Is $int_{0}^{infty}frac{x}{x^3+1} , , mathbb{d}x$ absolutely convergent or divergent? [on hold]












1












$begingroup$


We're asked to investigate whether this improper integral is (absolutely) convergent or divergent.
begin{equation}
int_{0}^{infty}frac{x}{x^3+1} , , mathbb{d}x
end{equation}

I've earlier worked on functions with logarithms and square roots that made it quite easy to use the Comparison Lemma or the Limit Criterion to go any further. But I think this expression needs to first be decomposed before I can operate on it and I can't seem to figure out how. Any hints/help would be appreciated










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put on hold as off-topic by Nosrati, RRL, Adrian Keister, Did, Lord_Farin yesterday


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, RRL, Adrian Keister, Did, Lord_Farin

If this question can be reworded to fit the rules in the help center, please edit the question.













  • $begingroup$
    $$int_{0}^{1}frac{x}{x^3+1},dx+int_{1}^{+infty}frac{x}{x^3+1},dx = int_{0}^{1}frac{x}{x^3+1},dx+int_{0}^{1}frac{1}{x^3+1},dx = int_{0}^{1}frac{dx}{x^2-x+1}$$ and since the discriminant of $x^2-x+1$ is negative, the function $frac{1}{x^2-x+1}$ is continuous on $[0,1]$, hence integrable. This also gives a straightforward way to compute the value of the given integral in terms of $pi$ and $sqrt{3}$.
    $endgroup$
    – Jack D'Aurizio
    yesterday
















1












$begingroup$


We're asked to investigate whether this improper integral is (absolutely) convergent or divergent.
begin{equation}
int_{0}^{infty}frac{x}{x^3+1} , , mathbb{d}x
end{equation}

I've earlier worked on functions with logarithms and square roots that made it quite easy to use the Comparison Lemma or the Limit Criterion to go any further. But I think this expression needs to first be decomposed before I can operate on it and I can't seem to figure out how. Any hints/help would be appreciated










share|cite|improve this question











$endgroup$



put on hold as off-topic by Nosrati, RRL, Adrian Keister, Did, Lord_Farin yesterday


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, RRL, Adrian Keister, Did, Lord_Farin

If this question can be reworded to fit the rules in the help center, please edit the question.













  • $begingroup$
    $$int_{0}^{1}frac{x}{x^3+1},dx+int_{1}^{+infty}frac{x}{x^3+1},dx = int_{0}^{1}frac{x}{x^3+1},dx+int_{0}^{1}frac{1}{x^3+1},dx = int_{0}^{1}frac{dx}{x^2-x+1}$$ and since the discriminant of $x^2-x+1$ is negative, the function $frac{1}{x^2-x+1}$ is continuous on $[0,1]$, hence integrable. This also gives a straightforward way to compute the value of the given integral in terms of $pi$ and $sqrt{3}$.
    $endgroup$
    – Jack D'Aurizio
    yesterday














1












1








1


1



$begingroup$


We're asked to investigate whether this improper integral is (absolutely) convergent or divergent.
begin{equation}
int_{0}^{infty}frac{x}{x^3+1} , , mathbb{d}x
end{equation}

I've earlier worked on functions with logarithms and square roots that made it quite easy to use the Comparison Lemma or the Limit Criterion to go any further. But I think this expression needs to first be decomposed before I can operate on it and I can't seem to figure out how. Any hints/help would be appreciated










share|cite|improve this question











$endgroup$




We're asked to investigate whether this improper integral is (absolutely) convergent or divergent.
begin{equation}
int_{0}^{infty}frac{x}{x^3+1} , , mathbb{d}x
end{equation}

I've earlier worked on functions with logarithms and square roots that made it quite easy to use the Comparison Lemma or the Limit Criterion to go any further. But I think this expression needs to first be decomposed before I can operate on it and I can't seem to figure out how. Any hints/help would be appreciated







calculus analysis convergence improper-integrals divergence






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edited 2 days ago









Asaf Karagila

302k32429760




302k32429760










asked 2 days ago









kareem bokaikareem bokai

555




555




put on hold as off-topic by Nosrati, RRL, Adrian Keister, Did, Lord_Farin yesterday


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, RRL, Adrian Keister, Did, Lord_Farin

If this question can be reworded to fit the rules in the help center, please edit the question.




put on hold as off-topic by Nosrati, RRL, Adrian Keister, Did, Lord_Farin yesterday


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, RRL, Adrian Keister, Did, Lord_Farin

If this question can be reworded to fit the rules in the help center, please edit the question.












  • $begingroup$
    $$int_{0}^{1}frac{x}{x^3+1},dx+int_{1}^{+infty}frac{x}{x^3+1},dx = int_{0}^{1}frac{x}{x^3+1},dx+int_{0}^{1}frac{1}{x^3+1},dx = int_{0}^{1}frac{dx}{x^2-x+1}$$ and since the discriminant of $x^2-x+1$ is negative, the function $frac{1}{x^2-x+1}$ is continuous on $[0,1]$, hence integrable. This also gives a straightforward way to compute the value of the given integral in terms of $pi$ and $sqrt{3}$.
    $endgroup$
    – Jack D'Aurizio
    yesterday


















  • $begingroup$
    $$int_{0}^{1}frac{x}{x^3+1},dx+int_{1}^{+infty}frac{x}{x^3+1},dx = int_{0}^{1}frac{x}{x^3+1},dx+int_{0}^{1}frac{1}{x^3+1},dx = int_{0}^{1}frac{dx}{x^2-x+1}$$ and since the discriminant of $x^2-x+1$ is negative, the function $frac{1}{x^2-x+1}$ is continuous on $[0,1]$, hence integrable. This also gives a straightforward way to compute the value of the given integral in terms of $pi$ and $sqrt{3}$.
    $endgroup$
    – Jack D'Aurizio
    yesterday
















$begingroup$
$$int_{0}^{1}frac{x}{x^3+1},dx+int_{1}^{+infty}frac{x}{x^3+1},dx = int_{0}^{1}frac{x}{x^3+1},dx+int_{0}^{1}frac{1}{x^3+1},dx = int_{0}^{1}frac{dx}{x^2-x+1}$$ and since the discriminant of $x^2-x+1$ is negative, the function $frac{1}{x^2-x+1}$ is continuous on $[0,1]$, hence integrable. This also gives a straightforward way to compute the value of the given integral in terms of $pi$ and $sqrt{3}$.
$endgroup$
– Jack D'Aurizio
yesterday




$begingroup$
$$int_{0}^{1}frac{x}{x^3+1},dx+int_{1}^{+infty}frac{x}{x^3+1},dx = int_{0}^{1}frac{x}{x^3+1},dx+int_{0}^{1}frac{1}{x^3+1},dx = int_{0}^{1}frac{dx}{x^2-x+1}$$ and since the discriminant of $x^2-x+1$ is negative, the function $frac{1}{x^2-x+1}$ is continuous on $[0,1]$, hence integrable. This also gives a straightforward way to compute the value of the given integral in terms of $pi$ and $sqrt{3}$.
$endgroup$
– Jack D'Aurizio
yesterday










1 Answer
1






active

oldest

votes


















8












$begingroup$

Hint



$$frac{x}{x^3color{red}{+1}} le frac{x}{x^3}=frac{1}{x^2}$$






share|cite|improve this answer









$endgroup$









  • 3




    $begingroup$
    Alright time to smash this keyboard. I've been trying to decompose this thing for the past couple hours and I didn't realize I could compare it in that sense. Damn. Thanks
    $endgroup$
    – kareem bokai
    2 days ago










  • $begingroup$
    No smashing, please! You're welcome :).
    $endgroup$
    – StackTD
    2 days ago






  • 2




    $begingroup$
    Small note. When using this comparison, one should change the lower limit of integration (WHY?), for example, go from 1 to infinity. Using the fact that integrating from 0 to 1 is a definite integral, the given comparison will do the rest. Also, if you are trying for hours to decompose the integral, well, use $a^3+b^3=(a+b)(a^2-ab+b^2)$. This integral can be evaluated, it is not too hard
    $endgroup$
    – imranfat
    2 days ago












  • $begingroup$
    I know, this was only a hint. The original function has no problem at $x=0$.
    $endgroup$
    – StackTD
    2 days ago






  • 1




    $begingroup$
    @StackTD. Oh yes, you are right, but your comparison function does have an issue at $0$. So one should not ignore that. Otherwise I would use the same comparison as you did
    $endgroup$
    – imranfat
    2 days ago


















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









8












$begingroup$

Hint



$$frac{x}{x^3color{red}{+1}} le frac{x}{x^3}=frac{1}{x^2}$$






share|cite|improve this answer









$endgroup$









  • 3




    $begingroup$
    Alright time to smash this keyboard. I've been trying to decompose this thing for the past couple hours and I didn't realize I could compare it in that sense. Damn. Thanks
    $endgroup$
    – kareem bokai
    2 days ago










  • $begingroup$
    No smashing, please! You're welcome :).
    $endgroup$
    – StackTD
    2 days ago






  • 2




    $begingroup$
    Small note. When using this comparison, one should change the lower limit of integration (WHY?), for example, go from 1 to infinity. Using the fact that integrating from 0 to 1 is a definite integral, the given comparison will do the rest. Also, if you are trying for hours to decompose the integral, well, use $a^3+b^3=(a+b)(a^2-ab+b^2)$. This integral can be evaluated, it is not too hard
    $endgroup$
    – imranfat
    2 days ago












  • $begingroup$
    I know, this was only a hint. The original function has no problem at $x=0$.
    $endgroup$
    – StackTD
    2 days ago






  • 1




    $begingroup$
    @StackTD. Oh yes, you are right, but your comparison function does have an issue at $0$. So one should not ignore that. Otherwise I would use the same comparison as you did
    $endgroup$
    – imranfat
    2 days ago
















8












$begingroup$

Hint



$$frac{x}{x^3color{red}{+1}} le frac{x}{x^3}=frac{1}{x^2}$$






share|cite|improve this answer









$endgroup$









  • 3




    $begingroup$
    Alright time to smash this keyboard. I've been trying to decompose this thing for the past couple hours and I didn't realize I could compare it in that sense. Damn. Thanks
    $endgroup$
    – kareem bokai
    2 days ago










  • $begingroup$
    No smashing, please! You're welcome :).
    $endgroup$
    – StackTD
    2 days ago






  • 2




    $begingroup$
    Small note. When using this comparison, one should change the lower limit of integration (WHY?), for example, go from 1 to infinity. Using the fact that integrating from 0 to 1 is a definite integral, the given comparison will do the rest. Also, if you are trying for hours to decompose the integral, well, use $a^3+b^3=(a+b)(a^2-ab+b^2)$. This integral can be evaluated, it is not too hard
    $endgroup$
    – imranfat
    2 days ago












  • $begingroup$
    I know, this was only a hint. The original function has no problem at $x=0$.
    $endgroup$
    – StackTD
    2 days ago






  • 1




    $begingroup$
    @StackTD. Oh yes, you are right, but your comparison function does have an issue at $0$. So one should not ignore that. Otherwise I would use the same comparison as you did
    $endgroup$
    – imranfat
    2 days ago














8












8








8





$begingroup$

Hint



$$frac{x}{x^3color{red}{+1}} le frac{x}{x^3}=frac{1}{x^2}$$






share|cite|improve this answer









$endgroup$



Hint



$$frac{x}{x^3color{red}{+1}} le frac{x}{x^3}=frac{1}{x^2}$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 2 days ago









StackTDStackTD

22.7k2049




22.7k2049








  • 3




    $begingroup$
    Alright time to smash this keyboard. I've been trying to decompose this thing for the past couple hours and I didn't realize I could compare it in that sense. Damn. Thanks
    $endgroup$
    – kareem bokai
    2 days ago










  • $begingroup$
    No smashing, please! You're welcome :).
    $endgroup$
    – StackTD
    2 days ago






  • 2




    $begingroup$
    Small note. When using this comparison, one should change the lower limit of integration (WHY?), for example, go from 1 to infinity. Using the fact that integrating from 0 to 1 is a definite integral, the given comparison will do the rest. Also, if you are trying for hours to decompose the integral, well, use $a^3+b^3=(a+b)(a^2-ab+b^2)$. This integral can be evaluated, it is not too hard
    $endgroup$
    – imranfat
    2 days ago












  • $begingroup$
    I know, this was only a hint. The original function has no problem at $x=0$.
    $endgroup$
    – StackTD
    2 days ago






  • 1




    $begingroup$
    @StackTD. Oh yes, you are right, but your comparison function does have an issue at $0$. So one should not ignore that. Otherwise I would use the same comparison as you did
    $endgroup$
    – imranfat
    2 days ago














  • 3




    $begingroup$
    Alright time to smash this keyboard. I've been trying to decompose this thing for the past couple hours and I didn't realize I could compare it in that sense. Damn. Thanks
    $endgroup$
    – kareem bokai
    2 days ago










  • $begingroup$
    No smashing, please! You're welcome :).
    $endgroup$
    – StackTD
    2 days ago






  • 2




    $begingroup$
    Small note. When using this comparison, one should change the lower limit of integration (WHY?), for example, go from 1 to infinity. Using the fact that integrating from 0 to 1 is a definite integral, the given comparison will do the rest. Also, if you are trying for hours to decompose the integral, well, use $a^3+b^3=(a+b)(a^2-ab+b^2)$. This integral can be evaluated, it is not too hard
    $endgroup$
    – imranfat
    2 days ago












  • $begingroup$
    I know, this was only a hint. The original function has no problem at $x=0$.
    $endgroup$
    – StackTD
    2 days ago






  • 1




    $begingroup$
    @StackTD. Oh yes, you are right, but your comparison function does have an issue at $0$. So one should not ignore that. Otherwise I would use the same comparison as you did
    $endgroup$
    – imranfat
    2 days ago








3




3




$begingroup$
Alright time to smash this keyboard. I've been trying to decompose this thing for the past couple hours and I didn't realize I could compare it in that sense. Damn. Thanks
$endgroup$
– kareem bokai
2 days ago




$begingroup$
Alright time to smash this keyboard. I've been trying to decompose this thing for the past couple hours and I didn't realize I could compare it in that sense. Damn. Thanks
$endgroup$
– kareem bokai
2 days ago












$begingroup$
No smashing, please! You're welcome :).
$endgroup$
– StackTD
2 days ago




$begingroup$
No smashing, please! You're welcome :).
$endgroup$
– StackTD
2 days ago




2




2




$begingroup$
Small note. When using this comparison, one should change the lower limit of integration (WHY?), for example, go from 1 to infinity. Using the fact that integrating from 0 to 1 is a definite integral, the given comparison will do the rest. Also, if you are trying for hours to decompose the integral, well, use $a^3+b^3=(a+b)(a^2-ab+b^2)$. This integral can be evaluated, it is not too hard
$endgroup$
– imranfat
2 days ago






$begingroup$
Small note. When using this comparison, one should change the lower limit of integration (WHY?), for example, go from 1 to infinity. Using the fact that integrating from 0 to 1 is a definite integral, the given comparison will do the rest. Also, if you are trying for hours to decompose the integral, well, use $a^3+b^3=(a+b)(a^2-ab+b^2)$. This integral can be evaluated, it is not too hard
$endgroup$
– imranfat
2 days ago














$begingroup$
I know, this was only a hint. The original function has no problem at $x=0$.
$endgroup$
– StackTD
2 days ago




$begingroup$
I know, this was only a hint. The original function has no problem at $x=0$.
$endgroup$
– StackTD
2 days ago




1




1




$begingroup$
@StackTD. Oh yes, you are right, but your comparison function does have an issue at $0$. So one should not ignore that. Otherwise I would use the same comparison as you did
$endgroup$
– imranfat
2 days ago




$begingroup$
@StackTD. Oh yes, you are right, but your comparison function does have an issue at $0$. So one should not ignore that. Otherwise I would use the same comparison as you did
$endgroup$
– imranfat
2 days ago



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