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I would like some help to find a closed form for the following integral:$$\int_0^1 \frac{\left(\log (1+x)\right)^3}{x}dx $$ I was told it could be calculated in a closed form. I've already proved that $$\int_0^1 \frac{\log (1+x)}{x}dx = \frac{\pi^2}{12}$$ using power series expansion.
Thank you.
Letting $u = \log(1+x)$,
$$ \begin{align} \int_{0}^{1} \frac{\log^{3}(1+x)}{x} \ dx &= \int_{0}^{\log 2} \frac{u^{3}}{e^{u}-1} e^{u} \ du \\ &= \int_{0}^{\log 2} \frac{u^{3}}{1-e^{-u}} \ du \\ &= \int_{0}^{\log 2} u^{3} \sum_{n=0}^{\infty} e^{-nu} \ du \\ &= \sum_{n=0}^{\infty} \int_{0}^{\log 2} u^{3} e^{-nu} \ du \\ &= \int_{0}^{\log 2} u^{3} \ du + \sum_{n=1}^{\infty}\int_{0}^{\log 2} u^{3} e^{-nu} \ du \\ &= \frac{\log^{4}(2)}{4} + \sum_{n=1}^{\infty}\int_{0}^{\log 2} u^{3} e^{-nu} \ du .\end{align}$$
Then integrating by parts 3 times,
$$\begin{align} \int_{0}^{1} \frac{\log^{3}(1+x)}{x} \ dx &= \frac{\log^{4}(2)}{4} -\sum_{n=1}^{\infty} e^{-nu} \left(\frac{6}{n^{4}} + \frac{6u}{n^{3}} + \frac{3 u^{2}}{n^{2}} + \frac{u^{3}}{n} \right)\Bigg|^{\log 2}_{0} \\ &= \frac{\log^{4}(2)}{4} - \sum_{n=1}^{\infty} \left[\frac{1}{2^{n}} \left(\frac{6}{n^{4}} + \frac{6 \log 2}{n^{3}} + \frac{3 \log^{2} (2)}{n^{2}} + \frac{\log^{3}(2)}{n}\right) - 6 \zeta(4) \right] \\ &= -\frac{3\log^{4}(2)}{4 } - 6 \text{Li}_{4} \left(\frac{1}{2} \right) - 6 \log (2) \ \text{Li}_{3} \left(\frac{1}{2} \right)-3 \log^{2}(2) \text{Li}_{2} \left(\frac{1}{2} \right) + 6 \zeta(4) \\ &\approx 0.1425141979 . \end{align}$$
The answer could of course be simplified using the known values of $\text{Li}_{2} \left(\frac{1}{2} \right)$, $\text{Li}_{3} \left( \frac{1}{2}\right) $, and $\zeta(4)$.
Hints:
Substitute $x=\frac{t}{1-t}$, followed by $t=1-u$, and then expand the denominator via partial fractions:
$$\begin{align} \mathcal{I} &=\int_{0}^{1}\frac{\ln^3{\left(1+x\right)}}{x}\,\mathrm{d}x\\ &=\int_{0}^{\frac12}\frac{(1-t)\ln^3{\left(\frac{1}{1-t}\right)}}{t}\,\frac{1}{(1-t)^2}\mathrm{d}t\\ &=-\int_{0}^{\frac12}\frac{\ln^3{\left(1-t\right)}}{t(1-t)}\,\mathrm{d}t\\ &=-\int_{\frac12}^{1}\frac{\ln^3{\left(u\right)}}{u(1-u)}\,\mathrm{d}u\\ &=-\int_{\frac12}^{1}\frac{\ln^3{\left(u\right)}}{u}\,\mathrm{d}u-\int_{\frac12}^{1}\frac{\ln^3{\left(u\right)}}{1-u}\,\mathrm{d}u\\ &=-\left[\frac{\ln^4{(u)}}{4}\right]_{\frac12}^{1}-\int_{\frac12}^{1}\frac{\ln^3{\left(u\right)}}{1-u}\,\mathrm{d}u\\ &=\frac{\ln^4{(2)}}{4}-\int_{\frac12}^{1}\frac{\ln^3{\left(u\right)}}{1-u}\,\mathrm{d}u.\\ \end{align}$$
The final integral $\int_{\frac12}^{1}\frac{\ln^3{\left(u\right)}}{1-u}\,\mathrm{d}u$ is hardly trivial, but neither is it terribly difficult.
First note that \begin{align} \int \frac{\ln^{2}(1+x)}{x} \, dx = - 2 \, Li_{3}(1+x) + 2 \, Li_{2}(1+x) \, \ln(1+x) + \ln(-x) \, \ln^{2}(1+x) \end{align} for which \begin{align} I_{2} = \int_{0}^{1} \frac{\ln^{2}(1+x)}{x} \, dx = \frac{\zeta(3)}{4}. \end{align}
Now, \begin{align} \int \frac{\ln^{3}(1+x)}{x} \, dx = \int \ln(1+x) \, \frac{\ln^{2}(1+x)}{x} \, dx \end{align} can be integrated by parts. This leads to \begin{align} \int \frac{\ln^{3}(1+x)}{x} \, dx &= \ln(1+x) \left( - 2 \, Li_{3}(1+x) + 2 \, Li_{2}(1+x) \, \ln(1+x) + \ln(-x) \, \ln^{2}(1+x) \right) \\ & - \int \frac{- 2 \, Li_{3}(1+x) + 2 \, Li_{2}(1+x) \, \ln(1+x) + \ln(-x) \, \ln^{2}(1+x)}{1+x} \, dx \\ &= \ln(1+x) \left( - 2 \, Li_{3}(1+x) + 2 \, Li_{2}(1+x) \, \ln(1+x) + \ln(-x) \, \ln^{2}(1+x) \right) \\ & \hspace{10mm} + 6 Li_{4}(1+x) + Li_{2}(1+x) \, \ln^{2}(1+x) - 4 Li_{3}(1+x) \, \ln(1+x) \end{align} This leads to \begin{align} I_{3} &= \int_{0}^{1} \frac{\ln^{3}(1+x)}{x} \, dx \\ &= 6 Li_{4}(2) + \ln^{2}(2) \, Li_{2}(2) - 4 \ln(2) \, Li_{3}(2) - 6 \zeta(4) - \frac{7}{4} \, \ln(2) \, \zeta(3) \\ &= \frac{\pi^{4}}{15} + \frac{\pi^{2}}{4} \, \ln^{2}(2) - \frac{21}{4} \, \zeta(3) \, \ln(2) - \frac{1}{4} \, \ln^{4}(2) - 6 Li_{4}\left(\frac{1}{2} \right) \end{align}
By using the generalization
$$\int_0^1\frac{\ln^n(1+x)}{x}dx=\frac{\ln^{n+1}(2)}{n+1}+n!\zeta(n+1)+\sum_{k=0}^n k!{n\choose k}\ln^{n-k}(2)\operatorname{Li}_{k+1}\left(\frac12\right)$$
setting $n=3$ and subbing the special values of $$\operatorname{Li}_2(1/2)=\frac12\zeta(2)-\frac12\ln^22$$ and $$\operatorname{Li}_3(1/2)=\frac78\zeta(3)-\frac12\ln2\zeta(2)+\frac16\ln^32$$
we get the desired closed form of the integral in the question.
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\int_{0}^{1}{\ln^{3}\pars{1 + x} \over x}\,\dd x} = \int_{1}^{2}{\ln^{3}\pars{x} \over x - 1}\,\dd x = \int_{1}^{1/2}{\ln^{3}\pars{1/x} \over 1/x - 1} \pars{-\,{\dd x \over x^{2}}} \\[5mm] = &\ -\int_{1/2}^{1}{\ln^{3}\pars{x} \over x\pars{1 - x}}\,\dd x = \overbrace{-\int_{1/2}^{1}{\ln^{3}\pars{x} \over x}\,\dd x} ^{\ds{{1 \over 4}\ln^{4}\pars{2}}}\ -\ \int_{1/2}^{1}{\ln^{3}\pars{x} \over 1 - x}\,\dd x \\[5mm] = &\ {1 \over 4}\ln^{4}\pars{2} - \bracks{\left.-\ln\pars{1 - x}\ln^{3}\pars{x}\right\vert_{\ 1/2}^{\ 1} - 3\int_{1/2}^{1}\mrm{Li}'_{2}\pars{x}\ln^{2}\pars{x}\,\dd x} \\[5mm] = &\ -\,{3 \over 4}\ln^{4}\pars{2} -3\,\mrm{Li}_{2}\pars{1 \over 2}\ln^{2}\pars{2} - 6\int_{1/2}^{1}\mrm{Li}'_{3}\pars{x}\ln\pars{x}\,\dd x \\[5mm] = &\ -\,{3 \over 4}\ln^{4}\pars{2} - 3\,\mrm{Li}_{2}\pars{1 \over 2}\ln^{2}\pars{2} - 6\,\mrm{Li}_{3}\pars{1 \over 2}\ln\pars{2} + 6\int_{1/2}^{1}\mrm{Li}'_{4}\pars{x}\,\dd x \\[5mm] = &\ -\,{3 \over 4}\ln^{4}\pars{2} - 3\,\mrm{Li}_{2}\pars{1 \over 2}\ln^{2}\pars{2} - 6\,\mrm{Li}_{3}\pars{1 \over 2}\ln\pars{2} \\[2mm] & \phantom{\,\,\,}+\ 6\,\mrm{Li}_{4}\pars{1} - 6\mrm{Li}_{4}\pars{1 \over 2} \end{align} $\ds{\mrm{Li}_{4}\pars{1} = \zeta\pars{4} = \pi^{4}/90}$ and values of $\ds{\mrm{Li}_{2}\pars{1/2}}$ and $\ds{\mrm{Li}_{3}\pars{1/2}}$ are already known such that the final result is reduced to: \begin{align} &\bbox[10px,#ffd]{\int_{0}^{1}{\ln^{3}\pars{1 + x} \over x}\,\dd x} \\[5mm] = &\ \bbx{{\pi^{4} \over 15} + {1 \over 4}\,\pi^{2}\ln^{2}\pars{2} - {1 \over 4}\,\ln^{4}\pars{2} - 6\,\mrm{Li}_{4}\pars{1 \over 2} - {21 \over 4}\,\ln\pars{2}\zeta\pars{3}} \\[5mm] \approx &\ 0.1425 \end{align}
\begin{align*}J&=\int_0^1 \frac{\ln^3(1+x)}{x}\,dx\\ &\overset{y=\frac{1}{1+x}}=-\int_{\frac{1}{2}}^1 \frac{\ln^3 y}{y(1-y)}\,dy\\ &=-\int_{\frac{1}{2}}^1 \frac{\ln^3 y}{y}\,dy-\int_{\frac{1}{2}}^1 \frac{\ln^3 y}{1-y}\,dy\\ &=\frac{1}{4}\ln^4 2-\left(\int_0^1 \frac{\ln^3 y}{1-y}\,dy-\int_0^{\frac{1}{2}} \frac{\ln^3 y}{1-y}\,dy\right)\\ &=\frac{1}{4}\ln^4 2+6\zeta(4)+\int_0^{\frac{1}{2}} \frac{\ln^3 y}{1-y}\,dy\\ &\overset{u=2y}=\frac{1}{4}\ln^4 2+6\zeta(4)+\frac{1}{2}\int_0^1 \frac{\ln^3\left(\frac{u}{2}\right)}{1-\frac{u}{2}}\,du\\ &=\frac{1}{4}\ln^4 2+6\zeta(4)+\\&\frac{1}{2}\left(\int_0^1 \frac{\ln^3 u}{1-\frac{u}{2}}\,du+\int_0^1 \frac{3\ln^2 2\ln u}{1-\frac{u}{2}}\,du-\int_0^1 \frac{3\ln 2\ln^2 u}{1-\frac{u}{2}}\,du-\int_0^1 \frac{\ln^3 2}{1-\frac{u}{2}}\,du\right)\\ &=\frac{1}{4}\ln^4 2+6\zeta(4)-6\text{ Li}_{4}\left(\frac{1}{2}\right)-3\ln^2 2 \text{ Li}_{2}\left(\frac{1}{2}\right)-6\ln 2\text{ Li}_{3}\left(\frac{1}{2}\right)-\ln^4 2\\ &=6\zeta(4)-6\text{ Li}_{4}\left(\frac{1}{2}\right)-\frac{3}{4}\ln^4 2-3\ln^2 2\left(\frac{\pi^2}{12}-\frac{1}{2}\ln^2 2\right)-\\&6\ln 2\left(\frac{1}{6}\ln^3 2-\frac{1}{12}\pi^2 \ln 2+\frac{7}{8}\zeta(3)\right)\\ &=\boxed{\frac{1}{4}\pi^2 \ln^2 2-\frac{21}{4}\zeta(3)\ln 2+\frac{1}{15}\pi^4-6\text{ Li}_{4}\left(\frac{1}{2}\right)-\frac{1}{4}\ln^4 2} \end{align*}
NB:
I assume that for $0<a\leq 1$, $r\geq 1$, integer,
\begin{align} \int_0^1 \frac{\ln^r x}{1-ax}dx&=\frac{(-1)^r r!}{a}\text{Li}_{r+1}(a)\\ \text{Li}_{2}\left(\frac{1}{2}\right)&=\frac{\pi^2}{12}-\frac{1}{2}\ln^2 2\\ \text{Li}_{3}\left(\frac{1}{2}\right)&=\frac{1}{6}\ln^3 2-\frac{1}{12}\pi^2 \ln 2+\frac{7}{8}\zeta(3)\\ \zeta(4)&=\frac{\pi^4}{90} \end{align}
do you know this results ?
$\int_0^1 \frac{\left(\log (1+x)\right)^2}{x}dx=(ζ(3))/4$
