Integral $\int_0^1\frac{\ln(1-x)}{x}\text{Li}_3\left({x} \right)\,dx$

Question

Is there a closed form for this integral?

$\displaystyle \int_0^1\frac{\ln(1-x)}{x}\text{Li}_3\left({x} \right)\,dx\\$

All I have been able to find, so far, is a numeric approximation of $-1.13348$

Answer 1

Yes there is: $$\mathcal I=\zeta(2)\zeta(3)-3\zeta(5).$$ See e.g. this answer.

Answer 2

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\begin{align} &\color{#f00}{\int_{0}^{1}{\ln\pars{1 - x} \over x}\,\Li{3}{x}\,\dd x} = \int_{0}^{1}{\ln\pars{1 - x} \over x}\, \sum_{n = 1}^{\infty}{x^{n} \over n^{3}}\,\dd x \\[5mm] = &\ \sum_{n = 1}^{\infty}{1 \over n^{3}}\int_{0}^{1}\ln\pars{1 - x}x^{n - 1}\,\dd x \label{1}\tag{1} \end{align}

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However, \begin{align} &\fbox{$\ds{\ \int_{0}^{1}\ln\pars{1 - x}x^{n - 1}\,\dd x\ }$} = \lim_{\mu \to 0}\partiald{}{\mu}\int_{0}^{1}\pars{1 - x}^{\mu}x^{n - 1}\,\dd x \\[5mm] = &\ \lim_{\mu \to 0}\partiald{}{\mu} \bracks{\Gamma\pars{\mu + 1}\Gamma\pars{n} \over \Gamma\pars{\mu + n + 1}} =\ \fbox{$\ds{\ -\,{H_{n} \over n}\ }$} \end{align}

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Then $\ds{\pars{~\mbox{see expression}\ \eqref{1}~}}$, $$ \color{#f00}{\int_{0}^{1}{\ln\pars{1 - x} \over x}\,\Li{3}{x}\,\dd x} = -\sum_{n = 1}^{\infty}{H_{n} \over n^{4}} = \color{#f00}{{1 \over 6}\,\pi^{2}\zeta\pars{3} - 3\zeta\pars{5}} $$

$\ds{\sum_{n = 1}^{\infty}{H_{n} \over n^{4}} = {3\zeta\pars{5} - {1 \over 6}\,\pi^{2}}\,\zeta\pars{3} - }$ is given as formula $\pars{20}$ in the MathWorld Harmonic Number page and it has been evaluated in a 1995 David Borwein and Jonathan Borwein paper.

Answer 3

\begin{align} I&=\int_0^1\frac{\ln(1-x)}{x}\operatorname{Li_3}(x)\ dx=\sum_{n=1}^\infty\frac1{n^3}\int_0^1x^{n-1}\ln(1-x)\ dx=-\sum_{n=1}^\infty\frac{H_n}{n^4}=\zeta(2)\zeta(3)-3\zeta(5) \end{align}

Answer 4

I recommend using following recursion:

$$\operatorname{Li}_{0}(z) =\frac{z}{1-z}; \quad \operatorname{Li}_{n+1}(z) = \int_0^z \frac{\operatorname{Li}_n(t)}{t}$$