Convergence of $int_0^infty frac{sin x}{x}dx$ [closed]
$begingroup$
$$
begin{align}
int_0^infty frac{sin x}{x}dx
end{align}
$$
Convergence at $x=0.$
Since, $lim_{xrightarrow 0}frac{sin x}{x} = 1$ so $0$ it is not a point of infinite discontinuity, hence the integrand is convergent at $0$.
Convergence at $x=infty.$
Since, $lim_{xrightarrow infty}frac{sin x}{x} = 0$ so $infty$ it is not a point of infinite discontinuity, hence the integrand is convergent at $infty$.
Will this be sufficient to prove that the integrand is convergent at $infty$?
real-analysis improper-integrals
asked Dec 13 '18 at 14:56
Ajay ChoudharyAjay Choudhary
888
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closed as unclear what you're asking by RRL, Saad, Rebellos, zhw., amWhy Dec 13 '18 at 20:07
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
$$
begin{align}
int_0^infty frac{sin x}{x}dx
end{align}
$$
Convergence at $x=0.$
Since, $lim_{xrightarrow 0}frac{sin x}{x} = 1$ so $0$ it is not a point of infinite discontinuity, hence the integrand is convergent at $0$.
Convergence at $x=infty.$
Since, $lim_{xrightarrow infty}frac{sin x}{x} = 0$ so $infty$ it is not a point of infinite discontinuity, hence the integrand is convergent at $infty$.
Will this be sufficient to prove that the integrand is convergent at $infty$?
real-analysis improper-integrals
asked Dec 13 '18 at 14:56
Ajay ChoudharyAjay Choudhary
888
$endgroup$
closed as unclear what you're asking by RRL, Saad, Rebellos, zhw., amWhy Dec 13 '18 at 20:07
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
@RRL I've edited the question. Please review it again.
$endgroup$
– Ajay Choudhary
Dec 14 '18 at 4:48
$begingroup$
Choudary: (1) In the real number system $infty$ is not a point so the "point of infinite continuity" terminology is not relevant -- particularly with respect to convergence of an improper integral. (2) If you are referring to the fact that $frac{sin x}{x} to 0$ as $x to infty$ which is obviously true, then this alone does not guarantee convergence of the improper integral. In this case the integral happens to converge.
$endgroup$
– RRL
Dec 14 '18 at 5:30
add a comment |
$begingroup$
$$
begin{align}
int_0^infty frac{sin x}{x}dx
end{align}
$$
Convergence at $x=0.$
Since, $lim_{xrightarrow 0}frac{sin x}{x} = 1$ so $0$ it is not a point of infinite discontinuity, hence the integrand is convergent at $0$.
Convergence at $x=infty.$
Since, $lim_{xrightarrow infty}frac{sin x}{x} = 0$ so $infty$ it is not a point of infinite discontinuity, hence the integrand is convergent at $infty$.
Will this be sufficient to prove that the integrand is convergent at $infty$?
real-analysis improper-integrals
asked Dec 13 '18 at 14:56
Ajay ChoudharyAjay Choudhary
888
$endgroup$
$$
begin{align}
int_0^infty frac{sin x}{x}dx
end{align}
$$
Convergence at $x=0.$
Since, $lim_{xrightarrow 0}frac{sin x}{x} = 1$ so $0$ it is not a point of infinite discontinuity, hence the integrand is convergent at $0$.
Convergence at $x=infty.$
Since, $lim_{xrightarrow infty}frac{sin x}{x} = 0$ so $infty$ it is not a point of infinite discontinuity, hence the integrand is convergent at $infty$.
Will this be sufficient to prove that the integrand is convergent at $infty$?
real-analysis improper-integrals
real-analysis improper-integrals
asked Dec 13 '18 at 14:56
Ajay ChoudharyAjay Choudhary
888
asked Dec 13 '18 at 14:56
Ajay ChoudharyAjay Choudhary
888
edited Dec 14 '18 at 4:33
Ajay Choudhary
asked Dec 13 '18 at 14:56
Ajay ChoudharyAjay Choudhary
888
asked Dec 13 '18 at 14:56
Ajay ChoudharyAjay Choudhary
888
asked Dec 13 '18 at 14:56
Ajay ChoudharyAjay Choudhary
888
888
closed as unclear what you're asking by RRL, Saad, Rebellos, zhw., amWhy Dec 13 '18 at 20:07
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by RRL, Saad, Rebellos, zhw., amWhy Dec 13 '18 at 20:07
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
@RRL I've edited the question. Please review it again.
$endgroup$
– Ajay Choudhary
Dec 14 '18 at 4:48
$begingroup$
Choudary: (1) In the real number system $infty$ is not a point so the "point of infinite continuity" terminology is not relevant -- particularly with respect to convergence of an improper integral. (2) If you are referring to the fact that $frac{sin x}{x} to 0$ as $x to infty$ which is obviously true, then this alone does not guarantee convergence of the improper integral. In this case the integral happens to converge.
$endgroup$
– RRL
Dec 14 '18 at 5:30
add a comment |
$begingroup$
@RRL I've edited the question. Please review it again.
$endgroup$
– Ajay Choudhary
Dec 14 '18 at 4:48
$begingroup$
Choudary: (1) In the real number system $infty$ is not a point so the "point of infinite continuity" terminology is not relevant -- particularly with respect to convergence of an improper integral. (2) If you are referring to the fact that $frac{sin x}{x} to 0$ as $x to infty$ which is obviously true, then this alone does not guarantee convergence of the improper integral. In this case the integral happens to converge.
$endgroup$
– RRL
Dec 14 '18 at 5:30
$begingroup$
@RRL I've edited the question. Please review it again.
$endgroup$
– Ajay Choudhary
Dec 14 '18 at 4:48
$begingroup$
@RRL I've edited the question. Please review it again.
$endgroup$
– Ajay Choudhary
Dec 14 '18 at 4:48
$begingroup$
Choudary: (1) In the real number system $infty$ is not a point so the "point of infinite continuity" terminology is not relevant -- particularly with respect to convergence of an improper integral. (2) If you are referring to the fact that $frac{sin x}{x} to 0$ as $x to infty$ which is obviously true, then this alone does not guarantee convergence of the improper integral. In this case the integral happens to converge.
$endgroup$
– RRL
Dec 14 '18 at 5:30
$begingroup$
Choudary: (1) In the real number system $infty$ is not a point so the "point of infinite continuity" terminology is not relevant -- particularly with respect to convergence of an improper integral. (2) If you are referring to the fact that $frac{sin x}{x} to 0$ as $x to infty$ which is obviously true, then this alone does not guarantee convergence of the improper integral. In this case the integral happens to converge.
$endgroup$
– RRL
Dec 14 '18 at 5:30
add a comment |
1 Answer
1
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oldest
votes
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No, because you could use the same argument when you replace $sin(x)$ with $frac{x}{x+1}$, but the statement is obviously false in this case.
answered Dec 13 '18 at 15:01
MindlackMindlack
4,760210
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1
$begingroup$
But then why that argument was sufficient for x=0 ?
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– Ajay Choudhary
Dec 13 '18 at 15:18
$begingroup$
Because $0$ is a “finite point”, that is, it has a neighborhood on which bounded functions are integrable.
$endgroup$
– Mindlack
Dec 13 '18 at 15:21
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No, because you could use the same argument when you replace $sin(x)$ with $frac{x}{x+1}$, but the statement is obviously false in this case.
answered Dec 13 '18 at 15:01
MindlackMindlack
4,760210
$endgroup$
1
$begingroup$
But then why that argument was sufficient for x=0 ?
$endgroup$
– Ajay Choudhary
Dec 13 '18 at 15:18
$begingroup$
Because $0$ is a “finite point”, that is, it has a neighborhood on which bounded functions are integrable.
$endgroup$
– Mindlack
Dec 13 '18 at 15:21
add a comment |
$begingroup$
No, because you could use the same argument when you replace $sin(x)$ with $frac{x}{x+1}$, but the statement is obviously false in this case.
answered Dec 13 '18 at 15:01
MindlackMindlack
4,760210
$endgroup$
1
$begingroup$
But then why that argument was sufficient for x=0 ?
$endgroup$
– Ajay Choudhary
Dec 13 '18 at 15:18
$begingroup$
Because $0$ is a “finite point”, that is, it has a neighborhood on which bounded functions are integrable.
$endgroup$
– Mindlack
Dec 13 '18 at 15:21
add a comment |
$begingroup$
No, because you could use the same argument when you replace $sin(x)$ with $frac{x}{x+1}$, but the statement is obviously false in this case.
answered Dec 13 '18 at 15:01
MindlackMindlack
4,760210
$endgroup$
No, because you could use the same argument when you replace $sin(x)$ with $frac{x}{x+1}$, but the statement is obviously false in this case.
answered Dec 13 '18 at 15:01
MindlackMindlack
4,760210
answered Dec 13 '18 at 15:01
MindlackMindlack
4,760210
answered Dec 13 '18 at 15:01
MindlackMindlack
4,760210
answered Dec 13 '18 at 15:01
MindlackMindlack
4,760210
4,760210
1
$begingroup$
But then why that argument was sufficient for x=0 ?
$endgroup$
– Ajay Choudhary
Dec 13 '18 at 15:18
$begingroup$
Because $0$ is a “finite point”, that is, it has a neighborhood on which bounded functions are integrable.
$endgroup$
– Mindlack
Dec 13 '18 at 15:21
add a comment |
1
$begingroup$
But then why that argument was sufficient for x=0 ?
$endgroup$
– Ajay Choudhary
Dec 13 '18 at 15:18
$begingroup$
Because $0$ is a “finite point”, that is, it has a neighborhood on which bounded functions are integrable.
$endgroup$
– Mindlack
Dec 13 '18 at 15:21
1
1
$begingroup$
But then why that argument was sufficient for x=0 ?
$endgroup$
– Ajay Choudhary
Dec 13 '18 at 15:18
$begingroup$
But then why that argument was sufficient for x=0 ?
$endgroup$
– Ajay Choudhary
Dec 13 '18 at 15:18
$begingroup$
Because $0$ is a “finite point”, that is, it has a neighborhood on which bounded functions are integrable.
$endgroup$
– Mindlack
Dec 13 '18 at 15:21
$begingroup$
Because $0$ is a “finite point”, that is, it has a neighborhood on which bounded functions are integrable.
$endgroup$
– Mindlack
Dec 13 '18 at 15:21
add a comment |
$begingroup$
@RRL I've edited the question. Please review it again.
$endgroup$
– Ajay Choudhary
Dec 14 '18 at 4:48
$begingroup$
Choudary: (1) In the real number system $infty$ is not a point so the "point of infinite continuity" terminology is not relevant -- particularly with respect to convergence of an improper integral. (2) If you are referring to the fact that $frac{sin x}{x} to 0$ as $x to infty$ which is obviously true, then this alone does not guarantee convergence of the improper integral. In this case the integral happens to converge.
$endgroup$
– RRL
Dec 14 '18 at 5:30