compute $lim_{n to infty} int_{[0,1]} frac{(ln x)^n}{sqrt{1-x^2}} , dx$












0












$begingroup$


$I = lim_{n to infty}I_n = lim_{n to infty} int_{[0,1]} frac{(ln x)^n}{sqrt{1-x^2}} , dx$



since we have $(ln x)^n leq x , forall x in [0,1], ngeq 1 $ then I could show that $I_n leq 1$



I wanted then to use the DCT but $ lim_{n to infty} (ln x)^n$ doesn't exist for positive $x$ less than $1$.



I'm a bit lost, anyone can give me hints or tell me what to do ?






integration limits measure-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    $ln x<0$ in $(0,1)$.
    $endgroup$
    – A.Γ.
    Dec 17 '18 at 13:41










  • $begingroup$
    @A.Γ. but that just implies that the L-1 norm is zero. not the limit itself
    $endgroup$
    – rapidracim
    Dec 17 '18 at 13:47
















0












$begingroup$


$I = lim_{n to infty}I_n = lim_{n to infty} int_{[0,1]} frac{(ln x)^n}{sqrt{1-x^2}} , dx$



since we have $(ln x)^n leq x , forall x in [0,1], ngeq 1 $ then I could show that $I_n leq 1$



I wanted then to use the DCT but $ lim_{n to infty} (ln x)^n$ doesn't exist for positive $x$ less than $1$.



I'm a bit lost, anyone can give me hints or tell me what to do ?






integration limits measure-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    $ln x<0$ in $(0,1)$.
    $endgroup$
    – A.Γ.
    Dec 17 '18 at 13:41










  • $begingroup$
    @A.Γ. but that just implies that the L-1 norm is zero. not the limit itself
    $endgroup$
    – rapidracim
    Dec 17 '18 at 13:47














0












0








0





$begingroup$


$I = lim_{n to infty}I_n = lim_{n to infty} int_{[0,1]} frac{(ln x)^n}{sqrt{1-x^2}} , dx$



since we have $(ln x)^n leq x , forall x in [0,1], ngeq 1 $ then I could show that $I_n leq 1$



I wanted then to use the DCT but $ lim_{n to infty} (ln x)^n$ doesn't exist for positive $x$ less than $1$.



I'm a bit lost, anyone can give me hints or tell me what to do ?






integration limits measure-theory







share|cite|improve this question









$endgroup$




$I = lim_{n to infty}I_n = lim_{n to infty} int_{[0,1]} frac{(ln x)^n}{sqrt{1-x^2}} , dx$



since we have $(ln x)^n leq x , forall x in [0,1], ngeq 1 $ then I could show that $I_n leq 1$



I wanted then to use the DCT but $ lim_{n to infty} (ln x)^n$ doesn't exist for positive $x$ less than $1$.



I'm a bit lost, anyone can give me hints or tell me what to do ?






integration limits measure-theory




integration limits measure-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 17 '18 at 13:35









rapidracimrapidracim

1,7241419




1,7241419












  • $begingroup$
    $ln x<0$ in $(0,1)$.
    $endgroup$
    – A.Γ.
    Dec 17 '18 at 13:41










  • $begingroup$
    @A.Γ. but that just implies that the L-1 norm is zero. not the limit itself
    $endgroup$
    – rapidracim
    Dec 17 '18 at 13:47


















  • $begingroup$
    $ln x<0$ in $(0,1)$.
    $endgroup$
    – A.Γ.
    Dec 17 '18 at 13:41










  • $begingroup$
    @A.Γ. but that just implies that the L-1 norm is zero. not the limit itself
    $endgroup$
    – rapidracim
    Dec 17 '18 at 13:47
















$begingroup$
$ln x<0$ in $(0,1)$.
$endgroup$
– A.Γ.
Dec 17 '18 at 13:41




$begingroup$
$ln x<0$ in $(0,1)$.
$endgroup$
– A.Γ.
Dec 17 '18 at 13:41












$begingroup$
@A.Γ. but that just implies that the L-1 norm is zero. not the limit itself
$endgroup$
– rapidracim
Dec 17 '18 at 13:47




$begingroup$
@A.Γ. but that just implies that the L-1 norm is zero. not the limit itself
$endgroup$
– rapidracim
Dec 17 '18 at 13:47










3 Answers
3






active

oldest

votes


















6












$begingroup$

Note that $-log xgeq 1-xgeq0$ on $[0,1]$. Thus your assertion that $I_n leq 1$ is wrong. We may write
$$
(-1)^nI_n = int_0^{frac{1}{e}}frac{(-log x)^n}{sqrt{1-x^2}}dx + int_{frac{1}{e}}^1frac{(-log x)^n}{sqrt{1-x^2}}dx .
$$ We can observe that for $xin (0,frac{1}{e})$, it holds $$
1leq (-log x)^n uparrow infty,
$$ and
$$
1geq (-log x)^n downarrow 0
$$ for $xin (frac{1}{e},1]$. By Lebesgue's dominated convergence theorem and monotone convergence theorem, it follows that
$$
lim_{ntoinfty} (-1)^n I_n = infty.
$$






share|cite|improve this answer











$endgroup$





















    4












    $begingroup$

    A shorter proof uses $int_0^1frac{(-ln x)^n dx}{sqrt{1-x^2}}geint_0^1(-ln x)^n dx=int_0^infty y^n e^{-y}dy=n!$.






    share|cite|improve this answer









    $endgroup$





















      2












      $begingroup$

      Graphing the initial few functions for integer-valued n, it is clear that the integral will change signs repeatedly, and be negative for odd n and positive for even n. As a result, I would suggest that the limit does not exist unless it converges at 0, which seems unlikely when plotting the graph for high n-values






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        I agree with you that either the limit is $0$ or the limit does not exists, but unfortunately, plotting a graph is not a proof.
        $endgroup$
        – nicomezi
        Dec 17 '18 at 13:49










      • $begingroup$
        @nicomezi That's an entirely valid point. I'll have a think about how to phrase a proper proof, but I think a much better written proof than I could come up with has been supplied in another answer.
        $endgroup$
        – M.M.
        Dec 17 '18 at 13:55











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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      6












      $begingroup$

      Note that $-log xgeq 1-xgeq0$ on $[0,1]$. Thus your assertion that $I_n leq 1$ is wrong. We may write
      $$
      (-1)^nI_n = int_0^{frac{1}{e}}frac{(-log x)^n}{sqrt{1-x^2}}dx + int_{frac{1}{e}}^1frac{(-log x)^n}{sqrt{1-x^2}}dx .
      $$       We can observe that for $xin (0,frac{1}{e})$, it holds $$
      1leq (-log x)^n uparrow infty,
      $$       and
      $$
      1geq (-log x)^n downarrow 0
      $$       for $xin (frac{1}{e},1]$. By Lebesgue's dominated convergence theorem and monotone convergence theorem, it follows that
      $$
      lim_{ntoinfty} (-1)^n I_n = infty.
      $$






      share|cite|improve this answer











      $endgroup$


















        6












        $begingroup$

        Note that $-log xgeq 1-xgeq0$ on $[0,1]$. Thus your assertion that $I_n leq 1$ is wrong. We may write
        $$
        (-1)^nI_n = int_0^{frac{1}{e}}frac{(-log x)^n}{sqrt{1-x^2}}dx + int_{frac{1}{e}}^1frac{(-log x)^n}{sqrt{1-x^2}}dx .
        $$         We can observe that for $xin (0,frac{1}{e})$, it holds $$
        1leq (-log x)^n uparrow infty,
        $$         and
        $$
        1geq (-log x)^n downarrow 0
        $$         for $xin (frac{1}{e},1]$. By Lebesgue's dominated convergence theorem and monotone convergence theorem, it follows that
        $$
        lim_{ntoinfty} (-1)^n I_n = infty.
        $$






        share|cite|improve this answer











        $endgroup$
















          6












          6








          6





          $begingroup$

          Note that $-log xgeq 1-xgeq0$ on $[0,1]$. Thus your assertion that $I_n leq 1$ is wrong. We may write
          $$
          (-1)^nI_n = int_0^{frac{1}{e}}frac{(-log x)^n}{sqrt{1-x^2}}dx + int_{frac{1}{e}}^1frac{(-log x)^n}{sqrt{1-x^2}}dx .
          $$           We can observe that for $xin (0,frac{1}{e})$, it holds $$
          1leq (-log x)^n uparrow infty,
          $$           and
          $$
          1geq (-log x)^n downarrow 0
          $$           for $xin (frac{1}{e},1]$. By Lebesgue's dominated convergence theorem and monotone convergence theorem, it follows that
          $$
          lim_{ntoinfty} (-1)^n I_n = infty.
          $$






          share|cite|improve this answer











          $endgroup$



          Note that $-log xgeq 1-xgeq0$ on $[0,1]$. Thus your assertion that $I_n leq 1$ is wrong. We may write
          $$
          (-1)^nI_n = int_0^{frac{1}{e}}frac{(-log x)^n}{sqrt{1-x^2}}dx + int_{frac{1}{e}}^1frac{(-log x)^n}{sqrt{1-x^2}}dx .
          $$           We can observe that for $xin (0,frac{1}{e})$, it holds $$
          1leq (-log x)^n uparrow infty,
          $$           and
          $$
          1geq (-log x)^n downarrow 0
          $$           for $xin (frac{1}{e},1]$. By Lebesgue's dominated convergence theorem and monotone convergence theorem, it follows that
          $$
          lim_{ntoinfty} (-1)^n I_n = infty.
          $$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 17 '18 at 13:54

























          answered Dec 17 '18 at 13:49









          SongSong

          18.3k21549




          18.3k21549























              4












              $begingroup$

              A shorter proof uses $int_0^1frac{(-ln x)^n dx}{sqrt{1-x^2}}geint_0^1(-ln x)^n dx=int_0^infty y^n e^{-y}dy=n!$.






              share|cite|improve this answer









              $endgroup$


















                4












                $begingroup$

                A shorter proof uses $int_0^1frac{(-ln x)^n dx}{sqrt{1-x^2}}geint_0^1(-ln x)^n dx=int_0^infty y^n e^{-y}dy=n!$.






                share|cite|improve this answer









                $endgroup$
















                  4












                  4








                  4





                  $begingroup$

                  A shorter proof uses $int_0^1frac{(-ln x)^n dx}{sqrt{1-x^2}}geint_0^1(-ln x)^n dx=int_0^infty y^n e^{-y}dy=n!$.






                  share|cite|improve this answer









                  $endgroup$



                  A shorter proof uses $int_0^1frac{(-ln x)^n dx}{sqrt{1-x^2}}geint_0^1(-ln x)^n dx=int_0^infty y^n e^{-y}dy=n!$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 17 '18 at 14:03









                  J.G.J.G.

                  30.8k23149




                  30.8k23149























                      2












                      $begingroup$

                      Graphing the initial few functions for integer-valued n, it is clear that the integral will change signs repeatedly, and be negative for odd n and positive for even n. As a result, I would suggest that the limit does not exist unless it converges at 0, which seems unlikely when plotting the graph for high n-values






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        I agree with you that either the limit is $0$ or the limit does not exists, but unfortunately, plotting a graph is not a proof.
                        $endgroup$
                        – nicomezi
                        Dec 17 '18 at 13:49










                      • $begingroup$
                        @nicomezi That's an entirely valid point. I'll have a think about how to phrase a proper proof, but I think a much better written proof than I could come up with has been supplied in another answer.
                        $endgroup$
                        – M.M.
                        Dec 17 '18 at 13:55
















                      2












                      $begingroup$

                      Graphing the initial few functions for integer-valued n, it is clear that the integral will change signs repeatedly, and be negative for odd n and positive for even n. As a result, I would suggest that the limit does not exist unless it converges at 0, which seems unlikely when plotting the graph for high n-values






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        I agree with you that either the limit is $0$ or the limit does not exists, but unfortunately, plotting a graph is not a proof.
                        $endgroup$
                        – nicomezi
                        Dec 17 '18 at 13:49










                      • $begingroup$
                        @nicomezi That's an entirely valid point. I'll have a think about how to phrase a proper proof, but I think a much better written proof than I could come up with has been supplied in another answer.
                        $endgroup$
                        – M.M.
                        Dec 17 '18 at 13:55














                      2












                      2








                      2





                      $begingroup$

                      Graphing the initial few functions for integer-valued n, it is clear that the integral will change signs repeatedly, and be negative for odd n and positive for even n. As a result, I would suggest that the limit does not exist unless it converges at 0, which seems unlikely when plotting the graph for high n-values






                      share|cite|improve this answer









                      $endgroup$



                      Graphing the initial few functions for integer-valued n, it is clear that the integral will change signs repeatedly, and be negative for odd n and positive for even n. As a result, I would suggest that the limit does not exist unless it converges at 0, which seems unlikely when plotting the graph for high n-values







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Dec 17 '18 at 13:44









                      M.M.M.M.

                      637




                      637












                      • $begingroup$
                        I agree with you that either the limit is $0$ or the limit does not exists, but unfortunately, plotting a graph is not a proof.
                        $endgroup$
                        – nicomezi
                        Dec 17 '18 at 13:49










                      • $begingroup$
                        @nicomezi That's an entirely valid point. I'll have a think about how to phrase a proper proof, but I think a much better written proof than I could come up with has been supplied in another answer.
                        $endgroup$
                        – M.M.
                        Dec 17 '18 at 13:55


















                      • $begingroup$
                        I agree with you that either the limit is $0$ or the limit does not exists, but unfortunately, plotting a graph is not a proof.
                        $endgroup$
                        – nicomezi
                        Dec 17 '18 at 13:49










                      • $begingroup$
                        @nicomezi That's an entirely valid point. I'll have a think about how to phrase a proper proof, but I think a much better written proof than I could come up with has been supplied in another answer.
                        $endgroup$
                        – M.M.
                        Dec 17 '18 at 13:55
















                      $begingroup$
                      I agree with you that either the limit is $0$ or the limit does not exists, but unfortunately, plotting a graph is not a proof.
                      $endgroup$
                      – nicomezi
                      Dec 17 '18 at 13:49




                      $begingroup$
                      I agree with you that either the limit is $0$ or the limit does not exists, but unfortunately, plotting a graph is not a proof.
                      $endgroup$
                      – nicomezi
                      Dec 17 '18 at 13:49












                      $begingroup$
                      @nicomezi That's an entirely valid point. I'll have a think about how to phrase a proper proof, but I think a much better written proof than I could come up with has been supplied in another answer.
                      $endgroup$
                      – M.M.
                      Dec 17 '18 at 13:55




                      $begingroup$
                      @nicomezi That's an entirely valid point. I'll have a think about how to phrase a proper proof, but I think a much better written proof than I could come up with has been supplied in another answer.
                      $endgroup$
                      – M.M.
                      Dec 17 '18 at 13:55


















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