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











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



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


















  • 1




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












  • 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 $$int_0^infty g(x)dx ?$$
    – Did
    Dec 4 at 20:46






  • 1




    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
    – SABOY
    Dec 4 at 20:49








  • 1




    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
    – Masacroso
    Dec 4 at 23:01






  • 1




    @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 $g=0$
    – Masacroso
    Dec 4 at 23:07

















up vote
1
down vote

favorite












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



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


















  • 1




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












  • 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 $$int_0^infty g(x)dx ?$$
    – Did
    Dec 4 at 20:46






  • 1




    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
    – SABOY
    Dec 4 at 20:49








  • 1




    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
    – Masacroso
    Dec 4 at 23:01






  • 1




    @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 $g=0$
    – Masacroso
    Dec 4 at 23:07















up vote
1
down vote

favorite









up vote
1
down vote

favorite











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



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













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



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 at 20:20









SABOY

502311




502311








  • 1




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












  • 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 $$int_0^infty g(x)dx ?$$
    – Did
    Dec 4 at 20:46






  • 1




    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
    – SABOY
    Dec 4 at 20:49








  • 1




    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
    – Masacroso
    Dec 4 at 23:01






  • 1




    @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 $g=0$
    – Masacroso
    Dec 4 at 23:07
















  • 1




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












  • 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 $$int_0^infty g(x)dx ?$$
    – Did
    Dec 4 at 20:46






  • 1




    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
    – SABOY
    Dec 4 at 20:49








  • 1




    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
    – Masacroso
    Dec 4 at 23:01






  • 1




    @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 $g=0$
    – Masacroso
    Dec 4 at 23:07










1




1




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






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














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 $$int_0^infty g(x)dx ?$$
– Did
Dec 4 at 20:46




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 $$int_0^infty g(x)dx ?$$
– Did
Dec 4 at 20:46




1




1




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
– SABOY
Dec 4 at 20:49






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
– SABOY
Dec 4 at 20:49






1




1




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
– Masacroso
Dec 4 at 23:01




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
– Masacroso
Dec 4 at 23:01




1




1




@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 $g=0$
– Masacroso
Dec 4 at 23:07






@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 $g=0$
– Masacroso
Dec 4 at 23:07












2 Answers
2






active

oldest

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up vote
2
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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




























    up vote
    1
    down vote













    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





















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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

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      active

      oldest

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      up vote
      2
      down vote













      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

























        up vote
        2
        down vote













        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























          up vote
          2
          down vote










          up vote
          2
          down vote









          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












          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 at 20:27









          p4sch

          4,800217




          4,800217






















              up vote
              1
              down vote













              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

























                up vote
                1
                down vote













                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























                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  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












                  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 at 20:27









                  Umberto P.

                  38.3k13063




                  38.3k13063






























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                      Bressuire

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                      Cabo Verde

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                        Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
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                      Gyllenstierna

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