# Compute $\int_0^{1/2}\frac{\left(\operatorname{Li}_2(x)\right)^2}{x}dx$ or $\sum_{n=1}^\infty \frac{H_n^{(2)}}{n^32^n}$

Prove that

I encountered this integral while working on the sum $$\displaystyle \sum_{n=1}^\infty \frac{H_n^{(2)}}{n^32^n}$$. Both of the integral and the sum were proposed by Cornel Valean:

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

$$\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^32^n}=-2\operatorname{Li}_5\left(\frac12\right)-3\ln2\operatorname{Li}_4\left(\frac12\right)+\frac{23}{64}\zeta(5)-\frac1{16}\ln2\zeta(4)+\frac{23}{16}\zeta(2)\zeta(3)\\-\frac{23}{16}\ln^22\zeta(3)+\frac7{12}\ln^32\zeta(2)-\frac{13}{120}\ln^52.$$

This sum itself is related, through the Cauchy product of $$\ln(1-x)\text{Li}_3(x)$$, to the following sum:

$$\sum_{n=1}^\infty \frac{H_n^{(3)}}{n^22^n}=4\operatorname{Li}_5\left(\frac12\right)+3\ln2\operatorname{Li}_4\left(\frac12\right)-\frac{81}{64}\zeta(5)+\frac5{16}\ln2\zeta(4)\nonumber\\ -\frac78\zeta(2)\zeta(3)+\frac{7}{8}\ln^22\zeta(3)-\frac5{12}\ln^32\zeta(2)+\frac{11}{120}\ln^52$$

The main integral is very related to the integral $$\int_0^1 \frac{\ln^3(1-x)\ln(1+x)}{x}dx$$ which I managed to solve using three tough results of alternating series, so again I am looking for a different approach that does not use these results ( mentioned in the link) to compute $$I$$.

Here is how the two integrals are related:

$$\int_0^{1/2}\frac{\left(\operatorname{Li}_2(x)\right)^2}{x}dx\overset{IBP}{=}\operatorname{Li}_2\left(\frac12\right)\operatorname{Li}_3\left(\frac12\right)-\ln2\operatorname{Li}_4\left(\frac12\right)-\operatorname{Li}_5\left(\frac12\right)+\sum_{n=1}^\infty\frac{H_n^{(4)}}{n2^n}$$

From this identity, we have $$\sum_{n=1}^\infty\frac{H_n^{(4)}}{n2^n}=-\frac16\int_0^1\frac{\ln^3(1-x)\ln(1+x)}{x}dx$$

Then

$$\int_0^{1/2}\frac{\left(\operatorname{Li}_2(x)\right)^2}{x}dx=\operatorname{Li}_2\left(\frac12\right)\operatorname{Li}_3\left(\frac12\right)-\ln2\operatorname{Li}_4\left(\frac12\right)-\operatorname{Li}_5\left(\frac12\right)\\-\frac16\int_0^1\frac{\ln^3(1-x)\ln(1+x)}{x}dx$$

So, any elegant way to solve any of these two integrals?

• Related Commented Jul 27, 2019 at 22:32
• @mrtaurho no its not related. the one you referring to is very easy and I did it before and the integral I posted here is completely different and challenging. Commented Jul 27, 2019 at 22:36
• I know, that you posted an answer there. I only mentioned that the other post is related as it is precisely the same integrand while, of course, the upper bound of one-half makes the whole thing a lot more complicated. However, for future (or current) reader the other post may be of interest aswell. I only linked the question, I did not marked yours as a duplicate (which it is definitely not!). Commented Jul 27, 2019 at 22:40
• oh got you now. Thanks anyway. Commented Jul 27, 2019 at 22:44
• That half in the upper limit makes a world of difference. Commented Jul 28, 2019 at 7:41

Considering the algebraic identity \begin{align*} &(a-b)^3b = a^3b - 3a^2b^2 + 3ab^3 - b^4 = -2a^3b +3(a^3b+ab^3) -3a^2b^2 -b^4\\ &\Longrightarrow \ \ \ 2a^3b = -{b^4 \over 2} -{b^4 + 6a^2b^2\over 2} + 3(a^3b+ab^3) - (a-b)^3b \end{align*} with $$a = \ln(1-x)$$ and $$b= \ln (1+x)$$ it follows that \begin{align*} 2\int_0^1 {\ln^3(1-x)\ln(1+x)\over x}dx =& - \frac 1 2\int_0^1 {\ln^4(1+x)\over x}d x \\ &-\frac 12 \int_0^1 \frac{\ln^4(1+x) + 6\ln^2(1-x)\ln^2(1+x)}{x}dx\\ &+3\int_0^1 \frac{\ln^3(1-x)\ln(1+x) + \ln(1-x)\ln^3(1+x)}{x}dx\\ &- \int_0^1 \frac{\ln^3\left(\frac{1-x}{1+x}\right)\ln(1+x)}{x}dx\\ =:& -I_1 - I_2 + I_3 -I_4. \end{align*}

For $$I_1$$, make substitution $$y = \frac x {1+x}$$ to get: \begin{align*} I_1 =& \frac 1 2 \int_0^{\frac 12} \frac{\ln^4(1-y)}{y(1-y)} dy \\ =& \frac 1 2\underbrace{ \int_0^{\frac 12} \frac{\ln^4(1-y)}{y} dy}_{z=1-y}+ \frac 1 2 \int_0^{\frac 12} \frac{\ln^4(1-y)}{1-y} dy\\ =& \frac 1 2 \int_{\frac 1 2 }^1 \frac{\ln^4 z} {1-z} dz + \frac {\ln^5 2}{10}\\ =& \frac 12 \sum_{n=1}^\infty \int_{\frac 1 2}^1 z^{n-1}\ln^4 z\ dz + \frac {\ln^5 2}{10}\\ =& \frac 12 \sum_{n=1}^\infty \frac{\partial^4}{\partial n^4}\left[\frac 1 n - \frac 1 {n2^n}\right] + \frac {\ln^5 2}{10}\\ =& \frac 12 \sum_{n=1}^\infty \left[\frac{24}{n^5} - \frac {24}{n^52^n} - \frac{24 \ln 2}{n^42^n}-\frac{12\ln^2 2}{n^3 2^n}-\frac{4\ln^3 2}{n^2 2^n} - \frac{\ln^4 2}{n2^n}\right] + \frac {\ln^5 2}{10}\\ =&12\zeta(5) - 12\text{Li}_5(1/2) - 12\ln 2 \text{Li}_4(1/2) -6\ln^2 2 \text{Li}_3(1/2) -2\ln^3 2\text{Li}_2(1/2)-\frac {2}{5}\ln^5 2\\ =&\boxed{-12\Big(\text{Li}_5(1/2) + \ln 2\text{Li}_4(1/2)-\zeta(5)\Big)-{21 \over 4}\zeta(3)\ln^2 2 +{1\over 3} \pi^2 \ln^3 2-{2 \over 5} \ln^5 2} \end{align*} where the well-known values \begin{align*}\text{Li}_2(1/2) = {\pi^2 \over 12}-{\ln^2 2\over 2} , \qquad \text{Li}_3(1/2) ={7\zeta(3) \over 8} -{\pi^2 \ln 2\over 12} + {\ln^3 2 \over 6} \end{align*} are used.

Actually, $$I_2$$ was already evaluated by the OP here using the algebraic identity $$b^4 + 6a^2b^2 = \frac {(a-b)^4} 2+\frac{(a+b)^4}{2} -a^4.$$ It holds that $$\boxed{I_2 = \frac {21}{8} \zeta(5).}$$

In fact, the value of $$I_3$$ can also be found in the previous answer of @Przemo's. For $$I_3$$, one can use the algebraic relation $$3(a^3b + ab^3) =\frac 3 8 \left[ (a+b)^4 - (a-b)^4\right]$$. This gives \begin{align*} I_3=& \underbrace{\frac 3 8 \int_0^1 \frac{\ln^4(1-x^2)}{x} dx}_{x^2 = y} - \underbrace{\frac 3 8 \int_0^1 \frac{\ln^4\left(\frac{1-x}{1+x}\right)}{x} dx}_{\frac{1-x}{1+x} = y}\\ =&\frac 3 {16}\underbrace{\int_0^1 \frac{\ln^4(1-y)}{y} dy }_{1-y\mapsto y}- \frac 3 4 \int_0^1 \frac{\ln^4 y}{1-y^2} dy\\ =&\frac 3 {16}\int_0^1 \frac{\ln^4 y}{1-y} dy - \frac 3 4 \sum_{n=0}^\infty \int_0^1 y^{2n} \ln^4 y \ dy\\ =&\frac 3 {16}\sum_{n=1}^\infty \int_0^1 y^{n-1}\ln^4 y \ dy - \frac 3 4 \sum_{n=0}^\infty \frac {24}{(2n+1)^5}\\ =&\frac 3 {16}\sum_{n=1}^\infty \frac{24}{n^5} - 18 \sum_{n=0}^\infty \frac {1}{(2n+1)^5}\\ =&\frac {9}{2} \zeta(5)- 18\cdot \frac {31}{32}\zeta(5)\\ =&\boxed{-\frac{207}{16}\zeta(5)} \end{align*} as can be found in @Przemo's answer.

For $$I_4$$, make substitution $$\frac{1-x}{1+x}\mapsto x$$ to get \begin{align*} I_4 = &2\int_0^1 \frac{\ln^3 x \ln\left(\frac 2 {1+x}\right)}{1-x^2} dx \\ =&2\ln 2 \int_0^1 \frac{\ln^3 x}{1-x^2} dx - \underbrace{2\int_0^1\frac{\ln^3 x \ln(1+x)}{1-x^2} dx }_{=:J}\\ =& 2\ln 2\sum_{n=0}^\infty \int_0^1 x^{2n} \ln^3 x\ dx - J\\ =& - 12\ln 2 \underbrace{\sum_{n=0}^\infty \frac 1 {(2n+1)^4}}_{\frac{15}{16}\zeta(4) = \frac{\pi^4}{96}} - J \\ =& -\frac{\pi^4 \ln 2}{8} - J. \end{align*} \begin{align*} J = &\int_0^1\frac{2\ln^3 x \ln(1+x)}{1-x^2} dx \\ =& \underbrace{\int_0^1 \frac{\ln^3 x \ln(1+x)}{1+x}dx}_{=:A} + \int_0^1 \frac{\ln^3 x \ln(1+x)}{1-x}dx\\ =& A + \int_0^1 \frac{\ln^3 x \ln(1-x^2)}{1-x}dx -\int_0^1 \frac{\ln^3 x \ln(1-x)}{1-x}dx\\ =&A + \int_0^1 \frac{(1+x)\ln^3 x \ln(1-x^2)}{1-x^2}dx -\int_0^1 \frac{\ln^3 x \ln(1-x)}{1-x}dx\\ =&A + \underbrace{\int_0^1 \frac{\ln^3 x \ln(1-x^2)}{1-x^2}dx }_{=:B}+\underbrace{\int_0^1 \frac{x\ln^3 x \ln(1-x^2)}{1-x^2}dx}_{x^2 \mapsto x}-\int_0^1 \frac{\ln^3 x \ln(1-x)}{1-x}dx\\ =&A + B - \underbrace{\frac {15}{16} \int_0^1 \frac{\ln^3 x \ln(1-x)}{1-x}dx}_{=:C}\\ =&A + B - C. \end{align*}

For $$A$$, we can use the McLaurin series of $$\frac{\ln (1+x)}{1+x} = \sum_{n=0}^\infty (-1)^{n-1}H_n x^n$$ ($$H_0= 0$$) to get \begin{align*} A = & \sum_{n=0}^\infty (-1)^{n-1}H_n \int_0^1 x^n\ln^3 x \ dx \\ =&6 \sum_{n=0}^\infty \frac{(-1)^{n}H_n}{(n+1)^4}\\ =&6 \sum_{n=0}^\infty \frac{(-1)^{n}H_{n+1}}{(n+1)^4} - 6\sum_{n=0}^\infty \frac{(-1)^{n}}{(n+1)^5}\\ =&6 \sum_{n=1}^\infty \frac{(-1)^{n-1}H_{n}}{n^4} - 6\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^5}\\ =& 6\left(\frac{59}{32}\zeta(5) - \frac{\pi^2\zeta(3)}{12}\right)-6\cdot \frac{15}{16}\zeta(5)\\ =& \frac{87}{16}\zeta(5) - \frac{\pi^2 \zeta(3)}{2}. \end{align*} Here, the known value of $$\sum_{n=1}^\infty (-1)^{n-1}{H_n \over n^4}$$ is used.

For $$B$$, make substitution $$u = x^2$$ to get \begin{align*} B =& \frac 1 {16} \int_0^1 \frac{\ln^3 u \ln(1-u)}{\sqrt u (1-u)} du \\ =& \frac 1 {16} \left[\frac{\partial^4}{\partial x^3\partial y} \text{B}(x,y)\right]_{x=\frac 1 2, y = 0^+} \end{align*} where $$\text{B}(\cdot,\cdot)$$ is Euler's Beta function. We can use the fact that \begin{align*} \lim_{y\to 0^+}\frac{\partial^2}{\partial x\partial y} \text{B}(x,y) = -\frac 1 2 \psi''(x) + \psi'(x) \big[\psi(x) + \gamma\big] \end{align*} to get \begin{align*} B =& \frac 1 {16}\frac{d^2}{dx^2}\left[-\frac 1 2 \psi''(x) + \psi'(x) \big[\psi(x) + \gamma\big]\right]_{x=\frac 1 2}\\ =&\frac 1 {16} \left[-\frac 1 2 \psi''''(1/2) + \psi'''(1/2)\big[\psi(1/2) + \gamma\big] + 3\psi'(1/2)\psi''(1/2)\right]\\ =& \frac 1 {16}\left[-21\pi^2 \zeta(3) + 372\zeta(5) - 2\pi^4 \ln 2\right] \end{align*} which can be evaluated using the series representations of polygamma functions $$\psi(x) +\gamma = - \frac 1 x +\sum_{n=1}^\infty \frac 1 n - \frac 1 { n+x},\\ \psi'(x) = \sum_{n=0}^\infty \frac 1 {(n+x)^2}$$ and the derived fact that $$\psi(\tfrac 1 2 )+\gamma = -2\ln 2$$ and $$\psi^{(k)}(\tfrac 1 2)=(-1)^{k+1}k!(2^{k+1}-1)\zeta(k+1)$$ for $$k\ge 1$$.

For $$C$$, we can use the same method as used in the evaluation of $$B$$. It holds that \begin{align*} C =& \frac {15}{16} \left[\frac{\partial^4}{\partial x^3\partial y} \text{B}(x,y)\right]_{x=1, y = 0^+}\\ =&\frac {15} {16}\left[-\frac 1 2 \psi''''(1) + \psi'''(1)\big[\psi(1) + \gamma\big] + 3\psi'(1)\psi''(1)\right]\\ =&\frac{15}{16}\left[12\zeta(5) -6\zeta(2)\zeta(3)\right]\\ =&\frac {45}{4}\zeta(5) -\frac {15\pi^2 \zeta(3)}{16} \end{align*} where $$\psi(1) +\gamma = 0$$, $$\psi'(1) = \zeta(2)$$, $$\psi''(1) = -2\zeta(3)$$ and $$\psi''''(1) = -24\zeta(5)$$ are used.

Combining $$A,B,C$$, we have that $$J =A+B-C= \frac{279}{16}\zeta(5) -\frac{7\pi^2\zeta(3)}{8} - \frac{\pi^4 \ln 2}{8}$$ and $$\boxed{I_4 = -\frac{\pi^4 \ln 2}{8} - J = -\frac{279}{16}\zeta(5)+\frac{7\pi^2\zeta(3)}{8}}$$

Finally, these evaluate $$\int_0^1 {\ln^3(1-x)\ln(1+x)\over x}dx =\frac 1 2\big[-I_1-I_2+I_3-I_4\big]$$ as follows.

\begin{align*} \int_0^1 {\ln^3(1-x)\ln(1+x)\over x}dx =&\ 6\text{Li}_5(1/2) + 6\ln 2\ \text{Li}_4(1/2)-\frac{81}{16}\zeta(5)-{7\pi^2 \over 16}\zeta(3)\\ &+\frac{21\ln^2 2}{8}\zeta(3)- \frac{1}{6}\pi^2\ln^3 2+\frac{1}{5}\ln^5 2. \end{align*}

Using the identity given in the OP, we get the desired integral $$I$$

\begin{align*} \int_0^{\frac 1 2}\frac{\text{Li}_2^2(x)}{x} dx = &-2\text{Li}_5(1/2) -2\ln 2\ \text{Li}_4(1/2)+\frac{27}{32}\zeta(5) +\frac{7\pi^2}{48}\zeta(3)-\frac{7\ln^2 2}{8}\zeta(3) \\ &-\frac{\pi^4\ln 2}{144} +\frac{\pi^2\ln^3 2}{12} - \frac{7\ln^5 2}{60}. \end{align*}

• Impressive! That was a really useful algebraic identity. Commented Aug 17, 2019 at 11:27
• @Zacky Thank you! I hope it was helpful :) Commented Aug 17, 2019 at 12:19
• @AliShather Glad you liked it :) and thank you for your comment on $J$. I'll be searching for an easier way while expecting yours! Commented Aug 17, 2019 at 14:05
• @Song such a beautiful work using an easy algebraic identity. By the way the integral J can be done in an simpler way and without using beta derivative I'll show you when I have time or you can give it a try. It's the same comment of few mins ago but just fixed some typos. Commented Aug 17, 2019 at 14:50
• @Song the integral $\int_0^1\frac{\ln^3x\ln(1+x)}{1-x}\ dx$ is calculated here math.stackexchange.com/q/3236084 Commented Aug 18, 2019 at 6:06

This is not a full solution to this problem but i believe it does provide useful insight and is not a cul de sac.

The following identities hold: $$\begin{eqnarray} \int\limits_0^1 \frac{\log(1-x)^3\cdot \log(1+x)}{x} dx + \int\limits_0^1 \frac{\log(1+x)^3\cdot \log(1-x)}{x} dx = -\frac{69}{16} \zeta(5) \quad (i) \\ \int\limits_0^1 \frac{\log(1-x)^2 \log(1+x)^2}{x} dx = 48 \text{Li}_5(2)-8 \text{Li}_2(2) \log ^3(2)+24 \text{Li}_3(2) \log ^2(2)-48 \text{Li}_4(2) \log (2)-\frac{75 \zeta (5)}{2}-2 i \pi \log ^4(2) \quad (ii) \end{eqnarray}$$

In[484]:= n = 4;
NIntegrate[Log[1 - x]^3/x Log[1 + x], {x, 0, 1}] +
NIntegrate[Log[1 + x]^3/x Log[1 - x], {x, 0, 1}]

-1/16 NIntegrate[Log[1 - x]^4/x, {x, 0, 1}] -
1/8 NIntegrate[Log[x]^4 (1/(1 + x)), {x, 0, 1}]
1/16 (Sum[
PolyLog[1 + p, 1] Binomial[n, p] p! (-1)^(p + 1), {p, n, n}]) -
1/8 NIntegrate[Log[x]^4 (1/(1 + x)), {x, 0, 1}]
val = 1/16 (PolyLog[1 + n, 1] n! (-1)^(n + 1)) -
1/8 (PolyLog[1 + n, -1] n! (-1)^(n + 1))
N[val, 50]

Out[485]= -4.47175

Out[486]= -4.47175

Out[487]= -4.47175

Out[488]= -((69 Zeta[5])/16)

Out[489]= -4.4717509440557828073040136603459598497461614653520

In[477]:= n = 4;
12 NIntegrate[Log[1 - x]^2/x Log[1 + x]^2, {x, 0, 1}]
(3/2 NIntegrate[Log[1 - x]^4/x, {x, 0, 1}] +
NIntegrate[Log[x]^4 (1/(1 + x)), {x, 0, 1}]) -
2 NIntegrate[Log[1 - x]^4/x, {x, 0, 1}] -
2 NIntegrate[Log[1 + x]^4/x, {x, 0, 1}]
(3/2 (PolyLog[1 + n, 1] n! (-1)^(n)) + (PolyLog[
1 + n, -1] n! (-1)^(n + 1))) -
2 (PolyLog[1 + n, 1] n! (-1)^(n)) -
2 NIntegrate[Log[1 + x]^4/x, {x, 0, 1}]
(3/2 (PolyLog[1 + n, 1] n! (-1)^(n)) + (PolyLog[
1 + n, -1] n! (-1)^(n + 1))) -
2 (PolyLog[1 + n, 1] n! (-1)^(n)) -
2 (Sum[ Log[1 + 1]^(n - p) PolyLog[1 + p, 1 + 1] Binomial[n,
p] p! (-1)^(p + 1), {p, 0, n}] -
PolyLog[1 + n, 1 + 0] Binomial[n, n] n! (-1)^(n + 1));

val = -2 I \[Pi] Log[2]^4 - 8 Log[2]^3 PolyLog[2, 2] +
24 Log[2]^2 PolyLog[3, 2] - 48 Log[2] PolyLog[4, 2] +
48 PolyLog[5, 2] - (75 Zeta[5])/2;
N[val, 50]

Out[478]= 10.7373

Out[479]= 10.7373

Out[480]= 10.7373

Out[483]= 10.7372609681247028385792813011310627400934758851668 +
0.*10^-50 I


We derived those identities in the following way. firstly we set $$(u,v)= (\log(1-x),\log(1+x))$$ and then we used the identity $$1/8((u+v)^4-(u-v)^4) = u^3 v + u v^3$$ in $$(i)$$ and the identity $$(u+v)^4 + (u-v)^4 = 2 u^4 + 12 u^2 v^2 + 2 v^4$$ in $$(ii)$$. After that we used change of variables and known anti-derivatives like the one below: $$$$\int \frac{\log(1-x)^n}{x} dx = \sum\limits_{p=0}^n \log(1-x)^{n-p} Li_{1+p}(1-x) \binom{n}{p} p! (-1)^{p+1}$$$$

Different approach

By Cauchy product we have

$$\operatorname{Li}_2^2(x)=\sum_{n=1}^\infty x^n\left(\frac{4H_n}{n^3}+\frac{2H_n^{(2)}}{n^2}-\frac{6}{n^4}\right)$$

Divide both sides by $$x$$ then $$\int_0^{1/2}$$ we get

$$\int_0^{1/2}\frac{\operatorname{Li}_2^2(x)}{x}\ dx=4\sum_{n=1}^\infty\frac{H_n}{n^42^n}+2\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^32^n}-6\operatorname{Li}_5\left(\frac12\right)$$

Substitute

\begin{align} \displaystyle\sum_{n=1}^{\infty}\frac{H_n}{ n^42^n}&=2\operatorname{Li_5}\left( \frac12\right)+\ln2\operatorname{Li_4}\left( \frac12\right)-\frac16\ln^32\zeta(2) +\frac12\ln^22\zeta(3)\\ &\quad-\frac18\ln2\zeta(4)- \frac12\zeta(2)\zeta(3)+\frac1{32}\zeta(5)+\frac1{40}\ln^52 \end{align}

we get

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

A solution by Cornel I. Valean to $$\displaystyle \int_0^{1/2}\frac{(\operatorname{Li}_2(x))^2}{x}\textrm{d}x$$ (based on a reduction strategy to weight $$5$$ alternating harmonic series by exploiting a Landen identity type in the series form)

We need the following identity presented in (Almost) Impossible Integrals, Sums, and Series (see page $$285$$, Sect. $$4.11$$), slightly rearranged, $$\frac{1}{6}\sum_{n=1}^{\infty} (-1)^{n-1}\frac{x^{n-1}}{n}(H_n^3+3H_n H_n^{(2)}+2H_n^{(3)})=\frac{1}{x}\operatorname{Li}_4\left(\frac{x}{x+1}\right). \tag1$$

Integrating both sides of $$(1)$$ from $$x=0$$ to $$x=1$$, then letting the variable change $$\displaystyle \frac{x}{x+1}\mapsto x$$ and integrating by parts two times in the right-hand side and then rearranging, we get that $$\int_0^{1/2}\frac{(\operatorname{Li}_2(x))^2}{x}\textrm{d}x$$ $$\small =-\frac{1}{12}\log^5(2)+\frac{1}{3}\log^3(2)\zeta(2)-\frac{7}{16}\log^2(2)\zeta(3)-\frac{5}{8}\log(2)\zeta(4)+\frac{7}{16}\zeta(2)\zeta(3)-\log(2)\operatorname{Li}_4\left(\frac{1}{2}\right)-\operatorname{Li}_5\left(\frac{1}{2}\right)$$ $$+\frac{1}{6}\sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_n^3}{n^2}+\frac{1}{2}\sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_n H_n^{(2)}}{n^2}+\frac{1}{3}\sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_n^{(3)}}{n^2}$$ $$=\frac{1}{2}\log^3(2)\zeta(2)-\frac{7}{8}\log^2(2)\zeta(3)-\frac{5}{8}\log(2)\zeta(4)+\frac{27}{32}\zeta(5)+\frac{7}{8}\zeta(2)\zeta(3)-\frac{7}{60}\log^5(2)\\-2\log(2)\operatorname{Li}_4\left(\frac{1}{2}\right)-2\operatorname{Li}_5\left(\frac{1}{2}\right),$$ where the remaining series are found and calculated in (Almost) Impossible Integrals, Sums, and Series, pages $$311$$-$$312$$.

End of story

• very nice (+1). Did you apply IBP? Commented Oct 23, 2021 at 22:24
• @AliShadhar Thank you! The substitution $x/(x+1)\mapsto x$ gives $\displaystyle \int_0^1\frac{1}{x}\operatorname{Li}_4\left(\frac{x}{x+1}\right)\textrm{d}x=\operatorname{Li}_5\left(\frac{1}{2}\right)+ \int_0^{1/2}\frac{\operatorname{Li}_4(x)}{1-x}\textrm{d}x$. In the remaining integral it is enough to integrate by parts two times to get the desired integral. Commented Oct 24, 2021 at 9:40