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This is a follow-up to my earlier question Closed form for ${\large\int}_0^1\frac{\ln^2x}{\sqrt{1-x+x^2}}dx$.

Is there a closed form for this integral? $$I=\int_0^1\frac{\ln^3x}{\sqrt{x^2-x+1}}dx\tag1$$

Mathematica and Maple cannot evaluate it directly. A numeric approximation for it is $$I\approx-6.1665252325192513801994672415450909679747097867356795481...\tag2$$ (click here to see more digits).

As I mentioned in the earlier question, Mathematica is able to find a closed form for a parameterized integral in terms of the Appell hypergeometric function: $$I(a)=\int_0^1\frac{x^a}{\sqrt{x^2-x+1}}dx\\=\frac1{a+1}F_1\left(a+1;\frac{1}{2},\frac{1}{2};a+2;(-1)^{\small1/3},-(-1)^{\small2/3}\right),\tag3$$ but taking a derivative from this looks a hard problem.

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    $\begingroup$ There is an equivalent rational integral, with no logs: $$\int_{1,1,1,1}^{\infty,\infty,\infty,\infty}\frac{96 \left(x_1^2 x_2^2+x_1 x_2+1\right) \left(x_1^2 x_2^2 x_3^2+x_1 x_2 x_3+1\right) \left(x_1^2 x_2^2 x_3^2 x_4^2+x_1 x_2 x_3 x_4+1\right)}{x_1 \left(x_1+2\right) x_2 \left(x_1^3 x_2^3+2 x_1^2 x_2^2-x_1 x_2-2\right) x_3 \left(x_1 x_2 x_3-1\right) \left(x_1 x_2 x_3+1\right) \left(x_1 x_2 x_3+2\right) x_4 \left(x_1 x_2 x_3 x_4-1\right) \left(x_1 x_2 x_3 x_4+1\right) \left(x_1 x_2 x_3 x_4+2\right)}$$ $\endgroup$
    – Kirill
    Sep 17, 2014 at 2:21
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    $\begingroup$ $$ 2\int_{0}^{1}\ln^{3}\left(\,1 - t^{2} \over 1 + 2t\,\right)\,{1 \over 1 + 2t}\,{\rm d}t $$ $\endgroup$
    – Felix Marin
    Sep 18, 2014 at 6:51
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    $\begingroup$ There is something bugging me here. Suppose that the approach to evaluate this integral is exactly the same as Mr. David & Tunk-Fey's answer in the previous problem, then we must evaluate it the integral of $\dfrac{\ln(1-x)\ln(1+x)\ln(1+2x)}{1+2x}$, your bounty question before this one, meanwhile that integral has no a closed-form which mean this integral has no a closed-form too $\endgroup$
    – Anastasiya-Romanova 秀
    Sep 18, 2014 at 16:20
  • $\begingroup$ @Anastasiya-Romanova Well, yes, it is possible to expand this integral to a sum of integrals, where one of the terms corresponds to my previous bounty question. But I still hope it might have a closed form. But even if it does not: Have you managed to evaluate all other terms? It's possible that "non-closed-form" part somehow cancels between them, leaving a closed form result for the integral in this question. $\endgroup$
    – Vladimir Reshetnikov
    Sep 18, 2014 at 17:17

1 Answer 1

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I followed the same approach as I used in an answer to another question, and expanded your integral in multiple polylogarithms of weight 4, then used some patterns in their values of weight 3 to guess terms that might appear in the integral. Then I used an integer relation algorithm to express your integral in terms of logs, zeta functions and polylogarithms of small rational arguments with a tolerance of about $10^{-200}$, and I found this value, which is correct to $3000$ digits.

There are $27$ polylogarithm terms there in total, and while I managed to simplify them somewhat, I never quite managed to evaluate some of them except by an integer relation algorithm.

Here it is: $$\textstyle\def\Li{\mathrm{Li}} -6 \Li_2(\frac{1}{3}) \zeta (2)+27 \Li_4(\frac{3}{4})+36 \Li_4(\frac{2}{3})-4 \Li_4(\frac{1}{2})-18 \Li_4(\frac{1}{3})-\frac{9}{2} \Li_4(\frac{1}{4})+6 \Li_2(\frac{1}{3}){}^2+6 \Li_2(\frac{1}{3}) \log ^2 3+24 \Li_2(\frac{1}{3}) \log ^2 2-48 \Li_3(\frac{2}{3}) \log3+96 \Li_3(\frac{2}{3}) \log2-48 \Li_3(\frac{1}{3}) \log3+72 \Li_3(\frac{1}{3}) \log2-24 \Li_2(\frac{1}{3}) \log2 \log3+78 \zeta (3) \log3-142 \zeta (3) \log2-\frac{151}{4} \zeta (4)-69 \zeta (2) \log ^2 3-122 \zeta (2) \log ^2 2+192 \zeta (2) \log2 \log3+\frac{73}{4} \log ^4 3+\frac{89}{6} \log ^4 2-70 \log2 \log ^3 3-56 \log ^3 2 \log3+93 \log ^2 2 \log ^2 3 $$

Here is the equivalent Mathematica expression to save people typing:

(-151*Pi^4)/360 - (61*Pi^2*Log[2]^2)/3 + (89*Log[2]^4)/6 + 32*Pi^2*Log[2]*Log[3] -  56*Log[2]^3*Log[3] - (23*Pi^2*Log[3]^2)/2 + 93*Log[2]^2*Log[3]^2 - 70*Log[2]*Log[3]^3 +  (73*Log[3]^4)/4 - Pi^2*PolyLog[2, 1/3] + 24*Log[2]^2*PolyLog[2, 1/3] -  24*Log[2]*Log[3]*PolyLog[2, 1/3] + 6*Log[3]^2*PolyLog[2, 1/3] + 6*PolyLog[2, 1/3]^2 +  72*Log[2]*PolyLog[3, 1/3] - 48*Log[3]*PolyLog[3, 1/3] + 96*Log[2]*PolyLog[3, 2/3] -  48*Log[3]*PolyLog[3, 2/3] - (9*PolyLog[4, 1/4])/2 - 18*PolyLog[4, 1/3] - 4*PolyLog[4, 1/2] +  36*PolyLog[4, 2/3] + 27*PolyLog[4, 3/4] - 142*Log[2]*Zeta[3] + 78*Log[3]*Zeta[3]
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    $\begingroup$ With all due respect, Sir, I don't think this is a proper way to answer this problem. I find an answer that comes from (human) mind is full of fascinating insights rather than output of machine. Please don't consider my comment as a rude expression Mr. Kirill but in my opinion, Mathematica & Maple, cs have ruined the art of solving integral & series like Deep Blue has ruined the art of playing chess. I hope someone here will come with legit proof for evaluating this integral. Peace... (>‿◠)✌ $\endgroup$
    – Anastasiya-Romanova 秀
    Sep 20, 2014 at 10:23
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    $\begingroup$ @Anastasiya-Romanova Don't be a Luddite :) Machines are very important tool to generate conjectures and verify manual computations. Many important known results would be very difficult or impossible to obtain without help of machines. $\endgroup$
    – Vladimir Reshetnikov
    Sep 20, 2014 at 17:42
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    $\begingroup$ @Anastasiya-Romanova While I sympathize with your point of view very much, I have to agree with Vladimir here. Part of being a good mathematician is making the best of the tools allotted to you, and this problem just looks too massive to be reasonably tackled by hand. $\endgroup$
    – David H
    Sep 20, 2014 at 18:33
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    $\begingroup$ @VladimirReshetnikov No, I'm not trying to be a Luddite. My point is where is the art of math if its problems i.e. the four color theorem or Riemann hypothesis is solved by a machine? Um, maybe I'm just a conservative being. I still rely on machines to do lots things in my life though. $\endgroup$
    – Anastasiya-Romanova 秀
    Sep 20, 2014 at 19:26
  • $\begingroup$ @DavidH You said too massive? How about this one: math.stackexchange.com/q/918821/133248? Still thinking too massive to be reasonably tackled by hand? You just looks the problem but not trying to solve it because you assume it's too massive to be reasonably tackled by hand. Or you're just unwilling to make the answer, I presume. (︺︹︺) $\endgroup$
    – Anastasiya-Romanova 秀
    Sep 20, 2014 at 19:33

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