Prove that $int_1^inftyfracsin^2(x)xdx$ diverges. [on hold]Convergence tests for improper multiple integralsShow that the improper integral $int_1^infty f(x) dx$ exists iff $sum_1^infty a_n$ converges.Does the $n$th term test work for integrals?Is $int_1^inftyfracx cos(x)^21+x^3$ convergent or divergent?Show that $int_1^infty fracln xleft(1+x^2right)^lambdamathrm dx$ is convergent only for $lambda > frac12$Prove that the function $f(x)=fraccos^2xsqrtx^4+1$ is improperly integrable on $(0,infty)$.How to prove convergence of $int_0^inftyfracsin(x)xdx$ without evaluating itf is $C^1$ and decreasing then $int_1^inftyf(x)sin(x)dx$ converges.Finding the derivative of $f(x)=int_1^inftyfrace^-xyy^2dy,:::xin(0,infty)$improper integral problem check

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Prove that $int_1^inftyfracsin^2(x)xdx$ diverges. [on hold]


Convergence tests for improper multiple integralsShow that the improper integral $int_1^infty f(x) dx$ exists iff $sum_1^infty a_n$ converges.Does the $n$th term test work for integrals?Is $int_1^inftyfracx cos(x)^21+x^3$ convergent or divergent?Show that $int_1^infty fracln xleft(1+x^2right)^lambdamathrm dx$ is convergent only for $lambda > frac12$Prove that the function $f(x)=fraccos^2xsqrtx^4+1$ is improperly integrable on $(0,infty)$.How to prove convergence of $int_0^inftyfracsin(x)xdx$ without evaluating itf is $C^1$ and decreasing then $int_1^inftyf(x)sin(x)dx$ converges.Finding the derivative of $f(x)=int_1^inftyfrace^-xyy^2dy,:::xin(0,infty)$improper integral problem check













0












$begingroup$


Def: $int_1^infty f(x)dx:= lim_brightarrowinftyint_1^b f(x)dx$.



Motivation: Just a routine homework problem to check the student's understanding of comparison tests for convergence of improper integrals.



I gave my own answer below.










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Z Kane is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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put on hold as off-topic by user21820, Cesareo, choco_addicted, Xander Henderson, Peter Foreman 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." – user21820, Cesareo, choco_addicted, Xander Henderson, Peter Foreman
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  • $begingroup$
    According to Wolfram Alpha, it diverges.
    $endgroup$
    – Viktor Glombik
    2 days ago















0












$begingroup$


Def: $int_1^infty f(x)dx:= lim_brightarrowinftyint_1^b f(x)dx$.



Motivation: Just a routine homework problem to check the student's understanding of comparison tests for convergence of improper integrals.



I gave my own answer below.










share|cite|improve this question









New contributor




Z Kane is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$



put on hold as off-topic by user21820, Cesareo, choco_addicted, Xander Henderson, Peter Foreman 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." – user21820, Cesareo, choco_addicted, Xander Henderson, Peter Foreman
If this question can be reworded to fit the rules in the help center, please edit the question.















  • $begingroup$
    According to Wolfram Alpha, it diverges.
    $endgroup$
    – Viktor Glombik
    2 days ago













0












0








0


0



$begingroup$


Def: $int_1^infty f(x)dx:= lim_brightarrowinftyint_1^b f(x)dx$.



Motivation: Just a routine homework problem to check the student's understanding of comparison tests for convergence of improper integrals.



I gave my own answer below.










share|cite|improve this question









New contributor




Z Kane is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




Def: $int_1^infty f(x)dx:= lim_brightarrowinftyint_1^b f(x)dx$.



Motivation: Just a routine homework problem to check the student's understanding of comparison tests for convergence of improper integrals.



I gave my own answer below.







integration analysis






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New contributor




Z Kane is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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share|cite|improve this question









New contributor




Z Kane is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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share|cite|improve this question




share|cite|improve this question








edited yesterday







Z Kane













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asked 2 days ago









Z KaneZ Kane

285




285




New contributor




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put on hold as off-topic by user21820, Cesareo, choco_addicted, Xander Henderson, Peter Foreman 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." – user21820, Cesareo, choco_addicted, Xander Henderson, Peter Foreman
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 user21820, Cesareo, choco_addicted, Xander Henderson, Peter Foreman 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." – user21820, Cesareo, choco_addicted, Xander Henderson, Peter Foreman
If this question can be reworded to fit the rules in the help center, please edit the question.











  • $begingroup$
    According to Wolfram Alpha, it diverges.
    $endgroup$
    – Viktor Glombik
    2 days ago
















  • $begingroup$
    According to Wolfram Alpha, it diverges.
    $endgroup$
    – Viktor Glombik
    2 days ago















$begingroup$
According to Wolfram Alpha, it diverges.
$endgroup$
– Viktor Glombik
2 days ago




$begingroup$
According to Wolfram Alpha, it diverges.
$endgroup$
– Viktor Glombik
2 days ago










3 Answers
3






active

oldest

votes


















3












$begingroup$

hint



$$sin^2(x)=frac1-cos(2x)2$$



with



$$int_1^+inftyfraccos(2x)2xdx$$ convergent by Dirichlet' test and



$$int_1^+inftyfracdx2x$$
divergent.
the sum is divergent near $+infty$.






share|cite|improve this answer











$endgroup$








  • 1




    $begingroup$
    @ZKane it does not decompose into two divergent improper integrals. The integral of cos(2x)/2x is convergent!
    $endgroup$
    – Tomislav Ostojich
    2 days ago










  • $begingroup$
    @hamam_Abdallah Ok, thank you. I need to learn about Dirichlet's test for improper integrals. Do you happen to know where a proof of this result can be found?
    $endgroup$
    – Z Kane
    2 days ago










  • $begingroup$
    @ZKane Sorry, no. But stackexchange or google could help.
    $endgroup$
    – hamam_Abdallah
    2 days ago


















3












$begingroup$

Hint: $sin^2(x)$ is large when $x = k pi + fracpi 2$, which implies that the integral majorates the harmonic series.



Centered at every number of the form $x_k = k pi + fracpi 2$, we can find an interval $I_k$ of size $2delta$ (independent of $k$) such that $sin^2(x) geq 0.99$ for $x in I_k$.



Then $$int_1^infty fracsin^2 xx geq delta sum_k = 0^infty frac0.99k pi + fracpi2 = infty$$






share|cite|improve this answer









$endgroup$








  • 1




    $begingroup$
    (+1) Same idea but maybe easier, as it is more concrete: $sin^2 xge 1/2$ for $xin [pi/4 +kpi, 3pi/4 + kpi]$.
    $endgroup$
    – peter a g
    2 days ago



















0












$begingroup$

I think of an elementary way to show the crucial convergence of $int_1^infty fraccos(x)xdx$ without referring to the Dirichlet's test (although the idea is similar, i.e., integration by parts).



$$int_1^infty fraccos(t)tdt=fracsin(t)tbigg|_2^infty+int_2^infty fracsin(t)t^2dt$$



And the first term on the right is finite since



$$0leq bigg|fracsin(t)tbigg|leqbigg|frac1tbigg|rightarrow0 text as t rightarrowinfty $$



And the second term on the right converges for a similar reason



$$bigg|fracsin(t)t^2bigg|leqbigg|frac1t^2bigg|quadRightarrowquad int_2^infty bigg|fracsin(t)t^2bigg|dtleqint_2^infty frac1t^2dt=frac12$$






share|cite|improve this answer










New contributor




Z Kane is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$



















    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    hint



    $$sin^2(x)=frac1-cos(2x)2$$



    with



    $$int_1^+inftyfraccos(2x)2xdx$$ convergent by Dirichlet' test and



    $$int_1^+inftyfracdx2x$$
    divergent.
    the sum is divergent near $+infty$.






    share|cite|improve this answer











    $endgroup$








    • 1




      $begingroup$
      @ZKane it does not decompose into two divergent improper integrals. The integral of cos(2x)/2x is convergent!
      $endgroup$
      – Tomislav Ostojich
      2 days ago










    • $begingroup$
      @hamam_Abdallah Ok, thank you. I need to learn about Dirichlet's test for improper integrals. Do you happen to know where a proof of this result can be found?
      $endgroup$
      – Z Kane
      2 days ago










    • $begingroup$
      @ZKane Sorry, no. But stackexchange or google could help.
      $endgroup$
      – hamam_Abdallah
      2 days ago















    3












    $begingroup$

    hint



    $$sin^2(x)=frac1-cos(2x)2$$



    with



    $$int_1^+inftyfraccos(2x)2xdx$$ convergent by Dirichlet' test and



    $$int_1^+inftyfracdx2x$$
    divergent.
    the sum is divergent near $+infty$.






    share|cite|improve this answer











    $endgroup$








    • 1




      $begingroup$
      @ZKane it does not decompose into two divergent improper integrals. The integral of cos(2x)/2x is convergent!
      $endgroup$
      – Tomislav Ostojich
      2 days ago










    • $begingroup$
      @hamam_Abdallah Ok, thank you. I need to learn about Dirichlet's test for improper integrals. Do you happen to know where a proof of this result can be found?
      $endgroup$
      – Z Kane
      2 days ago










    • $begingroup$
      @ZKane Sorry, no. But stackexchange or google could help.
      $endgroup$
      – hamam_Abdallah
      2 days ago













    3












    3








    3





    $begingroup$

    hint



    $$sin^2(x)=frac1-cos(2x)2$$



    with



    $$int_1^+inftyfraccos(2x)2xdx$$ convergent by Dirichlet' test and



    $$int_1^+inftyfracdx2x$$
    divergent.
    the sum is divergent near $+infty$.






    share|cite|improve this answer











    $endgroup$



    hint



    $$sin^2(x)=frac1-cos(2x)2$$



    with



    $$int_1^+inftyfraccos(2x)2xdx$$ convergent by Dirichlet' test and



    $$int_1^+inftyfracdx2x$$
    divergent.
    the sum is divergent near $+infty$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 2 days ago

























    answered 2 days ago









    hamam_Abdallahhamam_Abdallah

    38.2k21634




    38.2k21634







    • 1




      $begingroup$
      @ZKane it does not decompose into two divergent improper integrals. The integral of cos(2x)/2x is convergent!
      $endgroup$
      – Tomislav Ostojich
      2 days ago










    • $begingroup$
      @hamam_Abdallah Ok, thank you. I need to learn about Dirichlet's test for improper integrals. Do you happen to know where a proof of this result can be found?
      $endgroup$
      – Z Kane
      2 days ago










    • $begingroup$
      @ZKane Sorry, no. But stackexchange or google could help.
      $endgroup$
      – hamam_Abdallah
      2 days ago












    • 1




      $begingroup$
      @ZKane it does not decompose into two divergent improper integrals. The integral of cos(2x)/2x is convergent!
      $endgroup$
      – Tomislav Ostojich
      2 days ago










    • $begingroup$
      @hamam_Abdallah Ok, thank you. I need to learn about Dirichlet's test for improper integrals. Do you happen to know where a proof of this result can be found?
      $endgroup$
      – Z Kane
      2 days ago










    • $begingroup$
      @ZKane Sorry, no. But stackexchange or google could help.
      $endgroup$
      – hamam_Abdallah
      2 days ago







    1




    1




    $begingroup$
    @ZKane it does not decompose into two divergent improper integrals. The integral of cos(2x)/2x is convergent!
    $endgroup$
    – Tomislav Ostojich
    2 days ago




    $begingroup$
    @ZKane it does not decompose into two divergent improper integrals. The integral of cos(2x)/2x is convergent!
    $endgroup$
    – Tomislav Ostojich
    2 days ago












    $begingroup$
    @hamam_Abdallah Ok, thank you. I need to learn about Dirichlet's test for improper integrals. Do you happen to know where a proof of this result can be found?
    $endgroup$
    – Z Kane
    2 days ago




    $begingroup$
    @hamam_Abdallah Ok, thank you. I need to learn about Dirichlet's test for improper integrals. Do you happen to know where a proof of this result can be found?
    $endgroup$
    – Z Kane
    2 days ago












    $begingroup$
    @ZKane Sorry, no. But stackexchange or google could help.
    $endgroup$
    – hamam_Abdallah
    2 days ago




    $begingroup$
    @ZKane Sorry, no. But stackexchange or google could help.
    $endgroup$
    – hamam_Abdallah
    2 days ago











    3












    $begingroup$

    Hint: $sin^2(x)$ is large when $x = k pi + fracpi 2$, which implies that the integral majorates the harmonic series.



    Centered at every number of the form $x_k = k pi + fracpi 2$, we can find an interval $I_k$ of size $2delta$ (independent of $k$) such that $sin^2(x) geq 0.99$ for $x in I_k$.



    Then $$int_1^infty fracsin^2 xx geq delta sum_k = 0^infty frac0.99k pi + fracpi2 = infty$$






    share|cite|improve this answer









    $endgroup$








    • 1




      $begingroup$
      (+1) Same idea but maybe easier, as it is more concrete: $sin^2 xge 1/2$ for $xin [pi/4 +kpi, 3pi/4 + kpi]$.
      $endgroup$
      – peter a g
      2 days ago
















    3












    $begingroup$

    Hint: $sin^2(x)$ is large when $x = k pi + fracpi 2$, which implies that the integral majorates the harmonic series.



    Centered at every number of the form $x_k = k pi + fracpi 2$, we can find an interval $I_k$ of size $2delta$ (independent of $k$) such that $sin^2(x) geq 0.99$ for $x in I_k$.



    Then $$int_1^infty fracsin^2 xx geq delta sum_k = 0^infty frac0.99k pi + fracpi2 = infty$$






    share|cite|improve this answer









    $endgroup$








    • 1




      $begingroup$
      (+1) Same idea but maybe easier, as it is more concrete: $sin^2 xge 1/2$ for $xin [pi/4 +kpi, 3pi/4 + kpi]$.
      $endgroup$
      – peter a g
      2 days ago














    3












    3








    3





    $begingroup$

    Hint: $sin^2(x)$ is large when $x = k pi + fracpi 2$, which implies that the integral majorates the harmonic series.



    Centered at every number of the form $x_k = k pi + fracpi 2$, we can find an interval $I_k$ of size $2delta$ (independent of $k$) such that $sin^2(x) geq 0.99$ for $x in I_k$.



    Then $$int_1^infty fracsin^2 xx geq delta sum_k = 0^infty frac0.99k pi + fracpi2 = infty$$






    share|cite|improve this answer









    $endgroup$



    Hint: $sin^2(x)$ is large when $x = k pi + fracpi 2$, which implies that the integral majorates the harmonic series.



    Centered at every number of the form $x_k = k pi + fracpi 2$, we can find an interval $I_k$ of size $2delta$ (independent of $k$) such that $sin^2(x) geq 0.99$ for $x in I_k$.



    Then $$int_1^infty fracsin^2 xx geq delta sum_k = 0^infty frac0.99k pi + fracpi2 = infty$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 2 days ago









    rabotarabota

    14.2k32782




    14.2k32782







    • 1




      $begingroup$
      (+1) Same idea but maybe easier, as it is more concrete: $sin^2 xge 1/2$ for $xin [pi/4 +kpi, 3pi/4 + kpi]$.
      $endgroup$
      – peter a g
      2 days ago













    • 1




      $begingroup$
      (+1) Same idea but maybe easier, as it is more concrete: $sin^2 xge 1/2$ for $xin [pi/4 +kpi, 3pi/4 + kpi]$.
      $endgroup$
      – peter a g
      2 days ago








    1




    1




    $begingroup$
    (+1) Same idea but maybe easier, as it is more concrete: $sin^2 xge 1/2$ for $xin [pi/4 +kpi, 3pi/4 + kpi]$.
    $endgroup$
    – peter a g
    2 days ago





    $begingroup$
    (+1) Same idea but maybe easier, as it is more concrete: $sin^2 xge 1/2$ for $xin [pi/4 +kpi, 3pi/4 + kpi]$.
    $endgroup$
    – peter a g
    2 days ago












    0












    $begingroup$

    I think of an elementary way to show the crucial convergence of $int_1^infty fraccos(x)xdx$ without referring to the Dirichlet's test (although the idea is similar, i.e., integration by parts).



    $$int_1^infty fraccos(t)tdt=fracsin(t)tbigg|_2^infty+int_2^infty fracsin(t)t^2dt$$



    And the first term on the right is finite since



    $$0leq bigg|fracsin(t)tbigg|leqbigg|frac1tbigg|rightarrow0 text as t rightarrowinfty $$



    And the second term on the right converges for a similar reason



    $$bigg|fracsin(t)t^2bigg|leqbigg|frac1t^2bigg|quadRightarrowquad int_2^infty bigg|fracsin(t)t^2bigg|dtleqint_2^infty frac1t^2dt=frac12$$






    share|cite|improve this answer










    New contributor




    Z Kane is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






    $endgroup$

















      0












      $begingroup$

      I think of an elementary way to show the crucial convergence of $int_1^infty fraccos(x)xdx$ without referring to the Dirichlet's test (although the idea is similar, i.e., integration by parts).



      $$int_1^infty fraccos(t)tdt=fracsin(t)tbigg|_2^infty+int_2^infty fracsin(t)t^2dt$$



      And the first term on the right is finite since



      $$0leq bigg|fracsin(t)tbigg|leqbigg|frac1tbigg|rightarrow0 text as t rightarrowinfty $$



      And the second term on the right converges for a similar reason



      $$bigg|fracsin(t)t^2bigg|leqbigg|frac1t^2bigg|quadRightarrowquad int_2^infty bigg|fracsin(t)t^2bigg|dtleqint_2^infty frac1t^2dt=frac12$$






      share|cite|improve this answer










      New contributor




      Z Kane is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      $endgroup$















        0












        0








        0





        $begingroup$

        I think of an elementary way to show the crucial convergence of $int_1^infty fraccos(x)xdx$ without referring to the Dirichlet's test (although the idea is similar, i.e., integration by parts).



        $$int_1^infty fraccos(t)tdt=fracsin(t)tbigg|_2^infty+int_2^infty fracsin(t)t^2dt$$



        And the first term on the right is finite since



        $$0leq bigg|fracsin(t)tbigg|leqbigg|frac1tbigg|rightarrow0 text as t rightarrowinfty $$



        And the second term on the right converges for a similar reason



        $$bigg|fracsin(t)t^2bigg|leqbigg|frac1t^2bigg|quadRightarrowquad int_2^infty bigg|fracsin(t)t^2bigg|dtleqint_2^infty frac1t^2dt=frac12$$






        share|cite|improve this answer










        New contributor




        Z Kane is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.






        $endgroup$



        I think of an elementary way to show the crucial convergence of $int_1^infty fraccos(x)xdx$ without referring to the Dirichlet's test (although the idea is similar, i.e., integration by parts).



        $$int_1^infty fraccos(t)tdt=fracsin(t)tbigg|_2^infty+int_2^infty fracsin(t)t^2dt$$



        And the first term on the right is finite since



        $$0leq bigg|fracsin(t)tbigg|leqbigg|frac1tbigg|rightarrow0 text as t rightarrowinfty $$



        And the second term on the right converges for a similar reason



        $$bigg|fracsin(t)t^2bigg|leqbigg|frac1t^2bigg|quadRightarrowquad int_2^infty bigg|fracsin(t)t^2bigg|dtleqint_2^infty frac1t^2dt=frac12$$







        share|cite|improve this answer










        New contributor




        Z Kane is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.









        share|cite|improve this answer



        share|cite|improve this answer








        edited 2 days ago





















        New contributor




        Z Kane is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.









        answered 2 days ago









        Z KaneZ Kane

        285




        285




        New contributor




        Z Kane is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.





        New contributor





        Z Kane is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.






        Z Kane is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.













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