I am trying to calculate $$ I:=\int_0^1 dx \frac{\ln x \ln^2(1-x)\ln(1+x)}{x}$$ Note, the closed form is beautiful (yes beautiful) and is given by

$$ I=−\frac{3}{8}\zeta_2\zeta_3 -\frac{2}{3}\zeta_2\ln^3 2 +\frac{7}{4}\zeta_3\ln^2 2-\frac{7}{2}\zeta_5+4\ln 2 \operatorname{Li}_4\left(\frac{1}{2}\right)+\frac{2}{15}\ln^5 2+4\operatorname{Li}_5\left(\frac{1}{2}\right) $$ where $$ \zeta_s=\sum_{n=1}^\infty \frac{1}{n^{s}},\qquad \operatorname{Li}_s(z)=\sum_{n=1}^\infty \frac{z^n}{n^s},\qquad\text{for}\ |z|<1. $$ I succeeded in writing the integral as $$ I=-\sum_{i=0}^\infty \int_0^1 x^i\ln x\ln(1+x)\ln(1-x)\ dx, $$ but I am confused as to where to go from here. Possibly I was thinking of trying to use Mellin transforms or residues.

A reference to aid us is here. (Since somebody has asked for reference)

We can also write I as $$ I=\sum_{i=0}^\infty \sum_{j=1}^\infty \frac{1}{j}\sum_{k=1}^\infty \frac{1}{k} \int_0^1 x^{i+j+k} \ln x\ dx $$ using $$ \int_0^1 x^n \ln x\ dx= -\frac{1}{(n+1)^2}, $$ we can simplify this, but I am not sure then how to compute the triple sum. Thank you again.

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    $\begingroup$ Out of curiosity, what makes this closed form particularly beautiful? $\endgroup$ – Alex Becker Apr 20 '14 at 16:56
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    $\begingroup$ No doubt: "beautiful" is in the eyes of the beholder...If only the Riemann zeta function appeared I'd agree, but those$\;\color{red}L$ogarithmic $\;\color{red}i$ntegrals trash, imo, the beauty. $\endgroup$ – DonAntonio Apr 20 '14 at 16:58
  • $\begingroup$ @AlexBecker I would rather not discuss the form of beauty here, as we are not in a greek philosophy reading the Republic. However, this is a very nice integral from a mathematics journal that I found very beautiful. I apologize if you don't like it as much as I do. $\endgroup$ – Jeff Faraci Apr 20 '14 at 17:02
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    $\begingroup$ Since the polylogs are being evaluated at $\frac{1}{2}$, I don't feel that the beauty of the answer is ruined. $\endgroup$ – Random Variable Apr 20 '14 at 17:47
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    $\begingroup$ Sos440 gave an evaluation (which I helped simplify a bit) for the case where $\ln(1+x)$ is squared instead of $\ln(1-x)$. math.stackexchange.com/questions/560950/… $\endgroup$ – Random Variable Apr 20 '14 at 18:08

We will use similar approach as sos440's answer in I&S. Using the simple algebraic identity $$ ab^2=\frac{(a+b)^3+(a-b)^3-2a^3}{6}, $$ it follows that \begin{align} \int_0^1 \frac{\ln x\ln(1+x)\ln^2(1-x)}{x}\ dx &=\frac16I_1+\frac16I_2-\frac13I_3\ ,\tag1 \end{align} where \begin{align} I_1&=\int_0^1\frac{\ln x\ln^3(1-x^2)}{x}\ dx\\[12pt] I_2&=\int_0^1\frac{\ln x}{x}\ln^3\left(\frac{1+x}{1-x}\right)\ dx\\[12pt] I_3&=\int_0^1\frac{\ln x\ln^3(1+x)}{x}\ dx \end{align}

Evaluation of $I_1$

Setting $t=x^2$ followed by $t\mapsto1-t$, we have \begin{align} I_1&=\frac14\int_0^1\frac{\ln t\ln^3(1-t)}{t}\ dt\\ &=\frac14\int_0^1\frac{\ln (1-t)\ln^3t}{1-t}\ dt\\ \end{align} To evaluate the integral above, we can use multiple derivative of beta function

$$ I_1=\frac14\lim_{x\to1}\lim_{y\to0^+}\frac{\partial^4\text{B}(x,y)}{\partial x^3\partial y}=3\zeta(5)-\frac32\zeta(2)\zeta(3).\tag2 $$

Alternatively, we can use generating function for the harmonic numbers for $|z|<1$ $$ \sum_{n=1}^\infty H_n z^n=-\frac{\ln(1-z)}{1-z}, $$ identity of the harmonic numbers $$ H_{n+1}-H_n=\frac1{n+1}, $$ and $$ \int_0^1 x^\alpha \ln^n x\ dx=\frac{(-1)^n n!}{(\alpha+1)^{n+1}}, \qquad\text{for }\ n=0,1,2,\ldots\tag3 $$ We may refer to the following answer to see the complete approach for evaluating $I_1$.

Evaluation of $I_2$

$I_2$ has been evaluated by sos440 here and it is equal to

$$ I_2=\frac{21}{4}\zeta(2)\zeta(3)-\frac{93}{8}\zeta(5).\tag4 $$

Alternatively, we can use the following technique. Setting $t=\dfrac{1-x}{1+x}\ \color{red}{\Rightarrow}\ x=\dfrac{1-t}{1+t}$ and $dx=-\dfrac{2}{(1+t)^2}\ dt$, then \begin{align} I_2&=-\int_0^1\frac{\ln x}{x}\ln^3\left(\frac{1-x}{1+x}\right)\ dx\\ &=2\int_0^1\frac{\ln^3 t\ln(1+t)}{(1-t)(1+t)}\ dt-2\int_0^1\frac{\ln^3 t\ln(1-t)}{(1-t)(1+t)}\ dt.\tag5 \end{align} Using the fact that $$ \frac{2}{(1-t)(1+t)}=\frac1{1-t}+\frac1{1+t} $$ and $(5)$ can be evaluated by performing some tedious calculations involving series expansion (double summation or generating function for the harmonic numbers) of the form $\dfrac{\ln(1\pm t)}{1\pm t}$ and equation $(3)$.

Another alternative way to evaluate $I_2$ without using complex analysis and dividing integral into four separated integrals is substituting $t=\dfrac{1-x}{1+x}$ and $I_2$ turns out to be $$ I_2=-2\int_0^1\frac{\ln^3t}{1-t^2}\ln\left(\frac{1-t}{1+t}\right)\ dt,\tag6 $$ where $(6)$ has been evaluated by Omran Kouba (see evaluation of $K$).

Evaluation of $I_3$

$I_3$ has been evaluated here and it is equal to

\begin{align} I_3=&\ \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\\&-12\operatorname{Li}_4\left(\frac12\right)\ln2-12\operatorname{Li}_5\left(\frac12\right).\tag7 \end{align}

Thus, putting altogether we obtain

$$ I=\color{blue}{\small{\frac{2}{15}\ln^5 2-\frac{2}{3}\zeta(2)\ln^3 2 +\frac{7}{4}\zeta(3)\ln^2 2−\frac{3}{8}\zeta(2)\zeta(3) -\frac{7}{2}\zeta(5)+4\operatorname{Li}_4\left(\frac12\right)\ln2+4\operatorname{Li}_5\left(\frac{1}{2}\right)}}. $$

$$ \large\color{blue}{\text{# }\mathbb{Q.E.D.}\text{ #}} $$

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    $\begingroup$ This answer should be awarded a bounty. If no-one gives it, I'll reward it to you. +1 for your great answer! (>‿◠)✌ $\endgroup$ – Anastasiya-Romanova 秀 Sep 15 '14 at 13:08
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    $\begingroup$ @Anastasiya-Romanova秀 :0 no bounty was ever awarded! $\endgroup$ – Brevan Ellefsen Oct 25 '16 at 4:42

using the following identity proved by Cornel and can be found in his book, (Almost) Impossible Integrals, Sums and Series. $\quad\displaystyle\ln(1-x)\ln(1+x)=-\sum_{n=1}^\infty\left(\frac{H_{2n}-H_n}{n}+\frac1{2n^2}\right)x^{2n}$.

multiply both sides by $\frac{\ln(1-x)}{x}$ then integrate from $x=0$ to $1$, we get \begin{align} I&=-\sum_{n=1}^\infty\left(\frac{H_{2n}-H_n}{n}+\frac1{2n^2}\right)\int_0^1x^{2n-1}\ln(1-x)\ dx\\ &=-\sum_{n=1}^\infty\left(\frac{H_{2n}-H_n}{n}+\frac1{2n^2}\right)\left(-\frac{H_{2n}}{2n}\right)\\ &=2\sum_{n=1}^\infty\frac{H_{2n}^2}{(2n)^2}-\frac12\sum_{n=1}^\infty\frac{H_nH_{2n}}{n^2}+\sum_{n=1}^\infty\frac{H_{2n}}{(2n)^3}\\ &=\sum_{n=1}^\infty\frac{H_{n}^2}{n^2}+\sum_{n=1}^\infty\frac{(-1)^nH_{n}^2}{n^2}-\frac12\color{blue}{\sum_{n=1}^\infty\frac{H_{n}H_{2n}}{n^2}}+\sum_{n=1}^\infty\frac{H_{n}}{n^3}+\sum_{n=1}^\infty\frac{(-1)^nH_{n}}{n^3} \end{align} I managed here to prove $$\color{blue}{\sum_{n=1}^\infty\frac{H_nH_{2n}}{n^2}}=4\sum_{n=1}^\infty\frac{H_n^2}{n^2}+2\sum_{n=1}^\infty\frac{(-1)^nH_n^2}{n^2}+2\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}-4\sum_{n=1}^\infty\frac{H_n^{(3)}}{n2^n}-6\zeta(4)$$ which follows that $$\boxed{I=\sum_{n=1}^\infty\frac{H_n}{n^3}-\sum_{n=1}^\infty\frac{H_n^2}{n^2}+\sum_{n=1}^\infty\frac{H_n^{(3)}}{n2^n}+3\zeta(4)}$$ Plugging the following results: $$\displaystyle\sum_{n=1}^{\infty}\frac{H_n^2}{n^2}=6\zeta(4)-\frac74\zeta(4)=\frac{17}4\zeta(4)$$


The first sum can be found here, and the second sum can be found using the easy-to-prove generating function $\sum_{n=1}^\infty\frac{x^nH_n^{(3)}}{n}=\operatorname{Li_4}(x)-\ln(1-x)\operatorname{Li_3}(x)-\frac12\operatorname{Li_2}^2(x)$,substituting these two sums along with the well known value of $\sum_{n=1}^\infty\frac{H_n}{n^3}=3\zeta(5)-\zeta(2)\zeta(3)$, we get the desired closed form of $I$.

NOTE . Sorry guys I just noticed that my solution is for $\int_0^1\frac{\ln^2(1-x)\ln(1+x)}{x}\ dx$ without $\ln x$ in the numerator as in the original problem. I am keeping the solution as it was voted as useful.


Using the same start of Tunk-Fey's solution:

$$\begin{align} I=\int_0^1 \frac{\ln x\ln(1+x)\ln^2(1-x)}{x}\ dx &=\frac16I_1-\frac16I_2-\frac13I_3 \end{align}$$


\begin{align} I_1&=\int_0^1\frac{\ln x\ln^3(1-x^2)}{x}\ dx\\[12pt] I_2&=\int_0^1\frac{\ln x}{x}\ln^3\left(\frac{1-x}{1+x}\right)\ dx\\[12pt] I_3&=\int_0^1\frac{\ln x\ln^3(1+x)}{x}\ dx \end{align} What makes my solution different is that i am going to evaluate these integrals in different ways.

\begin{align} I_1&=\int_0^1\frac{\ln x\ln^3(1-x^2)}{x}\ dx=\frac14\int_0^1\frac{\ln x\ln^3(1-x)}{x}\ dx\\ &=\int_0^1\frac{\ln(1-x)\ln^3x}{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)=\boxed{12\zeta(5)-6\zeta(2)\zeta(3)=I_1} \end{align}

where we substituted $\sum_{n=1}^\infty\frac{H_n}{n^4}=3\zeta(5)-\zeta(2)\zeta(3)$

$$I_2=\small{\int_0^1\frac{\ln x}{x}\ln^3\left(\frac{1-x}{1+x}\right)\ dx\overset{IBP}{=}3\int_0^1\frac{\ln^2x}{1-x^2}\ln^2\left(\frac{1-x}{1+x}\right)\ dx\overset{\frac{1-x}{1+x}=y}{=}\frac32\int_0^1\frac{\ln^2x}{x}\ln^2\left(\frac{1-x}{1+x}\right)\ dx}$$

and by using $\ln^2\left(\frac{1-x}{1+x}\right)=-2\sum_{n=1}^\infty\frac{H_n-2H_{2n}}{n}x^{2n}$ (proved here) we get \begin{align} I_2&=-3\sum_{n=1}^\infty\frac{H_n-2H_{2n}}{n}\int_0^1x^{2n-1}\ln^2x\ dx=24\sum_{n=1}^\infty\frac{H_{2n}}{(2n)^4}-3\sum_{n=1}^\infty\frac{H_n}{n^4}\\ &=12\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}+\frac{45}4\sum_{n=1}^\infty\frac{H_n}{n^4}=\boxed{\frac{93}{8}\zeta(5)-\frac{21}{4}\zeta(2)\zeta(3)=I_2} \end{align} where we substituted $\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}$$=\frac12\zeta(2)\zeta(3)-\frac{59}{32}\zeta(5)\ $

$I_3$ was solved here in different ways but this solution is my favorite one which gives

$$\boxed{\small{I_3=-12\operatorname{Li}_5\left(\frac12\right)-12\ln2\operatorname{Li}_4\left(\frac12\right)+\frac{99}{16}\zeta(5)+3\zeta(2)\zeta(3)-\frac{21}4\ln^22\zeta(3)+2\ln^32\zeta(2)-\frac25\ln^52}\ \ }$$

Combine $I_1$, $I_2$ and $I_3$ to get

$$I=\small{4\operatorname{Li}_5\left(\frac{1}{2}\right)+4\ln2\operatorname{Li}_4\left(\frac12\right)-\frac{7}{2}\zeta(5)−\frac{3}{8}\zeta(2)\zeta(3)+\frac{7}{4}\ln^22\zeta(3)-\frac{2}{3}\ln^32\zeta(2)+\frac{2}{15}\ln^5 2}$$


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