Finding convergence (or divergence) of $int_0^{1/e} frac{dx}{x ln^2 x}$.












1















Compute or prove the divergence of $$int_0^{1/e} frac{dx}{x ln^2 x}$$




I have tried to find the usual derivative taking integration from α..(0<α<1/e)to 1/e..it turned out be log0..but the given solution is 1.






calculus integration convergence definite-integrals







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1















Compute or prove the divergence of $$int_0^{1/e} frac{dx}{x ln^2 x}$$




I have tried to find the usual derivative taking integration from α..(0<α<1/e)to 1/e..it turned out be log0..but the given solution is 1.






calculus integration convergence definite-integrals







share|cite|improve this question
























  • Please use MathJax in future. It is a type of $LaTeX$. A tutorial on it from this site is readily available online.
    – Shaun
    Nov 26 at 16:20














1












1








1








Compute or prove the divergence of $$int_0^{1/e} frac{dx}{x ln^2 x}$$




I have tried to find the usual derivative taking integration from α..(0<α<1/e)to 1/e..it turned out be log0..but the given solution is 1.






calculus integration convergence definite-integrals







share|cite|improve this question
















Compute or prove the divergence of $$int_0^{1/e} frac{dx}{x ln^2 x}$$




I have tried to find the usual derivative taking integration from α..(0<α<1/e)to 1/e..it turned out be log0..but the given solution is 1.






calculus integration convergence definite-integrals




calculus integration convergence definite-integrals






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share|cite|improve this question













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edited Nov 26 at 22:24









Key Flex

7,46841232




7,46841232










asked Nov 26 at 16:11









Kashmira

463




463












  • Please use MathJax in future. It is a type of $LaTeX$. A tutorial on it from this site is readily available online.
    – Shaun
    Nov 26 at 16:20


















  • Please use MathJax in future. It is a type of $LaTeX$. A tutorial on it from this site is readily available online.
    – Shaun
    Nov 26 at 16:20
















Please use MathJax in future. It is a type of $LaTeX$. A tutorial on it from this site is readily available online.
– Shaun
Nov 26 at 16:20




Please use MathJax in future. It is a type of $LaTeX$. A tutorial on it from this site is readily available online.
– Shaun
Nov 26 at 16:20










2 Answers
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$$int_0^{frac1e}dfrac{1}{xln^2x} dx$$



Use u-substitution: $u=ln ximplies du=dfrac{dx}{x}$



first compute without boundaries
$$=intdfrac{1}{u^2} du=dfrac{u^{-2+1}}{-2+1}=-left[dfrac{1}{ln x}right]_0^{frac1e}=1$$






share|cite|improve this answer

















  • 1




    Pardon me if this is a very stupid question, but wouldn't the integral diverge because of the fact that $$lim_{xto0^+}frac1{xlog^2x}=+infty$$
    – clathratus
    Nov 26 at 17:13










  • @clathratus Yes, it diverges because it has discontinuity at $x=0$ which gives improper bound
    – Key Flex
    Nov 26 at 17:16












  • So why does your answer say that $$int_0^{frac1e}frac{dx}{xlog^2x}=1$$
    – clathratus
    Nov 26 at 17:19










  • @clathratus I am evaluating the limit for $dfrac{1}{ln x}$ but not for $dfrac{1}{xln^2x}$
    – Key Flex
    Nov 26 at 17:22








  • 1




    Your answer to the first comment is not correct. The question in the comment is: doesn't the integral diverge since the integrand blows up at $0$? Answer: No
    – zhw.
    Nov 26 at 17:46



















0














HINT



Substitute $u = ln x$ with $du = dx/x$ to get $int u^{-2}du$...






share|cite|improve this answer





















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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2














    $$int_0^{frac1e}dfrac{1}{xln^2x} dx$$



    Use u-substitution: $u=ln ximplies du=dfrac{dx}{x}$



    first compute without boundaries
    $$=intdfrac{1}{u^2} du=dfrac{u^{-2+1}}{-2+1}=-left[dfrac{1}{ln x}right]_0^{frac1e}=1$$






    share|cite|improve this answer

















    • 1




      Pardon me if this is a very stupid question, but wouldn't the integral diverge because of the fact that $$lim_{xto0^+}frac1{xlog^2x}=+infty$$
      – clathratus
      Nov 26 at 17:13










    • @clathratus Yes, it diverges because it has discontinuity at $x=0$ which gives improper bound
      – Key Flex
      Nov 26 at 17:16












    • So why does your answer say that $$int_0^{frac1e}frac{dx}{xlog^2x}=1$$
      – clathratus
      Nov 26 at 17:19










    • @clathratus I am evaluating the limit for $dfrac{1}{ln x}$ but not for $dfrac{1}{xln^2x}$
      – Key Flex
      Nov 26 at 17:22








    • 1




      Your answer to the first comment is not correct. The question in the comment is: doesn't the integral diverge since the integrand blows up at $0$? Answer: No
      – zhw.
      Nov 26 at 17:46
















    2














    $$int_0^{frac1e}dfrac{1}{xln^2x} dx$$



    Use u-substitution: $u=ln ximplies du=dfrac{dx}{x}$



    first compute without boundaries
    $$=intdfrac{1}{u^2} du=dfrac{u^{-2+1}}{-2+1}=-left[dfrac{1}{ln x}right]_0^{frac1e}=1$$






    share|cite|improve this answer

















    • 1




      Pardon me if this is a very stupid question, but wouldn't the integral diverge because of the fact that $$lim_{xto0^+}frac1{xlog^2x}=+infty$$
      – clathratus
      Nov 26 at 17:13










    • @clathratus Yes, it diverges because it has discontinuity at $x=0$ which gives improper bound
      – Key Flex
      Nov 26 at 17:16












    • So why does your answer say that $$int_0^{frac1e}frac{dx}{xlog^2x}=1$$
      – clathratus
      Nov 26 at 17:19










    • @clathratus I am evaluating the limit for $dfrac{1}{ln x}$ but not for $dfrac{1}{xln^2x}$
      – Key Flex
      Nov 26 at 17:22








    • 1




      Your answer to the first comment is not correct. The question in the comment is: doesn't the integral diverge since the integrand blows up at $0$? Answer: No
      – zhw.
      Nov 26 at 17:46














    2












    2








    2






    $$int_0^{frac1e}dfrac{1}{xln^2x} dx$$



    Use u-substitution: $u=ln ximplies du=dfrac{dx}{x}$



    first compute without boundaries
    $$=intdfrac{1}{u^2} du=dfrac{u^{-2+1}}{-2+1}=-left[dfrac{1}{ln x}right]_0^{frac1e}=1$$






    share|cite|improve this answer












    $$int_0^{frac1e}dfrac{1}{xln^2x} dx$$



    Use u-substitution: $u=ln ximplies du=dfrac{dx}{x}$



    first compute without boundaries
    $$=intdfrac{1}{u^2} du=dfrac{u^{-2+1}}{-2+1}=-left[dfrac{1}{ln x}right]_0^{frac1e}=1$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 26 at 16:17









    Key Flex

    7,46841232




    7,46841232








    • 1




      Pardon me if this is a very stupid question, but wouldn't the integral diverge because of the fact that $$lim_{xto0^+}frac1{xlog^2x}=+infty$$
      – clathratus
      Nov 26 at 17:13










    • @clathratus Yes, it diverges because it has discontinuity at $x=0$ which gives improper bound
      – Key Flex
      Nov 26 at 17:16












    • So why does your answer say that $$int_0^{frac1e}frac{dx}{xlog^2x}=1$$
      – clathratus
      Nov 26 at 17:19










    • @clathratus I am evaluating the limit for $dfrac{1}{ln x}$ but not for $dfrac{1}{xln^2x}$
      – Key Flex
      Nov 26 at 17:22








    • 1




      Your answer to the first comment is not correct. The question in the comment is: doesn't the integral diverge since the integrand blows up at $0$? Answer: No
      – zhw.
      Nov 26 at 17:46














    • 1




      Pardon me if this is a very stupid question, but wouldn't the integral diverge because of the fact that $$lim_{xto0^+}frac1{xlog^2x}=+infty$$
      – clathratus
      Nov 26 at 17:13










    • @clathratus Yes, it diverges because it has discontinuity at $x=0$ which gives improper bound
      – Key Flex
      Nov 26 at 17:16












    • So why does your answer say that $$int_0^{frac1e}frac{dx}{xlog^2x}=1$$
      – clathratus
      Nov 26 at 17:19










    • @clathratus I am evaluating the limit for $dfrac{1}{ln x}$ but not for $dfrac{1}{xln^2x}$
      – Key Flex
      Nov 26 at 17:22








    • 1




      Your answer to the first comment is not correct. The question in the comment is: doesn't the integral diverge since the integrand blows up at $0$? Answer: No
      – zhw.
      Nov 26 at 17:46








    1




    1




    Pardon me if this is a very stupid question, but wouldn't the integral diverge because of the fact that $$lim_{xto0^+}frac1{xlog^2x}=+infty$$
    – clathratus
    Nov 26 at 17:13




    Pardon me if this is a very stupid question, but wouldn't the integral diverge because of the fact that $$lim_{xto0^+}frac1{xlog^2x}=+infty$$
    – clathratus
    Nov 26 at 17:13












    @clathratus Yes, it diverges because it has discontinuity at $x=0$ which gives improper bound
    – Key Flex
    Nov 26 at 17:16






    @clathratus Yes, it diverges because it has discontinuity at $x=0$ which gives improper bound
    – Key Flex
    Nov 26 at 17:16














    So why does your answer say that $$int_0^{frac1e}frac{dx}{xlog^2x}=1$$
    – clathratus
    Nov 26 at 17:19




    So why does your answer say that $$int_0^{frac1e}frac{dx}{xlog^2x}=1$$
    – clathratus
    Nov 26 at 17:19












    @clathratus I am evaluating the limit for $dfrac{1}{ln x}$ but not for $dfrac{1}{xln^2x}$
    – Key Flex
    Nov 26 at 17:22






    @clathratus I am evaluating the limit for $dfrac{1}{ln x}$ but not for $dfrac{1}{xln^2x}$
    – Key Flex
    Nov 26 at 17:22






    1




    1




    Your answer to the first comment is not correct. The question in the comment is: doesn't the integral diverge since the integrand blows up at $0$? Answer: No
    – zhw.
    Nov 26 at 17:46




    Your answer to the first comment is not correct. The question in the comment is: doesn't the integral diverge since the integrand blows up at $0$? Answer: No
    – zhw.
    Nov 26 at 17:46











    0














    HINT



    Substitute $u = ln x$ with $du = dx/x$ to get $int u^{-2}du$...






    share|cite|improve this answer


























      0














      HINT



      Substitute $u = ln x$ with $du = dx/x$ to get $int u^{-2}du$...






      share|cite|improve this answer
























        0












        0








        0






        HINT



        Substitute $u = ln x$ with $du = dx/x$ to get $int u^{-2}du$...






        share|cite|improve this answer












        HINT



        Substitute $u = ln x$ with $du = dx/x$ to get $int u^{-2}du$...







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 26 at 16:13









        gt6989b

        33k22452




        33k22452






























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