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
calculus analysis convergence improper-integrals divergence
edited 2 days ago
Asaf Karagila♦
302k32429760
asked 2 days ago
kareem bokaikareem bokai
555
$endgroup$
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.
add a comment |
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
calculus analysis convergence improper-integrals divergence
edited 2 days ago
Asaf Karagila♦
302k32429760
asked 2 days ago
kareem bokaikareem bokai
555
$endgroup$
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
add a comment |
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
calculus analysis convergence improper-integrals divergence
edited 2 days ago
Asaf Karagila♦
302k32429760
asked 2 days ago
kareem bokaikareem bokai
555
$endgroup$
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
calculus analysis convergence improper-integrals divergence
edited 2 days ago
Asaf Karagila♦
302k32429760
asked 2 days ago
kareem bokaikareem bokai
555
edited 2 days ago
Asaf Karagila♦
302k32429760
asked 2 days ago
kareem bokaikareem bokai
555
edited 2 days ago
Asaf Karagila♦
302k32429760
edited 2 days ago
Asaf Karagila♦
302k32429760
edited 2 days ago
Asaf Karagila♦
302k32429760
302k32429760
asked 2 days ago
kareem bokaikareem bokai
555
asked 2 days ago
kareem bokaikareem bokai
555
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
add a comment |
$begingroup$
$$int_{0}^{1}frac{x}{x^3+1},dx+int_{1}^{+infty}frac{x}{x^3+1},dx = int_{0}^{1}frac{x}{x^3+1},dx+int_{0}^{1}frac{1}{x^3+1},dx = int_{0}^{1}frac{dx}{x^2-x+1}$$ 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
add a comment |
1 Answer
1
active
oldest
votes
8
$begingroup$
Hint
$$frac{x}{x^3color{red}{+1}} le frac{x}{x^3}=frac{1}{x^2}$$
answered 2 days ago
StackTDStackTD
22.7k2049
$endgroup$
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
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
8
$begingroup$
Hint
$$frac{x}{x^3color{red}{+1}} le frac{x}{x^3}=frac{1}{x^2}$$
answered 2 days ago
StackTDStackTD
22.7k2049
$endgroup$
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
add a comment |
8
$begingroup$
Hint
$$frac{x}{x^3color{red}{+1}} le frac{x}{x^3}=frac{1}{x^2}$$
answered 2 days ago
StackTDStackTD
22.7k2049
$endgroup$
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
add a comment |
8
8
8
$begingroup$
Hint
$$frac{x}{x^3color{red}{+1}} le frac{x}{x^3}=frac{1}{x^2}$$
answered 2 days ago
StackTDStackTD
22.7k2049
$endgroup$
Hint
$$frac{x}{x^3color{red}{+1}} le frac{x}{x^3}=frac{1}{x^2}$$
answered 2 days ago
StackTDStackTD
22.7k2049
answered 2 days ago
StackTDStackTD
22.7k2049
answered 2 days ago
StackTDStackTD
22.7k2049
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
add a comment |
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
add a comment |
$begingroup$
$$int_{0}^{1}frac{x}{x^3+1},dx+int_{1}^{+infty}frac{x}{x^3+1},dx = int_{0}^{1}frac{x}{x^3+1},dx+int_{0}^{1}frac{1}{x^3+1},dx = int_{0}^{1}frac{dx}{x^2-x+1}$$ 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