20

How to prove $$\int_0^1x\ln^2(1+x)\ln\left(\frac{x^2}{1+x}\right)\frac{dx}{1+x^2}=-\frac{7}{32}\cdot\zeta{(3)}\ln2+\frac{3\pi^2}{128}\cdot\ln^22-\frac{1}{64}\cdot\ln^42-\frac{13\pi^4}{46080}$$ The substitution $$x=\frac{1-y}{1+y}$$ leads to calculate the integrals that are unknown: $$\int_0^1y\ln(1-y)\ln^2(1+y)\frac{dy}{1+y^2}, \int_0^1\frac{y\ln^3(1+y)}{1+y^2}dy$$ For the moment,I do not see how to calculate this integral.

Lizzi
  • 197
user178256
  • 5,467
  • 3
    Hi! Where did you found this integral? – Zacky Dec 25 '18 at 22:50
  • 2
    $$\ln\bigg(\frac{x^2}{x+1}\bigg)=\ln(x^2)-\ln(1+x)=2\ln(x)-\ln(1+x)$$ don't know if that helps – clathratus Dec 25 '18 at 23:08
  • Do any of you know the Maclaurin series for $\ln^2(1+x)$? Maybe we can solve it this way. – Larry Dec 25 '18 at 23:11
  • This integral has been proposed to me by friend. – user178256 Dec 25 '18 at 23:17
  • 5
    Here is an ideea, we have: $$a^2 b=\frac16\left((a+b)^3-(a-b)^3-2b^3\right)$$ Thus the integral is equal to: $$I=2\cdot \frac16\left( \int_0^1\frac{x\ln^3(1-x^2)}{1+x^2}dx+ \int_0^1\frac{x\ln^3\left(\frac{1-x}{1+x}\right)}{1+x^2}dx -2\int_0^1 \frac{x\ln^3(1-x)}{1+x^2}\right)-\int_0^1 \frac{x\ln^3(1+x)}{1+x^2}dx$$ Thus let $x^2=t$ for the first one and $\frac{1-x}{1+x}=t$ for the second one and try to simplify some. This way we don't have any product between the logarithms. – Zacky Dec 25 '18 at 23:22
  • 1
    @Zacky If it helps, making the substitutions $x\to -\frac{x}{1+x}$ and $x\to \frac{x-1}{2}$ shows that the integral in question is equal to $$\int_0^1 \ln^2\bigg(\frac{1+x}{2}\bigg)\ln\bigg(\frac{1}{2}\frac{(1-x)^2}{1+x}\bigg)\frac{x-1}{x+1}\frac{dx}{1+x^2}$$ Perhaps using your $a^2 b$ trick on this integral and combining it with the original representation can cause some sort of desirable cancellation to happen. – Franklin Pezzuti Dyer Dec 25 '18 at 23:30
  • Do you want an insightful proof or what do you expect? It is actually not too difficult to proof by using polylog identities, but i'm not sure thats what you want. – Diger Dec 27 '18 at 23:48
  • 1
    @Diger: I will be glad to see your solution using polylog identities. – FDP Dec 28 '18 at 10:32
  • Anyway, the proposed integral is the sum of two integrals expressible by rational linear combination of the constants $\zeta(3)\ln 2,\zeta(4),\pi^2\ln^2 2,\ln^4 2,\text{Li}_4\left(\frac{1}{2}\right)$ and the terms in $\text{Li}_4\left(\frac{1}{2}\right)$ are cancelling each other. (i don't have a proof, lindep of PARI GP rulez !) – FDP Dec 28 '18 at 11:10
  • What is canceling each other precisely? Regarding the polylog stuff I won't be able to write it here until end of next week earliest, since I only have mobile phone until then and I can assure you it is not "beautiful" just ugly algebra. – Diger Dec 28 '18 at 20:26
  • What I find far more interesting is the fact that each term can be expressed as a multiple of $\log^n(2) \zeta(4-n)$ for $n=0,1,2,4$. It is like there is a unit which needs to be conserved as in an underlying structure. – Diger Dec 28 '18 at 20:37
  • Numerically, \begin{align}\int_0^1 \frac{x\ln^2(1+x)\ln x}{1+x},dx=\frac{467}{256}\zeta(4)-\frac{5}{64}\ln^4 2+\frac{3}{32}\pi^2\ln^2 2-\frac{7}{4}\zeta(3)\ln 2-\frac{15}{8}\text{Li}_4\left(\frac{1}{2}\right)\end{align}

    \begin{align}\int_0^1 \frac{x\ln^3(1+x)}{1+x^2},dx=\frac{1881}{512}\zeta(4)-\frac{9}{64}\ln^4 2+\frac{21}{128}\pi^2\ln^2 2-\frac{105}{32}\zeta(3)\ln 2-\frac{15}{4}\text{Li}_4\left(\frac{1}{2}\right)\end{align} the terms in $\text{Li}_4\left(\frac{1}{2}\right)$ are killing each other.

     – FDP Dec 29 '18 at 10:37
  • @FDP your relationships are checked by Wolfram Alpha. – user178256 Dec 29 '18 at 22:03
  • Do you have we demonstrations? – user178256 Dec 29 '18 at 22:04
  • @user178256: of course not (yet). Maybe these relations are harder to obtain than the one in the question. I have supposed these integrals are rational combination of the constants $\zeta(4),\pi^2,\ln^4 2,\pi^2 \ln^2,\zeta(3)\ln 2,\text{Li_4}\left(\frac{1}{2}\right) $ and using GP PARI allow to guess if the supposition is right of not (but it's a proof of nothing for sure) – FDP Dec 29 '18 at 23:32
  • @zacky: I'm wondering how you obtained your result since I get $$\int_0^1 \frac{2x}{x^2+1} \log(x) \log^2(1+x) , {\rm d}x \ =\int_0^1 \frac{2(1-x)}{(1+x)(x^2+1)} \log\left( \frac{1-x}{1+x} \right) \log^2\left(\frac{2}{1+x}\right) , {\rm d}x , ,$$ different from the OP which you probably used. – Diger Dec 31 '18 at 22:26
  • @Diger You probably wanted to tag FDP, because I didn't obtain anything yet :D – Zacky Dec 31 '18 at 22:34
  • i just meant your "thus the integral is equal to: I=...". – Diger Dec 31 '18 at 22:39
  • Oh, I just used $a^2 b=\frac16\left((a+b)^3-(a-b)^3-2b^3\right)$. Where $a=\ln(1+x)$ and $b=\ln(1-x)$. Wait, I think I did a big mistake since I had $\frac{x}{1+x^2}$ – Zacky Dec 31 '18 at 22:49
  • So instead of the above we would then have $$ \int_0^1 \frac{2x}{x^2+1} \log(x) \log^2(1+x) , {\rm d}x \ = \frac{1}{3} \int_0^1 \frac{1-x}{1+x}\frac{1}{x^2+1} \Bigg{ \log^3\left( \frac{2(1-x)}{1+x} \right) - \log^3\left( \frac{2}{1-x^2} \right) - 2\log^3\left(1-x \right) \ - \log^3(2) + \log^3\left( \frac{2}{(1+x)^2} \right) + 2\log^3\left( 1+x \right) \Bigg} , {\rm d}x , , $$ of which only the second one $(\log^3\left( \frac{2}{1-x^2} \right))$ seems to be lacking an anti-derivative. – Diger Jan 03 '19 at 12:28
  • Just curios: Have you tried to use Mathematica to find an anti-derivative? – Diger Aug 01 '19 at 22:42

1 Answers1

5

Note that: $$\int _0^1\frac{x\ln ^3\left(1+x\right)}{1+x^2}\:dx=\frac{1}{2}\ln ^4\left(2\right)-\frac{3}{2}\int _0^1\frac{\ln ^2\left(1+x\right)\ln \left(1+x^2\right)}{1+x}\:dx$$

Now I'll evaluate that integral in large steps. $$\int _0^1\frac{\ln ^2\left(1+x\right)\ln \left(1+x^2\right)}{1+x}\:dx$$ $$=\int _{\frac{1}{2}}^1\frac{\ln ^2\left(x\right)\ln \left(1-2x+2x^2\right)}{x}\:dx-2\int _{\frac{1}{2}}^1\frac{\ln ^3\left(x\right)}{x}\:dx$$ $$=2\operatorname{\mathfrak{R}} \left\{\int _{\frac{1}{2}}^1\frac{\ln ^2\left(x\right)\ln \left(1-\left(1+i\right)x\right)}{x}\:dx\right\}+\frac{1}{2}\ln ^4\left(2\right)$$ $$=-2\operatorname{\mathfrak{R}} \left\{-2\operatorname{Li}_4\left(\frac{1+i}{2}\right)+2\operatorname{Li}_4\left(1+i\right)-2\ln \left(2\right)\operatorname{Li}_3\left(\frac{1+i}{2}\right)-\ln ^2\left(2\right)\operatorname{Li}_2\left(\frac{1+i}{2}\right)\right\}$$ $$+\frac{1}{2}\ln ^4\left(2\right)$$ The real part of these polylogarithm are well-known, using their closed-form yields: $$\int _0^1\frac{\ln ^2\left(1+x\right)\ln \left(1+x^2\right)}{1+x}\:dx$$ $$=-\frac{627}{256}\zeta \left(4\right)+\frac{5}{2}\operatorname{Li}_4\left(\frac{1}{2}\right)+\frac{35}{16}\ln \left(2\right)\zeta \left(3\right)-\frac{21}{32}\ln ^2\left(2\right)\zeta \left(2\right)+\frac{41}{96}\ln ^4\left(2\right)$$ Replacing this into the first line yields the closed-form for that other integral. $$\int _0^1\frac{x\ln ^3\left(1+x\right)}{1+x^2}\:dx$$ $$=\frac{1881}{512}\zeta \left(4\right)-\frac{15}{4}\operatorname{Li}_4\left(\frac{1}{2}\right)-\frac{105}{32}\ln \left(2\right)\zeta \left(3\right)+\frac{63}{64}\ln ^2\left(2\right)\zeta \left(2\right)-\frac{9}{64}\ln ^4\left(2\right)$$

Now for your next integral: $$\int _0^1\frac{x\ln \left(1-x\right)\ln ^2\left(1+x\right)}{1+x^2}\:dx$$ $$=\frac{1}{6}\int _0^1\frac{x\ln ^3\left(1-x^2\right)}{1+x^2}\:dx+\frac{1}{6}\int _0^1\frac{x\ln ^3\left(\frac{1-x}{1+x}\right)}{1+x^2}\:dx-\frac{1}{3}\int _0^1\frac{x\ln ^3\left(1-x\right)}{1+x^2}\:dx$$ $$=\frac{1}{12}\int _0^1\frac{\ln ^3\left(1-x\right)}{1+x}dx+\frac{1}{6}\int _0^1\frac{\ln ^3\left(x\right)}{1+x}\:dx-\frac{1}{6}\int _0^1\frac{x\ln ^3\left(x\right)}{1+x^2}\:dx$$ $$-\frac{1}{3}\operatorname{\mathfrak{R}} \left\{\frac{1-i}{2}\int _0^1\frac{\ln ^3\left(x\right)}{1-\frac{1-i}{2}x}\:dx\right\}$$ $$=-\frac{1}{2}\operatorname{Li}_4\left(\frac{1}{2}\right)-\frac{105}{128}\zeta \left(4\right)+2\operatorname{\mathfrak{R}} \left\{\operatorname{Li}_4\left(\frac{1-i}{2}\right)\right\}$$ Thus: $$\int _0^1\frac{x\ln \left(1-x\right)\ln ^2\left(1+x\right)}{1+x^2}\:dx=-\frac{77}{512}\zeta \left(4\right)+\frac{1}{8}\operatorname{Li}_4\left(\frac{1}{2}\right)-\frac{5}{64}\ln ^2\left(2\right)\zeta \left(2\right)+\frac{1}{48}\ln ^4\left(2\right)$$

Jorge Layja
  • 1,101