When is $int_a^b frac{1}{x}lnbigg(frac{x^3+1}{x^2+1}bigg)dx=0$?
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
20.7k638104
|
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
20.7k638104
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
20.7k638104
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
20.7k638104
asked Nov 21 at 15:16
Frpzzd
20.7k638104
edited Nov 23 at 16:03
asked Nov 21 at 15:16
Frpzzd
20.7k638104
asked Nov 21 at 15:16
Frpzzd
20.7k638104
asked Nov 21 at 15:16
Frpzzd
20.7k638104
20.7k638104
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
|
show 5 more comments
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3007841%2fwhen-is-int-ab-frac1x-ln-bigg-fracx31x21-biggdx-0%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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