# Challenging problem: Calculate $\int_0^{2\pi}x^2 \cos(x)\operatorname{Li}_2(\cos(x))dx$

The following problem is proposed by a friend:

$$\int_0^{2\pi}x^2 \cos(x)\operatorname{Li}_2(\cos(x))dx$$

$$=\frac{9}{8}\pi^4-2\pi^3-2\pi^2-8\ln(2)\pi-\frac12\ln^2(2)\pi^2+8\ln(2)\pi G+16\pi\Im\left\{\operatorname{Li}_3\left(\frac{1+i}{2}\right)\right\}$$

My only try is writing

$$\operatorname{Li}_2(\cos(x))=-\int_0^1\frac{\cos(x)\ln(y)}{1-\cos(x)y} \, dy$$

and have no idea how to continue with the double integral. I also tried $$\cos(x)=u$$, didn't do much.

Any help would be much appreciated.

• Is it ok if I offer a bounty for this question? I really want to drag more attention, very challenging problem.
– user809806
Oct 12, 2020 at 14:28
• @stefan sure go ahead. Oct 12, 2020 at 15:12
• What is $G$???? Oct 15, 2020 at 20:36
• @WhatsUp: G is the Catalan's constant, that is, $\displaystyle \text{G}=\int_0^1 \frac{\arctan x}{x}dx$
– FDP
Oct 16, 2020 at 5:24
• My long answer had almost no echo, is might be hard to digest, but while elaborating it i discovered a an unexpected, interesting, related relation, now posted as a new question... math.stackexchange.com/questions/3905908/… . It is one of the "clean relations" needed in between among the many other relations needed for a final answer, all of them show values of polylogs... Nov 13, 2020 at 14:24

Part I

This is a long answer, beyond the allowed maximal size = 30K characters. So i had to split it.

It's because of the question, but also because details for the performed steps are given, hoping that the text should be accesible, up to some complex analysis issues, to a wider circle of readers. The reader in hurry may want to skip long computations, if she or he knows the pattern. Also computer checks are provided, so that there is (i.e. was for me while typing) an immediate confirmation for the displayed results.

I will use sage for exact computations, and pari/gp for quick numerical checks. Most of the time for myself, to have an in between check, and further type with confidence. (Many arguments were done on paper days before, today i would maybe reshape 80 percent, but time... And maybe it is good to see a bloody computation, there is else too much refactored to fit in a few lines.)

I will use $$\operatorname{Li}_1$$ for the function $$x\to-\log(1-x)$$, which has the Taylor expansion $$\tag{1} \operatorname{Li}_1(x) := \frac x1+\frac {x^2}2+\frac{x^3}3+\dots$$ around zero, thus motivating the notation.

There is indeed an "idea" of computation, that can be isolated below:

Use the known primitive to integrate expressions like $$\frac 1{x-a}\log^2x$$ and $$\frac1{x-a}\log^3 x$$, then use "polarization" to obtain $$AB$$ from the squares $$A^2, B^2, (A-B)^2$$. (Doing the same with $$A^2B$$ and/or $$AB^2$$ to be obtained from the cubes $$A^3, B^3, (A\pm B)^3$$, yes, it is possible. But the integrals corresponding to $$(A\pm B)^3$$ are not in the same time easy.)

My feeling of progress decided only when to use a substitution or partial integration or something else, so that integrals of these functions show up. When they show up, we proceed almost algorithmically.

We define the level of complexity of an integral involving polylog factors like $$\operatorname{Li}_1(s)$$ (or $$-\log(1-s)$$), $$\operatorname{Li}_2(s)$$, $$\operatorname{Li}_3(s)$$, $$\operatorname{Li}_4(s)$$, and so on, by adding the "complexities" of the factors, which are $$1,2,3,4,$$ and so on. Instead of $$s$$ we may have an other argument, a rational function of $$s$$, usually $$\pm s$$ or $$1\pm s$$, et caetera.

As orientation, the following ideas to proceed (somehow) were applied.

• Dilogarithm identities are used. For instance: $$\tag{2}$$ \begin{aligned} \operatorname{Li}_2(x) + \operatorname{Li}_2(-x) &= \operatorname{Li}_2(x^2)\ ,\\ \operatorname{Li}_2(x) + \operatorname{Li}_2(1-x) &= -\log(x) \log(1-x)+\frac 16\pi^2\ ,\\ \end{aligned} (and combinations of them).

• Integrals involving trigonometric expressions in $$\sin x$$, $$\cos x$$ may be reshaped using the standard substitution $$t = \tan\frac x2$$. Then we have formally: $$\tag{3} \tan\frac x2 = t\ ,\ x = 2\arctan t \ ,\ dx =\frac{2\; dt}{1+t^2}\ ,\ \sin x=\frac {2t}{1+t^2}\ ,\ \cos x=\frac{1-t^2}{1+t^2}\ ,$$ (and so on).

• We would like to replace the factor $$\operatorname{Li}_2(\cos x)$$ of some integral, if possible, with the factor $$(\operatorname{Li}_2(\cos x)+\operatorname{Li}_2(-\cos x))$$. Then in case of a partial integration w.r.t. "other factors", there is a passage to $$\tag{4} \Big(\ \operatorname{Li}_2(\cos x)+\operatorname{Li}_2(-\cos x)\ \Big)' = -\sin x\cdot\Big(\ \operatorname{Li}_1(\cos x)-\operatorname{Li}_2(-\cos x)\ \Big) \\ = -\sin x\log\frac{1-\cos x}{1+\cos x}\ ,$$ and the last expression introduces a "simple $$\log$$ factor" using the above substitution, since $$\frac{1-\cos x}{1+\cos x} = \frac{(1+t^2)-(1-t^2)}{(1+t^2)+(1-t^2)}=t^2$$.

• We try to isolate expressions to be integrated like $$\frac 1{x-a}\log(x-b)\log(x-c)$$. In case of $$b=c$$ (and thus further without restriction $$b=c=0$$) there is for instance an explicit primitive function: $$\tag{5} G_a(x) = \int_0^x\frac{\log^2 t}{t-a}\; dt = -\log^2x\cdot\operatorname{Li}_1\left(\frac xa\right) +2\log x\cdot\operatorname{Li}_2\left(\frac xa\right) -2\operatorname{Li}_3\left(\frac xa\right)\ .$$ For different values $$b,c$$, we may use the "polarization" $$BC=\frac 12(B^2+C^2-(B-C)^2)$$, with $$B=\log(x-b)$$, $$C=\log(x-c)$$.

Note that in the formula for $$G_a$$, when $$x=0$$ is plugged in, each $$\log x$$ is joined with an $$O(x)$$-function, so the limit is zero. Also, if we plug in $$x=1$$, then the $$\log$$-terms are zero, so there is only a contribution from the trilogarithm.

• Not used, but it should be recorded here. There is a similar formula for the integral involving $$\log^3 t$$ (instead of $$\log^2t$$): $$\tag{6} \int_0^x\frac{\log^3 t}{t-a}\; dt = -\log^3x\cdot\operatorname{Li}_1\left(\frac xa\right) +3\log^2 x\cdot\operatorname{Li}_2\left(\frac xa\right) +6\log x\cdot\operatorname{Li}_3\left(\frac xa\right) +6\operatorname{Li}_4\left(\frac xa\right)\ .$$ To cover products of the shape $$B^2C$$ with $$B,C$$ as above, one can use $$3(B^2C+BC^2)=(B+C)^3-B^3-C^3$$ and $$3(B^2C-BC^2)=-(B-C)^3+B^3-C^3$$.

• The $$\arctan$$ is also a "kind of logarithm". To make this specific, write $$\frac 1{x^2+1}=\frac 1{2i}\left(\frac1{x-i}-\frac 1{x+i}\right)$$, and integrate. So, formally, $$\arctan x=\frac 1{2i}\log \frac {1+it}{1-it}$$. In particular, its "complexity" is also $$1$$, as the complexity of the $$\log$$.

• Not used, but it should be mentioned here: Parallel to the above relations, we record here:

$$\tag{7}$$ \begin{aligned} \int\frac{\log^2 (x-a)}{1+x^2}\; dx &= \log^2(x-a)\cdot(\arctan x-\arctan a) \\ &\qquad\qquad +i\log (x-a)\cdot\operatorname{Li}_2\left(\frac {a-x}{a+i}\right) -i\log (x-a)\cdot\operatorname{Li}_2\left(\frac {a-x}{a-i}\right) \\ &\qquad\qquad -i\operatorname{Li}_3\left(\frac {a-x}{a+i}\right) +i\operatorname{Li}_3\left(\frac {a-x}{a-i}\right) +C\ , \\ \int\frac{\log^3 (x-a)}{1+x^2}\; dx &= \log^3(x-a)\cdot(\arctan x-\arctan a) \\ &\qquad\qquad +\frac 32i\log^2 (x-a)\cdot\operatorname{Li}_2\left(\frac {a-x}{a+i}\right) -\frac 32i\log^2 (x-a)\cdot\operatorname{Li}_2\left(\frac {a-x}{a-i}\right) \\ &\qquad\qquad -3i\log(x-a)\cdot\operatorname{Li}_3\left(\frac {a-x}{a+i}\right) +3i\log(x-a)\cdot\operatorname{Li}_3\left(\frac {a-x}{a-i}\right) \\ &\qquad\qquad +3i\operatorname{Li}_4\left(\frac {a-x}{a+i}\right) -3i\operatorname{Li}_4\left(\frac {a-x}{a-i}\right) +C\ , \end{aligned}

• It is favorable to compute here formally the integral mentioned above in a relatively general case. We use the notation $$I_A^B(0,b;a)$$. The variables $$A,B;b;a$$ may be complex numbers. (In case an evaluation for specific values does not make sense, consider the limit for $$A,B$$ in the expression. The computation is formal since we do not care which branch of the one or other logarithm is taken. We use one branch, so that computations make sense. If i am careless, there is a discrete monodromy introduced, also for this reason, there is always a numerical check below.) $$\tag{8}$$ \begin{aligned} I_A^B(0,b;a) &:= \int_A^B\log(t-0)\log(t-b)\cdot\frac 1{t-a}\; dt \\ &= \frac 12\int_A^B\Big( \ \log^2 t + \log^2(t-b) - \log^2\frac t{t-b}\ Big)\cdot\frac 1{t-a}\; dt \ . \\ &\qquad\text{And now using the primitive function G_a above} \\ \int_A^B\frac {\log^2 t}{t-a}\; dt &= [\ G_a(t)\ ]_A^B=G_a(B)-G_a(A)\ , \\ \\ \int_A^B\frac {\log^2 (t-b)}{t-a}\; dt &= \int_A^B\log^2 u\cdot\frac 1{u-(a-b)}\; du \\ &=[\ G_{a-b}(t)\ ]_A^B=G_{a-b}(B)-G_{a-b}(A)\ , \\ \int_A^B\frac {\log^2 \frac t{t-b}}{t-a}\; dt &= \int_{A/(A-b)}^{B/(B-b)} \log^2 u\frac 1{\frac {ub}{u-1}-a}\cdot\frac{-b}{(u-1)^2}\; du \\ &= \int_{A/(A-b)}^{B/(B-b)} \log^2 u\frac 1{\frac {ub}{u-1}-a}\cdot\frac{-b}{(u-1)^2}\; du \\ &= \int_{A/(A-b)}^{B/(B-b)} \log^2 u\left(\frac 1{u-\frac a{a-b}} - \frac 1{u-1}\right)\; du \\ &= [\ G_{a/(a-b)}(t)\ ]_{A/(A-b)}^{B/(B-b)} - [\ G_1(t)\ ]_{A/(A-b)}^{B/(B-b)}\ . \end{aligned}

• The given integral can be reshaped as an integral on the one of the intervals $$[0,1]$$ or $$[-1,0]$$ or $$[0,\infty)$$ or... from functions having the complexity at most $$3$$. Integrals like $$\int_0^1R(t)\log t\; dt$$ with a rational function $$R$$ are considered to be "easy". Its complexity is one. Integrals like $$\int_0^1R(r)\log(1\pm t)\log t\; dt$$ or like $$\int_0^1R(r)\arctan t\log t\; dt$$ have complexity two, and using $$(8)$$, they are feasible. It turns out, that such integrals, together with the integral of felt complexity three $$K= \int_0^1\arctan^2 t\cdot \log t\cdot\frac2{1-t^2}\; dt$$ are enough to solve the issue.

• For the integral $$K$$ displayed above i had to use unusual ideas, so that it is rewritten to have complexity two. This is the only original point in this answer, else the story is part of the folklore.)

• Note that many "pieces" in the final answer are "polylogarithmic periods". For instance $$G=\Im\operatorname{Li}_2(i)$$. See for instance:

Catalan's constant

• In the same link, there is a trilogarithmic expression, mentioned to have a "simple answer": $$\frac 1{1^3}-\frac 1{3^3}+\frac 1{5^3}-\frac 1{7^3}+\dots =\Im\operatorname{Li}_3(i)=\frac 1{32}\pi^3\ .$$

We start the computation, and "do something" first. The substitution $$t=\arctan(x/2)$$ needs a smaller interval. So we shift as a first step the integration from $$[0,2\pi]$$ to the symmetric interval $$[-\pi,\pi]$$, breaking it into even and odd pieces, the odd part vanishes, the even part is twice the integral on $$[0,\pi]$$.

Let $$J$$ be the value of the integral to be computed. Then: $$\tag{9}$$ \begin{aligned} J &= \int_0^{2\pi}x^2\;\cos x\; \operatorname{Li}_2(\cos x)\;dx \\ &= \int_{-\pi}^{\pi}(x+\pi)^2\;\cos(x+\pi)\; \operatorname{Li}_2(\cos (x+\pi))\;dx \\ &= -\int_{-\pi}^{\pi}(x^2+\underbrace{2\pi x}_{\text{odd}}+\pi^2)\;\cos x\; \operatorname{Li}_2(-\cos x)\;dx \\ % &= % -\int_{-\pi}^{\pi}x^2\;\cos x\; \operatorname{Li}_2(-\cos x)\;dx % -\pi^2\int_{-\pi}^{\pi}\cos x\; \operatorname{Li}_2(-\cos x)\;dx % \\ &= 2 \underbrace{ \int_0^{\pi}-x^2\;\cos x\; \operatorname{Li}_2(-\cos x)\;dx }_{J_1} + 2\pi^2 \underbrace{ \int_0^{\pi}-\cos x\; \operatorname{Li}_2(-\cos x)\;dx }_{J_2} \\ &=2J_1+2\pi^2 J_2\ . \end{aligned} Here, after changing $$x\to\pi-x$$ we have $$\tag{10} J_2 = \int_0^{\pi}-\cos x\; \operatorname{Li}_2(-\cos x)\;dx = \int_0^{\pi}\cos x\; \operatorname{Li}_2(\cos x)\;dx =\frac 12\pi^2-\pi\ .$$ Thus $$J_1$$ gives the complexity. For $$J_2$$, explicitly: $$\tag{11}$$ \begin{aligned} J_2 &= \int_0^{\pi/2}\cos x\; \operatorname{Li}_2(\cos x)\;dx + \int_0^{\pi/2}\cos (\pi-x)\; \operatorname{Li}_2(\cos(\pi-x))\;dx \\ &= \Big[\sin x \operatorname{Li}_2(\cos x)\Big]_0^{\pi/2} - \int_0^{\pi/2}\sin x\cdot\frac 1{\cos x} \operatorname{Li}_1(\cos x)\cdot(-\sin x)\; dx \\ &\qquad -\Big[\sin x \operatorname{Li}_2(-\cos x)\Big]_0^{\pi/2} +\int_0^{\pi/2}\sin x\cdot\frac 1{-\cos x} \operatorname{Li}_1(-\cos x)\cdot(-\sin x)\; dx \\ &=\int_0^{\pi/2}\frac {\sin^2 x}{\cos x} \log\frac{1+\cos x}{1-\cos x}\; dx \\ &\qquad\text{and with t=\tan\frac x2, x=2\arctan t, dx=\frac2{1+t^2}\; dt, \sin x=\frac {2t}{1+t^2}, \cos x=\frac{1-t^2}{1+t^2}} \\ &= \int_0^1\frac{4t^2}{(1+t^2)^2}\cdot\frac{1+t^2}{1-t^2} \cdot\log\left(\frac{(1+t^2)+(1-t^2)}{(1+t^2)-(1-t^2)}\right) \cdot\frac 2{1+t^2}\; dt \\ &= -\int_0^1\frac{8t^2}{(1+t^2)^2(1-t^2)}\cdot\log t^2\; dt \\ &= 2\int_0^1\left(\frac {2t}{1+t^2}+\log(1-t) - \log(1+t)\right)' \log t\; dt \\ &= -2\int_0^1 \left(\frac {2t}{1+t^2}+\log(1-t) - \log(1+t)\right)\;\frac 1 t \; dt \\ &=-4\arctan\Big|_0^1 +2\Big[\operatorname{Li}_2(t) - \operatorname{Li}_2(-t)\Big]_0^1 \\ &=-\pi +\frac 12\pi^2\ . \end{aligned} Here, $$\operatorname{Li}_2(1)=\zeta(2)=\frac 16\pi^2$$.

And $$\operatorname{Li}_2(-1)=-\operatorname{Li}_2(1)+\frac 12\operatorname{Li}_2(1^2)= -\frac 1{12}\pi^2$$.

Computer check for the value of $$J_2$$. I will use pari/gp for this. There are some issues near $$0$$ and $$\pi$$, so i will integrate numerically on some interval $$[\epsilon, \pi-\epsilon]$$.

? \p 50
realprecision = 57 significant digits (50 digits displayed)
? eps = 0.000008;
? J2approx = intnum( x=eps, Pi-eps, cos(x)*dilog(cos(x)) )
%133 = 1.7931898077460863662640447913454517588781602204055
? Pi^2/2 - Pi
%134 = 1.7932095469548860709546021166585726834596803042453
? J2rewritten = intnum( x=0, Pi/2, sin(x)^2/cos(x) * log( (1+cos(x))/(1-cos(x)) ) )
%135 = 1.7932095469548860709546021166585726834596803042453
? J2rewritten2 = -intnum( t=0, 1, 16*t^2/(1+t^2)^2/(1-t^2) * log(t) )
%136 = 1.7932095469548860709546021166585726834596803042453

So the value $$\frac 12\pi^2-\pi$$ is numerically validated, it is the only way to check using pari/gp.

Using sage, we can "compute" / request an exact value:

sage: var('t');
sage: integral( -16*t^2 / (1+t^2)^2 / (1-t^2) * log(t), t, 0, 1 )
-pi + 1/2*pi^2
sage: integral( sin(x)^2/cos(x) * log( (1+cos(x))/(1-cos(x)) ), x, 0, pi/2 )
-pi + 1/2*pi^2

(Although for the initial form of the integral there are some maxima questions.)

So the integral $$J_1$$ is the issue. I will use for the dilog term the identity $$\operatorname{Li}_2(s)+ \operatorname{Li}_2(-s)= \frac 12\operatorname{Li}_2(s^2)$$, thus obtaining in part a similar grouping of $$\operatorname{Li}_2(\cos x)-\operatorname{Li}_2(-\cos x)$$, which is favorable. $$\tag{12}$$ \begin{aligned} J_1 &= \int_0^{\pi}-x^2\;\cos x\; \operatorname{Li}_2(-\cos x)\;dx \\ &= \frac 12 \int_0^{\pi}-x^2\;\cos x\; \operatorname{Li}_2(-\cos x)\;dx + \frac 12 \int_0^{\pi}-x^2\;\cos x\; \operatorname{Li}_2(-\cos x)\;dx \\ % &= % \frac 12 % \int_0^{\pi}x^2\;\cos x\; \operatorname{Li}_2(\cos x)\;dx % + % \frac 12 % \int_0^{\pi}-x^2\;\cos x\; \operatorname{Li}_2(-\cos x)\;dx % \\ % &\qquad\qquad % -\frac 14 % \int_0^{\pi}x^2\;\cos x\; \operatorname{Li}_2(\cos^2 x)\;dx % \\ &= \frac 12 \underbrace{ \int_0^{\pi}x^2\;\cos x\; \Big( \operatorname{Li}_2(\cos x) - \operatorname{Li}_2(-\cos x) \Big) \;dx}_{J_{11}} \\ &\qquad\qquad -\frac 14 \underbrace{ \int_0^{\pi}x^2\;\cos x\; \operatorname{Li}_2(\cos^2 x)\;dx }_{J_{12}} \\ &=\frac 12 J_{11}-\frac 14 J_{12}\ . \end{aligned} Here is a numerical check for the above equality.

eps = 0.000008;
J1  = intnum( x=eps, Pi-eps, -x^2 * cos(x) * dilog(-cos(x)) );
J11 = intnum( x=eps, Pi-eps,  x^2 * cos(x) * ( dilog(cos(x)) - dilog(-cos(x)) ) );
J12 = intnum( x=eps, Pi-eps, -x^2 * cos(x) * dilog( cos(x)^2 ) );

And with the above variables, the difference is in the range of the used precision...

? J1 - J11/2 - J12/4
%197 = 3.186183822264904554 E-58

Let us compute the "simpler" integral from above, $$J_{12}$$, first. We have: $$\tag{13}$$ \begin{aligned} J_{12} &= \int_0^\pi x^2\;\cos x\; \operatorname{Li}_2(\cos^2 x)\;dx \\ &= \int_0^\pi (\ 2x\cos x + (x^2-2)\sin x\ )'\; \operatorname{Li}_2(\cos^2 x)\;dx \\ &=2\pi\cos \pi\operatorname{Li}_2(1) - \int_0^\pi (\ 2x\cos x + (x^2-2)\sin x\ )\; \frac 1{\cos^2 x}\operatorname{Li}_1(\cos^2 x)\cdot (\cos^2 x)'\;dx \\ &=-\frac 13\pi^3 - 2\int_0^\pi (\ 2x\cos x + (x^2-2)\sin x\ )\; \frac {\sin x}{\cos x}\log(\sin^2 x)\;dx \\ &= -\frac 13\pi^3 - 8\underbrace{\int_0^\pi x\sin x\log \sin x\;dx}_{\pi(\log 2-1)} \\ &\qquad\qquad - 4\int_0^\pi x^2\frac {\sin^2 x}{\cos x}\log \sin x\;dx + 8\underbrace{ \int_0^\pi \frac {\sin^2 x}{\cos x}\log\sin x\;dx }_{0\text{ via }x\to\pi-x} \\ &= -\frac 13\pi^3 -8\pi\log 2+8\pi -2\int_0^\pi (x^2-(\pi-x)^2)\frac {\sin^2 x}{\cos x}\log \sin x\;dx \\ &= -\frac 13\pi^3 -8\pi\log 2+8\pi -4\pi\underbrace{\int_0^\pi x\frac {\sin^2 x}{\cos x}\log \sin x\;dx}_{J_{121}} \\ &= -\frac 13\pi^3 -8\pi\log 2+8\pi - 4\pi J_{121}\ ,\text{ where} \\[2mm] % J_{121} &:=\int_0^\pi x\;\frac {\sin^2 x}{\cos x}\;\log \sin x\;dx \\ &=-\int_{-\pi/2}^{\pi/2} \left(x+\frac\pi 2\right)\;\frac {\cos^2 x}{\sin^2 x}\;\log \cos x\;\cdot\;\sin x\;dx \\ &=-\int_{-\pi/2}^{\pi/2} x\;\frac {\cos^2 x}{\sin^2 x}\;\log \cos x\;\cdot\;\sin x\;dx \\ &=2\int_0^{\pi/2} x\;\frac {\cos^2 x}{\sin^2 x}\;\log \cos x\;\cdot\;d(\cos x) \\ &= 2\int_1^0 \arccos t\frac{t^2}{1-t^2}\log t\; dt \\ &= \int_0^1 2\arccos t\;\frac {(1-t^2)-1}{1-t^2}\;\log t\;dt \\ &= \underbrace{\int_0^1 2\arccos t\log t\;dt}_{2\log 2-4} - \int_0^1 \arccos t\left(\frac 1{1-t}+\frac 1{1+t}\right)\log t\;dt \\ &= 2\log2-4 - \int_0^1 \arccos t\; (\operatorname{Li}_2(1-t))'\; dt \\ &\qquad\qquad + \int_0^1 \arccos t\; (\operatorname{Li}_2(1+t))'\; dt - \log(-1) \int_0^1 \arccos t\cdot \frac 1{1+t}\; dt \\ &=2\log2-4 -\arccos 0\cdot\operatorname{Li}_2(1) \\ &\qquad\qquad - \int_0^1 \frac{\operatorname{Li}_2(1-t)}{\sqrt{1-t^2}}\; dt +\arccos 0\cdot\operatorname{Li}_2(1) + \Re\int_0^1 \frac{\operatorname{Li}_2(1+t)}{\sqrt{1-t^2}}\; dt \\ &=2\log2-4 - \int_0^{\pi/2} \frac{\operatorname{Li}_2(1-\cos u)}{\sin u}\; \sin u\;du + \Re\int_0^{\pi/2} \frac{\operatorname{Li}_2(1+\cos u)}{\sin u}\; \sin u\; du \\ &= 2\log2-4 - \int_0^{\pi/2} \operatorname{Li}_2(1-\cos u)\; du + \Re\int_0^{\pi/2} \operatorname{Li}_2(1+\cos u)\; du \\ &\qquad\text{ and with } \operatorname{Li}_2(1-c) = -\operatorname{Li}_2(c) + \frac 16\pi^2 -\log(c)\log(1-c)\ ,\\ \\ &\qquad\text{ and with } \operatorname{Li}_2(1+c) = -\operatorname{Li}_2(-c) + \frac 16\pi^2 -\log(-c)\log(1+c)\ ,\\ \\ &= 2\log2-4 + \underbrace{ \int_0^{\pi/2} \operatorname{Li}_2(\cos u)\; du - \int_0^{\pi/2} \operatorname{Li}_2(-\cos u)\; du }_{J_{1211}} \\ &\qquad\qquad + \underbrace {\int_0^{\pi/2} \log\cos u\cdot\log\frac {1-\cos u}{1+\cos u}\; du }_{J_{1212}} \ . \end{aligned} Well, $$J_{1212}$$ can be computed "algorithmically", so we eliminate this from the task list first. Recall, we have a formula to integrate expressions like $$\frac 1{x-a}\log^2 x$$. Using "polarization", products $$AB$$ of different logarithms $$A=\log(x-a)$$ and $$B=\log(x-b)$$, can be reshaped to products of the "same" log, use $$AB=\frac 12(A^2+B^2-(A-B)^2)$$. This gives: \begin{aligned} J_{1212} &=\int_0^{\pi/2} \log\cos u\cdot\log\frac {1-\cos u}{1+\cos u}\; du\\ &=\int_0^1 \log\frac{1-t^2}{1+t^2}\cdot\log\frac {(1+t^2)-(1-t^2)}{(1+t^2)+(1-t^2)}\; \frac 2{1+t^2}\; dt\\ &= 4\int_0^1 \frac{\log(1-t)\cdot\log t}{1+t^2}\; dt +4\int_0^1 \frac{\log(1+t)\cdot\log t}{1+t^2}\; dt -4\int_0^1 \frac{\log(1+t^2)\cdot\log t}{1+t^2}\; dt \\ &= 4\left( -\frac 1{128}\pi^3-\frac 1{32}\pi\log^2 2+\Im\operatorname{Li}_3\left(\frac {1+i}2\right) \right) + 4 \left( +\frac {11}{128}\pi^3+\frac 3{32}\pi\log^2 2-2G\log 2-3\Im\operatorname{Li}_3\left(\frac {1+i}2\right) \right) \\ &\qquad\qquad - 4\left( -\frac {2}{128}\pi^3-\frac 2{32}\pi\log^2 2-G\log 2+2\Im\operatorname{Li}_3\left(\frac {1+i}2\right) \right) \\ &= \frac 38\pi^3 +\frac 12\pi\log^2 2 -4G\log2 -16\Im\operatorname{Li}_3\left(\frac {1+i}2\right)\ . \end{aligned} Numerical check:

i = I; pi = Pi; G = imag(dilog(i));
J1212 = intnum( u=0, pi/2, log(cos(u)) * log( (1-cos(u)) / (1+cos(u)) ) );
J1212_claimed = 3/8 * pi^3 + 1/2*pi*log(2)^2 - 4*G*log(2) - 16*imag(polylog(3, (1+i)/2 ));

? J1212
%187 = 0.72121319477695937923367893878228892950489772911404
? J1212_claimed
%188 = 0.72121319477695937923367893878228892950489772911404

To see that we have played a "purely linear game" with (5), here are some computational details. We tacitly use $$\frac 1{t^2+1}=\frac 1{2i} \left(\frac1{t-i}-\frac 1{t+i}\right)$$. \begin{aligned} \int_0^1 \frac{\log^2 t}{1+t^2}\; dt &= \frac 1{2i}\left( \int_0^1 \frac{\log^2 t}{t-i}\; dt - \int_0^1 \frac{\log^2 t}{t+i}\; dt \right) \\ &=\frac 1{2i}\Big[\ G_i(t)-G_{-i}(t)\ \Big]_0^1 \\ &=\frac 1{2i}\Big[\ G_i(1)-G_{-i}(1)\ \Big] \\ &=\frac 1{2i}\left[\ -2\operatorname{Li}_3\left(\frac 1i\right) +2\operatorname{Li}_3\left(\frac 1{-i}\right) \ \right] \\ &=2\Im \operatorname{Li}_3(i) \\ &=2\cdot \frac 1{32}\pi^3=\frac 1{16}\pi^3\ . \end{aligned} Also: \begin{aligned} \int_0^1 \frac{\log^2 (1-t)}{1+t^2}\; dt &= \frac 1{2i}\left( \int_0^1 \frac{\log^2 t}{1-t-i}\; dt - \int_0^1 \frac{\log^2 t}{1-t+i}\; dt \right) \\ &= \frac 1{2i}\left( - \int_0^1 \frac{\log^2 t}{t-(1-i)}\; dt + \int_0^1 \frac{\log^2 t}{t-(1+i)}\; dt \right) \\ &=\frac 1{2i}\Big[\ G_{1+i}(t)-G_{1-i}(t)\ \Big]_0^1 \\ &=\frac 1{2i}\Big[\ G_{1+i}(1)-G_{1-i}(1)\ \Big]\ , \\ &=\frac 1{2i}\left[\ -2\operatorname{Li}_3\left(\frac 1{1+i}\right) +2\operatorname{Li}_3\left(\frac 1{1-i}\right) \ \right] \\ &=2\Im \operatorname{Li}_3\left(\frac {1+i}2\right) \ . \end{aligned} And finally, with the substitution $$u=t/(1-t)$$: \begin{aligned} &\!\!\!\int_0^1 \frac{\log^2 (t/(1-t))}{1+t^2}\; dt \\ &= \int_0^\infty \frac{\log^2 u}{1+\frac{u^2}{(1+u)^2}}\; \frac 1{(1+u)^2}\;du \\ &= \int_0^1 \frac{\log^2 u} {(1+u)^2+u^2}\;du + \int_1^\infty \frac{\log^2 u} {(1+u)^2+u^2}\;du \\ &= \int_0^1 \frac{\log^2 u} {(1+u)^2+u^2}\;du + \int_0^1 \frac{\log^2 u} {(u+1)^2+1^2}\;du \\ &= \int_0^1 \log^2 u\frac 1{2i}\left( \frac 1{u-\frac12(-1+i)} - \frac 1{u-\frac12(-1-i)} \right)\;du \\ &\qquad\qquad + \int_0^1 \log^2 u\frac 1{2i}\left( \frac 1{u-(-1+i)} - \frac 1{u-(-1-i)} \right)\;du \\ &= \frac 1{2i}\Big[\ G_{(-1+i)/2}(t) - G_{(-1-i)/2}(t)\ \Big]_0^1 + \frac 1{2i}\Big[\ G_{-1+i}(t) - G_{-1-i}(t)\ \Big]_0^1 \\ &= \frac 1{2i}\left( \ 2\operatorname{Li}_3\left(\frac 2{-1-i}\right) - 2\operatorname{Li}_3\left(\frac 2{-1+i}\right) + 2\operatorname{Li}_3\left(\frac 1{-1-i}\right) - 2\operatorname{Li}_3\left(\frac 1{-1+i}\right) \ \right) \\ &= 2\Im\operatorname{Li}_3\left(-1+i\right) + 2\Im\operatorname{Li}_3\left(\frac {-1+i}2\right) \qquad(z=1-i) \\ &= 2\Im\Big(\ \operatorname{Li}_3(-z) - \operatorname{Li}_3(-z^{-1})\ \Big) = 2\Im\left(-\frac16\log^3 z-\frac 16\pi^2\log z\right)=\dots \ . \end{aligned} (So $$\log 2$$ and $$\pi$$ show up soon.) Here is a quick numerical test for the above. (So that i can further type.)

? intnum( t=0, 1, log(t)^2 / (1+t^2) )
%231 = 1.9378922925187387609672696916938372001390805353678
? Pi^3/16
%232 = 1.9378922925187387609672696916938372001390805353678
? intnum( t=0, 1, log(1-t)^2 / (1+t^2) )
%233 = 1.1401548141775379563912195151801491021262916198375
? 2*imag( polylog(3, (1+i)/2) )
%234 = 1.1401548141775379563912195151801491021262916198375
? intnum( t=0, 1, log( t/(1-t) )^2 / (1+t^2) )
%235 = 2.5167020943309544685663530996649317514086075354493
? 2*imag( polylog(3, -1+i) + polylog(3, (-1+i)/2) )
%236 = 2.5167020943309544685663530996649317514086075354493
? z=1-I; 2*imag( -1/6*log(z)*(log(z)^2+pi^2) )
%237 = 2.5167020943309544685663530996649317514086075354493

In the following related post, pisco also computed these integrals using different methods. Please compare to have an alternative view.

Computation of integrals, math stackexchange question 3854736

This was $$J_{1212}$$. The remained integral $$J_{1211}$$ is not so simple. $$\tag{14}$$ \begin{aligned} J_{1211} &= \int_0^{\pi/2} u'\operatorname{Li}_2(\cos u)\; du - \int_0^{\pi/2} u'\operatorname{Li}_2(-\cos u)\; du \\ &= - \int_0^{\pi/2} u\cdot \frac{\sin u}{\cos u}\cdot\log\frac{1-\cos u}{1+\cos u}\; du \\ &= -\int_0^1 2\arctan t \;\frac{2t}{1-t^2}\; \log t^2\;\frac 2{1+t^2}\; dt \\ &= -4 \int_0^1 \left( \frac 1{1-t} - \frac 1{1+t} +\frac{2t}{1+t^2} \right)\cdot \arctan t\cdot \log t\; dt \\ &=-4(J_{1211a} - J_{1211b} + J_{1211c})\ . \end{aligned} Here, $$J_{1211a}$$, $$J_{1211b}$$, $$J_{1211c}$$ are the correspondingly integrals obtained by dissolving the parentheses.

One can show using either $$(8)$$, or the linked related computations, the formulas for the integrals indexed $$1211a$$, $$1211b$$, $$1211c$$: $$\tag{15}$$ \begin{aligned} J_{1211a} &= \frac 1{16}\left[\ -\pi^3-\pi\log^2 2+ 8G\log2 + 32\Im\operatorname{Li}_3\left(\frac {1+i}2\right)\ \right]\ ,\\ J_{1211b} &= \frac 1{64}\Big[\ -\pi^3 + 32G\log2\ \Big]\ ,\\ J_{1211c} &= \frac 1{16}\left[\ \pi^3+2\pi\log^2 2 -64\Im\operatorname{Li}_3\left(\frac {1+i}2\right)\ \right]\ , \\[3mm] J_{1211} &= \frac 1{16}\left[\ -\pi^3-4\pi\log^2 2 +128\Im\operatorname{Li}_3\left(\frac {1+i}2\right)\ \right]\ , \\ J_{1212} &= \frac 1{16}\left[\ 6\pi^3+8\pi\log^2 2 -64G\log 2 - 256\Im\operatorname{Li}_3\left(\frac {1+i}2\right)\ \right]\ , \\ J_{121} &= 2\log 2-4 + J_{1211} +J_{1212}\\ &= \frac 1{16}\left[\ 5\pi^3+4\pi\log^2 2 -64G\log 2 - 128\Im\operatorname{Li}_3\left(\frac {1+i}2\right)\ \right] +2\log 2-4 \ . \end{aligned} Numerical checks:

eps = 0.8e-5; pi = Pi; i = I;
G = imag( dilog(i) );
# J1211 = intnum( u=eps, pi/2, dilog(cos(u)) ) - intnum( u=eps, pi/2, dilog(-cos(u)) );
J1211  = intnum( t=0, 1, 4 * ( 1/(1-t) - 1/(1+t) +2*t/(1+t^2) ) * atan(t) * log(t) )
J1211a = intnum( t=0, 1, atan(t) * log(t) / (1-t) );
J1211b = intnum( t=0, 1, atan(t) * log(t) / (1+t) );
J1211c = intnum( t=0, 1, atan(t) * log(t) * 2*t / (1+t^2) );

J1211a - (-pi^3 -   pi*log(2)^2 +  8*G*log(2) + 32*imag(polylog(3, (1+i)/2))) / 16
J1211b - (-pi^3                 + 32*G*log(2)                               ) / 64
J1211c - (+pi^3 + 2*pi*log(2)^2               - 64*imag(polylog(3, (1+i)/2))) / 16

J1211
4*J1211a - 4*J1211b + 4*J1211c

Yes, the differences in the second block are covered by the precision used, and we have a final answer for $$J_{1211}$$.

To have an example of calculation: \begin{aligned} J_{1211b} &= \frac 1{2i} \int_0^1 \log t\cdot\log\frac {1+it}{1-it}\cdot\frac 1{t+1}\; dt\ , \\ \int_0^1 \frac {\log^2 t}{t+1}\; dt &= [\ G_{-1}(t)\ ]_0^1=-2\operatorname{Li}_3(-1)=\frac 32\zeta(3) \ ,\text{ (but not needed)} \\ \int_0^1 \frac {\log^2 (1+it)}{t+1}\; dt &= \int_1^{1+i} \log^2 u\;\cdot\frac 1{(u-1)+i}\; du \\ &= [ \ G_{1-i}(t)\ ]_1^{1+i} =G_{1-i}(1+i)-G_{1-i}(1) \\ &= -\log(1+i)^2\cdot\operatorname{Li}_1(i) +2\log(1+i)\cdot\operatorname{Li}_2(i) \ , \\ \int_0^1 \frac {\log^2 (1-it)}{t+1}\; dt &=\text{the complex conjugate of the above}\ , \\ \int_0^1 \frac {\log^2 \frac t{1+it}}{t+1}\; dt &= - \int_0^{1/(1+i)} \log^2 u\;\cdot\frac 1{\frac {iu}{u+i}+1}\; \frac 1{(u+i)^2}\; du % % u = t/(1+it), u + uit = t, t = u/(1-ui) = ui/(u+i) \\ &= \int_0^{1/(1+i)} \log^2 u\;\cdot\left(\frac 1{u+\frac 12(1+i)} - \frac 1{u+i}\right) \\ &=[\ G_{-(1+i)/2}(u)-G_{-i}(u)\ ]_0^{1/(1+i)} = G_{-(1+i)/2}\left(\frac{1-i}2\right)-G_{-i}\left(\frac{1-i}2\right) \\ &= -\log^2\frac{1-i}2\cdot \operatorname{Li}_1(i) +2\log\frac{1-i}2\cdot \operatorname{Li}_2(i) -2\operatorname{Li}_3(i) \\ &\qquad\qquad -\log^3\frac{1-i}2 - 2\log\frac{1-i}2 \cdot\operatorname{Li}_2\left(\frac{1+i}2\right) + 2\operatorname{Li}_3\left(\frac{1+i}2\right) \ . \\ \int_0^1 \frac{\log^2 \frac t{1-it}}{t+1}\; dt &=\text{the complex conjugate of the above value.} \\ J_{1211b} &= \frac 1{2i} \int_0^1 \log t\;\log\frac {1+it}{1-it}\cdot\frac 1{t+1}\; dt \\ &= \frac 1{4i} \int_0^1 \Bigg(\log^2(1+it)-\log^2(1-it) \\ &\qquad\qquad\qquad\qquad -\log^2\frac t{1+it} +\log^2\frac t{1-it}\Bigg)\cdot\frac 1{t+1}\; dt \ , \end{aligned} and the computation leads to the claimed result.

to be continued...

• If interested, please read arXiv $2007.03957$ for a systematic approach on these kind of integrals. Nov 14, 2020 at 15:21
• Thank you for the great efforts and solution. +1 Nov 15, 2020 at 17:57

Part II

Please look around for the first part, if this is incidentally first.

Finally, the most complicated integral, $$J_{11}$$. We have: \begin{aligned} J_{11} &= \int_0^{\pi}x^2\;\cos x\; \Big( \operatorname{Li}_2(\cos x) - \operatorname{Li}_2(-\cos x) \Big) \;dx \\ &= \int_0^{\pi/2}x^2\;\cos x\; \Big( \operatorname{Li}_2(\cos x) - \operatorname{Li}_2(-\cos x) \Big) \;dx \\ &\qquad\qquad + \int_0^{\pi/2}(\pi-x)^2\;\cos (\pi-x)\; \Big( \operatorname{Li}_2(\cos (\pi-x)) - \operatorname{Li}_2(-\cos (\pi-x)) \Big) \;dx \\ &= \int_0^{\pi/2}(x^2+(\pi-x)^2)\;\cos x\; \Big( \operatorname{Li}_2(\cos x) - \operatorname{Li}_2(-\cos x) \Big) \;dx \\ &= \int_0^{\pi/2}\Big( \ (4x-2\pi)\cos x + (2x^2-2\pi x +\pi^2-4) \sin x\ \Big)'\; \\ &\qquad\qquad\cdot \Big( \operatorname{Li}_2(\cos x) - \operatorname{Li}_2(-\cos x) \Big) \;dx \\ &= \frac 12\pi^3 - \int_0^{\pi/2} \Big( \ (4x-2\pi)\cos x + (2x^2-2\pi x +\pi^2-4) \sin x\ \Big) \\ &\qquad\qquad\cdot \left( \frac 1{\cos x}\operatorname{Li}_1(\cos x)\cdot (-\sin x) - \frac 1{-\cos x}\operatorname{Li}_1(-\cos x)\cdot \sin x \right)\; dx \\ &= \frac 12\pi^3 - \int_0^{\pi/2} \Big( \ (4x-2\pi)\cos x + (2x^2-2\pi x +\pi^2-4) \sin x\ \Big)\; \frac {\sin x}{\cos x}\log\frac{1-\cos x}{1+\cos x} \; dx \\ &= \frac 12\pi^3 - \int_0^{\pi/2} (4x-2\pi)\; \sin x\;\log\frac{1-\cos x}{1+\cos x} \; dx \\ &\qquad\qquad - \int_0^{\pi/2} (2x^2-2\pi x +\pi^2-4) \; \frac {\sin^2 x}{\cos x}\log\frac{1-\cos x}{1+\cos x} \; dx \\ &= \frac 12\pi^3 - \int_0^1 (8\arctan t-2\pi)\; \frac{2t}{1+t^2}\;\log t^2\;\frac2{1+t^2} \; dt \\ &\qquad\qquad - \int_0^1 (8\arctan^2 t-4\pi \arctan t +\pi^2-4) \; \frac {4t^2}{(1+t^2)(1-t^2)}\log t^2\;\frac2{1+t^2} \; dt \\ &= \frac 12\pi^3 - \underbrace{ \int_0^1 (8\arctan t-2\pi)\; \left( \frac{4t^2}{1+t^2}\log t - 2\log(1+t^2) \right)' \; dt} _{=8\pi\log 2- 4\pi} \\ &\qquad\qquad +2 \underbrace{ \int_0^1 (8\arctan^2 t-4\pi \arctan t +\pi^2-4) \; \left( \frac{2t}{1+t^2} + \log\frac{1-t}{1+t} \right)' \;\log t \; dt}_{=J_{111}} \ . \end{aligned} Some words on the value of the known integral above, the one with the value $$8\pi\log 2-4\pi$$. We use partial integration. Then the piece in $$(8\arctan t-2\pi)'=8/(1+t^2)$$ is rational, we use partial fraction decomposition over $$\Bbb C$$, and thus the integral is broken into pieces, each piece being of the form $$\frac 1{t-b}\log(t-a)$$ or $$\frac 1{(t-b)^2}\log(t-a)$$ with $$a,b$$ among $$0,\pm i$$. These integrals can be handled. Dilogarithms appear, we use than the dilogarithm identities.

A numerical check:

? 8*Pi*log(2) - 4*Pi
%41 = 4.8543181080696440901549376527829000419
? intnum( t=0, 1, (8*atan(t) - 2*Pi) * 4*t/(1+t^2)^2 * 2*log(t) )
%42 = 4.8543181080696440901549376527829000419
? intnum( x=0, Pi/2, (4*x - 2*Pi) * sin(x) * log( (1-cos(x)) / (1+cos(x)) ) )
%43 = 4.8543181080696440901549376527829000419

Now we look closer to the remained integral, $$J_{111}$$. Parts of it are rather easy. We have \begin{aligned} J_{111s} &:= \int_0^1 \left( \frac{2t}{1+t^2} + \log\frac{1-t}{1+t} \right)' \;\log t\; dt \\ &= \underbrace{\int_0^1 \left( \frac{2t}{1+t^2}\right)' \;\log t\; dt}_{=-2\arctan 1=-\pi/2} - \underbrace{ \int_0^1 \left( \frac 1{1-t}+\frac 1{1+t} \right) \;\log t\; dt}_{=-\pi^2/4} \ . \end{aligned} The integral with value $$-2\arctan 1$$ is done by partial integration, the other one using the (real part of the) primitive $$\operatorname{Li}_2(1\pm t)$$. The dilog vanishes in $$0$$, the two contributions in $$t=0$$ of $$\operatorname{Li}_2(1\pm t)=\operatorname{Li}_2(1\pm t)$$ cancel each other (different signs), and there remains $$-\Re\operatorname{Li}_2(1+1)=-\frac 14\pi^2$$.

The part in $$\arctan t\cdot \log t$$ from $$J_{111}$$ can also be computed. The parts derived from $$\left( \frac 1{1-t}+\frac 1{1+t} \right)$$ are considered in $$J_{1211a}$$, $$J_{1211b}$$. The remained part in $$\arctan t\cdot \log t$$ is \begin{aligned} J_{111t} &:= \int_0^1 \arctan t\; \left( \frac{2t}{1+t^2} \right)' \;\log t \; dt \\ &= - \int_0^1 \frac{2t}{1+t^2} \left( \frac 1{1+t^2}\cdot\log t + \arctan t\cdot\frac 1t \right) \; dt \\ &= \int_0^1 \left( \frac 1{1+t^2}-1 \right)' \cdot\log t \; dt - \int_0^1 \frac{2}{1+t^2} \arctan t \; dt \\ &=\frac 12\log 2-\frac 1{16}\pi^2\ . \end{aligned} So the main issue is $$\tag{16}$$ \begin{aligned} J_{111u} &:= \int_0^1 \arctan^2 t\; \left( \frac{2t}{1+t^2} + \log\frac{1-t}{1+t} \right)' \;\log t \; dt \\ &= \underbrace{ \int_0^1 \arctan^2 t\; \left( \frac{2t}{1+t^2} \right)' \;\log t \; dt}_{J_{111u1}} - \underbrace{ \int_0^1 \arctan^2 t\; \log t\; \left(\frac 1{1-t} +\frac 1{1+t}\right) \; dt}_{J_{111u2}} \ . \end{aligned} The term listed first is simpler. \begin{aligned} J_{111u1} &= - \int_0^1 \frac{2t}{1+t^2} \left( \arctan^2 t\cdot \frac 1t + 2\arctan t\cdot\log t\cdot\frac 1{1+t^2} \right) \; dt \\ &= - \left[\frac 23\arctan^3 t\right]_0^1 + 2\int_0^1 \left(\frac 1{1+t^2}\right)'\cdot\arctan t\cdot \log t\; dt \\ &= -\frac 23\cdot\frac 1{4^3}\pi^3 - 2\int_0^1 \frac 1{1+t^2}\; \left(\frac 1{1+t^2}\cdot \log t + \arctan t\cdot\frac 1t\right)\; dt \\ &= -\frac 1{96}\pi^3 -2\underbrace{\int_0^1\frac{\log t}{(1+t^2)^2}\; dt}_{-(\pi+4G)/8} -2\underbrace{\int_0^1 \frac 1t\;\arctan t\; dt}_{G} +\underbrace{\int_0^1 \frac {2t}{1+t^2}\;\arctan t\; dt}_{-\frac 14\pi\log 2+G} \\ &= -\frac 1{96}\pi^3 +\frac 14\pi -\frac 14\pi\log 2 \ . \end{aligned} For the integral of $$\frac 1t\arctan t$$ use maybe the Taylor expansion. For the integral with the $$\log t$$ use the partial fraction decomposition over $$\Bbb C$$ for $$\frac 1{(t^2-a^2)^2}=\frac 1{4a^3}\left(\frac 1{t+a}-\frac 1{t-a}\right)+\frac 1{4a^2}\left(\frac 1{(t-a)^2}+\frac 1{(t+a)^2}\right)$$, where $$a=i$$. For the remained integral, using partial integration we compute instead $$\int_0^1\frac{\log(1+t^2)}{1+t^2}\; dt$$. We split again in $$\log$$, and in partial fractions, using $$(t^2+1)=(t-i)(t+i)$$. Then $$\int_0^1\frac{\log(t+i)}{t+i}\; dt=\int_0^1\log(t+i)\cdot(\log(t+i))'\; dt$$, so we can integrate. And $$\int_0^1\frac{\log(t+i)}{t-i}\; dt$$ leads to a dilog value, $$\operatorname{Li}_2((1-i)/2)$$, and we need finally from it $$\Im\operatorname{Li}_2((1-i)/2)=-G+\frac 18\pi\log 2$$.

So up to mentioned white noise integrals, that can be handled, we are in position to start the solution. We show, using $$K$$ for a shorter name: $$\tag{17}$$ \begin{aligned} K:=J_{111u2} &:= \int_0^1 \arctan^2 t\cdot\log t\; \left( \frac 1{1-t} + \frac 1{1+t} \right) \; dt \\ &= \frac 3{128}\pi^4 + \frac 1{32}\pi^2\log^2 2 - \frac 12\pi G\log 2 - \pi\Im\operatorname{Li}_3\left(\frac{1+i}2\right)\ . \end{aligned} (This relation resisted to all standard attacks. I tried several ideas and tricks like partial integration, and the substitutions $$s=1/t$$ and $$t=(1-u)/(1+u)$$, and the deformation of the $$\arctan t$$ using the parameter $$a$$, thus replacing $$\arctan t=\int_0^1\frac {t\; da}{1+a^2t^2}$$ and/or replacing $$\arctan^2 t=\iint_{[0,1]^2}\frac {t\; da}{1+a^2t^2}\cdot \frac {t\; db}{1+b^2t^2}$$, and so on. The complexity of the resulted expressions was not reduced. Some of these lines are mentioned after the solution below, which is something experimentally found and new, that i never saw before. Well, when ideas and tricks do not work, only madness can help us...)

We will work using complex analysis and the first step is to write: $$\int_0^1=\int_0^i+\int_i^1\ .$$ Since $$K\in \Bbb R$$, we compute only the real part of the integrals in the R.H.S. above. Using $$\arctan t=\frac 1{2i}\log\frac{1+it}{1-it}$$ around zero we can compute by the parametrization $$t=iu$$, $$u\in[0,1)$$: \begin{aligned} K_1&:= \Re \int_0^i \arctan^2 t\cdot\log t \cdot\frac 2{1-t^2} \\ &= \Re\int_0^1 -\frac 14\cdot\log^2\frac{1+i\cdot iu}{1-i\cdot iu}\cdot\log(iu)\cdot\frac2{1+u^2}\; i\; du \\ &= \Re \int_0^1 -\frac 14\cdot\log^2\frac{1-u}{1+u}\cdot\left(i\frac \pi 2+\log u\right)\cdot\frac2{1+u^2}\; i\; du \\ &= \frac \pi 4 \int_0^1 \log^2\frac{1-u}{1+u}\cdot\frac 1{1+u^2}\; du \\ &= \frac \pi 4 \int_0^1 \log^2 s\cdot\frac 1{1+\left(\frac{1-s}{1+s}\right)^2}\; \frac 2{(1+s)^2}\;ds = \frac \pi 4 \int_0^1 \log^2 s\cdot\frac 2{(1+s)^2 + (1-s)^2}\; ds \\ &= \frac \pi 4 \int_0^1 \log^2 s\cdot\frac 1{2i}\left(\frac 1{s-i}-\frac 1{s+i}\right)\; ds \\ &= \frac \pi 4 \cdot\frac 1{2i} \left(-2\operatorname{Li}_3\left(\frac 1i\right) +2\operatorname{Li}_3\left(\frac 1{-i}\right) \right) =\frac \pi 4\cdot 2\cdot\frac{\pi^3}{32} \\ &=\frac{\pi^4}{64}\ . \end{aligned} (Similarly to the $$\log(1-t)$$ factor from the original integral, that may show some integration problem in $$u=1$$, but there isn't any, there also no issue for $$\log(1+iu)$$ in $$i$$.) Numerical check, pari/gp can do it also under such unusual circumstances:

? real(intnum( t=0, i, atan(t)^2 * log(t) *2/(1-t^2) ))
%290 = 1.5220170474062880818193801982610173632769935261357097139291853029682946165
? pi^4 / 64
%291 = 1.5220170474062880818193801982610173632769935261357097139291853029682946165

The other integral turns out to be also feasible by using a path from $$i$$ to $$1$$ on the unit circle, explicitly we use the parametrization "$$t=t(s)$$" with $$t(s) = \sin 2x + i\cos 2s=i\cdot e^{-2is}\ ,\qquad s\in[0,\ \pi/4]\ .$$ Then \begin{aligned} \frac {1+i\; t(s)} {1-i\; t(s)} &= \frac {1-e^{-2is}} {1+e^{-2is}} = \frac {(1-e^{-2is})(1+e^{+2is})} {(1+e^{-2is})(1+e^{+2is})} = \frac {2i\; \sin 2s} {2+2\cos 2s} \\ &=i\cdot\frac{2\sin s\cos s}{2\cos ^2 s} =i\;\tan s\ . \end{aligned} The other factors in the integrand can be reshaped in a similar way. So... \begin{aligned} K_2&:= \Re \int_i^1 \arctan^2 t\cdot\log t \cdot\frac 2{1-t^2} \\ &= \Re \int_0^{\pi/4} -\frac 14\cdot\underbrace{\log^2(i\tan s)}_{=\left(i\frac \pi2+\log\tan s\right)^2}\cdot i\;\left(\frac \pi 2-2s\right)\cdot\frac2{\cos 2s}\; ds \\ &= \int_0^{\pi/4} \frac 14\cdot2\cdot\frac \pi 2\cdot\log\tan s \cdot\left(\frac \pi 2-2s\right)\cdot\frac2{\cos 2s}\; ds\qquad(t=\tan s) \\ &= \frac \pi 2 \int_0^1 \log t \cdot\left(\frac \pi 2-2\arctan t\right)\cdot\frac{1+t^2}{1-t^2}\; \frac 1{1+t^2}\;dt \\ &= \frac{\pi^2}4 \cdot\frac 12 \int_0^1 \log t\left(\frac1{1-t}+\frac 1{1+t}\right)\; \;dt - \frac \pi \int_0^1 \log t\cdot \arctan t\cdot\frac 1{1-t^2}\;dt \ . \end{aligned} For the first integral above, note that $$\partial_t\operatorname{Li}_2(1-t) =\frac 1{1-t}\log t$$, and $$\partial_t\operatorname{Li}_2(1+t) =-\frac 1{1+t}\log (-t)$$. So its value is \begin{aligned} \Big[ \operatorname{Li}_2(1-t) \Big]_0^1 - \Re \Big[ \operatorname{Li}_2(1+t) \Big]_0^1 &= (\operatorname{Li}_2(0)-\operatorname{Li}_2(1)) - \Re (\operatorname{Li}_2(2)-\operatorname{Li}_2(1)) \\ &= -\Re\operatorname{Li}_2(2) =-\frac {\pi^2}4\ . \end{aligned} The second integral above was already computed as part of $$(15)$$, it is related to $$J_{1211a}$$ and $$J_{1211b}$$. This gives: \begin{aligned} K &= K_1+K_2 = \frac{\pi^4}{64} + K_2 \\ &=\frac{\pi^4}{64} + \frac{\pi^2}8\cdot\left(-\frac {\pi^2}4\right) -\frac\pi 2(J_{1211a}+ J_{1211b}) =\dots \end{aligned} and so on. We are in position to put all together. First of all $$\tag{18}$$ \begin{aligned} J_{111} &= 8(J_{111u1}-\underbrace{J_{111u2}}_{K}) -4\pi(J_{111t}-J_{1211a}-J_{1211b}) +(\pi^2-4)J_{111s} \\ &= -\frac{\pi^4}4 - \frac{\pi^2}4\log^2 2 +8 \pi G\log 2 +16\pi \operatorname{Li}_3\left(\frac {1+i}2\right) +4\pi - \pi^2 -\frac 13\pi^3 -4\pi\log 2 \ . \end{aligned} From $$(18)$$ and $$(15)$$ we get: \begin{aligned} J &= 2\underbrace{J_1}_{\frac 12J_{11}-\frac 14 J_{12}} + 2\pi^2\cdot \underbrace{J_2}_{\frac{\pi^2}2-\pi} \\ &={\color{blue}{J_{11}}} -\frac 12{\color{brown}{J_{12}}} +\pi^4-2\pi^3 \\ &= {\color{blue} {\left(\frac {\pi^3}2 -8\pi\log 2 +4\pi + 2J_{111}\right)}} -\frac 12 {\color{brown} {\left(-\frac {\pi^3}3 -8\pi\log 2 +8\pi -4\pi J_{121}\right)}} +\pi^4-2\pi^3 \\ &= \pi^4-\frac 43\pi^3-4\pi\log2 + 2J_{111} + 2\pi\cdot J_{121} \\ &= \pi^4-\frac 43\pi^3-4\pi\log2 \\ &\qquad +\frac 18\left[ -4\pi^4 - 8\pi^2\log^2 2 +128 \pi G\log 2 +256\pi \operatorname{Li}_3\left(\frac {1+i}2\right) -16 \pi^2 \right] -\frac 23\pi^3 -8\pi\log 2 + 8\pi \\ &\qquad + \frac 18\left[\ 5\pi^4+4\pi^2\log^2 2 -64\pi G\log 2 - 128\pi\Im\operatorname{Li}_3\left(\frac {1+i}2\right)\ \right] +4\pi\log 2-8\pi \\ &= \frac 98\pi^4-2\pi^3-2\pi^2 -\frac 12\pi^2\log^2 2 +8\pi G\log 2 + 16\pi\Im\operatorname{Li}_3\left(\frac {1+i}2\right) -8\pi\log 2 \ . \end{aligned}

$$\square$$

\begin{align} &\int_0^{2\pi}x^2 \cos(x)\text{Li}_2(\cos(x))dx \\[5mm] = &\ \int_0^{\pi}x^2 \cos(x)\text{Li}_2(\cos(x))dx+\int_{\pi}^{2\pi}x^2 \cos(x)\text{Li}_2(\cos(x))dx \\[5mm] = &\ 2\int_0^{\pi}x^2 \cos(x)\text{Li}_2(\cos(x))dx-4{\pi^2}(-\frac{\pi^2}{2}+{\pi}) \\[2mm] - &\ 4{\pi}\int_0^{\pi}x \cos(x)\text{Li}_2(\cos(x))dx \end{align} Here is an idea,but for me impossible to calculate these integrals. I prove $$\int_0^{\pi}x \cos(x)\text{Li}_2(\cos(x))dx=-\frac{\pi^3}{16}+4G\ln{2}-\frac{\pi}{4}\ln^22+6-4\ln2-\frac{7\pi^2}{12}+8\Im\left\{\text{Li}_3\left(\frac{1+i}{2}\right)\right\}$$ Put $$u=x\text{Li}_2(\cos(x)), v=\sin{x} (IBP)$$ $$\int_0^{\pi}x^2 \cos(x)\text{Li}_2(\cos(x))dx=-2\int_0^{\pi}x \sin(x)\text{Li}_2(\cos(x))dx-\int_0^{\pi}x^2 \sin(x)\tan(x)\ln(1-\cos(x))dx$$ $$\int_0^{\pi}x \sin(x)\text{Li}_2(\cos(x))dx=-\frac{\pi^3}{12}+\int_0^{\pi}\cos(x)\text{Li}_2(\cos(x))dx+\int_0^{\pi}x \sin(x)\ln(1-\cos(x))dx$$ $$\int_0^{\pi}\cos(x)\text{Li}_2(\cos(x))dx=\frac{\pi^2}{2}-{\pi},\int_0^{\pi}x \sin(x)\ln(1-\cos(x))dx=3{\pi}\ln2-2{\pi}$$ $$\int_0^{\pi}x \sin(x)\text{Li}_2(\cos(x))dx=-\frac{\pi^3}{12}+\frac{\pi^2}{2}-3{\pi}+3{\pi}\ln2$$ $$\int_0^{\pi}x^2 \cos(x)\text{Li}_2(\cos(x))dx=\frac{\pi^3}{6}-{\pi^2}+6{\pi}-6{\pi}\ln2-\int_0^{\pi}x^2 \sin(x)\tan(x)\ln(1-\cos(x))dx$$ $$\int_0^{\pi}x^2 \sin(x)\tan(x)\ln(1-\cos(x))dx=\int_{-1}^1{\operatorname{arccos}(y)}^2\sqrt{(1-y^2)}\ln(1-y)\frac{dy}{y}$$ put $$\cos{x}=y$$ $$\int_{-1}^1=\int_{-1}^0+\int_{0}^1{\operatorname{arccos}(y)}^2\sqrt{(1-y^2)}\ln(1-y)\frac{dy}{y}$$ $$\int_{-1}^0{\operatorname{arccos}(x)}^2\sqrt{(1-x^2)}\ln(1-x)\frac{dx}{x}=-\int_{0}^1{\operatorname{({\pi}-arccos}(y)})^2\sqrt{(1-y^2)}\ln(1-y)\frac{dy}{y}=-{\pi^2}\int_{0}^1\sqrt{(1-y^2)}\ln(1+y)\frac{dy}{y}+2{\pi}\int_{0}^1{\operatorname{arccos}(y)}\sqrt{(1-y^2)}\ln(1+y)\frac{dy}{y}-\int_{0}^1{\operatorname{arccos}(y)}^2\sqrt{(1-y^2)}\ln(1+y)\frac{dy}{y}$$ $$\int_{0}^1\sqrt{(1-y^2)}\ln(1+y)\frac{dy}{y}=\frac{\pi^2}{8}-\frac{\pi}{2}+1$$ $$\int_{-1}^1{\operatorname{arccos}(y)}^2\sqrt{(1-y^2)}\ln(1-y)\frac{dy}{y}=-{\pi^2}-\frac{\pi^4}{8}+\frac{\pi^3}{2}+2{\pi}A+\int_{0}^1{\operatorname{arccos}(y)}^2\sqrt{(1-y^2)}{\ln\left(\frac{1-y}{1+y}\right)}\frac{dy}{y}$$ $$A=\int_{0}^1{\operatorname{arccos}(y)}\sqrt{(1-y^2)}\ln(1+y)\frac{dy}{y}=\int_0^{\pi/2}x(\sin{x})^2\ln(1+\cos(x))\frac{dx}{\cos(x)}=\int_0^{\pi/2}x(\ln(1+\cos(x))\frac{dx}{\cos(x)}-\int_0^{\pi/2}x\cos(x)\ln(1+\cos(x))dx$$ $$\int_0^{\pi/2}x\cos(x)\ln(1+\cos(x))dx=2-\frac{\pi}{2}+\frac{\pi^2}{8}-2\ln2$$ $$\int_0^{\pi/2}x(\ln(1+\cos(x))\frac{dx}{\cos(x)}=4\int_0^1\frac{\arctan(y)}{1-y^2}\ln\left(\frac{2}{1+y^2}\right)dy$$ put $$x=2\arctan(y)$$ $$=2\int_0^1\frac{\arctan(y)}{1-y}\ln\left(\frac{2}{1+y^2}\right)dy+2\int_0^1\frac{\arctan(y)}{1+y}\ln\left(\frac{2}{1+y^2}\right)dy$$ IBP for this integral; we come across known integrals, $$\int_0^1\frac{\arctan(y)}{1-y}\ln\left(\frac{2}{1+y^2}\right)dy=-2\Im\left\{\text{Li}_3\left(\frac{1+i}{2}\right)\right\}+\frac{13\pi^3}{192}-G\ln2+\frac{\pi}{16}\ln^22$$ $$\int_0^1\frac{\arctan(y)}{1+y}\ln\left(\frac{2}{1+y^2}\right)dy=-2\Im\left\{\text{Li}_3\left(\frac{1+i}{2}\right)\right\}+\frac{11\pi^3}{192}+G\ln2+\frac{\pi}{16}\ln^22$$ $$\int_0^{\pi/2}x(\ln(1+\cos(x))\frac{dx}{\cos(x)}=-8\Im\left\{\text{Li}_3\left(\frac{1+i}{2}\right)\right\}+\frac{\pi^3}{4}-4G\ln2+\frac{\pi}{4}\ln^22$$ $$A=\int_{0}^1{\operatorname{arccos}(y)}\sqrt{(1-y^2)}\ln(1+y)\frac{dy}{y}=-8\Im\left\{\text{Li}_3\left(\frac{1+i}{2}\right)\right\}+\frac{\pi^3}{4}-4G\ln2+\frac{\pi}{4}\ln^22-2+\frac{\pi}{2}-\frac{\pi^2}{8}+2\ln2$$ $$\int_0^{\pi}x^2 \sin(x)\tan(x)\ln(1-\cos(x))dx=-16{\pi}\Im\left\{\text{Li}_3\left(\frac{1+i}{2}\right)\right\}-8{\pi}G\ln2+\frac{\pi^2}{2}\ln^22+\frac{3\pi^4}{8}+\frac{\pi^3}{4}-4{\pi}+4{\pi}\ln2+\int_{0}^1{\operatorname{arccos}(y)}^2\sqrt{(1-y^2)}{\ln\left(\frac{1-y}{1+y}\right)}\frac{dy}{y}$$ $$\int_{0}^1{\operatorname{arccos}(y)}^2\sqrt{(1-y^2)}{\ln\left(\frac{1-y}{1+y}\right)}\frac{dy}{y}=\int_0^{\pi/2}x^2(\sin{x})^2{\ln\left(\frac{1-\cos(x)}{1+\cos(x)}\right)}\frac{dx}{\cos(x)}=64\int_0^{1}u^2{(\arctan(u)})^2\ln(u)\frac{du}{(1+u^2)^2(1-u^2)}$$ $$\frac{u^2}{(1+u^2)^2(1-u^2)}=\frac{1}{4(1+u^2)}-\frac{1}{2(1+u^2)^2}+\frac{1}{4(1-u^2)}$$ $$\int_0^1=\frac{1}{4}\int_0^{1}{(\arctan(u)})^2\ln(u)\frac{du}{1+u^2}-\int_0^{1}{(\arctan(u)})^2\ln(u)\frac{du}{(1+u^2)^2}+\int_0^{1}{(\arctan(u)})^2\ln(u)\frac{du}{4(1-u^2)}$$

$$\int_0^{1}{(\arctan(u)})^2\ln(u)\frac{du}{1+u^2}=-\frac{\pi^2G}{16}+\frac{ \psi ^{(3)}\left(\frac{1}{4}\right)}{3072}-\frac{ \psi ^{(3)}\left(\frac{3}{4}\right)}{3072}$$ $$\int_0^{1}{(\arctan(u)})^2\ln(u)\frac{du}{1-u^2}=-\frac{\pi}{2}\Im\left\{\text{Li}_3\left(\frac{1+i}{2}\right)\right\}+\frac{3\pi^4}{256}-\frac{\pi}{4}G\ln2+\frac{\pi^2}{64}\ln^22$$ IBP$$\int_0^{1}{(\arctan(u)})^2\ln(u)\frac{du}{(1+u^2)^2}=-\frac{\pi}{16}\ln2+\frac{\pi}{16}-\frac{\pi^3}{384}-\frac{\pi^2G}{32}+\frac{ \psi ^{(3)}\left(\frac{1}{4}\right)}{6144}-\frac{ \psi ^{(3)}\left(\frac{3}{4}\right)}{6144}$$ $$\int_{0}^1{\operatorname{arccos}(y)}^2\sqrt{(1-y^2)}{\ln\left(\frac{1-y}{1+y}\right)}\frac{dy}{y}=-8{\pi}\Im\left\{\text{Li}_3\left(\frac{1+i}{2}\right)\right\}+\frac{3\pi^4}{16}-4{\pi}G\ln2+\frac{\pi^2}{4}\ln^22-2{\pi}+\frac{\pi^3}{12}+2{\pi}\ln2$$ In regards to,same process $$\int_0^{\pi}x \cos(x)\text{Li}_2(\cos(x))dx$$ We must calculate $$\int_{0}^1{\operatorname{arccos}(y)}\sqrt{(1-y^2)}\ln(1-y^2)\frac{dy}{y}=-8\Im\left\{\text{Li}_3\left(\frac{1+i}{2}\right)\right\}+\frac{3\pi^3}{16}-4G\ln2+\frac{\pi}{4}\ln^22-4+{\pi}+2\ln2$$

• $+1$. I didn't change anything. I was curious of your idea. So, I did it to clarify the whole structure. Thanks. Oct 15, 2020 at 19:00
• I think the equality is not true. Note that: $\displaystyle \int_0^{\pi}\cos x\text{Li}_2(\cos x)dx=\frac{\pi^2}{2}-\pi$
– FDP
Oct 16, 2020 at 11:10