Find the Values of $p$ and $q$ such that $int_0^{infty} x^pln(1+x)^q dx$ Converges or Diverges.












2












$begingroup$


The question is to find the value values of $p$ and $q$ such that the improper integral converges or diverges. My friend indeed found some values of $p$ and $q$ such that it converges, but after many trials, I am still getting divergent for all values. Via L'Hospital, we have the following:



$$lim_{x to 0} x^cln(1+x)^q = infty space text{when} space c < 0$$
$$lim_{x to 0} x^cln(1+x)^q = 0 space text{when} space c > 0$$
$$lim_{x to infty} x^cln(1+x)^q = infty space text{when} space c > 0$$
$$lim_{x to infty} x^cln(1+x)^q = 0 space text{when} space c < 0$$



Thus,
$$int_1^{infty} x^pln(1+x)^q dx = int_1^{infty} x^{p-c}x^cln(1+x)^q dx > int_1^{infty} x^{p-c} dx$$
which diverges when $p > -1$. Here, we set $c = frac{1}{2}(p+1) > 0$. On the other hand,
$$int_0^1 x^pln(1+x)^q dx = int_0^1 x^{p-c}x^cln(1+x)^q dx > int_0^1 x^{p-c} dx$$
which diverges when $p < -1$. Here, we set $c = frac{1}{2}(p+1) < 0$. Thus, the integral diverges whenever $p ne –1$. Then, let $p = -1$.



$$int_0^{infty} frac{1}{x}ln(1+x)^q dx = int_0^1 frac{1}{x}ln(1+x)^q dx + int_1^{infty} frac{1}{x}ln(1+x)^q dx $$
$$ ge int_0^1 frac{1}{x+1}ln(1+x)^q dx + int_1^{infty} frac{1}{x+1}ln(1+x)^q dx $$



The first summand diverges when $q < -1$ and the second diverges when $q > -1$. Finally, let $ p = -1$ and $q = -1$. The integral $int_0^{infty} frac{1}{xln(1+x)} dx ≥ int_0^{infty} frac{1}{(x+1)ln(1+x)} dx$ diverges on both ends. Did I make a mistake?






real-analysis improper-integrals







share|cite|improve this question









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    2












    $begingroup$


    The question is to find the value values of $p$ and $q$ such that the improper integral converges or diverges. My friend indeed found some values of $p$ and $q$ such that it converges, but after many trials, I am still getting divergent for all values. Via L'Hospital, we have the following:



    $$lim_{x to 0} x^cln(1+x)^q = infty space text{when} space c < 0$$
    $$lim_{x to 0} x^cln(1+x)^q = 0 space text{when} space c > 0$$
    $$lim_{x to infty} x^cln(1+x)^q = infty space text{when} space c > 0$$
    $$lim_{x to infty} x^cln(1+x)^q = 0 space text{when} space c < 0$$



    Thus,
    $$int_1^{infty} x^pln(1+x)^q dx = int_1^{infty} x^{p-c}x^cln(1+x)^q dx > int_1^{infty} x^{p-c} dx$$
    which diverges when $p > -1$. Here, we set $c = frac{1}{2}(p+1) > 0$. On the other hand,
    $$int_0^1 x^pln(1+x)^q dx = int_0^1 x^{p-c}x^cln(1+x)^q dx > int_0^1 x^{p-c} dx$$
    which diverges when $p < -1$. Here, we set $c = frac{1}{2}(p+1) < 0$. Thus, the integral diverges whenever $p ne –1$. Then, let $p = -1$.



    $$int_0^{infty} frac{1}{x}ln(1+x)^q dx = int_0^1 frac{1}{x}ln(1+x)^q dx + int_1^{infty} frac{1}{x}ln(1+x)^q dx $$
    $$ ge int_0^1 frac{1}{x+1}ln(1+x)^q dx + int_1^{infty} frac{1}{x+1}ln(1+x)^q dx $$



    The first summand diverges when $q < -1$ and the second diverges when $q > -1$. Finally, let $ p = -1$ and $q = -1$. The integral $int_0^{infty} frac{1}{xln(1+x)} dx ≥ int_0^{infty} frac{1}{(x+1)ln(1+x)} dx$ diverges on both ends. Did I make a mistake?






    real-analysis improper-integrals







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      The question is to find the value values of $p$ and $q$ such that the improper integral converges or diverges. My friend indeed found some values of $p$ and $q$ such that it converges, but after many trials, I am still getting divergent for all values. Via L'Hospital, we have the following:



      $$lim_{x to 0} x^cln(1+x)^q = infty space text{when} space c < 0$$
      $$lim_{x to 0} x^cln(1+x)^q = 0 space text{when} space c > 0$$
      $$lim_{x to infty} x^cln(1+x)^q = infty space text{when} space c > 0$$
      $$lim_{x to infty} x^cln(1+x)^q = 0 space text{when} space c < 0$$



      Thus,
      $$int_1^{infty} x^pln(1+x)^q dx = int_1^{infty} x^{p-c}x^cln(1+x)^q dx > int_1^{infty} x^{p-c} dx$$
      which diverges when $p > -1$. Here, we set $c = frac{1}{2}(p+1) > 0$. On the other hand,
      $$int_0^1 x^pln(1+x)^q dx = int_0^1 x^{p-c}x^cln(1+x)^q dx > int_0^1 x^{p-c} dx$$
      which diverges when $p < -1$. Here, we set $c = frac{1}{2}(p+1) < 0$. Thus, the integral diverges whenever $p ne –1$. Then, let $p = -1$.



      $$int_0^{infty} frac{1}{x}ln(1+x)^q dx = int_0^1 frac{1}{x}ln(1+x)^q dx + int_1^{infty} frac{1}{x}ln(1+x)^q dx $$
      $$ ge int_0^1 frac{1}{x+1}ln(1+x)^q dx + int_1^{infty} frac{1}{x+1}ln(1+x)^q dx $$



      The first summand diverges when $q < -1$ and the second diverges when $q > -1$. Finally, let $ p = -1$ and $q = -1$. The integral $int_0^{infty} frac{1}{xln(1+x)} dx ≥ int_0^{infty} frac{1}{(x+1)ln(1+x)} dx$ diverges on both ends. Did I make a mistake?






      real-analysis improper-integrals







      share|cite|improve this question









      $endgroup$




      The question is to find the value values of $p$ and $q$ such that the improper integral converges or diverges. My friend indeed found some values of $p$ and $q$ such that it converges, but after many trials, I am still getting divergent for all values. Via L'Hospital, we have the following:



      $$lim_{x to 0} x^cln(1+x)^q = infty space text{when} space c < 0$$
      $$lim_{x to 0} x^cln(1+x)^q = 0 space text{when} space c > 0$$
      $$lim_{x to infty} x^cln(1+x)^q = infty space text{when} space c > 0$$
      $$lim_{x to infty} x^cln(1+x)^q = 0 space text{when} space c < 0$$



      Thus,
      $$int_1^{infty} x^pln(1+x)^q dx = int_1^{infty} x^{p-c}x^cln(1+x)^q dx > int_1^{infty} x^{p-c} dx$$
      which diverges when $p > -1$. Here, we set $c = frac{1}{2}(p+1) > 0$. On the other hand,
      $$int_0^1 x^pln(1+x)^q dx = int_0^1 x^{p-c}x^cln(1+x)^q dx > int_0^1 x^{p-c} dx$$
      which diverges when $p < -1$. Here, we set $c = frac{1}{2}(p+1) < 0$. Thus, the integral diverges whenever $p ne –1$. Then, let $p = -1$.



      $$int_0^{infty} frac{1}{x}ln(1+x)^q dx = int_0^1 frac{1}{x}ln(1+x)^q dx + int_1^{infty} frac{1}{x}ln(1+x)^q dx $$
      $$ ge int_0^1 frac{1}{x+1}ln(1+x)^q dx + int_1^{infty} frac{1}{x+1}ln(1+x)^q dx $$



      The first summand diverges when $q < -1$ and the second diverges when $q > -1$. Finally, let $ p = -1$ and $q = -1$. The integral $int_0^{infty} frac{1}{xln(1+x)} dx ≥ int_0^{infty} frac{1}{(x+1)ln(1+x)} dx$ diverges on both ends. Did I make a mistake?






      real-analysis improper-integrals




      real-analysis improper-integrals






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      asked Oct 19 '14 at 3:36









      Andy TamAndy Tam

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          The result seems correct, but check the limits for $xto 0$.



          $int_1^{+infty} x^p log^q (1+x)dx$ diverges if $p>-1$,
          $int_0^{1} x^p log^q (1+x)dx$ diverges if $p<-1$, independently of $q$.



          It remains $p=-1$.



          $int_1^{+infty} x^{-1}log^q (1+x)dx$ diverges if $qgeq -1$. Note that for small $x$ the behavior of $log(1+x)sim x$, hence

          $$int_0^{1} x^{-1}log^q (1+x)dxsim int_0^{1} x^{-1}x^q dx,$$
          which diverges for any $qleq 0$.






          share|cite|improve this answer









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            $begingroup$

            The result seems correct, but check the limits for $xto 0$.



            $int_1^{+infty} x^p log^q (1+x)dx$ diverges if $p>-1$,
            $int_0^{1} x^p log^q (1+x)dx$ diverges if $p<-1$, independently of $q$.



            It remains $p=-1$.



            $int_1^{+infty} x^{-1}log^q (1+x)dx$ diverges if $qgeq -1$. Note that for small $x$ the behavior of $log(1+x)sim x$, hence

            $$int_0^{1} x^{-1}log^q (1+x)dxsim int_0^{1} x^{-1}x^q dx,$$
            which diverges for any $qleq 0$.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              The result seems correct, but check the limits for $xto 0$.



              $int_1^{+infty} x^p log^q (1+x)dx$ diverges if $p>-1$,
              $int_0^{1} x^p log^q (1+x)dx$ diverges if $p<-1$, independently of $q$.



              It remains $p=-1$.



              $int_1^{+infty} x^{-1}log^q (1+x)dx$ diverges if $qgeq -1$. Note that for small $x$ the behavior of $log(1+x)sim x$, hence

              $$int_0^{1} x^{-1}log^q (1+x)dxsim int_0^{1} x^{-1}x^q dx,$$
              which diverges for any $qleq 0$.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                The result seems correct, but check the limits for $xto 0$.



                $int_1^{+infty} x^p log^q (1+x)dx$ diverges if $p>-1$,
                $int_0^{1} x^p log^q (1+x)dx$ diverges if $p<-1$, independently of $q$.



                It remains $p=-1$.



                $int_1^{+infty} x^{-1}log^q (1+x)dx$ diverges if $qgeq -1$. Note that for small $x$ the behavior of $log(1+x)sim x$, hence

                $$int_0^{1} x^{-1}log^q (1+x)dxsim int_0^{1} x^{-1}x^q dx,$$
                which diverges for any $qleq 0$.






                share|cite|improve this answer









                $endgroup$



                The result seems correct, but check the limits for $xto 0$.



                $int_1^{+infty} x^p log^q (1+x)dx$ diverges if $p>-1$,
                $int_0^{1} x^p log^q (1+x)dx$ diverges if $p<-1$, independently of $q$.



                It remains $p=-1$.



                $int_1^{+infty} x^{-1}log^q (1+x)dx$ diverges if $qgeq -1$. Note that for small $x$ the behavior of $log(1+x)sim x$, hence

                $$int_0^{1} x^{-1}log^q (1+x)dxsim int_0^{1} x^{-1}x^q dx,$$
                which diverges for any $qleq 0$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Oct 19 '14 at 3:56









                MillyMilly

                2,628612




                2,628612






























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