13 '13 at 18:08
• We can get $$\sum_{n\geq 1} \frac{H^{(2)}_n}{n^3}=\frac{1}{2}\left(\pi^2 \zeta(3) - 9 \zeta(5) \right)$$ – Zaid Alyafeai Aug 13 '13 at 21:00
• @ZaidAlyafeai: it is not the same sum. – Mhenni Benghorbal Aug 14 '13 at 1:00
• @MhenniBenghorbal , yes . I just wanted to point that you can prove that the integral is equal to $$\frac{1}{6}\left(2 \pi^2 \zeta(3) - 27 \zeta(5) \right)$$ using the Euler sum you derived . – Zaid Alyafeai Aug 14 '13 at 2:54

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

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

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

\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 $$\color{blue}{\int^1_0\frac{\log{x} \ {\rm Li}_3(x)}{1-x}}=\zeta(5)-\frac{22\zeta(5)-8\zeta(2)\zeta(3)}{4}=\color{blue}{2\zeta(2)\zeta(3)-\frac{9}{2}\zeta(5)}$$

\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 $$I=2\zeta(2)\zeta(3)-\frac{9}{2}\zeta(5)$$

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