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.











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closed as off-topic by Michael Seifert, Decaf-Math, Gibbs, Jack D'Aurizio, Jyrki Lahtonen Dec 6 at 20:31


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  • You are integrating a positive and $2pi$-periodic function, hence the outcome is fairly obvious.
    – Jack D'Aurizio
    Dec 6 at 20:22
















-2















Determine whether $$int_0^infty sin(e^{sin{x}}), dx$$ converges or diverges.











share|cite|improve this question















closed as off-topic by Michael Seifert, Decaf-Math, Gibbs, Jack D'Aurizio, Jyrki Lahtonen Dec 6 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.













  • You are integrating a positive and $2pi$-periodic function, hence the outcome is fairly obvious.
    – Jack D'Aurizio
    Dec 6 at 20:22














-2












-2








-2








Determine whether $$int_0^infty sin(e^{sin{x}}), dx$$ converges or diverges.











share|cite|improve this question
















Determine whether $$int_0^infty sin(e^{sin{x}}), dx$$ converges or diverges.








calculus improper-integrals






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edited Dec 6 at 17:48









Brahadeesh

6,10242360




6,10242360










asked Dec 6 at 17:36









muhammath

12




12




closed as off-topic by Michael Seifert, Decaf-Math, Gibbs, Jack D'Aurizio, Jyrki Lahtonen Dec 6 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 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.












  • You are integrating a positive and $2pi$-periodic function, hence the outcome is fairly obvious.
    – Jack D'Aurizio
    Dec 6 at 20:22


















  • You are integrating a positive and $2pi$-periodic function, hence the outcome is fairly obvious.
    – Jack D'Aurizio
    Dec 6 at 20:22
















You are integrating a positive and $2pi$-periodic function, hence the outcome is fairly obvious.
– Jack D'Aurizio
Dec 6 at 20:22




You are integrating a positive and $2pi$-periodic function, hence the outcome is fairly obvious.
– Jack D'Aurizio
Dec 6 at 20:22










1 Answer
1






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4














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.






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    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4














    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.






    share|cite|improve this answer


























      4














      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.






      share|cite|improve this answer
























        4












        4








        4






        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.






        share|cite|improve this answer












        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.







        share|cite|improve this answer












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        answered Dec 6 at 17:41









        Chickenmancer

        3,309723




        3,309723















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