When is $int_a^b frac{1}{x}lnbigg(frac{x^3+1}{x^2+1}bigg)dx=0$?











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I would like to find positive, distinct, algebraic real numbers $a,bin mathbb R^+capmathbb A$ satisfying
$$int_a^b frac{1}{x}lnbigg(frac{x^3+1}{x^2+1}bigg)dx=0$$
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
$$frac{db}{b}lnbigg(frac{b^3+1}{b^2+1}bigg)=frac{da}{a}lnbigg(frac{a^3+1}{a^2+1}bigg)$$
...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
$$frac{text{Li}_2(-b^3)}{3}+frac{text{Li}_2(-b^2)}{2}=frac{text{Li}_2(-a^3)}{3}+frac{text{Li}_2(-a^2)}{2}$$






integration definite-integrals bounds-of-integration







share|cite|improve this question
























  • What is $mathbb{A}$?
    – Von Neumann
    Nov 21 at 15:20






  • 1




    If such $a,b$ do exist then $a$ belongs to $[0;1]$ and $b>1$. The integrand is negative on $[0;1]$ (see wolframalpha.com/input/… )
    – FDP
    Nov 22 at 10:52








  • 1




    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 at 14:46








  • 4




    @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 at 20:23








  • 1




    @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 at 23:19















up vote
12
down vote

favorite
3












I would like to find positive, distinct, algebraic real numbers $a,bin mathbb R^+capmathbb A$ satisfying
$$int_a^b frac{1}{x}lnbigg(frac{x^3+1}{x^2+1}bigg)dx=0$$
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
$$frac{db}{b}lnbigg(frac{b^3+1}{b^2+1}bigg)=frac{da}{a}lnbigg(frac{a^3+1}{a^2+1}bigg)$$
...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
$$frac{text{Li}_2(-b^3)}{3}+frac{text{Li}_2(-b^2)}{2}=frac{text{Li}_2(-a^3)}{3}+frac{text{Li}_2(-a^2)}{2}$$






integration definite-integrals bounds-of-integration







share|cite|improve this question
























  • What is $mathbb{A}$?
    – Von Neumann
    Nov 21 at 15:20






  • 1




    If such $a,b$ do exist then $a$ belongs to $[0;1]$ and $b>1$. The integrand is negative on $[0;1]$ (see wolframalpha.com/input/… )
    – FDP
    Nov 22 at 10:52








  • 1




    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 at 14:46








  • 4




    @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 at 20:23








  • 1




    @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 at 23:19













up vote
12
down vote

favorite
3









up vote
12
down vote

favorite
3






3





I would like to find positive, distinct, algebraic real numbers $a,bin mathbb R^+capmathbb A$ satisfying
$$int_a^b frac{1}{x}lnbigg(frac{x^3+1}{x^2+1}bigg)dx=0$$
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
$$frac{db}{b}lnbigg(frac{b^3+1}{b^2+1}bigg)=frac{da}{a}lnbigg(frac{a^3+1}{a^2+1}bigg)$$
...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
$$frac{text{Li}_2(-b^3)}{3}+frac{text{Li}_2(-b^2)}{2}=frac{text{Li}_2(-a^3)}{3}+frac{text{Li}_2(-a^2)}{2}$$






integration definite-integrals bounds-of-integration







share|cite|improve this question















I would like to find positive, distinct, algebraic real numbers $a,bin mathbb R^+capmathbb A$ satisfying
$$int_a^b frac{1}{x}lnbigg(frac{x^3+1}{x^2+1}bigg)dx=0$$
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
$$frac{db}{b}lnbigg(frac{b^3+1}{b^2+1}bigg)=frac{da}{a}lnbigg(frac{a^3+1}{a^2+1}bigg)$$
...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
$$frac{text{Li}_2(-b^3)}{3}+frac{text{Li}_2(-b^2)}{2}=frac{text{Li}_2(-a^3)}{3}+frac{text{Li}_2(-a^2)}{2}$$






integration definite-integrals bounds-of-integration




integration definite-integrals bounds-of-integration






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 23 at 16:03

























asked Nov 21 at 15:16









Frpzzd

21.2k639107




21.2k639107












  • What is $mathbb{A}$?
    – Von Neumann
    Nov 21 at 15:20






  • 1




    If such $a,b$ do exist then $a$ belongs to $[0;1]$ and $b>1$. The integrand is negative on $[0;1]$ (see wolframalpha.com/input/… )
    – FDP
    Nov 22 at 10:52








  • 1




    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 at 14:46








  • 4




    @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 at 20:23








  • 1




    @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 at 23:19


















  • What is $mathbb{A}$?
    – Von Neumann
    Nov 21 at 15:20






  • 1




    If such $a,b$ do exist then $a$ belongs to $[0;1]$ and $b>1$. The integrand is negative on $[0;1]$ (see wolframalpha.com/input/… )
    – FDP
    Nov 22 at 10:52








  • 1




    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 at 14:46








  • 4




    @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 at 20:23








  • 1




    @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 at 23:19
















What is $mathbb{A}$?
– Von Neumann
Nov 21 at 15:20




What is $mathbb{A}$?
– Von Neumann
Nov 21 at 15:20




1




1




If such $a,b$ do exist then $a$ belongs to $[0;1]$ and $b>1$. The integrand is negative on $[0;1]$ (see wolframalpha.com/input/… )
– FDP
Nov 22 at 10:52






If such $a,b$ do exist then $a$ belongs to $[0;1]$ and $b>1$. The integrand is negative on $[0;1]$ (see wolframalpha.com/input/… )
– FDP
Nov 22 at 10:52






1




1




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 at 14:46






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 at 14:46






4




4




@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 at 20:23






@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 at 20:23






1




1




@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 at 23:19




@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 at 23:19















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