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How to prove the following identity $$ \operatorname{Li}_{3}\left(\frac{1}{2} \right) = \sum_{n=1}^{\infty}\frac{1}{2^n n^3}= \frac{1}{24} \left( 21\zeta(3)+4\ln^3 (2)-2\pi^2 \ln2\right)\,?$$

Where $\operatorname{Li}_3 (x)$ is the trilogarithm also the result from above can be found here in $(2)$.

In particular for $x=\frac12$ we have:$$\operatorname{Li}_3\left(\frac12\right)=\sum_{n=1}^\infty \frac{1}{n^3}\frac{1}{2^n}=\frac14 \sum_{n=1}^\infty \frac{1}{2^{n-1}} \int_0^\frac12 x^{n-1}\ln^2 xdx=\frac14 \int_0^1 \frac{\ln^2 x}{1-\frac{x}{2}}dx$$

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    $\begingroup$ Maybe here? books.google.fr/… $\endgroup$ – Julien Feb 3 '13 at 16:44
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    $\begingroup$ @user97357329 This answer of mine may meet your standard. $\endgroup$ – Kemono Chen Jan 28 '20 at 6:24
  • $\begingroup$ @KemonoChen can you prepare a post here with the details of your approach? $\endgroup$ – user97357329 Feb 1 '20 at 9:25
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    $\begingroup$ @user97357329 Sure. I've completed filling the details. $\endgroup$ – Kemono Chen Feb 1 '20 at 14:17
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We can find the value similarly as seen here for $\operatorname{Li}_2\left(\frac12\right)=\frac{\pi^2}{12}-\frac{\ln^2 2}{2}$.

From the question we have: $$\operatorname{Li}_3\left(\frac12\right)=\sum_{n=1}^\infty \frac{1}{n^3}\frac{1}{2^n}=\frac12 \int_0^1 \frac{\ln^2 x}{2-x}dx\overset{\large \frac{x}{2-x}=t}=\frac12\int_0^1 \ln^2\left(\frac{2t}{1+t}\right)\frac{dt}{1+t}$$ $$=\frac12 \int_0^1\frac{\ln^2 (2t)-2\ln(2t)\ln(1+t)+\ln^2(1+t)}{1+t}dt$$

We can rewrite the first two terms as: $$\ln^2(2t)-2\ln(2t)\ln(1+t)=\color{blue}{\ln^2 2-2\ln 2 \ln(1+t)} +2\ln 2 \color{red}{\ln t}+\color{green}{\ln^2 t}-\color{orange}{2\ln t\ln(1+t)}$$ $$\small \color{blue}{\int_0^1 \frac{\ln^2 2-2\ln 2\ln(1+t)}{1+t}dt}=\ln 2 \int_0^1 \frac{\ln\left(\frac{2}{(1+t)^2}\right)}{1+t}dt\overset{\large t\to \frac{1-t}{1+t}}=-\ln 2 \int_0^1 \frac{\ln\left(\frac{2}{(1+t)^2}\right)}{1+t}dt=0$$ $$\color{red}{\int_0^1 \frac{\ln t}{1+t}dt}= \sum_{n=1}^\infty (-1)^{n-1} \int_0^1 t^{n-1} \ln t \, dt=-\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^2}=-\frac{\pi^2}{12}$$ $$\color{green}{\int_0^1 \frac{\ln^2 t}{1+t}dt}=\sum_{n=1}^\infty (-1)^{n-1} \int_0^1 t^{n-1} \ln^2 t \, dt=2\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^3}=\frac32 \zeta(3)$$ $$\color{orange}{\int_0^1 \frac{2\ln t \ln(1+t)}{1+t}dt}\overset{IBP}=-\int_0^1 \frac{\ln^2(1+t)}{t}dt=-\frac{\zeta(3)}{4}$$ For the latter integral look here or just let $m=1,n=0,q=1,p=0$ in the following relation: $$\small \int_0^1 \frac{[m\ln(1+x)+n\ln(1-x)][q\ln(1+x)+p\ln(1-x)]}{x}dx=\left(\frac{mq}{4}-\frac{5}{8}(mp+nq)+2np\right)\zeta(3)$$ The last term is easily found: $$\int_0^1 \frac{\ln^2(1+t)}{1+t}dt=\frac{\ln^3(1+t)}{3}\bigg|_0^1 =\frac{\ln^3 2}{3}$$ $$\Rightarrow \operatorname{Li}_3\left(\frac12\right)=\frac12 \left(-\frac{\pi^2}{6}\ln 2+\frac32\zeta(3)+\frac{\zeta(3)}{4}+\frac{\ln^3 2}{3}\right)=\boxed{\frac78\zeta(3)+\frac{\ln^3 2}{6}-\frac{\pi^2}{12}\ln 2}$$

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Another solution to both $\displaystyle \operatorname{Li}_2\left(\frac{1}{2}\right)$ and $\displaystyle \operatorname{Li}_3\left(\frac{1}{2}\right)$

Calculation of $\displaystyle \operatorname{Li}_2\left(\frac{1}{2}\right)$

If we derive and integrate back $\displaystyle \operatorname{Li}_2\left(\frac{x}{1+x}\right)$ from $x=0$ to $x=1$, we have $$\small \operatorname{Li}_2\left(\frac{1}{2}\right)=\int_0^1 \left(\operatorname{Li}_2\left(\frac{x}{1+x}\right)\right)'\textrm{d}x=\int_0^1\frac{\log(1+x)}{x}\textrm{d}x-\int_0^1\frac{\log(1+x)}{1+x}\textrm{d}x=\frac{\pi^2}{12}-\frac{\log^2(2)}{2},$$ and the calculation is finalized.

Good to mention that the value above can also be extracted immediately from the Dilogarithm Reflection formula which is easy to derive. Of course, one can use directly the strategy used in proving Dilogarithm Reflection formula for getting the desired result.

Calculation of $\displaystyle \operatorname{Li}_3\left(\frac{1}{2}\right)$

If we derive and integrate back $\displaystyle \operatorname{Li}_3\left(\frac{x}{1+x}\right)$ from $x=0$ to $x=1$, and then integrate by parts, we have $$\operatorname{Li}_3\left(\frac{1}{2}\right)=\int_0^1 \frac{\displaystyle \operatorname{Li}_2\left(\frac{x}{1+x}\right)}{x(1+x)}\textrm{d}x=-\log(2)\operatorname{Li}_2\left(\frac{1}{2}\right)-\int_0^1 \frac{\displaystyle \log(1+x)\log\left(\frac{x}{1+x}\right)}{x(1+x)}\textrm{d}x$$ $$=-\log(2)\operatorname{Li}_2\left(\frac{1}{2}\right)-\frac{1}{3}\log^3(2)-\underbrace{\int_0^1\frac{\log(x)\log(1+x)}{x}\textrm{d}x}_{\displaystyle -3/4\zeta(3)}+\frac{1}{2}\underbrace{\int_0^1\frac{\log^2(1+x)}{x}\textrm{d}x}_{\displaystyle 1/4\zeta(3)}$$ $$=\frac{7 }{8}\zeta (3)-\frac{1}{2}\log (2)\zeta(2)+\frac{1}{6}\log ^3(2),$$ where the first of the remaining integrals can be approached with polylogarithms or geometric series, and the second one can be approached easily with algebraic identities.

A first note: A nice approach (and possibly unexpected) involving a form of Beta function to the integral $\displaystyle \int_0^1\frac{\log^2(1+x)}{x}\textrm{d}x$ may be found in the book, (Almost) Impossible Integrals, Sums, and Series, pages $72$-$73$.

A second note: From the solution in the book (see previous note), we immediately have that

$$\int_0^1\frac{\log^2(1+x)}{x}\textrm{d}x-\int_0^1\frac{\log(x)\log(1+x)}{x}\textrm{d}x=\int_0^1 \frac{\log(x)\log(1-x)}{x}\textrm{d}x,$$

which can be perfectly used in the solution above to avoid the algebraic identities (if we want to).

A third note: Let's recall a well-known result in the work with polylogarithms,

$$\int_0^x \frac{\log^2(1+t)}{t} \textrm{d}t$$ $$=\log(x)\log^2(1+x)-\frac{2}{3}\log^3(1+x)-2\log(1+x) \operatorname{Li}_2\left(\frac{1}{1+x}\right)-2 \operatorname{Li}_3\left(\frac{1}{1+x}\right),$$

found in the famous book Polylogarithms and Associated Functions by Leonard Lewin. The nice thing about this method (involving the present generalization) is that in the right-hand side we get the desired trilogarithmic value with the opposite sign which assures the extraction, avoiding again the use of algebraic identities. The result is also met in the book, (Almost) Impossible Integrals, Sums, and Series, Section 1.4, page $3$.

A fourth note: perhaps no need to mention the third identity in $(1)$ here http://mathworld.wolfram.com/Trilogarithm.html

A fifth note: A fancy way of extracting the value $\displaystyle \operatorname{Li}_3\left(\frac{1}{2}\right)$ is based upon the identity

$$\small\int_0^1 \frac{\log (x) \log (1-x)}{1-a x} \textrm{d}x=\frac{\pi^2}{6}\frac{ \log (1-a)}{a}+\frac{1}{6}\frac{\log ^3(1-a)}{a}+\frac{1}{a}\operatorname{Li}_3(a)-\frac{1}{a}\operatorname{Li}_3\left(\frac{a}{a-1}\right), \ a<1,$$ used in this solution https://math.stackexchange.com/q/3503196. Setting $a=-1$ everything becomes clear in the right-hand side, and in the left-hand side the integral we have to deal with is already calculated elegantly in the paper A note presenting the generalization of aspecial logarithmic integral, that is $$\int_0^1 \frac{\log(x)\log(1-x)}{1+x}\textrm{d}x=\frac{13}{8}\zeta(3)-\frac{3}{2}\log(2)\zeta(2).$$

From another point of view, if we assume knowledge of the value of $\displaystyle \operatorname{Li}_3\left(\frac{1}{2}\right)$, we get another way of calculating the previously mentioned integral.

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Motivation of introducing the following integrals
These four integrals are strongly related to $\operatorname{Li}_3(1/2)$. With the evaluation of these integrals, we are able to deduce the polylogarithm value. $$\begin{aligned}I_1&=\int_0^{1/2}\frac{\ln^2(1-x)}x\mathrm {d}x\\ I_2&=\int_0^{1/2}\frac{\ln x\ln(1-x)}x\mathrm {d}x\\ I_3&=\int_0^{1/2}\frac{\ln^2 x}{1-x}\mathrm {d}x\\ I_4&=\int_0^{1/2}\frac{\ln x\ln(1-x)}{1-x}\mathrm {d}x\end{aligned}$$ Then, we find the relations between these integrals and hence evaluate them.

Relation 1

$$\begin{aligned}I_1+I_3&=\int_0^{1/2}\frac{\ln^2 x}{1-x}\mathrm{d}x+\int_{1/2}^1\frac{\ln^2 x}{1-x}\mathrm{d}x\\ &=\int_0^1\frac{\ln^2 x}{1-x}\mathrm {d}x\\ &=\int_0^1\sum_{n=0}^\infty x^n\ln^2x\mathrm{d}x\\ &=\sum_{n=0}^\infty \frac2{n^3}\\ &=2\zeta(3)\end{aligned}$$ Explanation: (1) Substitute $x\mapsto1-x$ in $I_1$
(2) Combine integration interval
(3) Expansion of $\frac1{1-x}$
(4) Change the order of summation and integration and integrate
(5) Definition of zeta function

Relation 2

$$\begin{aligned}I_2+I_4&=\int_0^{1/2}\frac{\ln x\ln(1-x)}x\mathrm {d}x+\int_{1/2}^1\frac{\ln x\ln(1-x)}x\mathrm {d}x\\ &=\int_0^1\frac{\ln x\ln(1-x)}x\mathrm{d}x\\ &=\left.\frac12\ln^2x\ln(1-x)\right|_0^1-\int_0^1-\frac12\frac{\ln^2x}{1-x}\mathrm{d}x\\ &=-\frac12(-2\zeta(3))\\ &=\zeta(3)\end{aligned}$$ Explanation: (1) Substitute $x\mapsto1-x$ in $I_4$
(2) Combine integration interval
(3) Integration by parts
(4) Use the result above

Relation 3

$$I_1=\ln^2(1-x)\ln x\big|_0^{1/2}+2\int_0^{1/2}\frac{\ln(1-x)\ln x}{1-x}\mathrm{d}x=-\ln^32+2I_4$$ Explanation: Integration by parts.

Relation 4

$$\begin{aligned}I_2-I_1&=\int_0^{1/2}\ln\frac x{1-x}\frac{\ln(1-x)}x\mathrm{d}x\\&=-\int_0^1\frac{\ln u\ln(1+u)}{u(1+u)}\mathrm{d}u\\ &=\int_0^1\ln u\sum_{n=1}^\infty(-1)^n\frac{H_n}nu^{n-1}\mathrm{d}u\\ &=\sum_{n=1}^\infty(-1)^{n-1}\frac{H_n}{n^2}\\ &=\frac58\zeta(3)\end{aligned}$$ Explaination: (1) Combine two integrands
(2) Use substitution $u=\frac{x}{1-x}$
(3) Use Taylor expansion of $\frac{\ln(1-z)}{1-z}$. This expansion can be deduced from Cauchy product of $\ln(1-z)$ and $(1-z)^{-1}$.
(4) Change the order of summation and integration and integrate
(5) See this answer.

Hence, we get an equation set: $$\left\{\begin{matrix}I_1+I_3&=&2\zeta(3)\\I_2+I_4&=&\zeta(3)\\2I_4-I_1&=&\ln^32\\I_2-I_1&=&\frac58\zeta(3)\end{matrix}\right.$$ Solving this gives $$I_1=\frac14\zeta(3)-\frac13\ln^32,\ I_2=\frac78\zeta(3)-\frac13\ln^32,$$ $$I_3=\frac74\zeta(3)+\frac13\ln^32,\ I_4=\frac18\zeta(3)+\frac13\ln^32.$$

Finally, we can calculate the value of $\operatorname{Li}_3(1/2)$. $$\operatorname{Li}_3\left(\frac12\right)=\int_0^{1/2}\frac{\operatorname{Li}_2(x)}x\mathrm{d}x\\ =\operatorname{Li}_2(x)\ln x\Big|_0^{1/2}+\int_0^{1/2}\frac{\ln(1-x)\ln x}x\mathrm{d}x\\ =-\frac1{12}\pi^2\ln2+\frac12\ln^32+\frac78\zeta(3)-\frac13\ln^32$$ Explanation: (1) Recurrence definition of polylogarithm
(2) Integration by parts
(3) Use the value of $\operatorname{Li}_2(1/2)$. See appendix for the deduction.

Appendix

$$\begin{aligned} \operatorname{Li}_2\left(\frac12\right)&=-\int_0^{1/2}\frac{\ln(1-x)}x\mathrm{d}x\\ &=-\ln^22-\int_0^{1/2}\frac{\ln x}{1-x}\mathrm{d}x\\ &=-\ln^22-\int_0^1\frac{\ln x}{1-x}\mathrm{d}x+\int_0^{1/2}\frac{\ln(1-x)}x\mathrm{d}x\\ &=-\ln^22+\frac16\pi^2-\operatorname{Li}_2\left(\frac12\right)\\ \Longrightarrow\operatorname{Li}_2\left(\frac12\right)&=\frac1{12}\pi^2-\frac12\ln^22 \end{aligned}$$

P.S.

This answer, which is requested to be created by user97357329, is the detailed version of my old answer.

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  • $\begingroup$ (+100 points) Excellent! $\endgroup$ – user97357329 Feb 1 '20 at 15:59
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    $\begingroup$ This is a canon. $\endgroup$ – Yuri Negometyanov Feb 1 '20 at 21:06
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\begin{align} K&=\int_0^{\frac{1}{2}}\left(\frac{\ln^2\left(\frac{x}{1-x}\right)}{x}-\frac{\ln^2 x}{x}\right)\,dx\\ &=\int_0^{\frac{1}{2}} \left(\frac{\ln^2(1-x)}{x}-\frac{2\ln x\ln(1-x)}{x}\right)\,dx\\ &=\int_0^{\frac{1}{2}} \frac{\ln^2(1-x)}{x}\,dx-\Big[\ln^2 x\ln(1-x)\Big]_0^{\frac{1}{2}}-\int_0^{\frac{1}{2}} \frac{\ln^2 x}{1-x}\,dx\\ &=\int_0^{\frac{1}{2}} \frac{\ln^2(1-t)}{t}\,dt+\ln^3 2-\int_0^{\frac{1}{2}} \frac{\ln^2 x}{1-x}\,dx\\ &\overset{x=1-t}=\int_{\frac{1}{2}}^1 \frac{\ln^2 x}{1-x}\,dx+\ln^3 2-\int_0^{\frac{1}{2}} \frac{\ln^2 x}{1-x}\,dx\\ &=\int_0^1 \frac{\ln^2 x}{1-x}\,dx+\ln^3 2-2\int_0^{\frac{1}{2}} \frac{\ln^2 t}{1-t}\,dt\\ &\overset{x=2t}=2\zeta(3)+\ln^3 2-\int_0^{1} \frac{\ln^2\left(\frac{x}{2}\right)}{1-\frac{x}{2}}\,dx\\ &=2\zeta(3)+\ln^3 2-\int_0^{1} \frac{\ln^2 x}{1-\frac{x}{2}}\,dx-\ln^2 2\int_0^{1} \frac{1}{1-\frac{x}{2}}\,dx+2\ln 2\int_0^1 \frac{\ln x}{1-\frac{x}{2}}\,dx\\ &=2\zeta(3)+\ln^3 2-4\text{Li}_3\left(\frac{1}{2}\right)-\ln^2 2\int_0^{1} \frac{1}{1-\frac{x}{2}}\,dx+2\ln 2\int_0^1 \frac{\ln x}{1-\frac{x}{2}}\,dx\\ &=2\zeta(3)-\ln^3 2-4\text{Li}_3\left(\frac{1}{2}\right)+2\ln 2\int_0^1 \frac{\ln x}{1-\frac{x}{2}}\,dx\\ \end{align}

On the other hand,

\begin{align} K&\overset{y=\frac{x}{1-x}}=\int_0^1 \left(\frac{\ln^2 x}{x(1+x)}-\frac{\ln^2\left(\frac{x}{1+x}\right)}{x(1+x)}\right)\,dx\\ &=\int_0^1 \left(\frac{2\ln x\ln(1+x)}{x(x+1)}-\frac{\ln^2(1+x)}{x(x+1)}\right)\,dx\\ &=2\int_0^1 \frac{\ln x\ln(1+x)}{x}\,dx-2\int_0^1 \frac{\ln x\ln(1+x)}{1+x}\,dx-\\ &\int_0^1 \frac{\ln^2(1+x)}{x}\,dx+\frac{1}{3}\ln^3 2\\ &=\Big(\Big[\ln^2 x\ln(1+x)\Big]_0^1-\int_0^1 \frac{\ln^2 x}{1+x}\,dx\Big)-2\int_0^1 \frac{\ln x\ln(1+x)}{1+x}\,dx-\\ &\Big(\Big[\ln x\ln^2(1+x)\Big]_0^1-2\int_0^1 \frac{\ln x\ln(1+x)}{1+x}\,dx\Big)+\frac{1}{3}\ln^3 2\\ &=-\int_0^1 \frac{\ln^2 x}{1+x}\,dx+\frac{1}{3}\ln^3 2\\ &=-\int_0^1 \frac{\ln^2 x}{1-x}\,dx+\int_0^1 \frac{2x\ln^2 t}{1-t^2}\,dt+\frac{1}{3}\ln^3 2\\ &\overset{x=t^2}=-\int_0^1 \frac{\ln^2 x}{1-x}\,dx+\frac{1}{4}\int_0^1 \frac{\ln^2 x}{1-x}\,dt+\frac{1}{3}\ln^3 2\\ &=\frac{1}{3}\ln^3 2-\frac{3}{4}\int_0^1 \frac{\ln^2 x}{1-x}\,dx\\ &=\frac{1}{3}\ln^3 2-\frac{3}{2}\zeta(3)\\ \end{align} Therefore, \begin{align} -4\text{Li}_3\left(\frac{1}{2}\right)&=\frac{4}{3}\ln^3 2-\frac{7}{2}\zeta(3)-2\ln 2\int_0^1 \frac{\ln x}{1-\frac{x}{2}}\,dx\\ \int_0^1 \frac{\ln t}{1-\frac{t}{2}}\,dt&\overset{x=\frac{t}{2-t}}=2\int_0^1 \frac{\ln\left(\frac{2x}{1+x}\right)}{1+x}\,dx\\ &=2\int_0^1 \frac{\ln x}{1+x}\,dx-\left[\ln^2\left(\frac{2}{1+x}\right)\right]_0^1\\ &=2\int_0^1 \frac{\ln x}{1+x}\,dx+\ln^2 2\\ &=2\int_0^1 \frac{\ln x}{1-x}\,dx-2\int_0^1 \frac{2t\ln t}{1-t^2}\,dt+\ln^2 2\\ &\overset{x=t^2}=2\int_0^1 \frac{\ln x}{1-x}\,dx-\int_0^1 \frac{\ln x}{1-x}\,dx+\ln^2 2\\ &=\int_0^1 \frac{\ln x}{1-x}\,dx+\ln^2 2\\ &=\ln^2 2-\zeta(2) \end{align} Therefore,

\begin{align} \text{Li}_3\left(\frac{1}{2}\right)&=\frac{1}{6}\ln^3 2+\frac{7}{8}\zeta(3)-\frac{1}{2}\zeta(2)\ln 2\\ &=\boxed{\frac{1}{6}\ln^3 2+\frac{7}{8}\zeta(3)-\frac{1}{12}\pi^2\ln 2} \end{align} NB: No need to compute $\displaystyle \int_0^1 \frac{\ln(1+x)\ln x}{1+x}\,dx$.

I assume: For $r\geq 1,0< a\leq 1,\displaystyle \int_0^1 \frac{\ln^r x }{1-ax}\,dx=\frac{(-1)^r r!}{a}\text{Li}_{r+1}(a)$

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Here is a partial solution:

Substitute $x=2\sin^2 \theta$ to obtain:

$$\int_0^{\frac{\pi}{4}} \left(\ln^2 2 + 4 \ln 2 \ln (\sin \theta) + 4 \ln^2 (\sin \theta)\right) \tan \theta \, \text{d}\theta$$

The $\ln^2 2$ part is trivial and evaluates to $\frac{\ln^3 2}{2}$.

The $\ln 2 \ln (\sin \theta)$ part is relatively straightforward.

Let $\displaystyle I = 4\ln 2 \int_0^\frac{\pi}{4} \ln (\sin \theta) \tan{\theta} \, \text{d}\theta$.

Integrate by parts to obtain $\displaystyle I = -\ln^3 2 + 4\ln 2 \int_0^\frac{\pi}{4} \ln (\cos \theta) \cot{\theta} \, \text{d}\theta$

Substitute $\theta \mapsto \frac{\pi}{2} - \theta$ to obtain $\displaystyle I = -\ln^3 2 + 4\ln 2 \int_\frac{\pi}{4}^\frac{\pi}{2} \ln (\sin \theta) \tan{\theta} \, \text{d}\theta$

which can then be averaged to obtain $\displaystyle I = -\frac{\ln^3 2}{2} + 2\ln 2 \int_0^\frac{\pi}{2} \ln (\sin \theta) \tan{\theta} \, \text{d}\theta$

Then use the substitution $u = \cos^2 \theta$ or $u = \sin^2 \theta$ to obtain the classic dilogarithm integral for $\zeta(2)$.

This simplifies down to $\displaystyle I = -\frac{\ln^3 2}{2} - \frac{\pi^2 \ln 2}{12}$

It then remains to show that $\displaystyle 4 \int_0^\frac{\pi}{4} \ln^2 (\sin \theta) \tan \theta = \frac{\ln^3 2}{6} + \frac{7}{8} \zeta(3)$.

I will continue attempting this integral and come back when/if I progress.

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  • $\begingroup$ (+1) For trying. $\endgroup$ – user97357329 Feb 1 '20 at 16:15

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