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
asked Nov 21 at 15:16
Frpzzd
21.2k639107
|
show 5 more comments
up vote
12
down vote
favorite
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
asked Nov 21 at 15:16
Frpzzd
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
|
show 5 more comments
up vote
12
down vote
favorite
up vote
12
down vote
favorite
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
asked Nov 21 at 15:16
Frpzzd
21.2k639107
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
asked Nov 21 at 15:16
Frpzzd
21.2k639107
asked Nov 21 at 15:16
Frpzzd
21.2k639107
edited Nov 23 at 16:03
asked Nov 21 at 15:16
Frpzzd
21.2k639107
asked Nov 21 at 15:16
Frpzzd
21.2k639107
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
|
show 5 more comments
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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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