# How to find ${\large\int}_0^1\frac{\ln^3(1+x)\ln x}x\mathrm dx$

Please help me to find a closed form for this integral: $$I=\int_0^1\frac{\ln^3(1+x)\ln x}x\mathrm dx\tag1$$ I suspect it might exist because there are similar integrals having closed forms: \begin{align}\int_0^1\frac{\ln^3(1-x)\ln x}x\mathrm dx&=12\zeta(5)-\pi^2\zeta(3)\tag2\\ \int_0^1\frac{\ln^2(1+x)\ln x}x\mathrm dx&=\frac{\pi^4}{24}-\frac16\ln^42+\frac{\pi^2}6\ln^22-\frac72\zeta(3)\ln2-4\operatorname{Li}_4\!\left(\tfrac12\right)\tag3\\ \int_0^1\frac{\ln^3(1+x)\ln x}{x^2}\mathrm dx&=\frac34\zeta(3)-\frac{63}4\zeta(3)\ln2+\frac{23\pi^4}{120}\\&-\frac34\ln^42-2\ln^32+\frac{3\pi^2}4\ln^22-18\operatorname{Li}_4\!\left(\tfrac12\right).\tag4\end{align} Thanks!

• Check out these techniques. Aug 25, 2014 at 4:39
• Aug 25, 2014 at 4:49
• I'm surprised there isn't a single mention in the answers below that the integral is just a special case of the Nielsen generalized polylogarithm so $$I = 6S_{2,3}(-1)$$ and the evaluation of $S_{n,p}(-1)$ is known for small $n,p$ such as in this post. Jun 1, 2019 at 6:32

Start with integration by parts (IBP) by setting $u=\ln^3(1+x)$ and $dv=\dfrac{\ln x}{x}\ dx$ yields \begin{align} I&=-\frac32\int_0^1\frac{\ln^2(1+x)\ln^2 x}{1+x}\ dx\\ &=-\frac32\int_1^2\frac{\ln^2x\ln^2 (x-1)}{x}\ dx\quad\Rightarrow\quad\color{red}{x\mapsto1+x}\\ &=-\frac32\int_{\large\frac12}^1\left[\frac{\ln^2x\ln^2 (1-x)}{x}-\frac{2\ln^3x\ln(1-x)}{x}+\frac{\ln^4x}{x}\right]\ dx\quad\Rightarrow\quad\color{red}{x\mapsto\frac1x}\\ &=-\frac32\int_{\large\frac12}^1\frac{\ln^2x\ln^2 (1-x)}{x}\ dx+3\int_{\large\frac12}^1\frac{\ln^3x\ln(1-x)}{x}\ dx-\left.\frac3{10}\ln^5x\right|_{\large\frac12}^1\\ &=-\frac32\color{red}{\int_{\large\frac12}^1\frac{\ln^2x\ln^2 (1-x)}{x}\ dx}+3\int_{\large\frac12}^1\frac{\ln^3x\ln(1-x)}{x}\ dx-\frac3{10}\ln^52. \end{align} Applying IBP again to evaluate the red integral by setting $u=\ln^2(1-x)$ and $dv=\dfrac{\ln^2 x}{x}\ dx$ yields \begin{align} \color{red}{\int_{\large\frac12}^1\frac{\ln^2x\ln^2 (1-x)}{x}\ dx}&=\frac13\ln^52+\frac23\color{blue}{\int_{\large\frac12}^1\frac{\ln^3x\ln (1-x)}{1-x}\ dx}. \end{align}

For the simplicity, let $$\color{blue}{\mathbf{H}_{m}^{(k)}(x)}=\sum_{n=1}^\infty \frac{H_{n}^{(k)}x^n}{n^m}\qquad\Rightarrow\qquad\color{blue}{\mathbf{H}(x)}=\sum_{n=1}^\infty H_{n}x^n,$$ Introduce a generating function for the generalized harmonic numbers for $|x|<1$ $$\color{blue}{\mathbf{H}^{(k)}(x)}=\sum_{n=1}^\infty H_{n}^{(k)}x^n=\frac{\operatorname{Li}_k(x)}{1-x}\qquad\Rightarrow\qquad\color{blue}{\mathbf{H}(x)}=-\frac{\ln(1-x)}{1-x}$$ and the following identity $$H_{n+1}^{(k)}-H_{n}^{(k)}=\frac1{(n+1)^k}\qquad\Rightarrow\qquad H_{n+1}-H_{n}=\frac1{n+1}$$

Let us integrating the indefinite form of the blue integral. \begin{align} \color{blue}{\int\frac{\ln^3x\ln (1-x)}{1-x}\ dx}=&-\int\sum_{n=1}^\infty H_nx^n\ln^3x\ dx\\ =&-\sum_{n=1}^\infty H_n\int x^n\ln^3x\ dx\\ =&-\sum_{n=1}^\infty H_n\frac{\partial^3}{\partial n^3}\left[\int x^n\ dx\right]\\ =&-\sum_{n=1}^\infty H_n\frac{\partial^3}{\partial n^3}\left[\frac{x^{n+1}}{n+1}\right]\\ =&-\sum_{n=1}^\infty H_n\left[\frac{x^{n+1}\ln^3x}{n+1}-\frac{3x^{n+1}\ln^2x}{(n+1)^2}+\frac{6x^{n+1}\ln x}{(n+1)^3}-\frac{6x^{n+1}}{(n+1)^4}\right]\\ =&-\ln^3x\sum_{n=1}^\infty \frac{H_{n+1}x^{n+1}}{n+1}+\ln^3x\sum_{n=1}^\infty \frac{x^{n+1}}{(n+1)^2}+3\ln^2x\sum_{n=1}^\infty \frac{H_{n+1}x^{n+1}}{(n+1)^2}\\&-3\ln^2x\sum_{n=1}^\infty \frac{x^{n+1}}{(n+1)^3}-6\ln x\sum_{n=1}^\infty \frac{H_{n+1}x^{n+1}}{(n+1)^3}+6\ln x\sum_{n=1}^\infty \frac{x^{n+1}}{(n+1)^4}\\&+6\sum_{n=1}^\infty \frac{H_{n+1}x^{n+1}}{(n+1)^4}-6\sum_{n=1}^\infty \frac{x^{n+1}}{(n+1)^5}\\ =&\ -\sum_{n=1}^\infty\left[\frac{H_nx^{n}\ln^3x}{n}-\frac{x^{n}\ln^3x}{n^2}-\frac{3H_nx^{n}\ln^2x}{n^2}+\frac{3x^{n}\ln^2x}{n^3}\right.\\& \left.\ +\frac{6H_nx^{n}\ln x}{n^3}-\frac{6x^{n}\ln x}{n^4}-\frac{6H_nx^{n}}{n^4}+\frac{6x^{n}}{n^5}\right]\\ =&\ -\color{blue}{\mathbf{H}_{1}(x)}\ln^3x+\operatorname{Li}_2(x)\ln^3x+3\color{blue}{\mathbf{H}_{2}(x)}\ln^2x-3\operatorname{Li}_3(x)\ln^2x\\&\ -6\color{blue}{\mathbf{H}_{3}(x)}\ln x+6\operatorname{Li}_4(x)\ln x+6\color{blue}{\mathbf{H}_{4}(x)}-6\operatorname{Li}_5(x). \end{align} Therefore \begin{align} \color{blue}{\int_{\Large\frac12}^1\frac{\ln^3x\ln (1-x)}{1-x}\ dx} =&\ 6\color{blue}{\mathbf{H}_{4}(1)}-6\operatorname{Li}_5(1)-\left[\color{blue}{\mathbf{H}_{1}\left(\frac12\right)}\ln^32-\operatorname{Li}_2\left(\frac12\right)\ln^32\right.\\&\left.\ +3\color{blue}{\mathbf{H}_{2}\left(\frac12\right)}\ln^22-3\operatorname{Li}_3\left(\frac12\right)\ln^22+6\color{blue}{\mathbf{H}_{3}\left(\frac12\right)}\ln 2\right.\\&\ -6\operatorname{Li}_4(x)\ln 2+6\color{blue}{\mathbf{H}_{4}(x)}-6\operatorname{Li}_5(x)\bigg]\\ =&\ 12\zeta(5)-\pi^2\zeta(3)+\frac{3}8\zeta(3)\ln^22-\frac{\pi^4}{120}\ln2-\frac{1} {4}\ln^52\\&\ -6\color{blue}{\mathbf{H}_{4}\left(\frac12\right)}+6\operatorname{Li}_4\left(\frac12\right)\ln 2+6\operatorname{Li}_5\left(\frac12\right). \end{align} Using the similar approach as calculating the blue integral, then \begin{align} \int\frac{\ln^3x\ln (1-x)}{x}\ dx&=-\int\sum_{n=1}^\infty \frac{x^{n-1}}{n}\ln^3x\ dx\\ &=-\sum_{n=1}^\infty \frac{1}{n}\int x^{n-1}\ln^3x\ dx\\ &=-\sum_{n=1}^\infty \frac{1}{n}\frac{\partial^3}{\partial n^3}\left[\int x^{n-1}\ dx\right]\\ &=-\sum_{n=1}^\infty \frac{1}{n}\frac{\partial^3}{\partial n^3}\left[\frac{x^{n}}{n}\right]\\ &=-\sum_{n=1}^\infty \frac{1}{n}\left[\frac{x^{n}\ln^3x}{n}-\frac{3x^{n}\ln^2x}{n^2}+\frac{6x^{n}\ln x}{n^3}-\frac{6x^{n}}{n^4}\right]\\ &=\sum_{n=1}^\infty \left[-\frac{x^{n}\ln^3x}{n^2}+\frac{3x^{n}\ln^2x}{n^3}-\frac{6x^{n}\ln x}{n^4}+\frac{6x^{n}}{n^5}\right]\\ &=6\operatorname{Li}_5(x)-6\operatorname{Li}_4(x)\ln x+3\operatorname{Li}_3(x)\ln^2x-\operatorname{Li}_2(x)\ln^3x. \end{align} Hence $$\int_{\large\frac{1}{2}}^1\frac{\ln^3x\ln (1-x)}{x}\ dx=\frac{\pi^2}{6}\ln^32-\frac{21}{8}\zeta(3)\ln^22-6\operatorname{Li}_4\left(\frac{1}{2}\right)\ln2-6\operatorname{Li}_5\left(\frac{1}{2}\right)+6\zeta(5).$$ Combining altogether, we have

\begin{align} I=&\ \frac{\pi^4}{120}\ln2-\frac{33}4\zeta(3)\ln^22+\frac{\pi^2}2\ln^32-\frac{11}{20}\ln^52+6\zeta(5)+\pi^2\zeta(3)\\ &\ +6\color{blue}{\mathbf{H}_{4}\left(\frac12\right)}-18\operatorname{Li}_4\left(\frac12\right)\ln2-24\operatorname{Li}_5\left(\frac12\right). \end{align}

Continuing my answer in: A sum containing harmonic numbers $\displaystyle\sum_{n=1}^\infty\frac{H_n}{n^3\,2^n}$, we have \begin{align} \color{blue}{\mathbf{H}_{3}\left(x\right)}=&\frac12\zeta(3)\ln x-\frac18\ln^2x\ln^2(1-x)+\frac12\ln x\left[\color{blue}{\mathbf{H}_{2}\left(x\right)}-\operatorname{Li}_3(x)\right]\\&+\operatorname{Li}_4(x)-\frac{\pi^2}{12}\operatorname{Li}_2(x)-\frac12\operatorname{Li}_3(1-x)\ln x+\frac{\pi^4}{60}.\tag1 \end{align} Dividing $(1)$ by $x$ and then integrating yields \small\begin{align} \color{blue}{\mathbf{H}_{4}\left(x\right)}=&\frac14\zeta(3)\ln^2 x-\frac18\int\frac{\ln^2x\ln^2(1-x)}x\ dx+\frac12\int\frac{\ln x}x\bigg[\color{blue}{\mathbf{H}_{2}\left(x\right)}-\operatorname{Li}_3(x)\bigg]\ dx\\&+\operatorname{Li}_5(x)-\frac{\pi^2}{12}\operatorname{Li}_3(x)-\frac12\int\frac{\operatorname{Li}_3(1-x)\ln x}x\ dx+\frac{\pi^4}{60}\ln x\\ =&\frac14\zeta(3)\ln^2 x+\frac{\pi^4}{60}\ln x+\operatorname{Li}_5(x)-\frac{\pi^2}{12}\operatorname{Li}_3(x)-\frac18\color{red}{\int\frac{\ln^2x\ln^2(1-x)}x\ dx}\\&+\frac12\left[\color{purple}{\sum_{n=1}^\infty\frac{H_{n}}{n^2}\int x^{n-1}\ln x\ dx}-\color{green}{\int\frac{\operatorname{Li}_3(x)\ln x}x\ dx}-\color{orange}{\int\frac{\operatorname{Li}_3(1-x)\ln x}x\ dx}\right].\tag2 \end{align} Evaluating the red integral using the same technique as the previous one yields \begin{align} \color{red}{\int\frac{\ln^2x\ln^2(1-x)}x\ dx}&=\frac13\ln^3x\ln^2(1-x)-\frac23\color{blue}{\int\frac{\ln(1-x)\ln^3 x}{1-x}\ dx}. \end{align} Evaluating the purple integral yields \begin{align} \color{purple}{\sum_{n=1}^\infty\frac{H_{n}}{n^2}\int x^{n-1}\ln x\ dx}&=\sum_{n=1}^\infty\frac{H_{n}}{n^2}\frac{\partial}{\partial n}\left[\int x^{n-1}\ dx\right]\\ &=\sum_{n=1}^\infty\frac{H_{n}}{n^2}\left[\frac{x^n\ln x}{n}-\frac{x^n}{n^2}\right]\\ &=\color{blue}{\mathbf{H}_{3}(x)}\ln x-\color{blue}{\mathbf{H}_{4}(x)}. \end{align} Evaluating the green integral using IBP by setting $u=\ln x$ and $dv=\dfrac{\operatorname{Li}_3(x)}{x}\ dx$ yields \begin{align} \color{green}{\int\frac{\operatorname{Li}_3(x)\ln x}x\ dx}&=\operatorname{Li}_4(x)\ln x-\int\frac{\operatorname{Li}_4(x)}x\ dx\\ &=\operatorname{Li}_4(x)\ln x-\operatorname{Li}_5(x). \end{align} Evaluating the orange integral using IBP by setting $u=\operatorname{Li}_3(1-x)$ and $dv=\dfrac{\ln x}{x}\ dx$ yields \begin{align} \color{orange}{\int\frac{\operatorname{Li}_3(1-x)\ln x}x\ dx}&=\frac12\operatorname{Li}_3(1-x)\ln^2 x+\frac12\color{maroon}{\int\frac{\operatorname{Li}_2(1-x)\ln^2 x}{1-x}\ dx}. \end{align} Applying IBP again to evaluate the maroon integral by setting $u=\operatorname{Li}_2(1-x)$ and $$dv=\dfrac{\ln^2 x}{1-x}\ dx\quad\Rightarrow\quad v=2\operatorname{Li}_3(x)-2\operatorname{Li}_2(x)\ln x-\ln(1-x)\ln^2x,$$ we have \small{\begin{align} \color{maroon}{\int\frac{\operatorname{Li}_2(1-x)\ln^2 x}{1-x}\ dx}=&\left[2\operatorname{Li}_3(x)-2\operatorname{Li}_2(x)\ln x-\ln(1-x)\ln^2x\right]\operatorname{Li}_2(1-x)\\ &-2\int\frac{\operatorname{Li}_3(x)\ln x}{1-x}\ dx+2\int\frac{\operatorname{Li}_2(x)\ln x}{1-x}\ dx+\color{blue}{\int\frac{\ln(1-x)\ln^3 x}{1-x}\ dx}. \end{align}}

We use the generating function for the generalized harmonic numbers evaluate the above integrals involving polylogarithm.

\begin{align} \int\frac{\operatorname{Li}_k(x)\ln x}{1-x}\ dx&=\sum_{n=1}^\infty H_{n}^{(k)}\int x^n\ln x\ dx\\ &=\sum_{n=1}^\infty H_{n}^{(k)}\frac{\partial}{\partial n}\left[\int x^n\ dx\right]\\ &=\sum_{n=1}^\infty H_{n}^{(k)}\left[\frac{x^{n+1}\ln x}{n+1}-\frac{x^{n+1}}{(n+1)^2}\right]\\ &=\sum_{n=1}^\infty\left[\frac{H_{n+1}^{(k)}x^{n+1}\ln x}{n+1}-\frac{x^{n+1}\ln x}{(n+1)^{k+1}}-\frac{H_{n+1}^{(k)}x^{n+1}}{(n+1)^2}+\frac{x^{n+1}}{(n+1)^{k+2}}\right]\\ &=\sum_{n=1}^\infty\left[\frac{H_{n}^{(k)}x^{n}\ln x}{n}-\frac{x^{n}\ln x}{n^{k+1}}-\frac{H_{n}^{(k)}x^{n}}{n^2}+\frac{x^{n}}{n^{k+2}}\right]\\ &=\color{blue}{\mathbf{H}_{1}^{(k)}(x)}\ln x-\operatorname{Li}_{k+1}(x)\ln x-\color{blue}{\mathbf{H}_{2}^{(k)}(x)}+\operatorname{Li}_{k+2}(x). \end{align}

Dividing generating function of $\color{blue}{\mathbf{H}^{(k)}(x)}$ by $x$ and then integrating yields

\begin{align} \sum_{n=1}^\infty \frac{H_{n}^{(k)}x^n}{n}&=\int\frac{\operatorname{Li}_k(x)}{x(1-x)}\ dx\\ \color{blue}{\mathbf{H}_{1}^{(k)}(x)}&=\int\frac{\operatorname{Li}_k(x)}{x}\ dx+\int\frac{\operatorname{Li}_k(x)}{1-x}\ dx\\ &=\operatorname{Li}_{k+1}(x)+\int\frac{\operatorname{Li}_k(x)}{1-x}\ dx. \end{align}

Repeating the process above yields

\begin{align} \sum_{n=1}^\infty \frac{H_{n}^{(k)}x^n}{n^2} &=\int\frac{\operatorname{Li}_{k+1}(x)}{x}\ dx+\int\frac{\operatorname{Li}_k(x)}{x(1-x)}\ dx\\ \color{blue}{\mathbf{H}_{2}^{(k)}(x)}&=\operatorname{Li}_{k+2}(x)+\operatorname{Li}_{k+1}(x)+\int\frac{\operatorname{Li}_k(x)}{1-x}\ dx, \end{align}

where it is easy to show by using IBP that

\begin{align} \int\frac{\operatorname{Li}_2(x)}{1-x}\ dx&=-\int\frac{\operatorname{Li}_2(1-x)}{x}\ dx\\ &=2\operatorname{Li}_3(x)-2\operatorname{Li}_2(x)\ln(x)-\operatorname{Li}_2(1-x)\ln x-\ln (1-x)\ln^2x \end{align}

and

$$\int\frac{\operatorname{Li}_3(x)}{1-x}\ dx=-\int\frac{\operatorname{Li}_3(1-x)}{x}\ dx=-\frac12\operatorname{Li}_2^2(1-x)-\operatorname{Li}_3(1-x)\ln x.$$

Now, all unknown terms have been obtained. Putting altogether to $(2)$, we have \small{\begin{align} \color{blue}{\mathbf{H}_{4}(x)} =&\ \frac1{10}\zeta(3)\ln^2 x+\frac{\pi^4}{150}\ln x-\frac{\pi^2}{30}\operatorname{Li}_3(x)-\frac1{60}\ln^3x\ln^2(1-x)+\frac65\operatorname{Li}_5(x)\\&-\frac15\left[\operatorname{Li}_3(x)-\operatorname{Li}_2(x)\ln x-\frac12\ln(1-x)\ln^2x\right]\operatorname{Li}_2(1-x)-\frac15\operatorname{Li}_4(x)\\&-\frac35\operatorname{Li}_4(x)\ln x+\frac15\operatorname{Li}_3(x)\ln x+\frac15\operatorname{Li}_3(x)\ln^2x-\frac1{10}\operatorname{Li}_3(1-x)\ln^2 x\\&-\frac1{15}\operatorname{Li}_2(x)\ln^3x-\frac15\color{blue}{\mathbf{H}_{2}^{(3)}(x)}+\frac15\color{blue}{\mathbf{H}_{2}^{(2)}(x)} +\frac15\color{blue}{\mathbf{H}_{1}^{(3)}(x)}\ln x\\&-\frac15\color{blue}{\mathbf{H}_{1}^{(2)}(x)}\ln x+\frac25\color{blue}{\mathbf{H}_{3}(x)}\ln x-\frac15\color{blue}{\mathbf{H}_{2}(x)}\ln^2x+\frac1{15}\color{blue}{\mathbf{H}_{1}(x)}\ln^3x+C.\tag3 \end{align}} The next step is finding the constant of integration. Setting $x=1$ to $(3)$ yields \small{\begin{align} \color{blue}{\mathbf{H}_{4}(1)} &=-\frac{\pi^2}{30}\operatorname{Li}_3(1)+\frac65\operatorname{Li}_5(1)-\frac15\operatorname{Li}_4(1)-\frac15\color{blue}{\mathbf{H}_{2}^{(3)}(1)}+\frac15\color{blue}{\mathbf{H}_{2}^{(2)}(1)}+C\\ 3\zeta(5)+\zeta(2)\zeta(3)&=-\frac{\pi^2}{30}\operatorname{Li}_3(1)+\frac{19}{30}\operatorname{Li}_5(1)+\frac{3}{5}\operatorname{Li}_3(1)+C\\ C&=\frac{\pi^4}{450}+\frac{\pi^2}{5}\zeta(3)-\frac35\zeta(3)+3\zeta(5). \end{align}} Thus \small{\begin{align} \color{blue}{\mathbf{H}_{4}(x)} =&\ \frac1{10}\zeta(3)\ln^2 x+\frac{\pi^4}{150}\ln x-\frac{\pi^2}{30}\operatorname{Li}_3(x)-\frac1{60}\ln^3x\ln^2(1-x)+\frac65\operatorname{Li}_5(x)\\&-\frac15\left[\operatorname{Li}_3(x)-\operatorname{Li}_2(x)\ln x-\frac12\ln(1-x)\ln^2x\right]\operatorname{Li}_2(1-x)-\frac15\operatorname{Li}_4(x)\\&-\frac35\operatorname{Li}_4(x)\ln x+\frac15\operatorname{Li}_3(x)\ln x+\frac15\operatorname{Li}_3(x)\ln^2x-\frac1{10}\operatorname{Li}_3(1-x)\ln^2 x\\&-\frac1{15}\operatorname{Li}_2(x)\ln^3x-\frac15\color{blue}{\mathbf{H}_{2}^{(3)}(x)}+\frac15\color{blue}{\mathbf{H}_{2}^{(2)}(x)} +\frac15\color{blue}{\mathbf{H}_{1}^{(3)}(x)}\ln x\\&-\frac15\color{blue}{\mathbf{H}_{1}^{(2)}(x)}\ln x+\frac25\color{blue}{\mathbf{H}_{3}(x)}\ln x-\frac15\color{blue}{\mathbf{H}_{2}(x)}\ln^2x+\frac1{15}\color{blue}{\mathbf{H}_{1}(x)}\ln^3x\\&+\frac{\pi^4}{450}+\frac{\pi^2}{5}\zeta(3)-\frac35\zeta(3)+3\zeta(5)\tag4 \end{align}} and setting $x=\frac12$ to $(4)$ yields \begin{align} \color{blue}{\mathbf{H}_{4}\left(\frac12\right)}=&\ \frac{\ln^52}{40}-\frac{\pi^2}{36}\ln^32+\frac{\zeta(3)}{2}\ln^22-\frac{\pi^2}{12}\zeta(3)\\&+\frac{\zeta(5)}{32}-\frac{\pi^4}{720}\ln2+\operatorname{Li}_4\left(\frac12\right)\ln2+2\operatorname{Li}_5\left(\frac12\right).\tag5 \end{align}

Finally, we obtain

\begin{align} \int_0^1\frac{\ln^3(1+x)\ln x}x\ dx=&\ \color{blue}{\frac{\pi^2}2\zeta(3)+\frac{99}{16}\zeta(5)-\frac25\ln^52+\frac{\pi^2}3\ln^32-\frac{21}4\zeta(3)\ln^22}\\&\color{blue}{-12\operatorname{Li}_4\left(\frac12\right)\ln2-12\operatorname{Li}_5\left(\frac12\right)}, \end{align}

References :

$[1]\$ Harmonic number

$[2]\$ Polylogarithm

• @Tunk-Fey Very impressive! Aug 28, 2014 at 23:23
• Consider going through your question and making the latex narrower is some places. Especially where you are aligning equal signs. Alot of whitespace is wasted there. It will make it clearer to read. Other than that this answer is amazing :) Aug 29, 2014 at 19:58
• @Aditya Considering your age, you're still young and one day when you go to college and major in math (physics, engineering, or science cs), you will learn something like these stuffs. For now, you can start to learn from Achille Hui, sos440, Felix Marin, Random Variable, Sasha, Vladimir Reshetnikov, Pranav Arora, Omran Kouba, Integrals and Series, Rob John, Olivier Oloa, Integrals, Jack D'Aurizio, SuperAbound, Raymond Manzoni, etc. Lots of users here are better than me at integration. And please, don't become like me. Just be yourself. $\ddot\smile$ Aug 30, 2014 at 13:46
• @JackD'Aurizio Indeed! This answer crashed my browser. And to Tunk, I'm very impressed with the overall organization. You're getting very good at these polylog integrals. +1 Aug 30, 2014 at 18:27
• Thanks @FelixMarin. of course you're one of my teachers in polylog integrals. $\ddot\smile$ Sep 2, 2014 at 7:25

Indeed, there is a closed form for this integral: $$I=\frac{\pi^2}3\ln^32-\frac25\ln^52+\frac{\pi^2}2\zeta(3)+\frac{99}{16}\zeta(5)-\frac{21}4\zeta(3)\ln^22\\-12\operatorname{Li}_4\left(\frac12\right)\ln2-12\operatorname{Li}_5\left(\frac12\right).$$

• Well, Cleo's at it again. Well done. Aug 25, 2014 at 4:33
• Notice that $\ln2$ acts here like a regularized value of $\zeta(1)$. Aug 25, 2014 at 6:33
• @BennetGardiner I agree. Cleo at its best! :-) Her answers make me chuckling and I'm pretty sure, that Ramanujan is her favorite. Nevertheless I hope that other users can provide additional helpful information. Best regards, Aug 25, 2014 at 7:05
• @Cleo Do you mind giving a slight hint as to how one should proceed with this integral? Thanks. Aug 26, 2014 at 5:52
• @Lucian What do you mean by a regularized value? Aug 27, 2014 at 0:56

This is an updated partial answer that is rather similar to Jack D'Aurizio's approach. (I really hope he doesn't mind.)

Step 1: Expressing the integral as a sum.

It is easy to derive the formula $$\left(\sum^{\infty}_{n=1}a_nx^n\right)\left(\sum^{\infty}_{n=1}b_nx^n\right)=\sum^\infty_{n=1}\sum^{n}_{k=1}a_kb_{n-k+1}x^{n+1}$$ We apply this formula to derive the Taylor series of $\ln^2(1+x)$. \begin{align} \ln^2(1+x) &=\left(\sum^{\infty}_{n=1}\frac{(-1)^{n-1}}{n}x^n\right)\left(\sum^{\infty}_{n=1}\frac{(-1)^{n-1}}{n}x^n\right)\\ &=\sum^\infty_{n=1}\sum^n_{k=1}\frac{(-1)^{k-1}(-1)^{n-k}}{k(n-k+1)}x^{n+1}\\ &=\sum^\infty_{n=1}\frac{(-1)^{n+1}}{n+1}\sum^n_{k=1}\left(\frac{1}{k}+\frac{1}{n-k+1}\right)x^{n+1}\\ &=\sum^\infty_{n=1}\frac{(-1)^{n+1}2H_n}{n+1}x^{n+1} \end{align} Apply this formula again to obtain the Taylor series of $\displaystyle\frac{\ln^2(1+x)}{1+x}$. \begin{align} \frac{\ln^2(1+x)}{1+x} &=\left(\sum^\infty_{n=1}\frac{(-1)^{n+1}2H_n}{n+1}x^{n+1}\right)\left(\sum^{\infty}_{n=1}(-1)^{n-1}x^{n-1}\right)\\ &=\sum^\infty_{n=1}\sum^n_{k=1}\frac{(-1)^{k+1}(-1)^{n-k}2H_k}{k+1}x^{n+1}\\ &=\sum^\infty_{n=1}2(-1)^{n+1}\sum^n_{k=1}\frac{H_k}{k+1}x^{n+1}\\ \end{align} The inner sum is \begin{align} \sum^n_{k=1}\frac{H_k}{k+1} &=\sum^n_{k=1}\frac{H_{k+1}}{k+1}-\sum^n_{k=1}\frac{1}{(k+1)^2}\\ &=\sum^{n+1}_{k=1}\frac{H_k}{k}-H_{n+1}^{(2)}\\ &=\sum^{n+1}_{k=1}\frac{1}{k}\sum^k_{j=1}\frac{1}{j}-H_{n+1}^{(2)}\\ &=\sum^{n+1}_{j=1}\frac{1}{j}\left(\sum^{n+1}_{k=1}\frac{1}{k}-\sum^{j-1}_{k=1}\frac{1}{k}\right)-H_{n+1}^{(2)}\\ &=H_{n+1}^2-\sum^{n+1}_{j=1}\frac{H_j}{j}\\ &=\frac{H_{n+1}^2-H_{n+1}^{(2)}}{2} \end{align} Hence $$\frac{\ln^2(1+x)}{1+x}=\sum^\infty_{n=1}(-1)^{n+1}\left(H_{n+1}^2-H_{n+1}^{(2)}\right)x^{n+1}$$ Pluck this into the integral. \begin{align} \int^1_0\frac{\ln^3(1+x)\ln{x}}{x}{\rm d}x &=-\frac{3}{2}\int^1_0\frac{\ln^2(1+x)\ln^2{x}}{1+x}{\rm d}x\\ &=-\frac{3}{2}\sum^\infty_{n=1}(-1)^{n+1}\left(H_{n+1}^2-H_{n+1}^{(2)}\right)\int^1_0x^{n+1}\ln^2{x} \ {\rm d}x\\ &=-3\sum^\infty_{n=1}\frac{(-1)^{n+1}\left(H_{n+1}^2-H_{n+1}^{(2)}\right)}{(n+2)^3}\\ &=3\sum^\infty_{n=1}\frac{(-1)^{n}\left(H_{n}^{(2)}-H_{n}^2\right)}{(n+1)^3}\\ \end{align}

Step 2: Evaluation of $\displaystyle\sum^\infty_{n=1}\frac{(-1)^nH_n^{(2)}}{(n+1)^3}$

We begin with some simple manipulations of the sum. \begin{align} \sum^\infty_{n=1}\frac{(-1)^nH_n^{(2)}}{(n+1)^3} &=\sum^\infty_{n=1}\frac{(-1)^nH_{n+1}^{(2)}}{(n+1)^3}-\sum^\infty_{n=1}\frac{(-1)^n}{(n+1)^5}\\ &=-\frac{15}{16}\zeta(5)-\underbrace{\sum^\infty_{n=1}\frac{(-1)^nH_n^{(2)}}{n^3}}_{S} \end{align} Consider the function $\displaystyle f(z)=\frac{\pi\csc(\pi z)\psi_1(-z)}{z^3}$. At the positive integers, \begin{align} {\rm Res}(f,n) &=\operatorname*{Res}_{z=n}\left[\frac{(-1)^n}{z^3(z-n)^3}+\frac{(-1)^n(H_n^{(2)}+2\zeta(2))}{z^3(z-n)}\right]\\ &=\frac{6(-1)^n}{n^5}+\frac{(-1)^nH_n^{(2)}}{n^3}+\frac{2(-1)^n\zeta(2)}{n^3} \end{align} Summing them up gives $$\sum^\infty_{n=1} {\rm Res}(f,n)=-\frac{45}{8}\zeta(5)+S-\frac{3}{2}\zeta(2)\zeta(3)$$ At the negative integers, \begin{align} {\rm Res}(f,-n) &=-\frac{(-1)^n\psi_1(n)}{n^3}\\ &=\frac{(-1)^nH_n^{(2)}}{n^3}-\frac{(-1)^n\zeta(2)}{n^3}-\frac{(-1)^n}{n^5} \end{align} Summing them up gives $$\sum^\infty_{n=1} {\rm Res}(f,-n)=S+\frac{3}{4}\zeta(2)\zeta(3)+\frac{15}{16}\zeta(5)$$ At $z=0$, \begin{align} {\rm Res}(f,0) &=[z^2]\left(\frac{1}{z}+\zeta(2)z\right)\left(\frac{1}{z^2}+\zeta(2)+2\zeta(3)z+3\zeta(4)z^2+4\zeta(5)z^3\right)\\ &=4\zeta(5)+2\zeta(2)\zeta(3) \end{align} Since the sum of the reisudes $=0$, $$\sum^\infty_{n=1}\frac{(-1)^nH_n^{(2)}}{(n+1)^3}=-\frac{41}{32}\zeta(5)+\frac{5}{8}\zeta(2)\zeta(3)$$

Step 3: Evaluation of $\displaystyle\sum^\infty_{n=1}\frac{(-1)^nH_n^{2}}{(n+1)^3}$

Formula $(45)$ in this page states that this sum is equal to $$4{\rm Li}_5\left(\frac{1}{2}\right)+4{\rm Li}_4\left(\frac{1}{2}\right)\ln{2}+\frac{2}{15}\ln^5{2}-\frac{107}{32}\zeta(5)+\frac{7}{4}\zeta(3)\ln^2{2}-\frac{2}{3}\zeta(2)\ln^2{2}-\frac{3}{8}\zeta(2)\zeta(3)$$ Using a previously derived result is really unsatisfactory for me. Nevertheless, I have not been able to derive this result, as contour integration fails here due to the power of the denominator being odd (which implies that the sum will vanish when I add the residues at the positive and negative integers up). It seems that Tunk-Fey's brilliant approach would be the most viable method to crack this last sum.

Step 4: Obtaining the final result

Combining our previous results, we get \begin{align} &\ \ \ \ \ \small{\int^1_0\frac{\ln^3(1+x)\ln{x}}{x}{\rm d}x}\\ &=\small{3\sum^\infty_{n=1}\frac{(-1)^n\left(H_{n}^{(2)}-H_n^2\right)}{(n+1)^3}}\\ &=\small{3\left(\frac{33}{16}\zeta(5)+\zeta(2)\zeta(3)-4{\rm Li}_5\left(\frac{1}{2}\right)-4{\rm Li}_4\left(\frac{1}{2}\right)\ln{2}-\frac{2}{15}\ln^5{2}-\frac{7}{4}\zeta(3)\ln^2{2}+\frac{2}{3}\zeta(2)\ln^3{2}\right)}\\ &=\small{\frac{99}{16}\zeta(5)+\frac{\pi^2}{2}\zeta(3)-12{\rm Li}_5\left(\frac{1}{2}\right)-12{\rm Li}_4\left(\frac{1}{2}\right)\ln{2}-\frac{2}{5}\ln^5{2}-\frac{21}{4}\zeta(3)\ln^2{2}+\frac{\pi^2}{3}\ln^3{2}} \end{align}

• Perhaps for the last integral you can use identities $$\sum_{n=1}^\infty H_nx^n=-\frac{\ln(1-x)}{1-x}$$ and $$\int_{1/2}^1\frac{\partial^3}{\partial n^3}x^n\ dx=\frac{\partial^3}{\partial n^3}\left[\frac1{n+1}-\frac1{2^{n+1}(n+1)}\right].$$ Although, it'll be tedious. Aug 25, 2014 at 13:35
• @Tunk-Fey Thank you for your suggestion. In fact, I have previously tried that method, however, the third derivative of the second term turned out to be very ugly. wolframalpha.com/input/… If all else fails though, I would probably revert to using this method. Aug 25, 2014 at 14:10
• nice approach @SuperAbound . $\displaystyle\sum^\infty_{n=1}\frac{(-1)^nH_n^{2}}{(n+1)^3}$ can be evaluated elegantly. i will post my solution to this integral and your sum in the right time. i have a different short approach. Apr 30, 2019 at 23:55

Just a partial answer for now.

We have: $$I = -\frac{3}{2}\int_{0}^{1}\frac{\log^2(1+x)\log^2 x}{1+x}\,dx$$ and since: $$\log(1+z)=\sum_{n=1}^{+\infty}\frac{(-1)^{n+1}}{n}z^n$$ it follows that: $$[z^N]\log^2(1+z)=(-1)^{N+1}\sum_{n=1}^{N-1}\frac{1}{n(N-n)}=(-1)^{N+1}\frac{2H_{N-1}}{N},$$ $$\log^2(1+z)=\sum_{n=1}^{+\infty}\frac{2(-1)^{n+1} H_{n-1}}{n}z^{n}.\tag{1}$$ Let we focus now on: $$J_n = \int_{0}^{1}\frac{x^n\log^2 x}{1+x}\,dx=\frac{\partial^2}{\partial n^2}\int_{0}^{1}\frac{x^n}{1+x}\,dx.$$ We have: $$J_n = \frac{1}{4}\left(H_{n/2}^{(3)}-H_{(n-1)/2}^{(3)}\right),$$ hence: $$\color{blue}{I = -\frac{3}{4}\sum_{n=1}^{+\infty}\frac{(-1)^{n+1}H_{n-1}\left(H_{n/2}^{(3)}-H_{(n-1)/2}^{(3)}\right)}{n}}.\tag{2}$$ or, by partial summation: $$\color{purple}{I=-\frac{3}{4}\sum_{n=1}^{+\infty}H_{n/2}^{(3)}(-1)^n\left(\frac{H_n}{n+1}+\frac{H_{n-1}}{n}\right).}\tag{3}$$ Another identity that follows from the Taylor series of $\log^3(1-z)$ is: $$\color{red}{I=3\sum_{n=1}^{+\infty}\frac{(-1)^{n+1}\left(H_n^2-H_n^{(2)}\right)}{(n+1)^3}.}\tag{4}$$

An alternate form of the answers given by @Cleo and @Tunk-Fey as sum of $1$ and $1/2$ argumented polylogarithm-products with rational coefficients:

$$I = \frac{99}{16}\operatorname{Li}_5(1)-12\operatorname{Li}_5\left(\frac{1}{2}\right) + 15\operatorname{Li}_1\left( \frac{1}{2} \right)\operatorname{Li}_4(1) - 12\operatorname{Li}_1\left(\frac{1}{2}\right)\operatorname{Li}_4\left(\frac{1}{2}\right) - 15\operatorname{Li}_2\left( \frac{1}{2} \right)\operatorname{Li}_3(1)-\frac{51}{4}\operatorname{Li}_1^2\left( \frac{1}{2} \right)\operatorname{Li}_3(1)+12\operatorname{Li}_2(1)\operatorname{Li}_3\left( \frac{1}{2} \right) - \frac{2}{5}\operatorname{Li}_1^5\left(\frac{1}{2}\right),$$

where $\operatorname{Li}_n$ is the polylogarithm function, and specifically

\begin{align} & \operatorname{Li}_5(1) \ \ \ = \zeta(5) \\ & \operatorname{Li}_5\left(\textstyle\frac{1}{2}\right) = \textstyle \sum_{k=1}^\infty {2^{-k} \over k^5} \\ & \operatorname{Li}_4(1) \ \ \ = \zeta(4) = \frac{\pi^4}{90} \\ & \operatorname{Li}_4\left(\textstyle\frac{1}{2}\right) = \textstyle \sum_{k=1}^\infty {2^{-k} \over k^4} \\ & \operatorname{Li}_3(1) \ \ \ = \zeta(3) \\ & \operatorname{Li}_3\left(\textstyle\frac{1}{2}\right) = \frac{7}{8} \zeta(3) - \frac{\pi^2}{12} \ln 2 + \frac{1}{6} \ln^3 2 \\ & \operatorname{Li}_2(1) \ \ \ = \zeta(2) = \frac{\pi^2}{6} \\ & \operatorname{Li}_2\left(\textstyle\frac{1}{2}\right) = \frac{\pi^2}{12} - \frac{1}{2} \ln^2 2 \\ & \operatorname{Li}_1\left(\textstyle\frac{1}{2}\right) = \ln2, \end{align} where $\zeta$ is the Riemann zeta function.

Lets start with letting $$x=(1-y)/y$$ we have: \begin{align} I&=\int_0^1 \frac{\ln^3(1+x)\ln x}{x}\ dx\\ &=\int_{1/2}^1\frac{\ln^4x}{x}\ dx+\int_{1/2}^1\frac{\ln^4x}{1-x}\ dx-\int_{1/2}^1\frac{\ln^3x\ln(1-x)}{x}\ dx-\int_{1/2}^1\frac{\ln^3x\ln(1-x)}{1-x}\ dx \end{align} Applying IBP for the second integral, we get \begin{align} I&=3\int_{1/2}^1\frac{\ln^3x\ln(1-x)}{x}\ dx-\int_{1/2}^1\frac{\ln^3x\ln(1-x)}{1-x}\ dx-\frac45\ln^52\\ &=4\int_{1/2}^1\frac{\ln^3x\ln(1-x)}{x}\ dx-\int_{1/2}^1\frac{\ln^3x\ln(1-x)}{x(1-x)}\ dx-\frac45\ln^52\\ &=4I_1-I_2-\frac45\ln^52 \end{align} Evaluating the first integral: \begin{align} I_1&=\int_{1/2}^1\frac{\ln^3x\ln(1-x)}{x}\ dx=-\sum_{n=1}^\infty\frac1n\int_{1/2}^1x^{n-1}\ln^3x\ dx\\ &=-\sum_{n=1}^\infty\frac1n\left(\frac{6}{n^42^n}+\frac{6\ln2}{n^32^n}+\frac{3\ln^22}{n^22^n}+\frac{\ln^32}{n2^n}-\frac{6}{n^4}\right)\\ &=-6\operatorname{Li_5}\left(\frac12\right)-6\ln2\operatorname{Li_4}\left(\frac12\right)-3\ln^22\operatorname{Li_3}\left(\frac12\right)-\ln^32\operatorname{Li_2}\left(\frac12\right)+6\zeta(5) \end{align} Evaluating the second integral \begin{align} I_2&=\int_{1/2}^1\frac{\ln^3x\ln(1-x)}{x(1-x)}\ dx=-\sum_{n=1}^\infty H_n\int_{1/2}^1 x^{n-1}\ln^3x\ dx\\ &=-\sum_{n=1}^\infty H_n\left(\frac{6}{n^42^n}+\frac{6\ln2}{n^32^n}+\frac{3\ln^22}{n^22^n}+\frac{\ln^32}{n2^n}-\frac{6}{n^4}\right)\\ &=-6\left(\color{blue}{\sum_{n=1}^\infty\frac{H_n}{n^42^n}+\ln2\sum_{n=1}^\infty\frac{H_n}{n^32^n}}\right)-3\ln^22\sum_{n=1}^\infty\frac{H_n}{n^22^n}-\ln^32\sum_{n=1}^\infty\frac{H_n}{n2^n}+6\sum_{n=1}^\infty\frac{H_n}{n^4} \end{align} I was able here to prove: $$\color{blue}{\sum_{n=1}^\infty\frac{H_n}{n^42^n}+\ln2\sum_{n=1}^\infty\frac{H_n}{n^32^n}} =-\frac12\ln^22\sum_{n=1}^{\infty}\frac{H_n}{n^22^n}-\frac16\ln^32\sum_{n=1}^{\infty}\frac{H_n}{n2^n}+\frac12\sum_{n=1}^{\infty}\frac{H_n}{n^4}-\frac{47}{32}\zeta(5) +\frac{1}{15}\ln^52+\frac{1}{3}\ln^32\operatorname{Li_2}\left( \frac12\right)+\ln^22\operatorname{Li_3}\left( \frac12\right)+2\ln2\operatorname{Li_4}\left( \frac12\right) +2\operatorname{Li_5}\left( \frac12\right)$$ which follows that: \begin{align*} I_2&=3\sum_{n=1}^{\infty}\frac{H_n}{n^4} -12\operatorname{Li_5}\left(\frac12\right)-12\ln2\operatorname{Li_4}\left( \frac12\right)-6\ln^22\operatorname{Li_3}\left( \frac12\right)\\ &\quad-2\ln^32\operatorname{Li_2}\left(\frac12\right)-\frac6{15}\ln^52+\frac{141}{16}\zeta(5) \end{align*} Grouping $$I_1$$ and $$I_2$$ we have: \begin{align} I&=-3\sum_{n=1}^\infty\frac{H_n}{n^4}-12\operatorname{Li_5}\left(\frac12\right)-12\ln2\operatorname{Li_4}\left( \frac12\right)-6\ln^22\operatorname{Li_3}\left( \frac12\right)\\ &\quad-2\ln^32\operatorname{Li_2}\left( \frac12\right)+\frac{243}{16}\zeta(5)-\frac25\ln^52 \end{align} Using the following common values: $$\sum_{n=1}^\infty \frac{H_n}{n^4}=3\zeta(5)-\zeta(2)\zeta(3)$$ $$\operatorname{Li_3}\left( \frac12\right)=\frac78\zeta(3)-\frac12\ln2\zeta(2)+\frac16\ln^32$$ $$\operatorname{Li_2}\left( \frac12\right) =\frac12\zeta(2)-\frac12\ln^22$$

Finally we get: \begin{align} I&=-12\operatorname{Li}_5\left(\frac12\right)-12\ln2\operatorname{Li}_4\left(\frac12\right)+\frac{99}{16}\zeta(5)+3\zeta(2)\zeta(3)\\ &\quad-\frac{21}4\ln^22\zeta(3)+2\ln^32\zeta(2)-\frac25\ln^52 \end{align}

UPDATE: The way below may be found in the preprint, A new perspective on the evaluation of the logarithmic integral, $$\int_0^1\frac{\log(x)\log^3(1+x)}{x}\textrm{d}x$$ by C.I.Valean.

A magical way proposed by Cornel Ioan Valean

We use the powerful form of the Beta function presented in the book, (Almost) Impossible Integrals, Sums, and Series, $$\displaystyle \int_0^1 \frac{x^{a-1}+x^{b-1}}{(1+x)^{a+b}} \textrm{d}x = \operatorname{B}(a,b)$$, (see pages $$72$$-$$73$$).

Here is the magic ...

By cleverly differentiating in two different ways to get rid of a nasty integral, we simply get the wonderful result

$$4\lim_{\substack{a\to0 \\ b \to 0}}\frac{\partial^{4}}{\partial a^3 \partial b}\operatorname{B}(a,b)-6\lim_{\substack{a\to0 \\ b \to 0}}\frac{\partial^{4}}{\partial a^2 \partial b^2}\operatorname{B}(a,b)$$ $$=8\int_0^1 \frac{\log(x)\log^3(1+x)}{x}\textrm{d}x-4\int_0^1 \frac{\log^3(x)\log(1+x)}{x}\textrm{d}x-4\int_0^1 \frac{\log^4(1+x)}{x}\textrm{d}x.$$ ... and we're wonderfully done!

A first note: a similar strategy has been used in this answer https://math.stackexchange.com/q/3531878.

A BIG BONUS (the extraction of the series $$\displaystyle \sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_n}{n^4}$$):

The extraction of the series $$\displaystyle \sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_n}{n^4}$$ is achieved immediately by observing that using the same Beta function limits, we arrive at

$$\lim_{\substack{a\to0 \\ b \to 0}}\frac{\partial^{4}}{\partial a^3 \partial b}\operatorname{B}(a,b)-\lim_{\substack{a\to0 \\ b \to 0}}\frac{\partial^{4}}{\partial a^2 \partial b^2}\operatorname{B}(a,b)$$ $$=\underbrace{\int_0^1 \frac{\log^2(x)\log^2(1+x)}{x}\textrm{d}x}_{\displaystyle 15/4\zeta(5)-4\sum_{n=1}^{\infty} (-1)^{n-1} H_n/n^4}-\int_0^1 \frac{\log^3(x)\log(1+x)}{x}\textrm{d}x,$$ which assures the desired extraction after turning the second integral into the series we want to calculate.

• Very useful technique for such integrals. (+1) Feb 3, 2020 at 2:08

Here is a simple approach that does not involve many results.

First, let $$x=(1-y)/y$$ to have: \begin{align} I&=\int_0^1 \frac{\ln^3(1+x)\ln x}{x}\ dx\\ &=\int_{1/2}^1\frac{\ln^4x}{x}\ dx+\int_{1/2}^1\frac{\ln^4x}{1-x}\ dx-\underbrace{\int_{1/2}^1\frac{\ln^3x\ln(1-x)}{x}\ dx}_{IBP}-\underbrace{\int_{1/2}^1\frac{\ln^3x\ln(1-x)}{1-x}\ dx}_{x\mapsto 1-x}\\ &=\frac15\ln^52+\int_{1/2}^1\frac{\ln^4x}{1-x}\ dx-\left(\frac14\ln^52+\frac14\int_{1/2}^1\frac{\ln^4x}{1-x}\ dx\right)-\underbrace{\int_{0}^{1/2}\frac{\ln^3(1-x)\ln x}{x}\ dx}_{\int_0^1-\int_{1/2}^1}\\ &=-\frac1{20}\ln^52+\frac34\int_{1/2}^1\frac{\ln^4x}{1-x}\ dx-\int_0^1\frac{\ln^3(1-x)\ln x}{x}\ dx+\color{blue}{\int_{1/2}^1\frac{\ln^3(1-x)\ln x}{x}\ dx} \end{align}

We have (proved below)

$$\color{blue}{\int_{1/2}^1\frac{\ln^3(1-x)\ln x}{x}\ dx}=\frac3{16}\zeta(5)+\frac3{20}\ln^52-\frac14\int_{1/2}^1\frac{\ln^4x}{1-x}\ dx+\frac12\int_0^1\frac{\ln^3(1-x)\ln x}{x}\ dx$$

Then we can write

$$I=\frac3{16}\zeta(5)+\frac1{10}\ln^52+\frac12\int_{1/2}^1\frac{\ln^4x}{1-x}\ dx-\frac12\int_0^1\frac{\ln^3(1-x)\ln x}{x}\ dx$$

Lets evaluate the first integral

$$\int_{1/2}^1\frac{\ln^4x}{1-x}\ dx=\sum_{n=1}^\infty\int_{1/2}^1 x^{n-1}\ln^4x\ dx$$ $$=\sum_{n=1}^\infty\left(\frac{24}{n^5}-\frac{24}{n^52^n}-\frac{24\ln2}{n^42^n}-\frac{12\ln^22}{n^32^n}-\frac{4\ln^32}{n^22^n}-\frac{\ln^42}{n2^n}\right)$$

$$=24\zeta(5)-24\operatorname{Li}_5\left(\frac12\right)-24\ln2\operatorname{Li}_4\left(\frac12\right)-12\ln^22\operatorname{Li}_3\left(\frac12\right)-4\ln^32\operatorname{Li}_2\left(\frac12\right)-\ln^52$$

$$=\boxed{4\ln^32\zeta(2)-\frac{21}2\ln^22\zeta(3)+24\zeta(5)-\ln^52-24\ln2\operatorname{Li}_4\left(\frac12\right)-24\operatorname{Li}_5\left(\frac12\right)}$$

where we used $$\operatorname{Li}_2\left(\frac12\right)=\frac12\zeta(2)-\frac12\ln^22$$ and $$\operatorname{Li}_3\left(\frac12\right)=\frac78\zeta(3)-\frac12\ln^22\zeta(2)+\frac16\ln^32$$

and the second integral

$$\int_0^1\frac{\ln^3(1-x)\ln x}{x}\ dx=\int_0^1\frac{\ln^3x\ln(1-x)}{1-x}\ dx$$ $$=-\sum_{n=1}^\infty H_n\int_0^1x^n\ln^3x\ dx=6\sum_{n=1}^\infty\frac{H_n}{(n+1)^4}$$ $$=6\sum_{n=1}^\infty\frac{H_n}{n^4}-6\zeta(5)=6\left(3\zeta(5)-\zeta(2)\zeta(3)\right)-6\zeta(5)=\boxed{12\zeta(5)-6\zeta(2)\zeta(3)}$$

Combining the boxed results gives

\begin{align} I&=-12\operatorname{Li}_5\left(\frac12\right)-12\ln2\operatorname{Li}_4\left(\frac12\right)+\frac{99}{16}\zeta(5)+3\zeta(2)\zeta(3)\\ &\quad-\frac{21}4\ln^22\zeta(3)+2\ln^32\zeta(2)-\frac25\ln^52 \end{align}

Proof of the blue integral: $$\color{blue}{A=\int_{1/2}^1\frac{\ln^3(1-x)\ln x}{x}\ dx}$$

We have the algebraic identity

$$4a^3b=a^4+b^4-(a-b)^4-4ab^3+6a^2b^2$$

set $$a=\ln(1-x)$$ and $$b=\ln x$$ and divide both sides by $$x$$ then integrate we get

$$\color{blue}{4A}=\underbrace{\int_{1/2}^1\frac{\ln^4(1-x)}{x}dx}_{x\mapsto1-x}+\underbrace{\int_{1/2}^1\frac{\ln^4x}{x}dx}_{\frac15\ln^52}-\underbrace{\int_{1/2}^1\frac1x\ln^4\left(\frac{1-x}{x}\right)dx}_{(1-x)/x= y}\\-4\underbrace{\int_{1/2}^1\frac{\ln(1-x)\ln^3x}{x}dx}_{IBP}+\underbrace{6\int_{1/2}^1\frac{\ln^2(1-x)\ln^2x}{x}dx}_{B}$$

$$=\underbrace{\int_0^{1/2}\frac{\ln^4x}{1-x}\ dx}_{\int_0^1-\int_{1/2}^1}+\frac15\ln^52-\underbrace{\int_0^1\frac{\ln^4x}{1+x}\ dx}_{\frac{45}2\zeta(5)}-4\left(\frac14\ln^52+\frac14\int_{1/2}^1\frac{\ln^4x}{1-x}\ dx\right)+B$$

$$=\int_0^1\frac{\ln^4x}{1-x}\ dx-2\int_{1/2}^1\frac{\ln^4x}{1-x}\ dx-\frac45\ln^52-\frac{45}2\zeta(5)+B$$

$$=24\zeta(5)-2\int_{1/2}^1\frac{\ln^4x}{1-x}\ dx-\frac45\ln^52-\frac{45}2\zeta(5)+B\tag{1}$$

Lets simplify the integral $$B$$

\begin{align} B&=6\int_{1/2}^1\frac{\ln^2(1-x)\ln^2x}{x}\ dx\overset{IBP}{=}2\ln^52+4\int_{1/2}^1\frac{\ln^3x\ln(1-x)}{1-x}\ dx\\ &\overset{x\mapsto1-x}{=}2\ln^52+4\underbrace{\int_{0}^{1/2}\frac{\ln^3(1-x)\ln x}{x}\ dx}_{\int_0^1-\int_{1/2}^1}\\ &=2\ln^52+4\int_{0}^{1}\frac{\ln^3(1-x)\ln x}{x}\ dx-\color{blue}{4A}\tag{2} \end{align}

Plugging (2) in (1) we have that

$$\color{blue}{8A}=\frac32\zeta(5)+\frac6{5}\ln^52-2\int_{1/2}^1\frac{\ln^4x}{1-x}\ dx+4\int_0^1\frac{\ln^3(1-x)\ln x}{x}\ dx$$

Or $$\boxed{\color{blue}{A}=\frac3{16}\zeta(5)+\frac3{20}\ln^52-\frac14\int_{1/2}^1\frac{\ln^4x}{1-x}\ dx+\frac12\int_0^1\frac{\ln^3(1-x)\ln x}{x}\ dx}$$

Here is a proof for $$\left(4\right)$$ since i couldn't find one: $$\int _0^1\frac{\ln ^3\left(1+x\right)\ln \left(x\right)}{x^2}\:dx$$ $$\overset{\operatorname{IBP}}=-\ln ^3\left(2\right)+3\int _0^1\frac{\ln ^2\left(1+x\right)}{x\left(1+x\right)}\:dx+3\int _0^1\frac{\ln \left(x\right)\ln ^2\left(1+x\right)}{x\left(1+x\right)}\:dx$$

$$3\underbrace{\int _0^1\frac{\ln ^2\left(1+x\right)}{x\left(1+x\right)}\:dx}_{x=\frac{1}{1+x}}=3\int _0^1\frac{\ln ^2\left(x\right)}{1-x}\:dx-3\int _0^{\frac{1}{2}}\frac{\ln ^2\left(x\right)}{1-x}\:dx$$ $$=6\sum _{k=1}^{\infty }\frac{1}{k^3}-6\sum _{k=1}^{\infty }\frac{1}{k^3\:2^k}-6\ln \left(2\right)\sum _{k=1}^{\infty }\frac{1}{k^2\:2^k}-3\ln ^3\left(2\right)$$ $$=6\zeta \left(3\right)-6\operatorname{Li}_3\left(\frac{1}{2}\right)-6\ln \left(2\right)\operatorname{Li}_2\left(\frac{1}{2}\right)-3\ln ^3\left(2\right)$$ $$=\frac{3}{4}\zeta \left(3\right)-\ln ^3\left(2\right)$$

$$3\underbrace{\int _0^1\frac{\ln \left(x\right)\ln ^2\left(1+x\right)}{x\left(1+x\right)}\:dx}_{x=\frac{1}{1+x}}$$ $$=3\int _0^{\frac{1}{2}}\frac{\ln \left(x\right)\ln ^2\left(1-x\right)}{x}\:dx-3\int _{\frac{1}{2}}^1\frac{\ln ^3\left(x\right)}{1-x}\:dx$$ $$=-6\sum _{k=1}^{\infty }\frac{H_k}{k^3\:2^k}-6\ln \left(2\right)\sum _{k=1}^{\infty }\frac{H_k}{k^2\:2^k}+6\sum _{k=1}^{\infty }\frac{1}{k^4\:2^k}+6\ln \left(2\right)\sum _{k=1}^{\infty }\frac{1}{k^3\:2^k}+18\sum _{k=1}^{\infty }\frac{1}{k^4}$$ $$-18\sum _{k=1}^{\infty }\frac{1}{k^4\:2^k}-18\ln \left(2\right)\sum _{k=1}^{\infty }\frac{1}{k^3\:2^k}-9\ln ^2\left(2\right)\sum _{k=1}^{\infty }\frac{1}{k^2\:2^k}-3\ln ^4\left(2\right)$$ $$=\frac{69}{4}\zeta \left(4\right)-18\operatorname{Li}_4\left(\frac{1}{2}\right)-\frac{63}{4}\ln \left(2\right)\zeta \left(3\right)+\frac{9}{2}\ln ^2\left(2\right)\zeta \left(2\right)-\frac{3}{4}\ln ^4\left(2\right)$$ Where $$\ln ^2\left(1-x\right)=2\sum _{k=1}^{\infty }\left(\frac{H_k}{k}-\frac{1}{k^2}\right)x^k$$ is used on the $$2$$nd line.

See here and here for the $$1$$st and $$2$$nd sum.

Collecting the results yields: $$\int _0^1\frac{\ln ^3\left(1+x\right)\ln \left(x\right)}{x^2}\:dx=\frac{69}{4}\zeta \left(4\right)+\frac{3}{4}\zeta \left(3\right)-18\operatorname{Li}_4\left(\frac{1}{2}\right)-\frac{63}{4}\ln \left(2\right)\zeta \left(3\right)$$ $$+\frac{9}{2}\ln ^2\left(2\right)\zeta \left(2\right)-2\ln ^3\left(2\right)-\frac{3}{4}\ln ^4\left(2\right)$$

• Your result gives $-0.1487$ while both usual numerical approximation and the use of Nielsen generalized polylogarithm function implemented by Mathematica by PolyLog[n,p,z] give $−0.0576$ Sep 21, 2020 at 9:08
• Wolfram|Alpha throws a numerical approximation of $-0.148706$ which coincides with my closed form. Sep 21, 2020 at 16:14
• Yes! But OP asked for ${\large\int}_0^1\frac{\ln^3(1+x)\ln x}x\mathrm dx$ Sep 21, 2020 at 16:32
• Oh sorry, i misunderstood, i just wanted to prove that other case given by the OP since the main question has already been answered in similar fashions. Sep 21, 2020 at 16:36
• You did a really nice work, compliments! +1 Sep 21, 2020 at 17:08

Related problems and techniques: (I), (II). Here is a different form of solution

$$I = -3\sum_{n=0}^{\infty} \sum_{k=0}^{n}\frac{(-1)^k{ n\brack k}k(k-1) }{(n+1)^3n!} ,$$

where ${n \brack k}$ is the Stirling numbers of the first kind.

• Interesting. But this is hardly a closed form. Aug 28, 2014 at 23:22