Determine whether $int_0^infty sin(e^{sin{x}}), dx$ converges or diverges. [closed]
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Determine whether $$int_0^infty sin(e^{sin{x}}), dx$$ converges or diverges.
calculus improper-integrals
edited Dec 6 '18 at 17:48
Brahadeesh
6,18742361
asked Dec 6 '18 at 17:36
muhammathmuhammath
12
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closed as off-topic by Michael Seifert, Decaf-Math, Gibbs, Jack D'Aurizio, Jyrki Lahtonen Dec 6 '18 at 20:31
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." – Michael Seifert, Decaf-Math, Gibbs, Jack D'Aurizio, Jyrki Lahtonen
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Determine whether $$int_0^infty sin(e^{sin{x}}), dx$$ converges or diverges.
calculus improper-integrals
edited Dec 6 '18 at 17:48
Brahadeesh
6,18742361
asked Dec 6 '18 at 17:36
muhammathmuhammath
12
$endgroup$
closed as off-topic by Michael Seifert, Decaf-Math, Gibbs, Jack D'Aurizio, Jyrki Lahtonen Dec 6 '18 at 20:31
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." – Michael Seifert, Decaf-Math, Gibbs, Jack D'Aurizio, Jyrki Lahtonen
If this question can be reworded to fit the rules in the help center, please edit the question.
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You are integrating a positive and $2pi$-periodic function, hence the outcome is fairly obvious.
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– Jack D'Aurizio
Dec 6 '18 at 20:22
add a comment |
$begingroup$
Determine whether $$int_0^infty sin(e^{sin{x}}), dx$$ converges or diverges.
calculus improper-integrals
edited Dec 6 '18 at 17:48
Brahadeesh
6,18742361
asked Dec 6 '18 at 17:36
muhammathmuhammath
12
$endgroup$
Determine whether $$int_0^infty sin(e^{sin{x}}), dx$$ converges or diverges.
calculus improper-integrals
calculus improper-integrals
edited Dec 6 '18 at 17:48
Brahadeesh
6,18742361
asked Dec 6 '18 at 17:36
muhammathmuhammath
12
edited Dec 6 '18 at 17:48
Brahadeesh
6,18742361
asked Dec 6 '18 at 17:36
muhammathmuhammath
12
edited Dec 6 '18 at 17:48
Brahadeesh
6,18742361
edited Dec 6 '18 at 17:48
Brahadeesh
6,18742361
edited Dec 6 '18 at 17:48
Brahadeesh
6,18742361
6,18742361
asked Dec 6 '18 at 17:36
muhammathmuhammath
12
asked Dec 6 '18 at 17:36
muhammathmuhammath
12
asked Dec 6 '18 at 17:36
muhammathmuhammath
12
12
closed as off-topic by Michael Seifert, Decaf-Math, Gibbs, Jack D'Aurizio, Jyrki Lahtonen Dec 6 '18 at 20:31
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." – Michael Seifert, Decaf-Math, Gibbs, Jack D'Aurizio, Jyrki Lahtonen
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Michael Seifert, Decaf-Math, Gibbs, Jack D'Aurizio, Jyrki Lahtonen Dec 6 '18 at 20:31
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." – Michael Seifert, Decaf-Math, Gibbs, Jack D'Aurizio, Jyrki Lahtonen
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
You are integrating a positive and $2pi$-periodic function, hence the outcome is fairly obvious.
$endgroup$
– Jack D'Aurizio
Dec 6 '18 at 20:22
add a comment |
$begingroup$
You are integrating a positive and $2pi$-periodic function, hence the outcome is fairly obvious.
$endgroup$
– Jack D'Aurizio
Dec 6 '18 at 20:22
$begingroup$
You are integrating a positive and $2pi$-periodic function, hence the outcome is fairly obvious.
$endgroup$
– Jack D'Aurizio
Dec 6 '18 at 20:22
$begingroup$
You are integrating a positive and $2pi$-periodic function, hence the outcome is fairly obvious.
$endgroup$
– Jack D'Aurizio
Dec 6 '18 at 20:22
add a comment |
1 Answer
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Since $sin(frac{1}{e})leq sin(e^{sin(x)})$ then the comparison theorem says that
$${int_0^infty sin(frac{1}{e})dx} leq int_0^inftysin(e^{sin(x)})dx.$$
Since the left hand integral diverges to $infty,$ then the right hand integral must diverge.
answered Dec 6 '18 at 17:41
ChickenmancerChickenmancer
3,564724
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Since $sin(frac{1}{e})leq sin(e^{sin(x)})$ then the comparison theorem says that
$${int_0^infty sin(frac{1}{e})dx} leq int_0^inftysin(e^{sin(x)})dx.$$
Since the left hand integral diverges to $infty,$ then the right hand integral must diverge.
answered Dec 6 '18 at 17:41
ChickenmancerChickenmancer
3,564724
$endgroup$
add a comment |
$begingroup$
Since $sin(frac{1}{e})leq sin(e^{sin(x)})$ then the comparison theorem says that
$${int_0^infty sin(frac{1}{e})dx} leq int_0^inftysin(e^{sin(x)})dx.$$
Since the left hand integral diverges to $infty,$ then the right hand integral must diverge.
answered Dec 6 '18 at 17:41
ChickenmancerChickenmancer
3,564724
$endgroup$
add a comment |
$begingroup$
Since $sin(frac{1}{e})leq sin(e^{sin(x)})$ then the comparison theorem says that
$${int_0^infty sin(frac{1}{e})dx} leq int_0^inftysin(e^{sin(x)})dx.$$
Since the left hand integral diverges to $infty,$ then the right hand integral must diverge.
answered Dec 6 '18 at 17:41
ChickenmancerChickenmancer
3,564724
$endgroup$
Since $sin(frac{1}{e})leq sin(e^{sin(x)})$ then the comparison theorem says that
$${int_0^infty sin(frac{1}{e})dx} leq int_0^inftysin(e^{sin(x)})dx.$$
Since the left hand integral diverges to $infty,$ then the right hand integral must diverge.
answered Dec 6 '18 at 17:41
ChickenmancerChickenmancer
3,564724
answered Dec 6 '18 at 17:41
ChickenmancerChickenmancer
3,564724
answered Dec 6 '18 at 17:41
ChickenmancerChickenmancer
3,564724
answered Dec 6 '18 at 17:41
ChickenmancerChickenmancer
3,564724
3,564724
add a comment |
add a comment |
$begingroup$
You are integrating a positive and $2pi$-periodic function, hence the outcome is fairly obvious.
$endgroup$
– Jack D'Aurizio
Dec 6 '18 at 20:22