Evaluation of $\int_0^1\frac{\ln ^2\left( 1-x \right) \text{Li}_2\left( \dfrac{1+x}{2} \right)}{x}\text{d}x$

Question

​show that \begin{align*} \int_0^1\frac{\ln ^2\left( 1-x \right) \text{Li}_2\left( \dfrac{1+x}{2} \right)}{x}\text{d}x&=\text{2Li}_5\left( \frac{1}{2} \right) +\text{2}\ln\text{2Li}_4\left( \frac{1}{2} \right) +\frac{81}{32}\zeta \left( 5 \right)\\ & -\frac{\ln ^22}{8}\zeta \left( 3 \right)+\frac{5\pi ^2}{16}\zeta \left( 3 \right) -\frac{\pi ^2\ln ^32}{18}-\frac{\pi ^4\ln 2}{15}\\ &+\frac{\ln ^52}{15} \end{align*}

$$I = \int_0^1 \frac{\ln^2(1-x) \text{Li}_2\left(\frac{1+x}{2}\right)}{x} \, dx = -\int_0^1 \frac{\ln^2(1-x)}{x} \int_0^{\frac{1+x}{2}} \frac{\ln(1-y)}{y} \, dy \, dx$$

Swapping the order of integration, we get:

\begin{align*} I = &-\int_0^{\frac{1}{2}} \frac{\ln(1-y)}{y} \int_0^1 \frac{\ln^2(1-x)}{x} \, dx \, dy \\ &- \int_{\frac{1}{2}}^1 \frac{\ln(1-y)}{y} \int_{2y-1}^1 \frac{\ln^2(1-x)}{x} \, dx \, dy \\ = &-\int_0^{\frac{1}{2}} \frac{\ln(1-y)}{y} \int_0^1 \frac{\ln^2(1-x)}{x} \, dx \, dy \\ &- 2\int_{\frac{1}{2}}^1 \frac{\ln(1-y) \text{Li}_3(2-2y)}{y} \, dy \\ &+ 2\int_{\frac{1}{2}}^1 \frac{\ln(1-y) \text{Li}_2(2-2y) \ln(2-2y)}{y} \, dy \\ &+ \int_{\frac{1}{2}}^1 \frac{\ln(1-y) \ln(2y-1) \ln^2(2-2y)}{y} \, dy \end{align*}

Answer 1

First of all, very subtle ways are possible here, and this will be supported by some future papers, but till then, let's see a super fast & elegant way.

A solution in very large steps by Cornel Ioan Valean

Step 1: By the Dilogarithm reflection formula here, we have $$\operatorname{Li}_2\left(\frac{1+x}{2}\right)=\frac{\pi ^2}{6}-\log \left(\frac{1-x}{2}\right) \log \left(\frac{1+x}{2}\right)-\operatorname{Li}_2\left(\frac{1-x}{2}\right),$$

and if we multiply both sides by $\log^2(1-x)/x$ and integrate from $x=0$ to $x=1$, we get the desired integral in the left-hand side, and in the right-hand side, if we expand, we get a sum of logarithmic integrals which are all known, plus another key integral $$\int_0^1 \frac{\log^2(1-x)}{x}\operatorname{Li}_2\left(\frac{1+x}{2}\right)\textrm{d}x$$ $$=\sum \text{known logarithmic integrals} -\int_0^1 \frac{\log^2(1-x)}{x}\operatorname{Li}_2\left(\frac{1-x}{2}\right)\textrm{d}x. \tag1$$

Step 2: Now, for the last resulting integral in the right-hand side, we let $x\mapsto 1-x$, and then use the Cauchy product to arrive at $$\int_0^1 \frac{\log^2(1-x)}{x}\operatorname{Li}_2\left(\frac{1-x}{2}\right)\textrm{d}x=\int_0^1 \frac{\log^2(x)}{1-x}\operatorname{Li}_2\left(\frac{x}{2}\right)\textrm{d}x$$ $$=2\zeta(3) \sum _{n=1}^{\infty } \frac{1}{n^2 2^n}-2 \sum _{n=1}^{\infty } \frac{H_n^{(3)}}{n^2 2^n}.$$

Step 3: At this point, we need to deal with an advanced harmonic series of weight $5$ with powers of $2$ in the denominator, where we want to exploit the simple fact that $\displaystyle \sum_{n=1}^{\infty} x^n \frac{H_n^{(3)}}{n}=-\frac{1}{2}(\operatorname{Li}_2(x))^2-\log(1-x)\operatorname{Li}_3(x)+\operatorname{Li}_4(x)$, and the needed series is obtained by dividing both sides of the generating function by $x$ and integrating from $x=0$ to $x=1/2$. The toughest resulting integrals can be extracted from this answer https://math.stackexchange.com/q/4284059, which you can also find it in More (Almost) Impossible Integrals, Sums, and Series (2023), page $63$, the sequel of (Almost) Impossible Integrals, Sums, and Series (2019).

End of story (Really, that simple!)

Answer 2

Used formulas

$$\text{Li}_2\left( \frac{1}{2} \right) =\frac{\pi ^2}{12}-\frac{\ln ^22}{2}$$

$$\text{Li}_3\left( \frac{1}{2} \right) =\frac{7}{8}\zeta \left( 3 \right) +\frac{\ln ^32}{6}-\frac{\pi ^2\ln 2}{12}$$

$$\sum_{n=1}^{\infty}{\frac{H_n}{2^nn^3}}=\text{Li}_4\left( \frac{1}{2} \right) +\frac{\pi ^4}{720}-\frac{\ln 2}{8}\zeta \left( 3 \right) +\frac{\ln ^42}{24}$$

$$\sum_{n=1}^{\infty}{\frac{H_{n}^{\left( 2 \right)}}{2^nn^2}}=\text{Li}_4\left( \frac{1}{2} \right) +\frac{\pi ^4}{1440}+\frac{\ln 2}{4}\zeta \left( 3 \right) -\frac{\pi ^2\ln ^22}{24}+\frac{\ln ^42}{24}$$

$$\sum_{n=1}^{\infty}{\frac{H_{n}^{\left( 3 \right)}}{2^nn}}=\text{Li}_4\left( \frac{1}{2} \right) -\frac{\pi ^4}{288}+\frac{\text{7}\ln 2}{8}\zeta \left( 3 \right) -\frac{\pi ^2\ln ^22}{24}+\frac{\ln ^42}{24}$$

\begin{align*} \sum_{n=1}^{\infty}{\frac{H_n}{2^nn^4}}&=\text{2Li}_5\left( \frac{1}{2} \right) +\ln\text{2Li}_4\left( \frac{1}{2} \right) +\frac{1}{32}\zeta \left( 5 \right) -\frac{\pi ^2}{12}\zeta \left( 3 \right)\\ & -\frac{\pi ^4\ln 2}{720}+\frac{\ln ^22}{2}\zeta \left( 3 \right) -\frac{\pi ^2\ln ^32}{36}+\frac{\ln ^52}{40} \end{align*}

\begin{align*} \sum_{n=1}^{\infty}{\frac{H_{n}^{\left( 2 \right)}}{2^nn^3}}&=-\text{2Li}_5\left( \frac{1}{2} \right) -\text{3}\ln\text{2Li}_4\left( \frac{1}{2} \right) +\frac{23}{64}\zeta \left( 5 \right) -\frac{23\pi ^2}{96}\zeta \left( 3 \right)\\ & -\frac{\pi ^4\ln 2}{1440}-\frac{\text{23}\ln ^22}{16}\zeta \left( 3 \right) +\frac{7\pi ^2\ln ^32}{72}-\frac{\text{13}\ln ^52}{120} \end{align*}

\begin{align*} \sum_{n=1}^{\infty}{\frac{H_{n}^{\left( 3 \right)}}{2^nn^2}}&=\text{4Li}_5\left( \frac{1}{2} \right) +\text{3}\ln\text{2Li}_4\left( \frac{1}{2} \right) -\frac{81}{64}\zeta \left( 5 \right) -\frac{7\pi ^2}{48}\zeta \left( 3 \right) \\ &+\frac{\pi ^4\ln 2}{288}+\frac{\text{7}\ln ^22}{8}\zeta \left( 3 \right) -\frac{5\pi ^2\ln ^32}{72}+\frac{\text{11}\ln ^52}{120} \end{align*}

\begin{align*} \sum_{n=1}^{\infty}{\frac{H_{n}^{\left( 4 \right)}}{2^nn}}&=-\text{Li}_5\left( \frac{1}{2} \right) -\ln\text{2Li}_4\left( \frac{1}{2} \right) +\frac{27}{32}\zeta \left( 5 \right) +\frac{7\pi ^2}{96}\zeta \left( 3 \right) \\ &-\frac{\text{7}\ln ^22}{16}\zeta \left( 3 \right) +\frac{\pi ^2\ln ^32}{36}-\frac{\ln ^52}{30} \end{align*}

Back to our topic

\begin{align*} I &= \int_0^1{\frac{\ln^2\left( 1-x \right)\text{Li}_2\left( \frac{1+x}{2} \right)}{x}\,dx} \\ &= -\int_0^1{\frac{\ln^2\left( 1-x \right)}{x}\int_0^{\frac{1+x}{2}}{\frac{\ln\left( 1-y \right)}{y}\,dy}\,dx} \end{align*}

\begin{align*} I=&-\int_0^{\frac{1}{2}}{\frac{\ln \left( 1-y \right)}{y}\int_0^1{\frac{\ln ^2\left( 1-x \right)}{x}\text{d}x\text{d}y}}-\int_{\frac{1}{2}}^1{\frac{\ln \left( 1-y \right)}{y}\int_{2y-1}^1{\frac{\ln ^2\left( 1-x \right)}{x}\text{d}x\text{d}y}}\\ =&-\int_0^{\frac{1}{2}}{\frac{\ln \left( 1-y \right)}{y}\int_0^1{\frac{\ln ^2\left( 1-x \right)}{x}\text{d}x\text{d}y}}-2\int_{\frac{1}{2}}^1{\frac{\ln \left( 1-y \right) \text{Li}_3\left( 2-2y \right)}{y}\text{d}y}\\ &+2\int_{\frac{1}{2}}^1{\frac{\ln \left( 1-y \right) \text{Li}_2\left( 2-2y \right) \ln \left( 2-2y \right)}{y}\text{d}y}+\int_{\frac{1}{2}}^1{\frac{\ln \left( 1-y \right) \ln \left( 2y-1 \right) \ln ^2\left( 2-2y \right)}{y}}\text{d}y\\ =&-\int_0^{\frac{1}{2}}{\frac{\ln \left( 1-y \right)}{y}\text{d}y\cdot \int_0^1{\frac{\ln ^2\left( 1-x \right)}{x}\text{d}x}}-2\int_0^1{\frac{\ln \left( \frac{t}{2} \right) \text{Li}_3\left( t \right)}{2-t}}\text{d}t+2\int_0^1{\frac{\ln \left( \frac{t}{2} \right) \text{Li}_2\left( t \right) \ln t}{2-t}}\text{d}t\\ &+\int_0^1{\frac{\ln \left( \frac{t}{2} \right) \ln \left( 1-t \right) \ln ^2t}{2-t}}\text{d}t\\ =&-\underset{I_1}{\underbrace{\int_0^{\frac{1}{2}}{\frac{\ln \left( 1-y \right)}{y}\text{d}y}}}\cdot \underset{I_2}{\underbrace{\int_0^1{\frac{\ln ^2\left( 1-x \right)}{x}\text{d}x}}}+2\underset{I_3}{\underbrace{\int_0^1{\frac{\ln t\text{Li}_3\left( t \right)}{2-t}}\text{d}t}}+\text{2}\ln 2\underset{I_4}{\underbrace{\int_0^1{\frac{\text{Li}_3\left( t \right)}{2-t}}\text{d}t}}\\ &-2\underset{I_5}{\underbrace{\int_0^1{\frac{\ln ^2t\text{Li}_2\left( t \right)}{2-t}}\text{d}t}}-\text{2}\ln 2\underset{I_6}{\underbrace{\int_0^1{\frac{\ln t\text{Li}_2\left( t \right)}{2-t}}\text{d}t}}+\underset{I_7}{\underbrace{\int_0^1{\frac{\ln ^3t\ln \left( 1-t \right)}{2-t}}\text{d}t}}-\ln 2\underset{I_8}{\underbrace{\int_0^1{\frac{\ln \left( 1-t \right) \ln ^2t}{2-t}}\text{d}t}} \end{align*}

$$I_1=-\text{Li}_2\left( \frac{1}{2} \right) =\frac{\ln ^22}{2}-\frac{\pi ^2}{12}$$

\begin{align*} I_2&=2\sum_{n=1}^{\infty}{\frac{H_n}{n+1}\int_0^1{x^n\text{d}x}}=2\sum_{n=1}^{\infty}{\frac{H_n}{\left( n+1 \right) ^2}}\\&=2\sum_{n=1}^{\infty}{\frac{H_{n+1}}{\left( n+1 \right) ^2}}-2\sum_{n=1}^{\infty}{\frac{1}{\left( n+1 \right) ^3}}=2\zeta \left( 3 \right) \end{align*}

\begin{align*} I_3&=\frac{1}{2}\int_0^1{\frac{\ln t\text{Li}_3\left( t \right)}{1-\frac{t}{2}}}\text{d}t=\frac{1}{2}\int_0^1{\sum_{k=0}^{\infty}{\frac{1}{2^k}\ln t\sum_{n=1}^{\infty}{\frac{t^{n+k}}{n^3}\text{d}t}}}=-\frac{1}{2}\sum_{k=0}^{\infty}{\frac{1}{2^k}\sum_{n=1}^{\infty}{\frac{1}{n^3\left( n+k+1 \right) ^2}}}\\ &=-\sum_{k=1}^{\infty}{\frac{1}{2^k}\sum_{n=1}^{\infty}{\frac{1}{n^3\left( n+k \right) ^2}}}=-\sum_{k=1}^{\infty}{\frac{1}{2^k}\frac{1}{k}\sum_{n=1}^{\infty}{\left[ \frac{1}{n^3\left( n+k \right)}-\frac{1}{n^2\left( n+k \right) ^2} \right]}}\\ &=-\sum_{k=1}^{\infty}{\frac{1}{2^k}\frac{1}{k}\sum_{n=1}^{\infty}{\frac{1}{n^3\left( n+k \right)}+}}\sum_{k=1}^{\infty}{\frac{1}{2^k}\frac{1}{k}\sum_{n=1}^{\infty}{\frac{1}{n^2\left( n+k \right) ^2}}}\\ &=-\sum_{k=1}^{\infty}{\frac{1}{2^k}\frac{1}{k^2}\sum_{n=1}^{\infty}{\left[ \frac{1}{n^3}-\frac{1}{n^2\left( n+k \right)} \right] +}}\sum_{k=1}^{\infty}{\frac{1}{2^k}\frac{1}{k^2}\sum_{n=1}^{\infty}{\left[ \frac{1}{n^2\left( n+k \right)}-\frac{1}{n\left( n+k \right) ^2} \right]}}\\ &=-\sum_{k=1}^{\infty}{\frac{1}{2^k}\frac{1}{k^2}\sum_{n=1}^{\infty}{\frac{1}{n^3}}}+\sum_{k=1}^{\infty}{\frac{1}{2^k}\frac{1}{k^2}\sum_{n=1}^{\infty}{\frac{1}{n^2\left( n+k \right)}}}+\sum_{k=1}^{\infty}{\frac{1}{2^k}\frac{1}{k^2}\sum_{n=1}^{\infty}{\frac{1}{n^2\left( n+k \right)}}}\\ &-\sum_{k=1}^{\infty}{\frac{1}{2^k}\frac{1}{k^2}\sum_{n=1}^{\infty}{\frac{1}{n\left( n+k \right) ^2}}}\\ &=-\zeta \left( 3 \right) \text{Li}_2\left( \frac{1}{2} \right) +2\sum_{k=1}^{\infty}{\frac{1}{2^k}\frac{1}{k^3}\sum_{n=1}^{\infty}{\left[ \frac{1}{n^2}-\frac{1}{n\left( n+k \right)} \right]}}\\ &-\sum_{k=1}^{\infty}{\frac{1}{2^k}\frac{1}{k^3}\sum_{n=1}^{\infty}{\left[ \frac{1}{n\left( n+k \right)}-\frac{1}{\left( n+k \right) ^2} \right]}}\\ &=-\zeta \left( 3 \right) \text{Li}_2\left( \frac{1}{2} \right) +2\sum_{k=1}^{\infty}{\frac{1}{2^k}\frac{1}{k^3}\sum_{n=1}^{\infty}{\frac{1}{n^2}-}}2\sum_{k=1}^{\infty}{\frac{1}{2^k}\frac{1}{k^3}\sum_{n=1}^{\infty}{\frac{1}{n\left( n+k \right)}}}\\&-\sum_{k=1}^{\infty}{\frac{1}{2^k}\frac{1}{k^3}\sum_{n=1}^{\infty}{\frac{1}{n\left( n+k \right)}}}+\sum_{k=1}^{\infty}{\frac{1}{2^k}\frac{1}{k^3}\sum_{n=1}^{\infty}{\frac{1}{\left( n+k \right) ^2}}}\\ &=-\zeta \left( 3 \right) \text{Li}_2\left( \frac{1}{2} \right) +\frac{\pi ^2}{2}\text{Li}_3\left( \frac{1}{2} \right) -3\sum_{k=1}^{\infty}{\frac{H_k}{2^kk^4}}-\sum_{k=1}^{\infty}{\frac{H_{k}^{\left( 2 \right)}}{2^kk^3}}\\ &=-\text{4Li}_5\left( \frac{1}{2} \right) +\frac{35\pi ^2}{96}\zeta \left( 3 \right) -\frac{29}{64}\zeta \left( 5 \right) +\frac{7\zeta \left( 3 \right) \ln ^22}{16}+\frac{\ln ^52}{30}\\ &+\frac{5\pi ^2\ln ^32}{72}-\frac{53\pi ^4\ln 2}{1440}\\ \end{align*}

\begin{align*} I_4&=\frac{1}{2}\int_0^1{\frac{\text{Li}_3\left( t \right)}{1-\frac{t}{2}}}\text{d}t=\frac{1}{2}\int_0^1{\sum_{k=0}^{\infty}{\frac{1}{2^k}\sum_{n=1}^{\infty}{\frac{t^{n+k}}{n^3}\text{d}t}}}=\frac{1}{2}\sum_{k=0}^{\infty}{\frac{1}{2^k}\sum_{n=1}^{\infty}{\frac{1}{n^3\left( n+k+1 \right)}}}\\ &=\sum_{k=1}^{\infty}{\frac{1}{2^k}\sum_{n=1}^{\infty}{\frac{1}{n^3\left( n+k \right)}}}=\sum_{k=1}^{\infty}{\frac{1}{2^k}\frac{1}{k}\sum_{n=1}^{\infty}{\left[ \frac{1}{n^3}-\frac{1}{n^2\left( n+k \right)} \right]}}\\ &=\sum_{k=1}^{\infty}{\frac{1}{2^k}\frac{1}{k}\sum_{n=1}^{\infty}{\frac{1}{n^3}}}-\sum_{k=1}^{\infty}{\frac{1}{2^k}\frac{1}{k}\sum_{n=1}^{\infty}{\frac{1}{n^2\left( n+k \right)}}}=\zeta \left( 3 \right) \ln 2-\sum_{k=1}^{\infty}{\frac{1}{2^k}\frac{1}{k^2}\sum_{n=1}^{\infty}{\left[ \frac{1}{n^2}-\frac{1}{n\left( n+k \right)} \right]}}\\ &=\zeta \left( 3 \right) \ln 2-\sum_{k=1}^{\infty}{\frac{1}{2^k}\frac{1}{k^2}\sum_{n=1}^{\infty}{\frac{1}{n^2}}}+\sum_{k=1}^{\infty}{\frac{1}{2^k}\frac{1}{k^2}\sum_{n=1}^{\infty}{\frac{1}{n\left( n+k \right)}}}\\ &=\zeta \left( 3 \right) \ln 2-\frac{\pi ^2}{6}\text{Li}_2\left( \frac{1}{2} \right) +\sum_{k=1}^{\infty}{\frac{H_k}{2^kk^3}}=\text{Li}_4\left( \frac{1}{2} \right) +\frac{\text{7}\ln 2}{8}\zeta \left( 3 \right) -\frac{\pi ^4}{80}+\frac{\ln ^42}{24}+\frac{\pi ^2\ln ^22}{12}\\ \end{align*}

\begin{align*} I_5&=\frac{1}{2}\int_0^1{\frac{\ln ^2t\text{Li}_2\left( t \right)}{1-\frac{t}{2}}}\text{d}t=\frac{1}{2}\int_0^1{\sum_{k=0}^{\infty}{\frac{1}{2^k}\ln ^2t\sum_{n=1}^{\infty}{\frac{t^{n+k}}{n^2}\text{d}t}}}=\sum_{k=0}^{\infty}{\frac{1}{2^k}\sum_{n=1}^{\infty}{\frac{1}{n^2\left( n+k+1 \right) ^3}}}\\ &=2\sum_{k=1}^{\infty}{\frac{1}{2^k}\sum_{n=1}^{\infty}{\frac{1}{n^2\left( n+k \right) ^3}}}=2\sum_{k=1}^{\infty}{\frac{1}{2^k}\frac{1}{k}\sum_{n=1}^{\infty}{\left[ \frac{1}{n^2\left( n+k \right) ^2}-\frac{1}{n\left( n+k \right) ^3} \right]}}\\ &=2\sum_{k=1}^{\infty}{\frac{1}{2^k}\frac{1}{k}\sum_{n=1}^{\infty}{\frac{1}{n^2\left( n+k \right) ^2}}}-2\sum_{k=1}^{\infty}{\frac{1}{2^k}\frac{1}{k}\sum_{n=1}^{\infty}{\frac{1}{n\left( n+k \right) ^3}}}\\ &=2\sum_{k=1}^{\infty}{\frac{1}{2^k}\frac{1}{k^2}\sum_{n=1}^{\infty}{\left[ \frac{1}{n^2\left( n+k \right)}-\frac{1}{n\left( n+k \right) ^2} \right]}}-2\sum_{k=1}^{\infty}{\frac{1}{2^k}\frac{1}{k^2}\sum_{n=1}^{\infty}{\left[ \frac{1}{n\left( n+k \right) ^2}-\frac{1}{\left( n+k \right) ^3} \right]}}\\ &=2\sum_{k=1}^{\infty}{\frac{1}{2^k}\frac{1}{k^2}\sum_{n=1}^{\infty}{\frac{1}{n^2\left( n+k \right)}}}-4\sum_{k=1}^{\infty}{\frac{1}{2^k}\frac{1}{k^2}\sum_{n=1}^{\infty}{\frac{1}{n\left( n+k \right) ^2}}}+2\sum_{k=1}^{\infty}{\frac{1}{2^k}\frac{1}{k^2}\sum_{n=1}^{\infty}{\frac{1}{\left( n+k \right) ^3}}}\\ &=2\sum_{k=1}^{\infty}{\frac{1}{2^k}\frac{1}{k^3}\sum_{n=1}^{\infty}{\left[ \frac{1}{n^2}-\frac{1}{n\left( n+k \right)} \right]}}-4\sum_{k=1}^{\infty}{\frac{1}{2^k}\frac{1}{k^3}\sum_{n=1}^{\infty}{\left[ \frac{1}{n\left( n+k \right)}-\frac{1}{\left( n+k \right) ^2} \right]}}\\ &+2\sum_{k=1}^{\infty}{\frac{1}{2^k}\frac{1}{k^2}\left[ \zeta \left( 3 \right) -H_{k}^{\left( 3 \right)} \right]}\\ &=2\sum_{k=1}^{\infty}{\frac{1}{2^k}\frac{1}{k^3}\sum_{n=1}^{\infty}{\frac{1}{n^2}}}-6\sum_{k=1}^{\infty}{\frac{1}{2^k}\frac{1}{k^3}\sum_{n=1}^{\infty}{\frac{1}{n\left( n+k \right)}}}+4\sum_{k=1}^{\infty}{\frac{1}{2^k}\frac{1}{k^3}\left[ \frac{\pi ^2}{6}-H_{k}^{\left( 2 \right)} \right]}\\ &+2\sum_{k=1}^{\infty}{\frac{1}{2^k}\frac{1}{k^2}\left[ \zeta \left( 3 \right) -H_{k}^{\left( 3 \right)} \right]}\\ &=\pi ^2\text{Li}_3\left( \frac{1}{2} \right) +2\zeta \left( 3 \right) \text{Li}_2\left( \frac{1}{2} \right) -6\sum_{k=1}^{\infty}{\frac{H_k}{2^kk^4}}-4\sum_{k=1}^{\infty}{\frac{H_{k}^{\left( 2 \right)}}{2^kk^3}}-2\sum_{k=1}^{\infty}{\frac{H_{k}^{\left( 3 \right)}}{2^kk^2}}\\ &=-\text{12Li}_5\left( \frac{1}{2} \right) +\frac{7\pi ^2}{8}\zeta \left( 3 \right) +\frac{29}{32}\zeta \left( 5 \right) +\frac{\ln ^52}{10}+\frac{\pi ^2\ln ^32}{12}-\frac{19\pi ^4\ln 2}{240} \end{align*}

\begin{align*} I_6&=\frac{1}{2}\int_0^1{\frac{\ln t\text{Li}_2\left( t \right)}{1-\frac{t}{2}}}\text{d}t=\frac{1}{2}\int_0^1{\sum_{k=0}^{\infty}{\frac{1}{2^k}\ln t\sum_{n=1}^{\infty}{\frac{t^{n+k}}{n^2}\text{d}t}}}=-\frac{1}{2}\sum_{k=0}^{\infty}{\frac{1}{2^k}\sum_{n=1}^{\infty}{\frac{1}{n^2\left( n+k+1 \right) ^2}}}\\ &=-\sum_{k=1}^{\infty}{\frac{1}{2^k}\sum_{n=1}^{\infty}{\frac{1}{n^2\left( n+k \right) ^2}}}=-\sum_{k=1}^{\infty}{\frac{1}{2^k}\frac{1}{k}\sum_{n=1}^{\infty}{\left[ \frac{1}{n^2\left( n+k \right)}-\frac{1}{n\left( n+k \right) ^2} \right]}}\\ &=-\sum_{k=1}^{\infty}{\frac{1}{2^k}\frac{1}{k}\sum_{n=1}^{\infty}{\frac{1}{n^2\left( n+k \right)}}}+\sum_{k=1}^{\infty}{\frac{1}{2^k}\frac{1}{k}\sum_{n=1}^{\infty}{\frac{1}{n\left( n+k \right) ^2}}}\\ &=-\sum_{k=1}^{\infty}{\frac{1}{2^k}\frac{1}{k^2}\sum_{n=1}^{\infty}{\left[ \frac{1}{n^2}-\frac{1}{n\left( n+k \right)} \right]}}+\sum_{k=1}^{\infty}{\frac{1}{2^k}\frac{1}{k^2}\sum_{n=1}^{\infty}{\left[ \frac{1}{n\left( n+k \right)}-\frac{1}{\left( n+k \right) ^2} \right]}}\\ &=-\sum_{k=1}^{\infty}{\frac{1}{2^k}\frac{1}{k^2}\sum_{n=1}^{\infty}{\frac{1}{n^2}}}+2\sum_{k=1}^{\infty}{\frac{1}{2^k}\frac{1}{k^2}\sum_{n=1}^{\infty}{\frac{1}{n\left( n+k \right)}}}\\ &-\sum_{k=1}^{\infty}{\frac{1}{2^k}\frac{1}{k^2}\sum_{n=1}^{\infty}{\frac{1}{\left( n+k \right) ^2}}}=-\frac{\pi ^2}{3}\text{Li}_2\left( \frac{1}{2} \right) +2\sum_{k=1}^{\infty}{\frac{H_k}{2^kk^3}}+\sum_{k=1}^{\infty}{\frac{H_{k}^{\left( 2 \right)}}{2^kk^2}}\\ &=\text{3Li}_4\left( \frac{1}{2} \right) -\frac{7\pi ^4}{288}+\frac{\ln ^42}{8}+\frac{\pi ^2\ln ^22}{8} \end{align*}

\begin{align*} I_7&=\frac{1}{2}\int_0^1{\frac{\ln ^3x\ln \left( 1-x \right)}{1-\frac{x}{2}}\text{d}x}=-\frac{1}{2}\sum_{k=0}^{\infty}{\frac{1}{2^k}}\sum_{n=1}^{\infty}{\frac{1}{n}}\int_0^1{x^{n+k}\ln ^3x\text{d}x}=3\sum_{k=0}^{\infty}{\frac{1}{2^k}}\sum_{n=1}^{\infty}{\frac{1}{n\left( n+k+1 \right) ^4}}\\ &=6\sum_{k=1}^{\infty}{\frac{1}{2^k}}\sum_{n=1}^{\infty}{\frac{1}{n\left( n+k \right) ^4}}=6\sum_{k=1}^{\infty}{\frac{1}{2^k}}\frac{1}{k}\sum_{n=1}^{\infty}{\left[ \frac{1}{n\left( n+k \right) ^3}-\frac{1}{\left( n+k \right) ^4} \right]}\\ &=6\sum_{k=1}^{\infty}{\frac{1}{2^k}}\frac{1}{k}\sum_{n=1}^{\infty}{\frac{1}{n\left( n+k \right) ^3}}-6\sum_{k=1}^{\infty}{\frac{1}{2^k}}\frac{1}{k}\sum_{n=1}^{\infty}{\left[ \frac{\pi ^4}{90}-H_{k}^{\left( 4 \right)} \right]}\\ &=6\sum_{k=1}^{\infty}{\frac{1}{2^k}}\frac{1}{k}\sum_{n=1}^{\infty}{\left[ \frac{1}{n\left( n+k \right) ^2}-\frac{1}{\left( n+k \right) ^3} \right]}\\ &-6\sum_{k=1}^{\infty}{\frac{1}{2^k}}\frac{1}{k}\sum_{n=1}^{\infty}{\left[ \frac{\pi ^4}{90}-H_{k}^{\left( 4 \right)} \right]}\\ &=6\sum_{k=1}^{\infty}{\frac{1}{2^k}}\frac{1}{k^2}\sum_{n=1}^{\infty}{\frac{1}{n\left( n+k \right) ^2}-}6\sum_{k=1}^{\infty}{\frac{1}{2^k}}\frac{1}{k^2}\sum_{n=1}^{\infty}{\left[ \zeta \left( 3 \right) -H_{k}^{\left( 3 \right)} \right]}-6\sum_{k=1}^{\infty}{\frac{1}{2^k}}\frac{1}{k}\sum_{n=1}^{\infty}{\left[ \frac{\pi ^4}{90}-H_{k}^{\left( 4 \right)} \right]}\\ &=6\sum_{k=1}^{\infty}{\frac{1}{2^k}}\frac{1}{k^3}\sum_{n=1}^{\infty}{\left[ \frac{1}{n\left( n+k \right)}-\frac{1}{\left( n+k \right) ^2} \right] -}6\sum_{k=1}^{\infty}{\frac{1}{2^k}}\frac{1}{k^2}\sum_{n=1}^{\infty}{\left[ \zeta \left( 3 \right) -H_{k}^{\left( 3 \right)} \right]}\\ &-6\sum_{k=1}^{\infty}{\frac{1}{2^k}}\frac{1}{k}\sum_{n=1}^{\infty}{\left[ \frac{\pi ^4}{90}-H_{k}^{\left( 4 \right)} \right]}\\ &=6\sum_{k=1}^{\infty}{\frac{1}{2^k}}\frac{1}{k^3}\sum_{n=1}^{\infty}{\frac{1}{n\left( n+k \right)}-6\sum_{k=1}^{\infty}{\frac{1}{2^k}}\frac{1}{k^3}\sum_{n=1}^{\infty}{\left[ \frac{\pi ^2}{6}-H_{k}^{\left( 2 \right)} \right]}-}6\sum_{k=1}^{\infty}{\frac{1}{2^k}}\frac{1}{k^2}\sum_{n=1}^{\infty}{\left[ \zeta \left( 3 \right) -H_{k}^{\left( 3 \right)} \right]}\\ &-6\sum_{k=1}^{\infty}{\frac{1}{2^k}}\frac{1}{k}\sum_{n=1}^{\infty}{\left[ \frac{\pi ^4}{90}-H_{k}^{\left( 4 \right)} \right]}\\ &=6\sum_{k=1}^{\infty}{\frac{1}{2^k}}\frac{H_k}{k^4}-6\sum_{k=1}^{\infty}{\frac{1}{2^k}}\frac{1}{k^3}\sum_{n=1}^{\infty}{\left[ \frac{\pi ^2}{6}-H_{k}^{\left( 2 \right)} \right]}-6\sum_{k=1}^{\infty}{\frac{1}{2^k}}\frac{1}{k^2}\sum_{n=1}^{\infty}{\left[ \zeta \left( 3 \right) -H_{k}^{\left( 3 \right)} \right]}\\ &-6\sum_{k=1}^{\infty}{\frac{1}{2^k}}\frac{1}{k}\sum_{n=1}^{\infty}{\left[ \frac{\pi ^4}{90}-H_{k}^{\left( 4 \right)} \right]}\\ &=6\sum_{k=1}^{\infty}{\frac{H_k}{2^kk^4}}-\pi ^2\sum_{k=1}^{\infty}{\frac{1}{2^kk^3}}+6\sum_{k=1}^{\infty}{\frac{H_{k}^{\left( 2 \right)}}{2^kk^3}}-6\zeta \left( 3 \right) \sum_{k=1}^{\infty}{\frac{1}{2^kk^2}}\\ &+6\sum_{k=1}^{\infty}{\frac{H_{k}^{\left( 3 \right)}}{2^kk^2}}-\frac{\pi ^4}{15}\sum_{k=1}^{\infty}{\frac{1}{2^kk}}+6\sum_{k=1}^{\infty}{\frac{H_{k}^{\left( 4 \right)}}{2^kk}}\\ &=\text{18Li}_5\left( \frac{1}{2} \right) -\frac{7\pi ^2}{8}\zeta \left( 3 \right) -\frac{3}{16}\zeta \left( 5 \right) -\frac{\text{3}\ln ^52}{20}+\frac{\pi ^4\ln 2}{40} \end{align*}

\begin{align*} I_8&=\frac{1}{2}\int_0^1{\frac{\ln ^2x\ln \left( 1-x \right)}{1-\frac{x}{2}}\text{d}x}=-\frac{1}{2}\sum_{k=0}^{\infty}{\frac{1}{2^k}}\sum_{n=1}^{\infty}{\frac{1}{n}}\int_0^1{x^{n+k}\ln ^2x\text{d}x}=-\sum_{k=0}^{\infty}{\frac{1}{2^k}}\sum_{n=1}^{\infty}{\frac{1}{n\left( n+k+1 \right) ^3}}\\ &=-2\sum_{k=1}^{\infty}{\frac{1}{2^k}}\sum_{n=1}^{\infty}{\frac{1}{n\left( n+k \right) ^3}}=-2\sum_{k=1}^{\infty}{\frac{1}{2^k}}\frac{1}{k}\sum_{n=1}^{\infty}{\left[ \frac{1}{n\left( n+k \right) ^2}-\frac{1}{\left( n+k \right) ^3} \right]}\\ &=-2\sum_{k=1}^{\infty}{\frac{1}{2^k}}\frac{1}{k}\sum_{n=1}^{\infty}{\frac{1}{n\left( n+k \right) ^2}}+2\sum_{k=1}^{\infty}{\frac{1}{2^k}}\frac{1}{k}\sum_{n=1}^{\infty}{\left[ \zeta \left( 3 \right) -H_{k}^{\left( 3 \right)} \right]}\\ &=-2\sum_{k=1}^{\infty}{\frac{1}{2^k}}\frac{1}{k^2}\sum_{n=1}^{\infty}{\left[ \frac{1}{n\left( n+k \right)}-\frac{1}{\left( n+k \right) ^2} \right]}+2\sum_{k=1}^{\infty}{\frac{1}{2^k}}\frac{1}{k}\sum_{n=1}^{\infty}{\left[ \zeta \left( 3 \right) -H_{k}^{\left( 3 \right)} \right]}\\ &=-2\sum_{k=1}^{\infty}{\frac{H_k}{2^kk^3}}+2\sum_{k=1}^{\infty}{\frac{1}{2^k}}\frac{1}{k^2}\left[ \frac{\pi ^2}{6}-H_{k}^{\left( 2 \right)} \right] +2\sum_{k=1}^{\infty}{\frac{1}{2^k}}\frac{1}{k}\sum_{n=1}^{\infty}{\left[ \zeta \left( 3 \right) -H_{k}^{\left( 3 \right)} \right]}\\ &=-2\sum_{k=1}^{\infty}{\frac{H_k}{2^kk^3}}+\frac{\pi ^2}{3}\text{Li}_2\left( \frac{1}{2} \right) -2\sum_{k=1}^{\infty}{\frac{H_{k}^{\left( 2 \right)}}{2^kk^2}}+\text{2}\ln 2\zeta \left( 3 \right) -2\sum_{k=1}^{\infty}{\frac{H_{k}^{\left( 3 \right)}}{2^kk}}\\ &=-\text{6Li}_4\left( \frac{1}{2} \right) +\frac{11\pi ^4}{360}-\frac{\ln ^42}{4} \end{align*}

\begin{align*} I&=\text{2Li}_5\left( \frac{1}{2} \right) +\text{2}\ln\text{2Li}_4\left( \frac{1}{2} \right) +\frac{81}{32}\zeta \left( 5 \right) -\frac{\ln ^22}{8}\zeta \left( 3 \right) \\ &+\frac{5\pi ^2}{16}\zeta \left( 3 \right) -\frac{\pi ^2\ln ^32}{18}-\frac{\pi ^4\ln 2}{15}+\frac{\ln ^52}{15} \end{align*}