Calculating the integral $int_0^infty fracarctan xx(ln(x+2))^2 dx$ The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Using Comparison test to determine if $int_0^infty fracarctan x 2+e^x dx$ convergesWhy is the integral $int_0^infty frac arctan x x^3/2, dx$ convergent?Does $int_0^infty fraccos x1+x$ absolutly converge?Convergence test of the following improper integral $int_0^infty frac e^-1/x-1 x^2/3dx$Closed form of $int_0^pi/2 fracarctan^2 (sin^2 theta)sin^2 theta,dtheta$Proving convergence of a generalized integral $int_0^infty fracarctan(x)1+x^adx$Does $int _1^infty left(arctanleft(e^xright)-fracpi 2:right):dx$ converge?Closed form of $int_0^infty (fracarctan(x)x)^ndx$Integral $int_0^infty fracarctan(x^2)x^4+x^2+1dx$Convergence for improper integral $int_0^infty x^re^-x dx$

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Calculating the integral $int_0^infty fracarctan xx(ln(x+2))^2 dx$



The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Using Comparison test to determine if $int_0^infty fracarctan x 2+e^x dx$ convergesWhy is the integral $int_0^infty frac arctan x x^3/2, dx$ convergent?Does $int_0^infty fraccos x1+x$ absolutly converge?Convergence test of the following improper integral $int_0^infty frac e^-1/x-1 x^2/3dx$Closed form of $int_0^pi/2 fracarctan^2 (sin^2 theta)sin^2 theta,dtheta$Proving convergence of a generalized integral $int_0^infty fracarctan(x)1+x^adx$Does $int _1^infty left(arctanleft(e^xright)-fracpi 2:right):dx$ converge?Closed form of $int_0^infty (fracarctan(x)x)^ndx$Integral $int_0^infty fracarctan(x^2)x^4+x^2+1dx$Convergence for improper integral $int_0^infty x^re^-x dx$










0












$begingroup$


I'm trying to prove that the integral of the following function converges:
$$int_0^infty fracarctan xx(ln(x+2))^2 dx$$
it's easy to prove that for each integral starting from $1$ using Dirichlet but i didn't manage to proof for the integral starting at $0$



Can I have a hint?
Thank you










share|cite|improve this question











$endgroup$











  • $begingroup$
    The title is a bit misleading since calculating an indefinite integral is usually much more difficult than proving its convergence only.
    $endgroup$
    – Jack D'Aurizio
    Mar 25 at 15:29















0












$begingroup$


I'm trying to prove that the integral of the following function converges:
$$int_0^infty fracarctan xx(ln(x+2))^2 dx$$
it's easy to prove that for each integral starting from $1$ using Dirichlet but i didn't manage to proof for the integral starting at $0$



Can I have a hint?
Thank you










share|cite|improve this question











$endgroup$











  • $begingroup$
    The title is a bit misleading since calculating an indefinite integral is usually much more difficult than proving its convergence only.
    $endgroup$
    – Jack D'Aurizio
    Mar 25 at 15:29













0












0








0


1



$begingroup$


I'm trying to prove that the integral of the following function converges:
$$int_0^infty fracarctan xx(ln(x+2))^2 dx$$
it's easy to prove that for each integral starting from $1$ using Dirichlet but i didn't manage to proof for the integral starting at $0$



Can I have a hint?
Thank you










share|cite|improve this question











$endgroup$




I'm trying to prove that the integral of the following function converges:
$$int_0^infty fracarctan xx(ln(x+2))^2 dx$$
it's easy to prove that for each integral starting from $1$ using Dirichlet but i didn't manage to proof for the integral starting at $0$



Can I have a hint?
Thank you







integration






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 25 at 9:19









mathreadler

15.5k72263




15.5k72263










asked Mar 25 at 9:18









MyNickMyNick

214




214











  • $begingroup$
    The title is a bit misleading since calculating an indefinite integral is usually much more difficult than proving its convergence only.
    $endgroup$
    – Jack D'Aurizio
    Mar 25 at 15:29
















  • $begingroup$
    The title is a bit misleading since calculating an indefinite integral is usually much more difficult than proving its convergence only.
    $endgroup$
    – Jack D'Aurizio
    Mar 25 at 15:29















$begingroup$
The title is a bit misleading since calculating an indefinite integral is usually much more difficult than proving its convergence only.
$endgroup$
– Jack D'Aurizio
Mar 25 at 15:29




$begingroup$
The title is a bit misleading since calculating an indefinite integral is usually much more difficult than proving its convergence only.
$endgroup$
– Jack D'Aurizio
Mar 25 at 15:29










3 Answers
3






active

oldest

votes


















2












$begingroup$

$$lim_xto 0fracarctan(x)xln^2(x+2)=frac1ln^2(2)$$ and thus it's prolongeable by continuity at $0$.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    you supposed to get 0/0, what am i missing?
    $endgroup$
    – MyNick
    Mar 25 at 10:01






  • 1




    $begingroup$
    The very standard limit $lim_xto 0fracarctan(x)x=1.$ @MyNick
    $endgroup$
    – user657324
    Mar 25 at 10:04



















2












$begingroup$

$int_2^infty frac 1 x(log, x)^2, dx <infty$ as seen easily buy the substitution $y=log, x$. Now just use the fact that $arctan$ is bounded. Integrability near $0$ is clear.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    i dont see how it's clear, can you give anotgher hint please?
    $endgroup$
    – MyNick
    Mar 25 at 10:02






  • 1




    $begingroup$
    $frac arctan, x x to 1$ as $ x to 0$ (by L'Hopital's Rule) so the integrand is bounded on $(0,2)$. Is this the part where you had difficulty?
    $endgroup$
    – Kavi Rama Murthy
    Mar 25 at 10:05











  • $begingroup$
    didn't know that limit. Thank you :)
    $endgroup$
    – MyNick
    Mar 25 at 10:06


















1












$begingroup$

There is no problem around $x=0$ since, by Taylor expansion,
$$fractan ^-1(x)x log ^2(x+2)=frac1log ^2(2)-fracxlog ^3(2)+Oleft(x^2right)$$



I do not finish since Kavi Rama Murthy explained what is going on for infinite values of $x$.






share|cite|improve this answer









$endgroup$













    Your Answer








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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    $$lim_xto 0fracarctan(x)xln^2(x+2)=frac1ln^2(2)$$ and thus it's prolongeable by continuity at $0$.






    share|cite|improve this answer











    $endgroup$












    • $begingroup$
      you supposed to get 0/0, what am i missing?
      $endgroup$
      – MyNick
      Mar 25 at 10:01






    • 1




      $begingroup$
      The very standard limit $lim_xto 0fracarctan(x)x=1.$ @MyNick
      $endgroup$
      – user657324
      Mar 25 at 10:04
















    2












    $begingroup$

    $$lim_xto 0fracarctan(x)xln^2(x+2)=frac1ln^2(2)$$ and thus it's prolongeable by continuity at $0$.






    share|cite|improve this answer











    $endgroup$












    • $begingroup$
      you supposed to get 0/0, what am i missing?
      $endgroup$
      – MyNick
      Mar 25 at 10:01






    • 1




      $begingroup$
      The very standard limit $lim_xto 0fracarctan(x)x=1.$ @MyNick
      $endgroup$
      – user657324
      Mar 25 at 10:04














    2












    2








    2





    $begingroup$

    $$lim_xto 0fracarctan(x)xln^2(x+2)=frac1ln^2(2)$$ and thus it's prolongeable by continuity at $0$.






    share|cite|improve this answer











    $endgroup$



    $$lim_xto 0fracarctan(x)xln^2(x+2)=frac1ln^2(2)$$ and thus it's prolongeable by continuity at $0$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Mar 25 at 9:35

























    answered Mar 25 at 9:30









    user657324user657324

    59510




    59510











    • $begingroup$
      you supposed to get 0/0, what am i missing?
      $endgroup$
      – MyNick
      Mar 25 at 10:01






    • 1




      $begingroup$
      The very standard limit $lim_xto 0fracarctan(x)x=1.$ @MyNick
      $endgroup$
      – user657324
      Mar 25 at 10:04

















    • $begingroup$
      you supposed to get 0/0, what am i missing?
      $endgroup$
      – MyNick
      Mar 25 at 10:01






    • 1




      $begingroup$
      The very standard limit $lim_xto 0fracarctan(x)x=1.$ @MyNick
      $endgroup$
      – user657324
      Mar 25 at 10:04
















    $begingroup$
    you supposed to get 0/0, what am i missing?
    $endgroup$
    – MyNick
    Mar 25 at 10:01




    $begingroup$
    you supposed to get 0/0, what am i missing?
    $endgroup$
    – MyNick
    Mar 25 at 10:01




    1




    1




    $begingroup$
    The very standard limit $lim_xto 0fracarctan(x)x=1.$ @MyNick
    $endgroup$
    – user657324
    Mar 25 at 10:04





    $begingroup$
    The very standard limit $lim_xto 0fracarctan(x)x=1.$ @MyNick
    $endgroup$
    – user657324
    Mar 25 at 10:04












    2












    $begingroup$

    $int_2^infty frac 1 x(log, x)^2, dx <infty$ as seen easily buy the substitution $y=log, x$. Now just use the fact that $arctan$ is bounded. Integrability near $0$ is clear.






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      i dont see how it's clear, can you give anotgher hint please?
      $endgroup$
      – MyNick
      Mar 25 at 10:02






    • 1




      $begingroup$
      $frac arctan, x x to 1$ as $ x to 0$ (by L'Hopital's Rule) so the integrand is bounded on $(0,2)$. Is this the part where you had difficulty?
      $endgroup$
      – Kavi Rama Murthy
      Mar 25 at 10:05











    • $begingroup$
      didn't know that limit. Thank you :)
      $endgroup$
      – MyNick
      Mar 25 at 10:06















    2












    $begingroup$

    $int_2^infty frac 1 x(log, x)^2, dx <infty$ as seen easily buy the substitution $y=log, x$. Now just use the fact that $arctan$ is bounded. Integrability near $0$ is clear.






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      i dont see how it's clear, can you give anotgher hint please?
      $endgroup$
      – MyNick
      Mar 25 at 10:02






    • 1




      $begingroup$
      $frac arctan, x x to 1$ as $ x to 0$ (by L'Hopital's Rule) so the integrand is bounded on $(0,2)$. Is this the part where you had difficulty?
      $endgroup$
      – Kavi Rama Murthy
      Mar 25 at 10:05











    • $begingroup$
      didn't know that limit. Thank you :)
      $endgroup$
      – MyNick
      Mar 25 at 10:06













    2












    2








    2





    $begingroup$

    $int_2^infty frac 1 x(log, x)^2, dx <infty$ as seen easily buy the substitution $y=log, x$. Now just use the fact that $arctan$ is bounded. Integrability near $0$ is clear.






    share|cite|improve this answer









    $endgroup$



    $int_2^infty frac 1 x(log, x)^2, dx <infty$ as seen easily buy the substitution $y=log, x$. Now just use the fact that $arctan$ is bounded. Integrability near $0$ is clear.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Mar 25 at 9:36









    Kavi Rama MurthyKavi Rama Murthy

    74.6k53270




    74.6k53270











    • $begingroup$
      i dont see how it's clear, can you give anotgher hint please?
      $endgroup$
      – MyNick
      Mar 25 at 10:02






    • 1




      $begingroup$
      $frac arctan, x x to 1$ as $ x to 0$ (by L'Hopital's Rule) so the integrand is bounded on $(0,2)$. Is this the part where you had difficulty?
      $endgroup$
      – Kavi Rama Murthy
      Mar 25 at 10:05











    • $begingroup$
      didn't know that limit. Thank you :)
      $endgroup$
      – MyNick
      Mar 25 at 10:06
















    • $begingroup$
      i dont see how it's clear, can you give anotgher hint please?
      $endgroup$
      – MyNick
      Mar 25 at 10:02






    • 1




      $begingroup$
      $frac arctan, x x to 1$ as $ x to 0$ (by L'Hopital's Rule) so the integrand is bounded on $(0,2)$. Is this the part where you had difficulty?
      $endgroup$
      – Kavi Rama Murthy
      Mar 25 at 10:05











    • $begingroup$
      didn't know that limit. Thank you :)
      $endgroup$
      – MyNick
      Mar 25 at 10:06















    $begingroup$
    i dont see how it's clear, can you give anotgher hint please?
    $endgroup$
    – MyNick
    Mar 25 at 10:02




    $begingroup$
    i dont see how it's clear, can you give anotgher hint please?
    $endgroup$
    – MyNick
    Mar 25 at 10:02




    1




    1




    $begingroup$
    $frac arctan, x x to 1$ as $ x to 0$ (by L'Hopital's Rule) so the integrand is bounded on $(0,2)$. Is this the part where you had difficulty?
    $endgroup$
    – Kavi Rama Murthy
    Mar 25 at 10:05





    $begingroup$
    $frac arctan, x x to 1$ as $ x to 0$ (by L'Hopital's Rule) so the integrand is bounded on $(0,2)$. Is this the part where you had difficulty?
    $endgroup$
    – Kavi Rama Murthy
    Mar 25 at 10:05













    $begingroup$
    didn't know that limit. Thank you :)
    $endgroup$
    – MyNick
    Mar 25 at 10:06




    $begingroup$
    didn't know that limit. Thank you :)
    $endgroup$
    – MyNick
    Mar 25 at 10:06











    1












    $begingroup$

    There is no problem around $x=0$ since, by Taylor expansion,
    $$fractan ^-1(x)x log ^2(x+2)=frac1log ^2(2)-fracxlog ^3(2)+Oleft(x^2right)$$



    I do not finish since Kavi Rama Murthy explained what is going on for infinite values of $x$.






    share|cite|improve this answer









    $endgroup$

















      1












      $begingroup$

      There is no problem around $x=0$ since, by Taylor expansion,
      $$fractan ^-1(x)x log ^2(x+2)=frac1log ^2(2)-fracxlog ^3(2)+Oleft(x^2right)$$



      I do not finish since Kavi Rama Murthy explained what is going on for infinite values of $x$.






      share|cite|improve this answer









      $endgroup$















        1












        1








        1





        $begingroup$

        There is no problem around $x=0$ since, by Taylor expansion,
        $$fractan ^-1(x)x log ^2(x+2)=frac1log ^2(2)-fracxlog ^3(2)+Oleft(x^2right)$$



        I do not finish since Kavi Rama Murthy explained what is going on for infinite values of $x$.






        share|cite|improve this answer









        $endgroup$



        There is no problem around $x=0$ since, by Taylor expansion,
        $$fractan ^-1(x)x log ^2(x+2)=frac1log ^2(2)-fracxlog ^3(2)+Oleft(x^2right)$$



        I do not finish since Kavi Rama Murthy explained what is going on for infinite values of $x$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 25 at 9:41









        Claude LeiboviciClaude Leibovici

        125k1158135




        125k1158135



























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