# Compute $\int_0^1\frac{\ln^2(1+x)\operatorname{Li}_2(-x)}{x}dx$

How to prove

$$\int_0^1\frac{\ln^2(1+x)\operatorname{Li}_2(-x)}{x}dx=4\operatorname{Li}_5\left(\frac12\right)+4\ln2\operatorname{Li}_4\left(\frac12\right)-\frac{125}{32}\zeta(5)-\frac{1}{8}\zeta(2)\zeta(3)\\+\frac{7}{4}\ln^22\zeta(3)-\frac2{3}\ln^32\zeta(2)+\frac{2}{15}\ln^52$$

This integral was nicely computed by Cornel here in page $$5$$ using tricky manipulation.

Another form of the integral, after subbing and applying integration by parts is

$$\int_0^1\frac{\ln^2(1-x)\operatorname{Li}_2(x/2)}{x}dx$$

My question is how to evaluate any of these integrals in a different way?

Thanks

• \begin{align}\int_0^1\frac{\ln^2(1+x)\operatorname{Li}_2(-x)}{x}dx&=\frac{1}{2}\int_0^1\frac{\left(\operatorname{Li}_2(-x)\right)^2}{1+x}dx-\frac{1}{8}\zeta(2)^2\ln 2\end{align}
– FDP
Commented Nov 30, 2019 at 5:15
• Then what? You just applied IBP I dont see any progress ? Commented Nov 30, 2019 at 5:17
• That's the reason i have posted it as a comment. This formula is nice for me.
– FDP
Commented Nov 30, 2019 at 5:22
• I see , I thought you had an idea. Commented Nov 30, 2019 at 5:23
• Do you like this one? I hope it's not obvious. \begin{align}\int_0^1\frac{\ln^2(1+x)\operatorname{Li}_2(-x)}{x}dx=-\int_0^1\frac{\operatorname{Li}_2\left(\frac{x}{1+x}\right)\ln^2(1+x)}{x}\,dx-\frac{1}{2}\int_0^1 \frac{\ln^4(1+x)}{x}\,dx\end{align}
– FDP
Commented Dec 2, 2019 at 12:46

Finally I got the idea:

Starting with $$\frac{1}{1+x}\mapsto x$$ then using Landens identity $$\operatorname{Li}_2\left(\frac{x-1}{x}\right)=-\frac12\ln^2x-\operatorname{Li}_2(1-x)$$ we obtain

$$\mathcal{I}=\int_0^1\frac{\ln^2(1+x)\operatorname{Li}_2(-x)}{x}dx=\int_{1/2}^1\frac{\ln^2x\operatorname{Li}_2\left(\frac{x-1}{x}\right)}{x(1-x)}dx$$

$$=-\frac12\underbrace{\int_{1/2}^1\frac{\ln^4x}{x(1-x)}dx}_{\mathcal{\large J}}-\underbrace{\int_{1/2}^1\frac{\ln^2x\operatorname{Li}_2(1-x)}{x}dx}_{ IBP}-\underbrace{\int_{1/2}^1\frac{\ln^2x\operatorname{Li}_2(1-x)}{1-x}dx}_{ IBP}$$

$$\text{Note for the third integral that} \int\frac{\ln x}{1-x}dx=\operatorname{Li}_2(1-x)$$

$$=-\frac12\mathcal{J}-\frac13\ln^32\operatorname{Li}_2(1/2)+\frac13\underbrace{\int_{1/2}^1\frac{\ln^4x}{1-x}dx}_{\frac1{1-x}=\frac1{x(1-x)}-\frac1x}-\frac12\ln2\operatorname{Li}_2^2(1/2)+\frac12\underbrace{\int_{1/2}^1\frac{\operatorname{Li}_2^2(1-x)}{x}dx}_{\mathcal{\large K}}$$

and the integral simplifies into

$$\mathcal{I}=\frac12\mathcal{K}-\frac16\mathcal{J}+\frac1{12}\ln^32\zeta(2)-\frac5{16}\ln2\zeta(4)-\frac1{40}\ln^52$$

where we substituted $$\operatorname{Li}_2(1/2)=\frac12\zeta(2)-\frac12\ln^22$$

The integral $$\mathcal{J}$$ is classical and can be done using the generalization

$$(-1)^n\int_{1/2}^1\frac{\ln^nx}{x(1-x)}dx=\frac{\ln^{n+1}(2)}{n+1}+n!\zeta(n+1)+\sum_{k=0}^n k!{n\choose k}\ln^{n-k}(2)\operatorname{Li}_{k+1}\left(\frac12\right)$$

so

$$\boxed{\mathcal{J}=24\zeta(5)-\frac{21}2\ln^22\zeta(3)+4\ln^32\zeta(2)-\frac45\ln^52-24\ln2\operatorname{Li}_4\left(\frac12\right)-24\operatorname{Li}_5\left(\frac12\right)}$$

where we substituted $$\operatorname{Li}_3(1/2)=\frac78\zeta(3)-\frac12\ln2\zeta(2)+\frac16\ln^32$$

For the integral $$\mathcal{K}$$, we can just use the dilogarithm reflection formula: $$\operatorname{Li}_2(1-x)=\zeta(2)-\ln x\ln(1-x)-\operatorname{Li}_2(x)$$

upon expanding $$\operatorname{Li}_2^2(1-x)$$ we get

$$\mathcal{K}=\underbrace{\int_{1/2}^1\frac{\zeta^2(2)-2\zeta(2)\operatorname{Li}_2(x)}{x}dx}_{\mathcal{\large {K_1}}}-2\zeta(2)\underbrace{\int_{1/2}^1\frac{\ln x\ln(1-x)}{x}dx}_{\mathcal{\large {K_2}}}+\underbrace{\int_{1/2}^1\frac{\ln^2x\ln^2(1-x)}{x}dx}_{\mathcal{\large {K_3}}}\\+2\underbrace{\int_{1/2}^1\frac{\ln x\ln(1-x)\operatorname{Li}_2(x)}{x}dx}_{\mathcal{\large {K_4}}}+\underbrace{\int_{1/2}^1\frac{\operatorname{Li}_2^2(x)}{x}dx}_{\mathcal{\large {K_5}}}$$

$$\mathcal{K_1}=\frac52\ln2\zeta(4)-2\zeta(2)[\zeta(3)-\operatorname{Li}_3(1/2)]$$

$$\boxed{\mathcal{K_1}=\frac13\ln^32\zeta(2)-\frac14\zeta(2)\zeta(3)}$$

$$\mathcal{K_2}\overset{IBP}{=}-\ln2\operatorname{Li}_2(1/2)+\int_{1/2}^1\frac{\operatorname{Li}_2(x)}{x}dx$$ $$=-\ln2\operatorname{Li}_2(1/2)+\zeta(3)-\operatorname{Li}_3(1/2)$$ $$\boxed{\mathcal{K_2}=\frac18\zeta(3)+\frac13\ln^32}$$

$$\mathcal{K_3}=\int_0^{1/2}\frac{\ln^2x\ln^2(1-x)}{x}dx\overset{IBP}{=}\frac23{\int_{0}^{1/2}\frac{\ln^3x\ln(1-x)}{1-x}}dx=\frac23\color{blue}{\int_{1/2}^1\frac{\ln^3(1-x)\ln x}{x}dx}$$

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

where

$$\int_{1/2}^1\frac{\ln^4x}{1-x}=\mathcal{J}-\int_{1/2}^1\frac{\ln^4x}{x}dx$$

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

and

$$\int_0^1\frac{\ln^3(1-x)\ln x}{x}\ dx=\int_0^1\frac{\ln^3x\ln(1-x)}{1-x}\ dx=-\sum_{n=1}^\infty H_n\int_0^1x^n\ln^3x=6\sum_{n=1}^\infty\frac{H_n}{(n+1)^4}$$

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

combine the two integrals to get

$$\color{blue}{\int_{1/2}^1\frac{\ln^3(1-x)\ln x}{x}\ dx}\\=\frac3{16}\zeta(5)-3\zeta(2)\zeta(3)+\frac{21}8\ln^22\zeta(3)-\ln^32\zeta(2)+\frac25\ln^52+6\ln2\operatorname{Li}_4\left(\frac12\right)+6\operatorname{Li}_5\left(\frac12\right)$$

which gives

$$\boxed{\mathcal{K_3}=\frac1{8}\zeta(5)-2\zeta(2)\zeta(3)+\frac{7}4\ln^22\zeta(3)-\frac23\ln^32\zeta(2)+\frac4{15}\ln^52+4\ln2\operatorname{Li}_4\left(\frac12\right)+4\operatorname{Li}_5\left(\frac12\right)}$$

$$\mathcal{K_4}\overset{IBP}{=}-\frac12\ln2\operatorname{Li}_2^2(1/2)+\frac12\underbrace{\int_{1/2}^1\frac{\operatorname{Li}_2^2(x)}{x}dx}_{\mathcal{\large{K_5}}}$$

so

$$2\mathcal{K_4}+\mathcal{K_5}=-\ln2\operatorname{Li}_2^2(1/2)+2\int_{1/2}^1\frac{\operatorname{Li}_2^2(x)}{x}dx$$

where

$$\int_{1/2}^1\frac{\operatorname{Li}_2^2(x)}{x}dx=\int_0^{1}\frac{\operatorname{Li}_2^2(x)}{x}dx-\int_0^{1/2}\frac{\operatorname{Li}_2^2(x)}{x}dx$$

we have

$$\int_0^{1}\frac{\operatorname{Li}_2^2(x)}{x}dx=\sum_{n=1}^\infty\frac1{n^2}\int_0^1 x^{n-1}\operatorname{Li}_2(x)dx=\sum_{n=1}^\infty\frac1{n^2}\left(\frac{\zeta(2)}{n}-\frac{H_n}{n^2}\right)$$

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

and @Song proved here

$$\int_0^{1/2}\frac{\operatorname{Li}_2^2(x)}{x}dx=\frac12\ln^32\zeta(2)-\frac78\ln^22\zeta(3)-\frac58\ln2\zeta(4)+\frac{27}{32}\zeta(5)+\frac78\zeta(2)\zeta(3)\\-\frac{7}{60}\ln^52-2\ln2\operatorname{Li}_4\left(\frac12\right)-2\operatorname{Li}_5\left(\frac12\right)$$

giving us

$$\int_{1/2}^1\frac{\operatorname{Li}_2^2(x)}{x}dx=-\frac12\ln^32\zeta(2)+\frac78\ln^22\zeta(3)+\frac58\ln2\zeta(4)-\frac{123}{32}\zeta(5)+\frac98\zeta(2)\zeta(3)\\+\frac{7}{60}\ln^52+2\ln2\operatorname{Li}_4\left(\frac12\right)+2\operatorname{Li}_5\left(\frac12\right)$$

consequently

$$\boxed{2\mathcal{K_4}+\mathcal{K_5}=-\frac12\ln^32\zeta(2)+\frac74\ln^22\zeta(3)+\frac{5}8\ln2\zeta(4)-\frac{123}{16}\zeta(5)+\frac94\zeta(2)\zeta(3)\\ \qquad\qquad\qquad-\frac{1}{60}\ln^52+4\ln2\operatorname{Li}_4\left(\frac12\right)+4\operatorname{Li}_5\left(\frac12\right)}$$

Finally combine the boxed results we obtain

$$\small{\mathcal{I}=4\operatorname{Li}_5\left(\frac12\right)+4\ln2\operatorname{Li}_4\left(\frac12\right)-\frac{125}{32}\zeta(5)-\frac{1}{8}\zeta(2)\zeta(3)+\frac{7}{4}\ln^22\zeta(3)-\frac2{3}\ln^32\zeta(2)+\frac{2}{15}\ln^52}$$

• $-\int_{1/2}^1\frac{\ln^2x\operatorname{Li}_2(1-x)}{1-x}dx$ can computed using $\int_{1/2}^1 \frac{(\operatorname{Li}_2(x))^2}{x}dx$ the other is the sister of your "first integral".
– FDP
Commented Dec 4, 2019 at 21:28
• Yes this integral appeared along with other hard integrals but i did them all. I will complete my solution soon Commented Dec 4, 2019 at 23:23

Incomplete solution is following,

\begin{align} J&=\int_0^1\frac{\ln^2(1+x)\operatorname{Li}_2(-x)}{x}dx\\ &=-\int_0^1 \int_0^1\frac{\ln(1+tx)\ln^2(1+x)}{tx}\,dt\,dx\\ &\overset{u=\frac{1-t}{1+tx}}=-\int_0^1 \int_0^1 \frac{(1+x)\ln^2(1+x)\ln\left(\frac{1+x}{1+ux}\right)}{x(1-u)(1+ux)}\,du\,dx\\ &=-\int_0^1 \frac{(1+x)\ln^2(1+x)}{x}\left(\int_0^1\frac{x\ln\left(\frac{1+x}{1+ux}\right)}{(1+x)(1+ux)}+\frac{\ln\left(\frac{1+x}{1+ux}\right)}{(1-u)(1+x)}\,du\right)dx\\ &=\frac{1}{2}\int_0^1 \frac{(1+x)\ln^2(1+x)}{x}\left[\frac{\ln^2 \left(\frac{1+x}{1+ux}\right)}{1+x}\right]_{u=0}^{u=1} dx-\int_0^1\int_0^1 \frac{\ln^2(1+x)\ln\left(\frac{1+x}{1+ux}\right)}{x(1-u)}dudx\\ &=-\frac{1}{2}\int_0^1 \frac{\ln^4(1+x)}{x}\,dx-\int_0^1\int_0^1 \frac{\ln^2(1+x)\ln\left(\frac{1+x}{1+ux}\right)}{x(1-u)}dudx\\ \end{align} Let $$0<\alpha<1$$, \begin{align} A(\alpha)&=\int_0^1\frac{\ln^2(1+x)}{x}\left(\int_0^\alpha \frac{\ln\left(\frac{1+x}{1+ux}\right)}{1-u}du\right)dx\\ &=-\ln\left(1-\alpha\right)\left(\int_0^1 \frac{\ln^3(1+x)}{x}\,dx\right)-\int_0^1 \frac{\ln^2(1+x)}{x}\left(\int_0^{\alpha}\frac{\ln(1+ux)}{1-u}du\right)dx\\ &=-\ln\left(1-\alpha\right)\left(\int_0^1 \frac{\ln^3(1+x)}{x}\,dx\right)+\\ &\int_0^1 \frac{\ln^2(1+x)}{x}\left[\ln(1+ux)\ln\left(\frac{(1-u)x}{1+x}\right)+\operatorname{Li}_2\left(\frac{1+ux}{1+x}\right)\right]_{u=0}^{u=\alpha}\,dx\\ &=\ln(1-\alpha)\int_0^1 \frac{\ln^2(1+x)\ln\left(\frac{1+\alpha x}{1+x}\right)}{x}\,dx-\int_0^1 \frac{\ln^3(1+x)\ln(1+\alpha x)}{x}\,dx+\\ &\int_0^1 \frac{\ln^2(1+x)\ln(1+ux)\ln x}{x}\,dx+\int_0^1 \frac{\ln^2(1+x)\left(\operatorname{Li}_2\left(\frac{1+\alpha x}{1+x}\right)-\operatorname{Li}_2(1)\right)}{x}\,dx\\ \end{align} Therefore, \begin{align} &\int_0^1\int_0^1 \frac{\ln^2(1+x)\ln\left(\frac{1+x}{1+ux}\right)}{x(1-u)}dudx\\ &=\lim_{\alpha \rightarrow 1}J(\alpha)\\ &=\int_0^1 \frac{\ln^3(1+x)\ln\left(\frac{x}{1+x}\right)}{x}\,dx-\int_0^1 \frac{\ln^2(1+x)\left(\operatorname{Li}_2\left(\frac{1}{1+x}\right)-\operatorname{Li}_2(1)\right)}{x}\,dx\\ R&=\int_0^1 \frac{\ln^2(1+x)\left(\operatorname{Li}_2\left(\frac{1}{1+x}\right)-\operatorname{Li}_2(1)\right)}{x}\,dx\\ &\overset{y=\frac{1}{1+x}}=\int_{\frac{1}{2}}^1 \frac{\ln^2 x\big(\operatorname{Li}_2\left(x\right)-\operatorname{Li}_2(1)\big)}{x(1-x)}\,dx\\ &=\int_{\frac{1}{2}}^1 \frac{\ln^2 x\big(\operatorname{Li}_2\left(x\right)-\operatorname{Li}_2(1)\big)}{x}\,dx+\int_{\frac{1}{2}}^1 \frac{\ln^2 x\big(\operatorname{Li}_2\left(x\right)-\operatorname{Li}_2(1)\big)}{1-x}\,dx\\ &\overset{\text{IBP}}=\frac{1}{3}\ln^3 2\left(\operatorname{Li}_2\left(\frac{1}{2}\right)-\operatorname{Li}_2(1)\right)+\frac{1}{3}\int_{\frac{1}{2}}^1 \frac{\ln^3 x\ln(1-x)}{x}\,dx+\\ &\int_{\frac{1}{2}}^1 \frac{\ln^2 x\big(\operatorname{Li}_2\left(x\right)-\operatorname{Li}_2(1)\big)}{1-x}\,dx\\ &\overset{\text{IBP}}=\frac{1}{3}\ln^3 2\left(\operatorname{Li}_2\left(\frac{1}{2}\right)-\operatorname{Li}_2(1)\right)+\frac{1}{3}\int_{\frac{1}{2}}^1 \frac{\ln^3 x\ln(1-x)}{x}\,dx+\\ &\left[\left(\int_0^x \frac{\ln^2 t}{1-t}\,dt\right)\left(\operatorname{Li}_2\left(x\right)-\operatorname{Li}_2(1)\right)\right]_{\frac{1}{2}}^1+\\&\int_{\frac{1}{2}}^1 \left(-\ln(1-x)\ln^2 x+2\int_0^x \frac{\ln (1-t)\ln t}{t}\,dt\right)\frac{\ln(1-x)}{x}\,dx\\ &=\left(\frac{1}{3}\ln^3 2-\left(\int_0^{\frac{1}{2}} \frac{\ln^2 t}{1-t}\,dt\right)\right)\left(\operatorname{Li}_2\left(\frac{1}{2}\right)-\operatorname{Li}_2(1)\right)+\frac{1}{3}\int_{\frac{1}{2}}^1 \frac{\ln^3 x\ln(1-x)}{x}\,dx-\\ &\int_{\frac{1}{2}}^1 \frac{\ln^2 x\ln^2(1-x)}{x}\,dx+2\int_{\frac{1}{2}}^1 \frac{\ln(1-x)}{x}\left(\int_0^x\frac{\ln (1-t)\ln t}{t}\,dt\right)\,dx\\ &\overset{\text{IBP}}=\left(\frac{1}{3}\ln^3 2-\left(\int_0^{\frac{1}{2}} \frac{\ln^2 t}{1-t}\,dt\right)\right)\left(\operatorname{Li}_2\left(\frac{1}{2}\right)-\operatorname{Li}_2(1)\right)+\frac{1}{3}\int_{\frac{1}{2}}^1 \frac{\ln^3 x\ln(1-x)}{x}\,dx-\\ &\int_{\frac{1}{2}}^1 \frac{\ln^2 x\ln^2(1-x)}{x}\,dx+2\left[-\operatorname{Li}_2(x)\left(\int_0^x\frac{\ln (1-t)\ln t}{t}\,dt\right)\right]_{\frac{1}{2}}^1+\\ &2\int_0^{\frac{1}{2}} \frac{\operatorname{Li}_2(x)\ln (1-x)\ln x}{x}\,dx\\ &=\left(\frac{1}{3}\ln^3 2-\left(\int_0^{\frac{1}{2}} \frac{\ln^2 t}{1-t}\,dt\right)\right)\left(\operatorname{Li}_2\left(\frac{1}{2}\right)-\operatorname{Li}_2(1)\right)+\frac{1}{3}\int_{\frac{1}{2}}^1 \frac{\ln^3 x\ln(1-x)}{x}\,dx-\\ &\int_{\frac{1}{2}}^1 \frac{\ln^2 x\ln^2(1-x)}{x}\,dx+2\operatorname{Li}_2\left(\frac{1}{2}\right)\left(\int_0^{\frac{1}{2}}\frac{\ln (1-t)\ln t}{t}\,dt\right)-\\ &2\operatorname{Li}_2\left(1\right)\left(\int_0^{1}\frac{\ln (1-t)\ln t}{t}\,dt\right)+2\int_{\frac{1}{2}}^1 \frac{\operatorname{Li}_2(x)\ln (1-x)\ln x}{x}\,dx\\ \end{align} Since, \begin{align} \frac{\partial}{\partial x}\left(\operatorname{Li}_2(x)\right)^2&=-2\operatorname{Li}_2(x)\frac{\ln(1-x)}{x} \end{align} then, \begin{align} R&\overset{\text{IBP}}=\left(\frac{1}{3}\ln^3 2-\left(\int_0^{\frac{1}{2}} \frac{\ln^2 t}{1-t}\,dt\right)\right)\left(\operatorname{Li}_2\left(\frac{1}{2}\right)-\operatorname{Li}_2(1)\right)+\frac{1}{3}\int_{\frac{1}{2}}^1 \frac{\ln^3 x\ln(1-x)}{x}\,dx-\\ &\int_{\frac{1}{2}}^1 \frac{\ln^2 x\ln^2(1-x)}{x}\,dx+2\operatorname{Li}_2\left(\frac{1}{2}\right)\left(\int_0^{\frac{1}{2}}\frac{\ln (1-t)\ln t}{t}\,dt\right)-\\ &2\operatorname{Li}_2\left(1\right)\left(\int_0^{1}\frac{\ln (1-t)\ln t}{t}\,dt\right)-\left(\operatorname{Li}_2\left(\frac{1}{2}\right)\right)^2\ln 2+ \int_{\frac{1}{2}}^1 \frac{\left(\operatorname{Li}_2(x)\right)^2}{x}\,dx\\ \end{align} Therefore, \begin{align} J&=-\frac{1}{2}\int_0^1 \frac{\ln^4(1+x)}{x}\,dx-\int_0^1 \frac{\ln^3(1+x)\ln\left(\frac{x}{1+x}\right)}{x}\,dx+\\ &\left(\frac{1}{3}\ln^3 2-\left(\int_0^{\frac{1}{2}} \frac{\ln^2 t}{1-t}\,dt\right)\right)\left(\operatorname{Li}_2\left(\frac{1}{2}\right)-\operatorname{Li}_2(1)\right)+\frac{1}{3}\int_{\frac{1}{2}}^1 \frac{\ln^3 x\ln(1-x)}{x}\,dx-\\ &\int_{\frac{1}{2}}^1 \frac{\ln^2 x\ln^2(1-x)}{x}\,dx+2\operatorname{Li}_2\left(\frac{1}{2}\right)\left(\int_0^{\frac{1}{2}}\frac{\ln (1-t)\ln t}{t}\,dt\right)-\\ &2\operatorname{Li}_2\left(1\right)\left(\int_0^{1}\frac{\ln (1-t)\ln t}{t}\,dt\right)-\left(\operatorname{Li}_2\left(\frac{1}{2}\right)\right)^2\ln 2+ \int_{\frac{1}{2}}^1 \frac{\left(\operatorname{Li}_2(x)\right)^2}{x}\,dx \end{align} For the very last integral see:

Compute $\int_0^{1/2}\frac{\left(\operatorname{Li}_2(x)\right)^2}{x}dx$

Definite Dilogarithm integral $\int^1_0 \frac{\operatorname{Li}_2^2(x)}{x}\, dx$

• Very nice (+1) i appreciate your efforts. all these sub integrals at the end are manageable but $\int_0^1\frac{\ln^3(1+x)\ln\left(\frac{x}{1+x}\right)}{x}dx$ but i will give it a try. Commented Dec 3, 2019 at 16:54
• $\displaystyle \int_0^1\frac{\ln^3(1+x)\ln x}{x}dx$ is the problem here. Equivalently $\displaystyle \int_0^1\frac{\ln^2(1+x)\ln^2 x}{1+x}dx$
– FDP
Commented Dec 3, 2019 at 17:03
• ah i missed that. so your solution is complete why you mentioned its not? Commented Dec 3, 2019 at 17:21
• I haven't computed most of the integrals at the end. For now, i think one of these integrals could be not straightforward to compute. That's $\displaystyle \int_0^1\frac{\ln^3(1+x)\ln x}{x}dx$
– FDP
Commented Dec 3, 2019 at 17:43
• Actually you have computed it: math.stackexchange.com/questions/908108/…
– FDP
Commented Dec 3, 2019 at 17:49