121
$\begingroup$

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!

$\endgroup$
145
+200
$\begingroup$

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}

which again matches @Cleo's answer.


References :

$[1]\ $ Harmonic number

$[2]\ $ Polylogarithm

| cite | improve this answer | |
$\endgroup$
  • 9
    $\begingroup$ @Tunk-Fey Very impressive! $\endgroup$ – Oksana Gimmel Aug 28 '14 at 23:23
  • 5
    $\begingroup$ 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 :) $\endgroup$ – Ali Caglayan Aug 29 '14 at 19:58
  • 12
    $\begingroup$ @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. $\ddot\smile$ $\endgroup$ – Tunk-Fey Aug 30 '14 at 13:46
  • 5
    $\begingroup$ @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 $\endgroup$ – David H Aug 30 '14 at 18:27
  • 2
    $\begingroup$ Thanks @FelixMarin. of course you're one of my teachers in polylog integrals. $\ddot\smile$ $\endgroup$ – Tunk-Fey Sep 2 '14 at 7:25
77
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
  • 11
    $\begingroup$ Well, Cleo's at it again. Well done. $\endgroup$ – Bennett Gardiner Aug 25 '14 at 4:33
  • 5
    $\begingroup$ Notice that $\ln2$ acts here like a regularized value of $\zeta(1)$. $\endgroup$ – Lucian Aug 25 '14 at 6:33
  • 25
    $\begingroup$ @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, $\endgroup$ – Markus Scheuer Aug 25 '14 at 7:05
  • 13
    $\begingroup$ @Cleo Do you mind giving a slight hint as to how one should proceed with this integral? Thanks. $\endgroup$ – SuperAbound Aug 26 '14 at 5:52
  • 4
    $\begingroup$ @Lucian What do you mean by a regularized value? $\endgroup$ – Vladimir Reshetnikov Aug 27 '14 at 0:56
23
$\begingroup$

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}

| cite | improve this answer | |
$\endgroup$
  • 4
    $\begingroup$ 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. $\endgroup$ – Tunk-Fey Aug 25 '14 at 13:35
  • $\begingroup$ @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. $\endgroup$ – SuperAbound Aug 25 '14 at 14:10
  • $\begingroup$ 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. $\endgroup$ – Ali Shather Apr 30 '19 at 23:55
21
$\begingroup$

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: $$\color{red}{I=3\sum_{n=1}^{+\infty}\frac{(-1)^{n+1}\left(H_n^2-H_n^{(2)}\right)}{(n+1)^3}.}\tag{4}$$

| cite | improve this answer | |
$\endgroup$
8
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
4
$\begingroup$

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}

| cite | improve this answer | |
$\endgroup$
4
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
4
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Very useful technique for such integrals. (+1) $\endgroup$ – Ali Shather Feb 3 at 2:08
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • 8
    $\begingroup$ Interesting. But this is hardly a closed form. $\endgroup$ – Oksana Gimmel Aug 28 '14 at 23:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.