Find $lim_{nto infty} int_{mathbb R}frac{n log^{4}(x)}{n+nx+x^2}chi_{[0,infty[}dlambda(x)$












1












$begingroup$


$lim_{nto infty} int_{mathbb R}frac{n log^{4}(x)}{n+nx+x^2}chi_{[0,infty[}dlambda(x)$



This seems like an apt situation to utilize dominating convergence.



$f_{n}(x):=frac{n log^{4}(x)}{n+nx+x^2}$ is continuous on $]0,infty[$ and therefore measurable $forall n in mathbb N$. But I am struggling to show $int_{[0,infty[}frac{n log^{4}(x)}{n+nx+x^2}dlambda(x)=int_{]0,infty[}frac{n log^{4}(x)}{n+nx+x^2}dlambda(x)<infty$ to ensure $(f_{n})_{n}subseteqmathcal{L}^{1}(mu)$.



$|frac{n log^{4}(x)}{n+nx+x^2}|leq...?$



Any hints?






real-analysis integration measure-theory convergence







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Hint: The integrand monotone-increases to $(log x)^4/(1+x) mathbf{1}_{[0,infty)}(x)$ as $ntoinfty$.
    $endgroup$
    – Sangchul Lee
    Dec 4 '18 at 20:27












  • $begingroup$
    Is there a reason why you repeatedly use, for various functions $g$, the incorrect formula $$int_{mathbb R}g(x)chi_{[0,infty[}dlambda(x)$$ rather than the correct $$int_0^infty g(x)dx ?$$
    $endgroup$
    – Did
    Dec 4 '18 at 20:46






  • 1




    $begingroup$
    I was of the view $int_{mathbb R}g(x)chi_{[0,infty[}dlambda(x)=int_{[0,infty[}g(x)dlambda(x)$ as the lebesgue integral rather than $int^{infty}_{0}g(x)dx$ which we defined as the Riemann integral
    $endgroup$
    – SABOY
    Dec 4 '18 at 20:49








  • 1




    $begingroup$
    maybe you mean that you are searching an estimate for the integral $int_{[0,infty)}frac{(log x)^4}{1+x},lambda(dx)$? You have one in the answer of @p4sch, just fill the details, find a rough estimation of the integral in the set $[0,e]$, just note that the integrand is positive and that $lim_{xto 0^+}frac{(log x)^4}{1+x}=infty$, at least the integral is positive here
    $endgroup$
    – Masacroso
    Dec 4 '18 at 23:01






  • 1




    $begingroup$
    @SABOY I dont know your version of Fatou's lemma, I just know the version of above. Probably your version of Fatou's lemma is extended to a measure space $(X,mu,overline{Bbb R})$ instead of $(X,mu,overline{Bbb R}^+)$. In this case the condition $gle f_n$ seems to be added to ensure that, by the dominated convergence theorem, the integral $lim_nint_X f_n^-,dmu=int_Xlim f_n^-, dmu$, where $f_n^-$ is the negative part of $f_n$. Anyway Im just guessing, Idk really the reasons on your version of Fatou's lemma. In you case you can choose $g=0$
    $endgroup$
    – Masacroso
    Dec 4 '18 at 23:07


















1












$begingroup$


$lim_{nto infty} int_{mathbb R}frac{n log^{4}(x)}{n+nx+x^2}chi_{[0,infty[}dlambda(x)$



This seems like an apt situation to utilize dominating convergence.



$f_{n}(x):=frac{n log^{4}(x)}{n+nx+x^2}$ is continuous on $]0,infty[$ and therefore measurable $forall n in mathbb N$. But I am struggling to show $int_{[0,infty[}frac{n log^{4}(x)}{n+nx+x^2}dlambda(x)=int_{]0,infty[}frac{n log^{4}(x)}{n+nx+x^2}dlambda(x)<infty$ to ensure $(f_{n})_{n}subseteqmathcal{L}^{1}(mu)$.



$|frac{n log^{4}(x)}{n+nx+x^2}|leq...?$



Any hints?






real-analysis integration measure-theory convergence







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Hint: The integrand monotone-increases to $(log x)^4/(1+x) mathbf{1}_{[0,infty)}(x)$ as $ntoinfty$.
    $endgroup$
    – Sangchul Lee
    Dec 4 '18 at 20:27












  • $begingroup$
    Is there a reason why you repeatedly use, for various functions $g$, the incorrect formula $$int_{mathbb R}g(x)chi_{[0,infty[}dlambda(x)$$ rather than the correct $$int_0^infty g(x)dx ?$$
    $endgroup$
    – Did
    Dec 4 '18 at 20:46






  • 1




    $begingroup$
    I was of the view $int_{mathbb R}g(x)chi_{[0,infty[}dlambda(x)=int_{[0,infty[}g(x)dlambda(x)$ as the lebesgue integral rather than $int^{infty}_{0}g(x)dx$ which we defined as the Riemann integral
    $endgroup$
    – SABOY
    Dec 4 '18 at 20:49








  • 1




    $begingroup$
    maybe you mean that you are searching an estimate for the integral $int_{[0,infty)}frac{(log x)^4}{1+x},lambda(dx)$? You have one in the answer of @p4sch, just fill the details, find a rough estimation of the integral in the set $[0,e]$, just note that the integrand is positive and that $lim_{xto 0^+}frac{(log x)^4}{1+x}=infty$, at least the integral is positive here
    $endgroup$
    – Masacroso
    Dec 4 '18 at 23:01






  • 1




    $begingroup$
    @SABOY I dont know your version of Fatou's lemma, I just know the version of above. Probably your version of Fatou's lemma is extended to a measure space $(X,mu,overline{Bbb R})$ instead of $(X,mu,overline{Bbb R}^+)$. In this case the condition $gle f_n$ seems to be added to ensure that, by the dominated convergence theorem, the integral $lim_nint_X f_n^-,dmu=int_Xlim f_n^-, dmu$, where $f_n^-$ is the negative part of $f_n$. Anyway Im just guessing, Idk really the reasons on your version of Fatou's lemma. In you case you can choose $g=0$
    $endgroup$
    – Masacroso
    Dec 4 '18 at 23:07
















1












1








1





$begingroup$


$lim_{nto infty} int_{mathbb R}frac{n log^{4}(x)}{n+nx+x^2}chi_{[0,infty[}dlambda(x)$



This seems like an apt situation to utilize dominating convergence.



$f_{n}(x):=frac{n log^{4}(x)}{n+nx+x^2}$ is continuous on $]0,infty[$ and therefore measurable $forall n in mathbb N$. But I am struggling to show $int_{[0,infty[}frac{n log^{4}(x)}{n+nx+x^2}dlambda(x)=int_{]0,infty[}frac{n log^{4}(x)}{n+nx+x^2}dlambda(x)<infty$ to ensure $(f_{n})_{n}subseteqmathcal{L}^{1}(mu)$.



$|frac{n log^{4}(x)}{n+nx+x^2}|leq...?$



Any hints?






real-analysis integration measure-theory convergence







share|cite|improve this question









$endgroup$




$lim_{nto infty} int_{mathbb R}frac{n log^{4}(x)}{n+nx+x^2}chi_{[0,infty[}dlambda(x)$



This seems like an apt situation to utilize dominating convergence.



$f_{n}(x):=frac{n log^{4}(x)}{n+nx+x^2}$ is continuous on $]0,infty[$ and therefore measurable $forall n in mathbb N$. But I am struggling to show $int_{[0,infty[}frac{n log^{4}(x)}{n+nx+x^2}dlambda(x)=int_{]0,infty[}frac{n log^{4}(x)}{n+nx+x^2}dlambda(x)<infty$ to ensure $(f_{n})_{n}subseteqmathcal{L}^{1}(mu)$.



$|frac{n log^{4}(x)}{n+nx+x^2}|leq...?$



Any hints?






real-analysis integration measure-theory convergence




real-analysis integration measure-theory convergence






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 4 '18 at 20:20









SABOYSABOY

614311




614311








  • 1




    $begingroup$
    Hint: The integrand monotone-increases to $(log x)^4/(1+x) mathbf{1}_{[0,infty)}(x)$ as $ntoinfty$.
    $endgroup$
    – Sangchul Lee
    Dec 4 '18 at 20:27












  • $begingroup$
    Is there a reason why you repeatedly use, for various functions $g$, the incorrect formula $$int_{mathbb R}g(x)chi_{[0,infty[}dlambda(x)$$ rather than the correct $$int_0^infty g(x)dx ?$$
    $endgroup$
    – Did
    Dec 4 '18 at 20:46






  • 1




    $begingroup$
    I was of the view $int_{mathbb R}g(x)chi_{[0,infty[}dlambda(x)=int_{[0,infty[}g(x)dlambda(x)$ as the lebesgue integral rather than $int^{infty}_{0}g(x)dx$ which we defined as the Riemann integral
    $endgroup$
    – SABOY
    Dec 4 '18 at 20:49








  • 1




    $begingroup$
    maybe you mean that you are searching an estimate for the integral $int_{[0,infty)}frac{(log x)^4}{1+x},lambda(dx)$? You have one in the answer of @p4sch, just fill the details, find a rough estimation of the integral in the set $[0,e]$, just note that the integrand is positive and that $lim_{xto 0^+}frac{(log x)^4}{1+x}=infty$, at least the integral is positive here
    $endgroup$
    – Masacroso
    Dec 4 '18 at 23:01






  • 1




    $begingroup$
    @SABOY I dont know your version of Fatou's lemma, I just know the version of above. Probably your version of Fatou's lemma is extended to a measure space $(X,mu,overline{Bbb R})$ instead of $(X,mu,overline{Bbb R}^+)$. In this case the condition $gle f_n$ seems to be added to ensure that, by the dominated convergence theorem, the integral $lim_nint_X f_n^-,dmu=int_Xlim f_n^-, dmu$, where $f_n^-$ is the negative part of $f_n$. Anyway Im just guessing, Idk really the reasons on your version of Fatou's lemma. In you case you can choose $g=0$
    $endgroup$
    – Masacroso
    Dec 4 '18 at 23:07
















  • 1




    $begingroup$
    Hint: The integrand monotone-increases to $(log x)^4/(1+x) mathbf{1}_{[0,infty)}(x)$ as $ntoinfty$.
    $endgroup$
    – Sangchul Lee
    Dec 4 '18 at 20:27












  • $begingroup$
    Is there a reason why you repeatedly use, for various functions $g$, the incorrect formula $$int_{mathbb R}g(x)chi_{[0,infty[}dlambda(x)$$ rather than the correct $$int_0^infty g(x)dx ?$$
    $endgroup$
    – Did
    Dec 4 '18 at 20:46






  • 1




    $begingroup$
    I was of the view $int_{mathbb R}g(x)chi_{[0,infty[}dlambda(x)=int_{[0,infty[}g(x)dlambda(x)$ as the lebesgue integral rather than $int^{infty}_{0}g(x)dx$ which we defined as the Riemann integral
    $endgroup$
    – SABOY
    Dec 4 '18 at 20:49








  • 1




    $begingroup$
    maybe you mean that you are searching an estimate for the integral $int_{[0,infty)}frac{(log x)^4}{1+x},lambda(dx)$? You have one in the answer of @p4sch, just fill the details, find a rough estimation of the integral in the set $[0,e]$, just note that the integrand is positive and that $lim_{xto 0^+}frac{(log x)^4}{1+x}=infty$, at least the integral is positive here
    $endgroup$
    – Masacroso
    Dec 4 '18 at 23:01






  • 1




    $begingroup$
    @SABOY I dont know your version of Fatou's lemma, I just know the version of above. Probably your version of Fatou's lemma is extended to a measure space $(X,mu,overline{Bbb R})$ instead of $(X,mu,overline{Bbb R}^+)$. In this case the condition $gle f_n$ seems to be added to ensure that, by the dominated convergence theorem, the integral $lim_nint_X f_n^-,dmu=int_Xlim f_n^-, dmu$, where $f_n^-$ is the negative part of $f_n$. Anyway Im just guessing, Idk really the reasons on your version of Fatou's lemma. In you case you can choose $g=0$
    $endgroup$
    – Masacroso
    Dec 4 '18 at 23:07










1




1




$begingroup$
Hint: The integrand monotone-increases to $(log x)^4/(1+x) mathbf{1}_{[0,infty)}(x)$ as $ntoinfty$.
$endgroup$
– Sangchul Lee
Dec 4 '18 at 20:27






$begingroup$
Hint: The integrand monotone-increases to $(log x)^4/(1+x) mathbf{1}_{[0,infty)}(x)$ as $ntoinfty$.
$endgroup$
– Sangchul Lee
Dec 4 '18 at 20:27














$begingroup$
Is there a reason why you repeatedly use, for various functions $g$, the incorrect formula $$int_{mathbb R}g(x)chi_{[0,infty[}dlambda(x)$$ rather than the correct $$int_0^infty g(x)dx ?$$
$endgroup$
– Did
Dec 4 '18 at 20:46




$begingroup$
Is there a reason why you repeatedly use, for various functions $g$, the incorrect formula $$int_{mathbb R}g(x)chi_{[0,infty[}dlambda(x)$$ rather than the correct $$int_0^infty g(x)dx ?$$
$endgroup$
– Did
Dec 4 '18 at 20:46




1




1




$begingroup$
I was of the view $int_{mathbb R}g(x)chi_{[0,infty[}dlambda(x)=int_{[0,infty[}g(x)dlambda(x)$ as the lebesgue integral rather than $int^{infty}_{0}g(x)dx$ which we defined as the Riemann integral
$endgroup$
– SABOY
Dec 4 '18 at 20:49






$begingroup$
I was of the view $int_{mathbb R}g(x)chi_{[0,infty[}dlambda(x)=int_{[0,infty[}g(x)dlambda(x)$ as the lebesgue integral rather than $int^{infty}_{0}g(x)dx$ which we defined as the Riemann integral
$endgroup$
– SABOY
Dec 4 '18 at 20:49






1




1




$begingroup$
maybe you mean that you are searching an estimate for the integral $int_{[0,infty)}frac{(log x)^4}{1+x},lambda(dx)$? You have one in the answer of @p4sch, just fill the details, find a rough estimation of the integral in the set $[0,e]$, just note that the integrand is positive and that $lim_{xto 0^+}frac{(log x)^4}{1+x}=infty$, at least the integral is positive here
$endgroup$
– Masacroso
Dec 4 '18 at 23:01




$begingroup$
maybe you mean that you are searching an estimate for the integral $int_{[0,infty)}frac{(log x)^4}{1+x},lambda(dx)$? You have one in the answer of @p4sch, just fill the details, find a rough estimation of the integral in the set $[0,e]$, just note that the integrand is positive and that $lim_{xto 0^+}frac{(log x)^4}{1+x}=infty$, at least the integral is positive here
$endgroup$
– Masacroso
Dec 4 '18 at 23:01




1




1




$begingroup$
@SABOY I dont know your version of Fatou's lemma, I just know the version of above. Probably your version of Fatou's lemma is extended to a measure space $(X,mu,overline{Bbb R})$ instead of $(X,mu,overline{Bbb R}^+)$. In this case the condition $gle f_n$ seems to be added to ensure that, by the dominated convergence theorem, the integral $lim_nint_X f_n^-,dmu=int_Xlim f_n^-, dmu$, where $f_n^-$ is the negative part of $f_n$. Anyway Im just guessing, Idk really the reasons on your version of Fatou's lemma. In you case you can choose $g=0$
$endgroup$
– Masacroso
Dec 4 '18 at 23:07






$begingroup$
@SABOY I dont know your version of Fatou's lemma, I just know the version of above. Probably your version of Fatou's lemma is extended to a measure space $(X,mu,overline{Bbb R})$ instead of $(X,mu,overline{Bbb R}^+)$. In this case the condition $gle f_n$ seems to be added to ensure that, by the dominated convergence theorem, the integral $lim_nint_X f_n^-,dmu=int_Xlim f_n^-, dmu$, where $f_n^-$ is the negative part of $f_n$. Anyway Im just guessing, Idk really the reasons on your version of Fatou's lemma. In you case you can choose $g=0$
$endgroup$
– Masacroso
Dec 4 '18 at 23:07












2 Answers
2






active

oldest

votes


















2












$begingroup$

Note that for $x > 0$ we have
$$f_n(x):= frac{log(x)^4}{1+x+x^2/n}.$$
Thus for all $x >0$
$$f_n(x) le f_{n+1}(x) le f(x):=frac{log(x)^4}{1+x}$$
and pointwise $f_n rightarrow f$. Therefore, we can apply the monotone convergence theorem. However, the limes $f$ is not integrable, because
$$int_{e}^infty f(x) , dx ge int_{e}^infty frac{1}{1+x} , dx =infty,$$
and we cannot apply the dominated convergence theorem.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    To apply LDCT you need an $L^1$ majorant $g$ satisfying $|f_n(x)| le g(x)$ almost everywhere for all $n$. If $f_n to f$ almost everywhere then you have $|f(x)| le g(x)$ too.



    You can write $$f_n(x) = frac{n log^4 x}{n + nx + x^2} = frac{log^4 x}{1 + x + x^2/n}$$ so that $$f_n(x) to frac{log^4 x}{1+x}.$$
    This function is not integrable, so you won't find an integrable majorant $g$ with $|f_n| le g$ for all $n$. It looks like LDCT may not be the way to go.






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Note that for $x > 0$ we have
      $$f_n(x):= frac{log(x)^4}{1+x+x^2/n}.$$
      Thus for all $x >0$
      $$f_n(x) le f_{n+1}(x) le f(x):=frac{log(x)^4}{1+x}$$
      and pointwise $f_n rightarrow f$. Therefore, we can apply the monotone convergence theorem. However, the limes $f$ is not integrable, because
      $$int_{e}^infty f(x) , dx ge int_{e}^infty frac{1}{1+x} , dx =infty,$$
      and we cannot apply the dominated convergence theorem.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        Note that for $x > 0$ we have
        $$f_n(x):= frac{log(x)^4}{1+x+x^2/n}.$$
        Thus for all $x >0$
        $$f_n(x) le f_{n+1}(x) le f(x):=frac{log(x)^4}{1+x}$$
        and pointwise $f_n rightarrow f$. Therefore, we can apply the monotone convergence theorem. However, the limes $f$ is not integrable, because
        $$int_{e}^infty f(x) , dx ge int_{e}^infty frac{1}{1+x} , dx =infty,$$
        and we cannot apply the dominated convergence theorem.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          Note that for $x > 0$ we have
          $$f_n(x):= frac{log(x)^4}{1+x+x^2/n}.$$
          Thus for all $x >0$
          $$f_n(x) le f_{n+1}(x) le f(x):=frac{log(x)^4}{1+x}$$
          and pointwise $f_n rightarrow f$. Therefore, we can apply the monotone convergence theorem. However, the limes $f$ is not integrable, because
          $$int_{e}^infty f(x) , dx ge int_{e}^infty frac{1}{1+x} , dx =infty,$$
          and we cannot apply the dominated convergence theorem.






          share|cite|improve this answer









          $endgroup$



          Note that for $x > 0$ we have
          $$f_n(x):= frac{log(x)^4}{1+x+x^2/n}.$$
          Thus for all $x >0$
          $$f_n(x) le f_{n+1}(x) le f(x):=frac{log(x)^4}{1+x}$$
          and pointwise $f_n rightarrow f$. Therefore, we can apply the monotone convergence theorem. However, the limes $f$ is not integrable, because
          $$int_{e}^infty f(x) , dx ge int_{e}^infty frac{1}{1+x} , dx =infty,$$
          and we cannot apply the dominated convergence theorem.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 4 '18 at 20:27









          p4schp4sch

          5,120217




          5,120217























              1












              $begingroup$

              To apply LDCT you need an $L^1$ majorant $g$ satisfying $|f_n(x)| le g(x)$ almost everywhere for all $n$. If $f_n to f$ almost everywhere then you have $|f(x)| le g(x)$ too.



              You can write $$f_n(x) = frac{n log^4 x}{n + nx + x^2} = frac{log^4 x}{1 + x + x^2/n}$$ so that $$f_n(x) to frac{log^4 x}{1+x}.$$
              This function is not integrable, so you won't find an integrable majorant $g$ with $|f_n| le g$ for all $n$. It looks like LDCT may not be the way to go.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                To apply LDCT you need an $L^1$ majorant $g$ satisfying $|f_n(x)| le g(x)$ almost everywhere for all $n$. If $f_n to f$ almost everywhere then you have $|f(x)| le g(x)$ too.



                You can write $$f_n(x) = frac{n log^4 x}{n + nx + x^2} = frac{log^4 x}{1 + x + x^2/n}$$ so that $$f_n(x) to frac{log^4 x}{1+x}.$$
                This function is not integrable, so you won't find an integrable majorant $g$ with $|f_n| le g$ for all $n$. It looks like LDCT may not be the way to go.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  To apply LDCT you need an $L^1$ majorant $g$ satisfying $|f_n(x)| le g(x)$ almost everywhere for all $n$. If $f_n to f$ almost everywhere then you have $|f(x)| le g(x)$ too.



                  You can write $$f_n(x) = frac{n log^4 x}{n + nx + x^2} = frac{log^4 x}{1 + x + x^2/n}$$ so that $$f_n(x) to frac{log^4 x}{1+x}.$$
                  This function is not integrable, so you won't find an integrable majorant $g$ with $|f_n| le g$ for all $n$. It looks like LDCT may not be the way to go.






                  share|cite|improve this answer









                  $endgroup$



                  To apply LDCT you need an $L^1$ majorant $g$ satisfying $|f_n(x)| le g(x)$ almost everywhere for all $n$. If $f_n to f$ almost everywhere then you have $|f(x)| le g(x)$ too.



                  You can write $$f_n(x) = frac{n log^4 x}{n + nx + x^2} = frac{log^4 x}{1 + x + x^2/n}$$ so that $$f_n(x) to frac{log^4 x}{1+x}.$$
                  This function is not integrable, so you won't find an integrable majorant $g$ with $|f_n| le g$ for all $n$. It looks like LDCT may not be the way to go.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 4 '18 at 20:27









                  Umberto P.Umberto P.

                  38.9k13064




                  38.9k13064






























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