Measure theory integral












4














This is a qualifying exam practice question - so not being graded for homework purposes, just studying!



Calculate $lim_{n rightarrow infty} int_0^infty frac{x^n}{ x^{(n+3)}+1} dx$



I tried the following:



$lim_{n rightarrow infty} int_0^infty frac{x^n}{ x^{(n+3)}+1} , dx$ = $frac{d}{dn}int_0^infty int_0^inftyfrac{x^n}{ x^{(n+3)}+1}dn , dx$ = -$frac{d}{dn} int_0^infty frac{ln(x^3+1}{x^3 ln(x)} , dx$



Not really sure where to go from here, any advice would be appreciated!






real-analysis measure-theory convergence lebesgue-integral lebesgue-measure







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  • 3




    I'm trying to understand your thinking when you went from $lim_{ntoinfty}$ to $frac{d}{dn}int_0^infty dn$. I don't see how that could work.
    – Teepeemm
    yesterday










  • (Another take on Teepeemm's comment:) $frac{mathrm{d}}{mathrm{d}n} langle text{expression containing no "$n$"s} rangle = 0$.
    – Eric Towers
    yesterday
















4














This is a qualifying exam practice question - so not being graded for homework purposes, just studying!



Calculate $lim_{n rightarrow infty} int_0^infty frac{x^n}{ x^{(n+3)}+1} dx$



I tried the following:



$lim_{n rightarrow infty} int_0^infty frac{x^n}{ x^{(n+3)}+1} , dx$ = $frac{d}{dn}int_0^infty int_0^inftyfrac{x^n}{ x^{(n+3)}+1}dn , dx$ = -$frac{d}{dn} int_0^infty frac{ln(x^3+1}{x^3 ln(x)} , dx$



Not really sure where to go from here, any advice would be appreciated!






real-analysis measure-theory convergence lebesgue-integral lebesgue-measure







share|cite|improve this question




















  • 3




    I'm trying to understand your thinking when you went from $lim_{ntoinfty}$ to $frac{d}{dn}int_0^infty dn$. I don't see how that could work.
    – Teepeemm
    yesterday










  • (Another take on Teepeemm's comment:) $frac{mathrm{d}}{mathrm{d}n} langle text{expression containing no "$n$"s} rangle = 0$.
    – Eric Towers
    yesterday














4












4








4


3





This is a qualifying exam practice question - so not being graded for homework purposes, just studying!



Calculate $lim_{n rightarrow infty} int_0^infty frac{x^n}{ x^{(n+3)}+1} dx$



I tried the following:



$lim_{n rightarrow infty} int_0^infty frac{x^n}{ x^{(n+3)}+1} , dx$ = $frac{d}{dn}int_0^infty int_0^inftyfrac{x^n}{ x^{(n+3)}+1}dn , dx$ = -$frac{d}{dn} int_0^infty frac{ln(x^3+1}{x^3 ln(x)} , dx$



Not really sure where to go from here, any advice would be appreciated!






real-analysis measure-theory convergence lebesgue-integral lebesgue-measure







share|cite|improve this question















This is a qualifying exam practice question - so not being graded for homework purposes, just studying!



Calculate $lim_{n rightarrow infty} int_0^infty frac{x^n}{ x^{(n+3)}+1} dx$



I tried the following:



$lim_{n rightarrow infty} int_0^infty frac{x^n}{ x^{(n+3)}+1} , dx$ = $frac{d}{dn}int_0^infty int_0^inftyfrac{x^n}{ x^{(n+3)}+1}dn , dx$ = -$frac{d}{dn} int_0^infty frac{ln(x^3+1}{x^3 ln(x)} , dx$



Not really sure where to go from here, any advice would be appreciated!






real-analysis measure-theory convergence lebesgue-integral lebesgue-measure




real-analysis measure-theory convergence lebesgue-integral lebesgue-measure






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday









md2perpe

7,72111028




7,72111028










asked yesterday









Math Lady

1146




1146








  • 3




    I'm trying to understand your thinking when you went from $lim_{ntoinfty}$ to $frac{d}{dn}int_0^infty dn$. I don't see how that could work.
    – Teepeemm
    yesterday










  • (Another take on Teepeemm's comment:) $frac{mathrm{d}}{mathrm{d}n} langle text{expression containing no "$n$"s} rangle = 0$.
    – Eric Towers
    yesterday














  • 3




    I'm trying to understand your thinking when you went from $lim_{ntoinfty}$ to $frac{d}{dn}int_0^infty dn$. I don't see how that could work.
    – Teepeemm
    yesterday










  • (Another take on Teepeemm's comment:) $frac{mathrm{d}}{mathrm{d}n} langle text{expression containing no "$n$"s} rangle = 0$.
    – Eric Towers
    yesterday








3




3




I'm trying to understand your thinking when you went from $lim_{ntoinfty}$ to $frac{d}{dn}int_0^infty dn$. I don't see how that could work.
– Teepeemm
yesterday




I'm trying to understand your thinking when you went from $lim_{ntoinfty}$ to $frac{d}{dn}int_0^infty dn$. I don't see how that could work.
– Teepeemm
yesterday












(Another take on Teepeemm's comment:) $frac{mathrm{d}}{mathrm{d}n} langle text{expression containing no "$n$"s} rangle = 0$.
– Eric Towers
yesterday




(Another take on Teepeemm's comment:) $frac{mathrm{d}}{mathrm{d}n} langle text{expression containing no "$n$"s} rangle = 0$.
– Eric Towers
yesterday










2 Answers
2






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7














Hint:



You have $$lim_n int_0^infty frac{x^n}{x^{n+3}+1} = lim_n int_0^1 frac{x^n}{x^{n+3}+1}+lim_n int_1^infty frac{x^n}{x^{n+3}+1}$$



Use on each term the dominated convergence theorem to get the limits inside. If $0<x<1$ then $x^nrightarrow 0$ so the first term is easy to calculate.



The second converges to $frac{1}{x^3}$ and then you need to evaluate $int_1^infty frac{1}{x^3} dx$.






share|cite|improve this answer





























    2














    Hint. Note that
    $$int_0^{infty}frac{x^n}{ x^{n+3}+1},dx=int_0^{1}frac{x^n}{ x^{n+3}+1},dx+int_1^{infty}frac{dx}{x^3}-int_1^{infty}frac{dx}{x^3( x^{n+3}+1)},$$
    where
    $$0leq int_0^1frac{x^n}{ x^{n+3}+1},dx leq int_0^1 x^n ,dx=left[frac{x^{n+1}}{n+1}right]_0^{1}=frac{1}{n+1},$$
    and
    $$0leq int_1^{infty}frac{dx}{x^3( x^{n+3}+1)}leq
    int_1^{infty}frac{1}{ x^{n+3}},dx=left[-frac{1}{(n+2)x^{n+2}}right]_1^{infty}=frac{1}{n+2}.$$






    share|cite|improve this answer























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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






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      active

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      7














      Hint:



      You have $$lim_n int_0^infty frac{x^n}{x^{n+3}+1} = lim_n int_0^1 frac{x^n}{x^{n+3}+1}+lim_n int_1^infty frac{x^n}{x^{n+3}+1}$$



      Use on each term the dominated convergence theorem to get the limits inside. If $0<x<1$ then $x^nrightarrow 0$ so the first term is easy to calculate.



      The second converges to $frac{1}{x^3}$ and then you need to evaluate $int_1^infty frac{1}{x^3} dx$.






      share|cite|improve this answer


























        7














        Hint:



        You have $$lim_n int_0^infty frac{x^n}{x^{n+3}+1} = lim_n int_0^1 frac{x^n}{x^{n+3}+1}+lim_n int_1^infty frac{x^n}{x^{n+3}+1}$$



        Use on each term the dominated convergence theorem to get the limits inside. If $0<x<1$ then $x^nrightarrow 0$ so the first term is easy to calculate.



        The second converges to $frac{1}{x^3}$ and then you need to evaluate $int_1^infty frac{1}{x^3} dx$.






        share|cite|improve this answer
























          7












          7








          7






          Hint:



          You have $$lim_n int_0^infty frac{x^n}{x^{n+3}+1} = lim_n int_0^1 frac{x^n}{x^{n+3}+1}+lim_n int_1^infty frac{x^n}{x^{n+3}+1}$$



          Use on each term the dominated convergence theorem to get the limits inside. If $0<x<1$ then $x^nrightarrow 0$ so the first term is easy to calculate.



          The second converges to $frac{1}{x^3}$ and then you need to evaluate $int_1^infty frac{1}{x^3} dx$.






          share|cite|improve this answer












          Hint:



          You have $$lim_n int_0^infty frac{x^n}{x^{n+3}+1} = lim_n int_0^1 frac{x^n}{x^{n+3}+1}+lim_n int_1^infty frac{x^n}{x^{n+3}+1}$$



          Use on each term the dominated convergence theorem to get the limits inside. If $0<x<1$ then $x^nrightarrow 0$ so the first term is easy to calculate.



          The second converges to $frac{1}{x^3}$ and then you need to evaluate $int_1^infty frac{1}{x^3} dx$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered yesterday









          Yanko

          6,394727




          6,394727























              2














              Hint. Note that
              $$int_0^{infty}frac{x^n}{ x^{n+3}+1},dx=int_0^{1}frac{x^n}{ x^{n+3}+1},dx+int_1^{infty}frac{dx}{x^3}-int_1^{infty}frac{dx}{x^3( x^{n+3}+1)},$$
              where
              $$0leq int_0^1frac{x^n}{ x^{n+3}+1},dx leq int_0^1 x^n ,dx=left[frac{x^{n+1}}{n+1}right]_0^{1}=frac{1}{n+1},$$
              and
              $$0leq int_1^{infty}frac{dx}{x^3( x^{n+3}+1)}leq
              int_1^{infty}frac{1}{ x^{n+3}},dx=left[-frac{1}{(n+2)x^{n+2}}right]_1^{infty}=frac{1}{n+2}.$$






              share|cite|improve this answer




























                2














                Hint. Note that
                $$int_0^{infty}frac{x^n}{ x^{n+3}+1},dx=int_0^{1}frac{x^n}{ x^{n+3}+1},dx+int_1^{infty}frac{dx}{x^3}-int_1^{infty}frac{dx}{x^3( x^{n+3}+1)},$$
                where
                $$0leq int_0^1frac{x^n}{ x^{n+3}+1},dx leq int_0^1 x^n ,dx=left[frac{x^{n+1}}{n+1}right]_0^{1}=frac{1}{n+1},$$
                and
                $$0leq int_1^{infty}frac{dx}{x^3( x^{n+3}+1)}leq
                int_1^{infty}frac{1}{ x^{n+3}},dx=left[-frac{1}{(n+2)x^{n+2}}right]_1^{infty}=frac{1}{n+2}.$$






                share|cite|improve this answer


























                  2












                  2








                  2






                  Hint. Note that
                  $$int_0^{infty}frac{x^n}{ x^{n+3}+1},dx=int_0^{1}frac{x^n}{ x^{n+3}+1},dx+int_1^{infty}frac{dx}{x^3}-int_1^{infty}frac{dx}{x^3( x^{n+3}+1)},$$
                  where
                  $$0leq int_0^1frac{x^n}{ x^{n+3}+1},dx leq int_0^1 x^n ,dx=left[frac{x^{n+1}}{n+1}right]_0^{1}=frac{1}{n+1},$$
                  and
                  $$0leq int_1^{infty}frac{dx}{x^3( x^{n+3}+1)}leq
                  int_1^{infty}frac{1}{ x^{n+3}},dx=left[-frac{1}{(n+2)x^{n+2}}right]_1^{infty}=frac{1}{n+2}.$$






                  share|cite|improve this answer














                  Hint. Note that
                  $$int_0^{infty}frac{x^n}{ x^{n+3}+1},dx=int_0^{1}frac{x^n}{ x^{n+3}+1},dx+int_1^{infty}frac{dx}{x^3}-int_1^{infty}frac{dx}{x^3( x^{n+3}+1)},$$
                  where
                  $$0leq int_0^1frac{x^n}{ x^{n+3}+1},dx leq int_0^1 x^n ,dx=left[frac{x^{n+1}}{n+1}right]_0^{1}=frac{1}{n+1},$$
                  and
                  $$0leq int_1^{infty}frac{dx}{x^3( x^{n+3}+1)}leq
                  int_1^{infty}frac{1}{ x^{n+3}},dx=left[-frac{1}{(n+2)x^{n+2}}right]_1^{infty}=frac{1}{n+2}.$$







                  share|cite|improve this answer














                  share|cite|improve this answer



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                  edited yesterday

























                  answered yesterday









                  Robert Z

                  93.7k1061132




                  93.7k1061132






























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