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I would like to find positive, distinct, algebraic real numbers $a,b\in \mathbb R^+\cap\mathbb A$ satisfying $$\int_a^b \frac{1}{x}\ln\bigg(\frac{x^3+1}{x^2+1}\bigg)dx=0$$ Does anyone know of a systematic way to go about solving this problem? Calculating a definite integral is one thing, but solving for the values of its bounds is something that I have no experience with. If we let $a,b$ be numbers satisfying the above relation, then we know that $$\frac{db}{b}\ln\bigg(\frac{b^3+1}{b^2+1}\bigg)=\frac{da}{a}\ln\bigg(\frac{a^3+1}{a^2+1}\bigg)$$ ...but this is not useful since the chance of an antiderivative existing is slim.

Can anyone find such $a,b$?

Inspired by this question.

NOTE: Because the antiderivative of the integrand can be expressed in terms of dilogarithms, the problem is equivalent to finding distinct real algebraic numbers $a,b$ satisfying $$\frac{\text{Li}_2(-b^3)}{3}+\frac{\text{Li}_2(-b^2)}{2}=\frac{\text{Li}_2(-a^3)}{3}+\frac{\text{Li}_2(-a^2)}{2}$$

Franklin Pezzuti Dyer
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    What is $\mathbb{A}$? – Enrico M. Nov 21 '18 at 15:20
  • @VonNeumann Algebraic numbers. – Franklin Pezzuti Dyer Nov 21 '18 at 15:23
  • $\ln(\frac{x^3+1}{x^2+1})=\ln(x^2-x+1)-\ln(x^2+1)+\ln(x+1)$ if that is remotely helpful... – abiessu Nov 21 '18 at 15:23
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    If such $a,b$ do exist then $a$ belongs to $[0;1]$ and $b>1$. The integrand is negative on $[0;1]$ (see https://www.wolframalpha.com/input/?i=plot+log((x%5E3%2B1)%2F(x%5E2%2B1))%2Fx ) – FDP Nov 22 '18 at 10:52
  • Do you have numerics suggesting values of $a$,$b$? It should be possible to use an arbitrary precision package to minimize the square of the integral over all intervals to high precision. – asd Nov 27 '18 at 05:00
  • @asd No, I have not found any values that seem to satisfy this equality. – Franklin Pezzuti Dyer Nov 27 '18 at 14:07
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    Mathematica finds the antiderivative of your integral by $ln((x^3+1)/(x^2+1))=ln(x^3+1)-ln(x^2+1)$ and then writing the argument in a power expansion like $(x^2+1)=(1+i x)*(1-i x)$. Findroot calculates the zero findings for a=0 and b=1.8464. – stocha Nov 27 '18 at 14:46
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    @asd It would be useless: given the integrand changes sign, it's easy to prove there are infinitely many distinct real $a,b$ such that the integral vanishes. The problem is to find algebraic ones. Numerically you have no way to tell the difference. – Jean-Claude Arbaut Nov 28 '18 at 20:23
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    @Jean-ClaudeArbaut Nominally useless. There is a detail missing in the comment above that one can minimize the square of the integral starting with some $a$ algebraic so that one finds to high accuracy $b$ which satisfies the integral being zero. If $b$ in fact is algebraic (and satisfies a polynomial of low degree) then it is possible using algdep in Pari/GP to help find this value exactly from the numerics. – asd Nov 29 '18 at 23:19
  • $$\int_0^1 \frac{1}{x}\ln\bigg(\frac{x^3+1}{x^2+1}\bigg)dx=\frac{-\pi^2}{72}$$ not sure if this helps – quantus14 Nov 30 '18 at 14:27
  • The fact that the integrand is negative in $(0,1)$ is inconvenient as it deprives usage of the identity $\operatorname{Li}_2(z)+\operatorname{Li}_2(1-z)=\zeta(2)-\log z\log(1-z)$. – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Dec 25 '19 at 20:43
  • Though if we assume $\exists a,b\in\Bbb A^+$ s.t. $b^2(1+a^2)=1$ then a simple dilogarithm identity gives $$\operatorname{Li}_2\left(-\frac1{(1+a^2)^{3/2}}\right)-\operatorname{Li}_2(-a^3)=\frac34\log^2(1+a^2)-3\log(1+a^2)\log a-\frac{\pi^2}4.$$ From Desmos, we know such $a,b$ exist in $\Bbb R^+$ (and there are in fact two of them). However, I'm not sure what to do with the cubic terms, or how to show that they are even algebraic. – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Dec 25 '19 at 20:52

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Just an extended comment...

A figure might be helpful. Using the Rubi package in Mathematica one finds the following:

Get["Rubi`"]
integral = Int[(Log[x^3 + 1] - Log[x^2 + 1])/x, {x, a, b}]

$$\frac{1}{6} \left(2 \text{Li}_2\left(-a^3\right)-3 \text{Li}_2\left(-a^2\right)\right)+\frac{1}{6} \left(3 \text{Li}_2\left(-b^2\right)-2 \text{Li}_2\left(-b^3\right)\right)$$

as shown by the OP. A contour plot shows the contours of zero:

ContourPlot[integral, {a, -1, 2}, {b, -1, 2}, Contours -> {0},
 PlotPoints -> 100, ContourShading -> None, AspectRatio -> 1]

Contour plot of level 0

JimB
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  • Can Mathematica atleast aproximate such numbers $a$ and $b$? I highly believe that $b$ might be $\phi =\frac{1+\sqrt 5}{2}$ – Zacky Jan 10 '19 at 17:35
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    @Zacky Using FindRoot[integral /. b -> (1 + Sqrt[5])/2, {a, -0.3}, WorkingPrecision -> 30] Mathematica gets {a -> -0.311511113597363354426784499497}. – JimB Jan 10 '19 at 17:41
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    @Zacky Sorry, I forgot the other root around 0.4: sol = FindRoot[integral /. b -> (1 + Sqrt[5])/2, {a, 0.4}, WorkingPrecision -> 50] which results in {a -> 0.40710884690013078787253080882807595702108710161649}. – JimB Jan 10 '19 at 18:17