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I am trying to evaluate $\displaystyle \int _0^1\frac{\text{Li}_2\left(-x\right)\ln \left(1-x\right)}{1+x}\:dx$

I first tried using the series expansion for the dilogarithm like this $$\sum _{n=1}^{\infty }\frac{\left(-1\right)^n}{n^2}\int _0^1\frac{x^n\ln \left(1-x\right)}{1+x}\:dx$$ Then I used integration by parts but this lead to nothing useful.

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    $\begingroup$ Just for answer-checking, the result is $$-4 \text{Li}_4\left(\frac{1}{2}\right)-\frac{21}{8} \zeta (3) \log (2)+\frac{\pi ^4}{24}-\frac{1}{6} \log ^4(2)+\frac{1}{8} \pi ^2 \log ^2(2)$$ $\endgroup$ – pisco Aug 4 '20 at 9:25
  • $\begingroup$ This question is a duplicate. It was already asked and answered here math.stackexchange.com/questions/3528761/… $\endgroup$ – user97357329 Aug 5 '20 at 17:23
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Lets start with the following integral

$$\int_0^1\frac{\text{Li}_2(-x)\ln(1-x)}{1+x}dx=\sum_{n=0}^\infty (-1)^nH_n^{(2)}\int_0^1 x^{n}\ln(1-x)dx$$

$$=-\sum_{n=0}^\infty \frac{(-1)^nH_n^{(2)}H_{n+1}}{n+1}=\sum_{n=1}^\infty \frac{(-1)^nH_{n-1}^{(2)}H_{n}}{n}$$

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

where

$$\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}=2\operatorname{Li_4}\left(\frac12\right)-\frac{11}4\zeta(4)+\frac74\ln2\zeta(3)-\frac12\ln^22\zeta(2)+\frac{1}{12}\ln^42$$

The other sum was already evaluated by Cornel in this solution and I am writing it in more details:

First relation:

From here we have

$$\int_0^1x^{n-1}\ln^3(1-x)\ dx=-\frac{H_n^3+3H_nH_n^{(2)}+2H_n^{(3)}}{n}$$

Multiply both sides by $(-1)^{n-1}$ then $\sum_{n=1}^\infty$ we have

$$\sum_{n=1}^\infty (-1)^n\frac{H_n^3+3H_nH_n^{(2)}+2H_n^{(3)}}{n}=\int_0^1\ln^3(1-x)\sum_{n=1}^\infty (-x)^{n-1}dx=\int_0^1\frac{\ln^3(1-x)}{1+x}dx$$

$$=\int_0^1\frac{\ln^3x}{2-x}dx=\sum_{n=1}^\infty\frac1{2^n}\int_0^1 x^{n-1}\ln^3xdx=-6\sum_{n=1}^\infty\frac{1}{2^nn^4}=-6\text{Li}_4\left(\frac12\right)\tag1$$

Second relation:

From here we have

$$\sum_{n=1}^\infty\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)x^n=-\frac{\ln^3(1-x)}{1-x}$$

Replace $x$ by $-x$ then divide both sides by $x$ and $\int_0^1$ we get

$$\sum_{n=1}^\infty (-1)^n\frac{H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}}{n}=-\int_0^1\frac{\ln^3(1+x)}{x(1+x)}dx$$ $$\overset{x=\frac{1-y}{y}}{=}\int_{1/2}^1\frac{\ln^3x}{1-x}dx=\sum_{n=1}^\infty \int_{1/2}^1 x^{n-1}\ln^3xdx$$

$$=\sum_{n=1}^\infty\left(\frac{6}{n^42^n}-\frac{6}{n^4}+\frac{6\ln2}{n^32^n}+\frac{3\ln^22}{n^22^n}+\frac{\ln^32}{n2^n}\right)$$

$$=6\text{Li}_4\left(\frac12\right)-6\zeta(4)+6\ln2\text{Li}_3\left(\frac12\right)+3\ln^22\text{Li}_2\left(\frac12\right)+\ln^42$$

$$=6\text{Li}_4\left(\frac12\right)-6\zeta(4)+\frac{21}{4}\ln2\zeta(3)-\frac32\ln^22\zeta(2)+\frac12\ln^42\tag2$$

Thus, $(1)-(2)$ gives

$$\sum_{n=1}^\infty\frac{(-1)^nH_nH_n^{(2)}}{n}=-2\text{Li}_4\left(\frac12\right)+\zeta(4)-\frac{7}{8}\ln2\zeta(3)+\frac14\ln^22\zeta(2)-\frac1{12}\ln^42$$

Combining the two sums we finally get

$$\int_0^1\frac{\text{Li}_2(-x)\ln(1-x)}{1+x}dx=-4 \text{Li}_4\left(\frac{1}{2}\right)+\frac{15}{4}\zeta(4)-\frac{21}{8}\ln2\zeta(3)+\frac34\ln^22\zeta(2)-\frac16\ln^42$$


Bonus:

By writing $\text{Li}_2(-x)=\int_0^1\frac{x\ln y}{1+xy}dy$ then changing the order of integration we have

$$\int_0^1\frac{\text{Li}_2(-x)\ln(1-x)}{1+x}dx=\int_0^1\ln y\left(\int_0^1\frac{x\ln(1-x)}{(1+x)(1+xy)}dx\right)dy$$

$$=\int_0^1 \ln y\left(\frac{\zeta(2)-\ln^22}{2(1-y)}-\frac{\text{Li}_2\left(\frac{y}{1+y}\right)}{y(1-y)}\right)dy$$

$$=-\frac12(\zeta(2)-\ln^22)\zeta(2)-\int_0^1\frac{\ln y}{y(1-y)}\text{Li}_2\left(\frac{y}{1+y}\right)dx$$

Or

$$\int_0^1\frac{\text{Li}_2(-x)\ln(1-x)}{1+x}dx+\int_0^1\frac{\ln x}{x(1-x)}\text{Li}_2\left(\frac{x}{1+x}\right)dx=\frac12\ln^22\zeta(2)-\frac54\zeta(4)$$

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  • $\begingroup$ This question has already been answered using such a strategy at this link math.stackexchange.com/q/3528838. $\endgroup$ – user97357329 Aug 6 '20 at 3:44
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    $\begingroup$ @user97357329 ah you are right I didnt noticed that. So this question is a duplicate. $\endgroup$ – Ali Shadhar Aug 6 '20 at 5:22
  • $\begingroup$ My bad i didnt know,still, thanks for your effort in such great answer. $\endgroup$ – user806882 Aug 6 '20 at 6:38

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