I am trying to see if Mathematica can calculate:
$$\int_0^1\frac{\ln(x)\ln(1-x)\ln(1+x)}{x}dx,$$
which has a closed form found here. So I tried
Integrate[Log[x]Log[1-x]Log[1+x]/x,{x,0,1}] but Mathematica failed to give an answer. Is there a special command for evaluating this integral?
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1$\begingroup$ "a well-known closed form" - for the more ignorant among us, what is this well-known closed form? $\endgroup$– MarcoBMar 3, 2021 at 2:52
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$\begingroup$ @MarcoB I edited my question adding the link where you can see many solutions to that integral and also you will find a generalization among these solutions. My point of saying " well-known" is that I dont need to calculate the integral, I am just curious why mathematica could not evaluate it. $\endgroup$– Ali ShadharMar 3, 2021 at 3:17
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2$\begingroup$ The linked answer is not a closed-form expression, but an analytic expression. All that is art for arts sake. All we need in most cases is a number. $\endgroup$– user64494Mar 3, 2021 at 12:28
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2$\begingroup$ I am voting to close this question because it is nonsense to hope that there a command that helps Mathematica. Ask a professional consultant for help. I also agree with @user64494 that this is art for arts sake. Especially asking for a black box solution does not bring any progress or deepens any understanding. $\endgroup$– yarchikMar 4, 2021 at 6:33
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$\begingroup$ @yarchik Chill no need to be so angry. There are some professional people here who know more than we both do. Wait and see and learn and dont complain. $\endgroup$– Ali ShadharMar 4, 2021 at 15:54
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1 Answer
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In the math.stackexchange post that you have linked, P. Teruo Nagasava shows that the integral can be reduced to two simpler ones:
$$I_1=\int_0^1 \frac{\log ^2(1+x) \log (x)}{x} \, dx,\\ I_2=\int_0^1 \frac{\log ^2(1-x) \log (x)}{x} \, dx. $$
Both of them MA computes without any problems
i1=Integrate[Log[1 + x]^2 Log[x]/x, {x, 0, 1}]
$$I_1=-4 \text{Li}_4\big(\tfrac{1}{2}\big)-\frac{7}{2} \zeta (3) \log (2)+\frac{\pi ^4}{24}-\frac{1}{6} \log ^4(2)+\frac{1}{6} \pi ^2 \log ^2(2);$$
i2=Integrate[Log[1 - x]^2 Log[x]/x, {x, 0, 1}]
$$I_2=-\frac{\pi ^4}{180}.$$
Human insight is still needed to perform rather trivial manipulations to reduce the integral to the linear combination of $I_1$ and $I_2$. But that is good, isn't it?
Actually even this step can be realized with MA. First compute the original integral numerically with a high precision:
i = NIntegrate[Log[x] Log[1 - x] Log[1 + x]/x, {x, 0, 1},
WorkingPrecision -> 100];
Next, find an integer relation between the three values:
FindIntegerNullVector[{i1, i2, i}]
(* {-4, -3, -8} *)
Thus, we have
$$I=-\frac12 I_1-\frac38 I_2.$$
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$\begingroup$ Thank you but thats not my question. Of course we can manipulate many integrals. I am asking if there is a smart command for evaluating the main integral by Mathematica without doing any integral manipulation. $\endgroup$ Mar 4, 2021 at 3:55
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2$\begingroup$ So what does your question mean? You've hoped that MA has some special command for a narrow circle of privileged which would allow to compute any integra,l or that someone will just run
Integratefor longer than 20 min because you got bored to wait? In my opinion you should just delete your question in both cases because these are just nonsense reasons. Voting to close. $\endgroup$– yarchikMar 4, 2021 at 6:31 -
$\begingroup$ I see people like you on MSE, on Tex and here they just enjoy fighting and arguing. I think there is no way avoiding negative people in this world. Thanks for your efforts though. $\endgroup$ Mar 4, 2021 at 15:58
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2$\begingroup$ (Hoping all concerned have calmed a bit...) I cannot see how this answer misses the question, at least as it is worded. This shows a smart way of manipulating the problem to get a good result. Are you asking for a way to do so that does not use mathematical knowledge about the problem e.g. to transform to a different integral? That would be asking for a lot. An automated mathematical assistant, so to speak. Such things are perhaps viable, down the road. Certainly there are research projects in that direction. But they are not on the immediate horizon (to the best of my knowledge). $\endgroup$ Mar 5, 2021 at 14:28