Why does this lead to zero when integrating a non-negative function?

Question

When I try to integrate this $\int_0^{1} \frac{\log ^2(x+1) \log ^2(1-x)}{x} \, dx$, it leads to 0.

And when I use NIntegrate, it turns into a positive number, which seems to be right.

The codes are shown below.

Clear["Global`*"];
Plot[Log[1 + x]^2 Log[1 - x]^2/x, {x, 0, 1}]
Integrate[Log[1 + x]^2 Log[1 - x]^2/x, {x, 0, 1}]
NIntegrate[Log[1 + x]^2 Log[1 - x]^2/x, {x, 0, 1}]

Answer 1

I cannot answer why, but I can provide an analytical solution with MA. We have

$$\int_0^1 \frac{\log ^2(1-x) \log ^2(x+1)}{x} \, dx= \frac{1}{12} \int_0^1 \frac{\log ^4\left(1-x^2\right)}{x} \, dx +\frac{1}{12} \int_0^1 \frac{\log ^4\left(\frac{1-x}{x+1}\right)}{x} \, dx -\frac{1}{6} \int_0^1 \frac{\log ^4(1-x)}{x} \, dx -\frac{1}{6} \int_0^1 \frac{\log ^4(x+1)}{x} \, dx.$$

Each of them can be correctly computed with MA

i1 = 1/12 Integrate[Log[1 - x^2]^4/x, {x, 0, 1}]
(*Zeta[5]*)
i2 = 1/12 Integrate[Log[(1 - x)/(1 + x)]^4/x, {x, 0, 1}]
(*(31 Zeta[5])/8*)
i3 = -1/6 Integrate[Log[1 - x]^4/x, {x, 0, 1}]
(*-4 Zeta[5]*)
i4 = -1/6 Integrate[Log[1 + x]^4/x, {x, 0, 1}]
(*1/6 (-(2/3) \[Pi]^2 Log[2]^3 + (4 Log[2]^5)/5 + 21/2 Log[2]^2 Zeta[3] +
   24 (Log[2] PolyLog[4, 1/2] + PolyLog[5, 1/2] - Zeta[5]))*)

Combining together we obtain

i = i1 + i2 + i3 + i4

$$4 \text{Li}_5\left(\frac{1}{2}\right)+\text{Li}_4\left(\frac{1}{2}\right) \log (16)-\frac{25 \zeta (5)}{8}+\frac{7}{4} \zeta (3) \log ^2(2)+\frac{2 \log ^5(2)}{15}-\frac{1}{9} \pi ^2 \log ^3(2).$$

Answer 2

Not an answer for the question only how easy compute integral

This such form integrals we can compute with MultipleZetaValues-1.1.0 package.

<< MultipleZetaValues`
MZIntegrate[Log[1 + x]^2 Log[1 - x]^2/x, {x, 0, 1}]

(*-(1/9) \[Pi]^2 Log[2]^3 + (2 Log[2]^5)/15 + 
4 Log[2] PolyLog[4, 1/2] + 4 PolyLog[5, 1/2] + 
7/4 Log[2]^2 Zeta[3] - (25 Zeta[5])/8*)

Answer 3

I am the OP. The current situation is like this. I am running it in my 14.0.0 version:

AbsoluteTiming[Integrate[Log[1 + x]^2*(Log[1 - x]^2/x), {x, 0, 1}]]
$Version

The first result is like this: which shows that after more than 3 minutes, Mathematica returns an unevaluated output.

---

After sending the bug to Wolfram's Product Feedback team on 2021-09-24, they didn't reply after confirming the issue. Then I asked them again yesterday and got the following reply(personal information has been erased):

Hello ████ ███,

Thank you for your email. The issue has been fixed since Mathematica 13.3. There is no closed-form solution to the given integral, so Mathematica will return an unevaluated output.

In[1]:= AbsoluteTiming[Integrate[Log[1+x]^2*(Log[1-x]^2/x),{x,0,1}]]
Out[1]= {131.705, Integrate[(Log[1 - x]^2*Log[1 + x]^2)/x, {x, 0, 1}]}
In[2]:= $Version
Out[2]= 14.0.0 for Mac OS X ARM (64-bit) (December 13, 2023)

Please feel free to let me know if you have any further questions.

Sincerely,

██ ███, PhD

Wolfram Technical Support

Wolfram Research, Inc.

http://support.wolfram.com

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@yarchik's answer is an analytical form for the integration, and there is indeed no closed-form solution to the given integral. Thanks to @user64494 for reminding me.