# Evaluating $\int^1_0 \frac{\log(1+x)\log(1-x) \log(x)}{x}\, \mathrm dx$

a friend posted the following integral

$$I=\int^1_0 \frac{\log(1+x)\log(1-x) \log(x)}{x}\, \mathrm dx$$

The best we could do is expressing it in terms of Euler sums

$$I=-\frac{\zeta^2(2)}{2}+ \sum_{n\geq 1}\frac{(-1)^{n-1}}{n^2} H_{n}^{(2)}+\sum_{n\geq 1}\frac{(-1)^{n-1}}{n^3}H_{n}$$

I am wondering if the approach I followed made the integral complicated ? What approach would you follow to solve the integral?, can we find a better solution ?

• By the way, why are you interested in Euler sums? Aug 14, 2013 at 3:19
• @Mhenni Benghorbal, I was working with polylogarithm function and I found an intimate relation with Euler sums . Aug 14, 2013 at 3:25
• @mhenniBenghorbal, sorry I don't get what you are saying ? Aug 16, 2013 at 16:53

The values of the two Euler Sums are

$$\displaystyle \sum_{n=1}^\infty (-1)^{n-1} \frac{H_n}{n^{3}} = \frac{11\pi^4}{360}-2\text{Li}_4 \left(\frac{1}{2} \right)-\frac{7}{4}\log(2) \zeta(3)+\frac{\pi^2}{12}\log^2(2)-\frac{1}{12}\log^4(2)$$ $$\displaystyle \sum_{n=1}^\infty (-1)^{n-1} \frac{H_n^{(2)}}{n^{2}} =-\frac{17}{480}\pi^4 +4 \text{Li}_4 \left(\frac{1}{2} \right)+\frac{7}{2}\log(2) \zeta(3)-\frac{\pi^2 \log^2(2)}{6}+\frac{\log^4(2)}{6}$$

Therefore the integral evaluates to

\begin{align*} \int_0^1 \frac{\log(1-x)\log(x)\log(1+x)}{x}dx &=-\frac{3 \pi^4}{160}+\frac{7\log(2)}{4}\zeta(3)-\frac{\pi^2 \log^2(2)}{12} +\frac{\log^4(2)}{12} \\ &\quad+ 2 \text{Li}_4 \left(\frac{1}{2} \right) \sim 0.290721 \end{align*}

• The following relation is useful for your calculations. Aug 17, 2013 at 19:24

using an identity devoloped by Cornel Ioan Valean and it can be found in his book " Almost impossible integrals, sums, and series": $$\ln(1-x)\ln(1+x)=-\sum_{n=1}^\infty\left(\frac{H_{2n}-H_n}{n}+\frac1{2n^2}\right)x^{2n}$$ we get: \begin{align} I&=\int_0^1\frac{\ln(1-x)\ln(1+x)\ln x}{x}\ dx=-\sum_{n=1}^\infty\left(\frac{H_{2n}-H_n}{n}+\frac1{2n^2}\right)\int_0^1x^{2n-1}\ln x\ dx\\ &=-\sum_{n=1}^\infty\left(\frac{H_{2n}-H_n}{n}+\frac1{2n^2}\right)\left(-\frac1{(2n)^2}\right)=2\sum_{n=1}^\infty\frac{H_{2n}}{(2n)^3}-\frac14\sum_{n=1}^\infty\frac{H_n}{n^3}+\frac18\zeta(4)\\ &=\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}+\frac34\sum_{n=1}^\infty\frac{H_n}{n^3}+\frac18\zeta(4)\\ &=2\operatorname{Li}_4\left(\frac12\right)-\frac12\ln^22\zeta(2)+\frac74\ln2\zeta(3)-\frac{27}{16}\zeta(4)+\frac1{12}\ln^42 \end{align}

where we used the value of the first sum proved here and the common value of the second sum which is $$\frac54\zeta(4)$$.

A solution by Cornel Ioan Valean (the nice, interesting part about this solution is that we calculate the integral without using harmonic series, Beta function at all)

Proof. Multiplying both sides of $$i)$$ (see below) by $$1/(1-a)$$ and integrating from $$a=0$$ to $$a=1/2$$, we obtain $$\int_0^{1/2}\left(\int_0^1 \frac{\log (x) \log (1-x)}{(1-a x)(1-a)} \textrm{d}x\right)\textrm{d}a=\int_0^1\left(\int_0^{1/2} \frac{\log (x) \log (1-x)}{(1-a x)(1-a)} \textrm{d}a\right)\textrm{d}x$$ $$=\int_0^1 \frac{\log (x) \log (1-x) \log (2-x)}{1-x} \textrm{d}x=\int_0^1 \frac{\log (1-x) \log (x) \log (1+x)}{x} \textrm{d}x$$ $$=\frac{\pi^2}{6}\int_0^{1/2}\frac{ \log (1-a)}{a(1-a)}\textrm{d}a+\frac{1}{6}\int_0^{1/2}\frac{\log ^3(1-a)}{a(1-a)}\textrm{d}a+\int_0^{1/2}\frac{\operatorname{Li}_3(a)}{a(1-a)}\textrm{d}a$$ $$-\int_0^{1/2}\frac{1}{a(1-a)}\operatorname{Li}_3\left(\frac{a}{a-1}\right) \textrm{d}a$$ $$=\frac{1}{12}\log^4(2)-\frac{1}{2}\log^2(2)\zeta(2)+\frac{7}{4}\log(2)\zeta(3)-\frac{27}{16}\zeta(4)+2\operatorname{Li}_4\left(\frac{1}{2}\right),$$ and the solution is complete.

In the calculations we needed the following results:

Let $$a<1$$ be a real number. The following equality holds: $$i) \ \int_0^1 \frac{\log (x) \log (1-x)}{1-a x} \textrm{d}x=\frac{\pi^2}{6}\frac{ \log (1-a)}{a}+\frac{1}{6}\frac{\log ^3(1-a)}{a}+\frac{1}{a}\operatorname{Li}_3(a)-\frac{1}{a}\operatorname{Li}_3\left(\frac{a}{a-1}\right),$$ where $$\operatorname{Li}_3$$ is the Trilogarithm function. The result is stated and proved in the paper A special way of extracting the real part of the Trilogarithm, $$\operatorname{Li}_3\left(\frac{1\pm i}{2}\right)$$ by Cornel Ioan Valean.

$$ii) \ \int_0^a \frac{\log (1-x)}{x (1-x)}\textrm{d}x=-\frac{1}{2} \log ^2(1-a)-\operatorname{Li}_2(a).$$

$$iii) \ \int_0^a \frac{\log ^3(1-x)}{x (1-x)} \textrm{d}x$$ $$=6 \operatorname{Li}_4(1-a)-6 \operatorname{Li}_3(1-a) \log (1-a)+3 \operatorname{Li}_2(1-a) \log ^2(1-a)$$ $$-\frac{1}{4} \log ^4(1-a)+\log (a) \log ^3(1-a)-\frac{\pi^4}{15},$$

which is straightforward with integration by parts.

$$iv) \ \int_0^a \frac{\operatorname{Li}_3(x)}{1-x} \textrm{d}x=-\frac{1}{2}(\operatorname{Li}_2(a))^2-\operatorname{Li}_3(a) \log (1-a),$$

and it's straightforward with integration by parts.

$$v) \ \int_0^a \frac{1}{x(1-x)}\operatorname{Li}_3\left(\frac{x}{x-1}\right)\textrm{d}x =\operatorname{Li}_4\left(\frac{a}{a-1}\right).$$

$$vi) \ \operatorname{Li}_2\left(\frac{1}{2}\right)=\frac{1}{2}(\zeta(2)-\log^2(2)).$$

$$vii) \ \operatorname{Li}_3\left(\frac{1}{2}\right)=\frac{7}{8}\zeta(3)-\frac{1}{2}\log(2)\zeta(2)+\frac{1}{6}\log^3(2).$$

A first note: A generalization of the present integral with $$\log^{2n}(x)$$ is given in the book (Almost) Impossible Integrals, Sums, and Series (see page 6),

$$\int_0^1\frac{\log(1-x)\log^{2n}(x)\log(1+x)}{x} \textrm{d}x =\frac{1}{2}(2n)!\left(1-\frac{1}{2^{2n+1}}\right)\sum_{k=1}^{2n} \zeta(k+1)\zeta(2n-k+2)$$ $$-(2n)!\sum_{k=1}^{n}\left(1-\frac{1}{2^{2k-1}}\right)\zeta(2k)\zeta(2n-2k+3) + \frac{1}{2^{2n+3}} (2n+3-2^{2n+3})(2n)!\zeta(2n+3),$$ and the result is obtained by exploiting the series representation of $$\log(1-x)\log(1+x)$$.

A second note: By the strategy presented above more interesting integrals may be calculated (without Beta function, harmonic series). Another nice example is

$$\int _0^1 \frac{\log(1-x) \log(x)\log(1+x)}{1-x}\textrm{d}x$$ $$=\frac{17 }{16}\zeta(4)-\frac{1}{4} \log ^2(2)\zeta(2)+\frac{7}{8} \log (2)\zeta (3) -\frac{1}{12} \log ^4(2)-2 \text{Li}_4\left(\frac{1}{2}\right).$$

A second solution in large steps (we circumvent the use of harmonic series)

If we use the algebraic identities $$(a+b)^2=a^2+2ab+b^2$$ and $$(a-b)^2=a^2-2ab+b^2$$ which we combine with clever rearrangements of the resulting integrals, we arrive at

$$\int _0^1 \frac{\log(1-x) \log(x)\log(1+x)}{x}\textrm{d}x$$ $$=\frac{1}{2}\int_0^1 \frac{\log^2(x)\log(1+x)}{1+x}\textrm{d}x-\frac{3}{8}\underbrace{\int_0^1 \frac{\log(1-x)\log^2(x)}{1-x}\textrm{d}x}_{\text{Beta function}},$$ where the first integral is calculated in the book, (Almost) Impossible Integrals, Sums, and Series, on pages $$503-505$$, and the second integral is a form of Beta function.

• This is magic. (+1). Jan 14, 2020 at 18:32
• @AliShather Thanks! Jan 15, 2020 at 14:38

$$I=\int_0^1\frac{\log(1+x)\log(1-x)\log(x)}{x}dx$$

Let's rewrite the integral using the following fact:

$$\color{blue}{\log(1+x)\log(1-x)=\frac{\log^2(1-x^2)-\log^2(1-x)-\log^2(1+x)}{2}}$$

$$I=\frac{1}{2}\int_0^1\underbrace{\frac{\log^2(1-x^2)\log(x)}{x}}_{x^2\rightarrow x}dx-\frac{1}{2}\int_0^1\frac{\log^2(1-x)\log(x)}{x}dx-\frac{1}{2}\int_0^1\underbrace{\frac{\log^2(1+x)\log(x)}{x}}_{1+x\rightarrow x}dx$$

$$I=-\frac{3}{8}\int_0^1\underbrace{\frac{\log^2(1-x)\log(x)}{x}}_{x\rightarrow 1-x}dx-\frac{1}{2}\int_1^2\underbrace{\frac{\log^2(x)\log(x-1)}{x-1}}_{x\rightarrow \frac{1}{x}}dx$$

$$I=\underbrace{-\frac{3}{8}\int_0^1\frac{\log^2(x)\log(1-x)}{1-x}dx}_{I_1}\underbrace{-\frac{1}{2}\int_{1/2}^1\frac{\log^2(x)\log(1-x)-\log^3(x)}{x(1-x)}dx}_{I_2}$$

To solve $$I_1$$ let's use Differentiation Under the Integral Sign, then let's switch the order of integration and apply partial fractions:

$$I_1=\frac{3}{8}\int_0^1\frac{1}{1-y}\int_0^1\left[\frac{\log^2(x)}{1-x}-\frac{\log^2(x)}{1-yx}\right]dxdy=\frac{3}{4}\int_0^1\left[\frac{Li_3(1)-Li_3(y)}{1-y}-\frac{Li_3(y)}{y}\right]dy$$

$$I_1=\frac{3}{4}\left[-\left(Li_3(1)-Li_3(y)\right)\log(1-y)-Li_4(y)\right]^1_0-\frac{3}{4}\int_0^1\frac{Li_2(y)\log(1-y)}{y}dy$$

$$I_1=-\frac{3}{4}Li_4(1)+\frac{3}{8}Li^2_2(1)=\frac{3}{16}\zeta(4)$$

$$\color{red}{I_1=\frac{3}{16}\zeta(4)}$$

To solve $$I_2$$ let's apply partial fractions and then solve the resulting integrals. Most of them are straightfoward, but one of them requires some algebric work:

$$I_2=-\frac{1}{2}\int_{1/2}^1\left[\frac{\log^2{\left(x\right)}\log{\left(1-x\right)}-\log^3{\left(x\right)}}{x}+\frac{\log^2{\left(x\right)}\log{\left(1-x\right)}-\log^3{\left(x\right)}}{1-x}\right]dx$$

$$I_2=\frac{1}{2}\left[Li_2(x)\log^2(x)-2Li_3(x)\log(x)+2Li_4(x)+\frac{\log^4(x)}{4}-\log^3(x)\log(1-x)-3Li_2(x)\log^2(x)+6Li_3(x)\log(x)-6Li_4(x)\right]_{1/2}^1\underbrace{-\frac{1}{2}\int_{1/2}^1\frac{\log^2{\left(x\right)}\log{\left(1-x\right)}}{1-x}dx}_{I_{2a}}$$

$$I_2=\frac{1}{2}\left[-4\zeta(4)+4Li_4\left(\frac{1}{2}\right)+4Li_3\left(\frac{1}{2}\right)\log(2)+2Li_2\left(\frac{1}{2}\right)\log^2(2)+\frac{3}{4}\log^4(2)\right]+I_{2a}$$

$$I_2=-2\zeta(4)+\frac{7\log(2)}{4}\zeta(3)-\frac{\log^2(2)}{2}\zeta(2)+2Li_4\left(\frac{1}{2}\right)+\frac{5\log^4(2)}{24}+I_{2a}$$

Instead of applying Differentiation Under the Integral Sign again to evaluate $$I_{2a}$$, let's do the following:

$$\left\{ \begin{array}{c} A+B=\int_0^{1/2}\frac{\log^2(x)\log(1-x)}{1-x}dx+\int_{1/2}^1\frac{\log^2(x)\log(1-x)}{1-x}dx \\ A-B=\int_0^{1/2}\underbrace{\frac{\log^2(x)\log(1-x)}{1-x}}_{x\rightarrow 1-x}dx-\int_{1/2}^1\frac{\log^2(x)\log(1-x)}{1-x}dx \end{array} \right.$$

$$\left\{ \begin{array}{c} A+B=\int_0^{1}\frac{\log^2(x)\log(1-x)}{1-x}dx=-\frac{\zeta(4)}{2} (From\ I_1) \\ A-B=\int_{1/2}^1\underbrace{\frac{\log^2(1-x)\log(x)}{x}}_{IBP}dx-\int_{1/2}^1\frac{\log^2(x)\log(1-x)}{1-x}dx \end{array} \right.$$

$$\left\{ \begin{array}{c} A+B=-\frac{\zeta(4)}{2} \\ A-B=-\frac{\log^4(2)}{2}+\int_{1/2}^1\frac{\log^2(x)\log(1-x)}{1-x}dx -\int_{1/2}^1\frac{\log^2(x)\log(1-x)}{1-x}dx \end{array} \right.$$

Thus $$B=\int_{1/2}^1\frac{\log^2(x)\log(1-x)}{1-x}dx=\frac{\log^4(2)-\zeta(4)}{4}$$ $$I_{2a}=-\frac{B}{2}=\frac{\zeta(4)-\log^4(2)}{8}$$

Hence, gathering all results, it's possible to conclude that: $$I=\int_0^1\frac{\log(1+x)\log(1-x)\log(x)}{x}dx=$$ $$-\frac{27}{16}\zeta(4)+\frac{7log(2)}{4}\zeta(3)-\frac{\log^2(2)}{2}\zeta(2)+2Li_4\left(\frac{1}{2}\right)+\frac{\log^4(2)}{12}$$

• Thank you :) I just eddited the post following your advices. I hope now that the overbrace thing is no longer an issue. Jan 15, 2020 at 14:30
• (+100 points) Excellent! Thanks! Jan 15, 2020 at 14:35
• OMG, thank you so much for the bounty reward :) Jan 15, 2020 at 14:37

Related problems: (I). You can have the following solution

$$\frac{3\gamma}{4}\,\zeta( 3 )+{\frac {7\pi^4}{360}}+\sum _{m=1}^{\infty }{\frac { \left( -1 \right) ^{m-1}\psi \left( m \right) }{{m}^{3}}}+\sum _{m=1}^{\infty }-{\frac { \left( -1 \right) ^{m-1}\psi' \left( m \right) }{{m}^{2}}}\sim 0.2907212779,$$

which you might be able to simplify it further.

Note: If you use the identity

$$\frac{\pi^4}{90}=\zeta(4),$$

in the above expression, then you will have the form

$$\frac{3\gamma}{4}\,\zeta( 3 )+{\frac {7}{4}}\zeta(4)+\sum _{m=1}^{\infty }{\frac { \left( -1 \right) ^{m-1}\psi \left( m \right) }{{m}^{3}}}+\sum _{m=1}^{\infty }-{\frac { \left( -1 \right) ^{m-1}\psi' \left( m \right) }{{m}^{2}}}.$$

\begin{align*} J&=\int_0^1 \frac{\ln x\ln(1-x)\ln(1+x)}{x}\,dx\\ &\overset{IBP}=\frac{1}{2}\Big[\ln^2 x\ln(1-x)\ln(1+x)\Big]_0^1 -\frac{1}{2}\int_0^1 \ln^2 x\left(\frac{\ln(1-x)}{1+x}-\frac{\ln(1+x)}{1-x}\right)\,dx\\ &=\frac{1}{2}\int_0^1 \ln^2 x\left(\frac{\ln(1+x)}{1-x}-\frac{\ln(1-x)}{1+x}\right)\,dx\\ K&=\int_0^1 \frac{\ln^2 x\ln(1+x)}{1-x}\,dx,L=\int_0^1 \frac{\ln^2 x\ln(1-x)}{1+x}\,dx,M=\int_0^1\frac{\ln(1+x)\ln^2 x}{1+x}\,dx\\ &\overset{IBP}=\left[\left(\int_0^x \frac{\ln^2 t}{1-t}\,dt\right)\ln(1+x)\right]_0^1-\int_0^1 \frac{1}{1+x}\left(\int_0^x \frac{\ln^2 t}{1-t}\,dt\right)\,dx\\ &\overset{u(t)=xt}=2\zeta(3)\ln 2-\int_0^1 \int_0^1 \frac{x\ln^2(tx)}{(1-tx)(1+x)}\,dt\,dx\\ &=2\zeta(3)\ln 2-\frac{1}{2}\left(\int_0^1 \int_0^1 \frac{x\ln^2(tx)}{(1-tx)(1+x)}\,dt\,dx+\int_0^1 \int_0^1 \frac{t\ln^2(tx)}{(1-tx)(1+t)}\,dt\,dx\right)\\ &=2\zeta(3)\ln 2-\frac{1}{2}\left(\int_0^1 \int_0^1 \frac{\ln^2(tx)}{1-tx}\,dt\,dx-\int_0^1 \int_0^1 \frac{\ln^2(tx)}{(1+t)(1+x)}\,dt\,dx\right)\\ &=2\zeta(3)\ln 2+\int_0^1 \frac{\ln^2 x+\ln t\ln x}{(1+t)(1+x)}\,dt\,dx-\frac{1}{2}\int_0^1\int_0^1 \frac{\ln^2 (tx)}{1-tx}\,dt\,dx\\ &=\frac{7}{2}\zeta(3)\ln 2+\frac{\pi^4}{144}-\frac{1}{2}\int_0^1\int_0^1 \frac{\ln^2 (tx)}{1-tx}\,dt\,dx\\ &\overset{u=tx}=\frac{7}{2}\zeta(3)\ln 2+\frac{\pi^4}{144}-\frac{1}{2}\int_0^1 \frac{1}{x}\left(\int_0^x \frac{\ln^2 u}{1-u}\,du\right)\,dx\\ &\overset{IBP}=\frac{7}{2}\zeta(3)\ln 2+\frac{\pi^4}{144}-\frac{1}{2}\left[\ln x\left(\int_0^x \frac{\ln^2 u}{1-u}\,du\right)\right]_0^1+\frac{1}{2}\int_0^1 \frac{\ln^3 x}{1-x}\,dx\\ &=\frac{7}{2}\zeta(3)\ln 2+\frac{\pi^4}{144}+\frac{1}{2}\int_0^1 \frac{\ln^3 x}{1-x}\,dx\\ &=\frac{7}{2}\zeta(3)\ln 2+\frac{\pi^4}{144}+\frac{1}{2}\times -\frac{\pi^4}{15}\\ &=\boxed{\frac{7}{2}\zeta(3)\ln 2-\frac{19\pi^4}{720}} \end{align*} \begin{align*} 0& \begin{align*} U&=\int_0^1 \frac{\ln^3\left(\frac{x}{1+x}\right)}{1+x}\,dx\\ &\overset{y=\frac{x}{1+x}}=\int_0^{\frac{1}{2}}\frac{\ln^3 x}{1-x}\,dx\\ U&=\int_0^1 \frac{\ln^3 x}{1+x}\,dx-\int_0^1 \frac{\ln^3(1+x)}{1+x}\,dx-3\int_0^1 \frac{\ln^2 x\ln(1+x)}{1+x}\,dx+3\int_0^1 \frac{\ln^2(1+x)\ln x}{1+x}\,dx\\ &=\int_0^1 \frac{\ln^3 x}{1+x}\,dx-\frac{1}{4}\ln^4 2-3M+\Big[\ln^3(1+x)\ln x\Big]_0^1-\int_0^1 \frac{\ln^3(1+t)}{t}\,dt\\ &\overset{x=\frac{1}{1+t}}=\int_0^1 \frac{\ln^3 x}{1+x}\,dx-\frac{1}{4}\ln^4 2-3M+\int_{\frac{1}{2}}^1 \frac{\ln^3 x}{x(1-x)}\,dx\\ &=\int_0^1 \frac{\ln^3 x}{1+x}\,dx-\frac{1}{4}\ln^4 2-3M+\int_{\frac{1}{2}}^1 \frac{\ln^3 x}{x}\,dx-\int_{\frac{1}{2}}^1 \frac{\ln^3 x}{1-x}\,dx\\ &=2\int_0^1 \frac{\ln^3 x}{1-x^2}\,dx-\frac{1}{2}\ln^4 2-3M-\int_0^{\frac{1}{2}} \frac{\ln^3 x}{1-x}\,dx\\ &=\left(2\int_0^1 \frac{\ln^3 x}{1-x}\,dx-\int_0^1 \frac{2t\ln^3 t}{1-t}\,dt\right)-\frac{1}{2}\ln^4 2-3M-\int_0^{\frac{1}{2}} \frac{\ln^3 x}{1-x}\,dx\\ &\overset{x=t^2}=\frac{15}{8}\int_0^1 \frac{\ln^3 x}{1-x}\,dx-\frac{1}{2}\ln^4 2-3M-\int_0^{\frac{1}{2}} \frac{\ln^3 x}{1-x}\,dx\\ \end{align*}

Therefore, \begin{align*} M&=\frac{5}{8}\int_0^1 \frac{\ln^3 x}{1-x}\,dx-\frac{1}{6}\ln^4 2-\frac{2}{3}\int_0^{\frac{1}{2}} \frac{\ln^3 x}{1-x}\,dx\\ \int_0^{\frac{1}{2}} \frac{\ln^3 x}{1-x}\,dx&\overset{y=2x}=\frac{1}{2}\int_0^1 \frac{\ln^3\left(\frac{1}{2}x\right)}{1-\frac{1}{2}x}\,dx\\ &=\frac{1}{2}\int_0^1 \frac{\ln^3 x}{1-\frac{1}{2}x}\,dx-\frac{\ln^3 2}{2}\int_0^1 \frac{1}{1-\frac{1}{2}x}\,dx-\\ &\frac{3\ln 2}{2}\int_0^1 \frac{\ln^2 x}{1-\frac{1}{2}x}dx+\frac{3\ln^2 2}{2}\int_0^1 \frac{\ln x}{1-\frac{1}{2}x}dx\\ &=-6\text{Li}_4\left(\frac{1}{2}\right)-\ln^4 2-6\ln 2\text{Li}_3\left(\frac{1}{2}\right)-3\ln^2 2 \text{Li}_2\left(\frac{1}{2}\right)\\ &=-6\text{Li}_4\left(\frac{1}{2}\right)-\frac{21\zeta(3)}{4}\ln 2+\frac{\pi^2 \ln^2 2 }{4}-\frac{\ln^4 2}{2}\\ M&=4\text{Li}_4\left(\frac{1}{2}\right)-\frac{\pi^4}{24}+\frac{7\zeta(3)\ln 2}{2}-\frac{\pi^2 \ln^2 2}{6}+\frac{\ln^4 2}{6}\\ L&=\boxed{\frac{\pi^4}{90}-4\text{Li}_4\left(\frac{1}{2}\right)+\frac{\pi^2 \ln^2 2}{6}-\frac{\ln^4 2}{6}}\\ J&=\frac{1}{2}\left(K-L\right)\\ &=\boxed{2\text{Li}_4\left(\frac{1}{2}\right)-\frac{3\pi^4}{160}+\frac{7\zeta(3)\ln 2}{4}-\frac{\pi^2 \ln^2 2}{12}+\frac{\ln^4 2}{12}} \end{align*}

NB: I assume, $$r\geq 1,0< a\leq 1$$, integers \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{\ln^2 2}{2},\text{Li}_2(1)=\zeta(2)=\frac{\pi^2}{6}\\ \text{Li}_3(1)&=\zeta(3),\text{Li}_3\left(\frac{1}{2}\right)=\frac{7\zeta(3)}{8}+\frac{\ln^3 2}{6}-\frac{\pi^2\ln 2}{12},\text{Li}_4(1)=\zeta(4)=\frac{\pi^4}{90} \end{align*}

• (+1) Looks nice. Jan 15, 2020 at 22:55

I would do the following variable change.

$$x=e^{-t}$$ Then we can represent the integral as follows:

$$I=-\int_{0}^{\infty}t\ln(1+e^{-t})\ln(1-e^{-t})\;dt$$ Now, apply the Taylor expansion of the logarithm:

$$\ln(1+x)=\sum_{i=1}^{\infty}(-1)^{i-1}\frac{x^i}{i}$$

$$I=\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}\frac{(-1)^{i-1}}{ij}\int_{0}^{\infty}te^{-(i+j)t}dt=$$

$$=\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}\frac{(-1)^{i-1}}{ij(i+j)^2}$$

$$-\frac{\partial^2}{\partial s\partial t}\left[B(s+1,t+1)\;_3 F_2(1,1,s+1;2,s+t+2;-1)\right]_{s=t=0}$$

It may be that the Hypergeometric function is summable. In this case, the differentiation is trivial. (B denotes Euler's beta function.)

• Your $LaTeX$ got messed up. Try to fix it. Aug 12, 2013 at 4:59