Measure theory integral












4












$begingroup$


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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$endgroup$








  • 3




    $begingroup$
    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.
    $endgroup$
    – Teepeemm
    Jan 5 at 15:35










  • $begingroup$
    (Another take on Teepeemm's comment:) $frac{mathrm{d}}{mathrm{d}n} langle text{expression containing no "$n$"s} rangle = 0$.
    $endgroup$
    – Eric Towers
    Jan 5 at 22:58
















4












$begingroup$


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











$endgroup$








  • 3




    $begingroup$
    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.
    $endgroup$
    – Teepeemm
    Jan 5 at 15:35










  • $begingroup$
    (Another take on Teepeemm's comment:) $frac{mathrm{d}}{mathrm{d}n} langle text{expression containing no "$n$"s} rangle = 0$.
    $endgroup$
    – Eric Towers
    Jan 5 at 22:58














4












4








4


3



$begingroup$


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











$endgroup$




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 Jan 5 at 15:02









md2perpe

7,80111028




7,80111028










asked Jan 5 at 14:01









Math LadyMath Lady

1196




1196








  • 3




    $begingroup$
    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.
    $endgroup$
    – Teepeemm
    Jan 5 at 15:35










  • $begingroup$
    (Another take on Teepeemm's comment:) $frac{mathrm{d}}{mathrm{d}n} langle text{expression containing no "$n$"s} rangle = 0$.
    $endgroup$
    – Eric Towers
    Jan 5 at 22:58














  • 3




    $begingroup$
    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.
    $endgroup$
    – Teepeemm
    Jan 5 at 15:35










  • $begingroup$
    (Another take on Teepeemm's comment:) $frac{mathrm{d}}{mathrm{d}n} langle text{expression containing no "$n$"s} rangle = 0$.
    $endgroup$
    – Eric Towers
    Jan 5 at 22:58








3




3




$begingroup$
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.
$endgroup$
– Teepeemm
Jan 5 at 15:35




$begingroup$
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.
$endgroup$
– Teepeemm
Jan 5 at 15:35












$begingroup$
(Another take on Teepeemm's comment:) $frac{mathrm{d}}{mathrm{d}n} langle text{expression containing no "$n$"s} rangle = 0$.
$endgroup$
– Eric Towers
Jan 5 at 22:58




$begingroup$
(Another take on Teepeemm's comment:) $frac{mathrm{d}}{mathrm{d}n} langle text{expression containing no "$n$"s} rangle = 0$.
$endgroup$
– Eric Towers
Jan 5 at 22:58










2 Answers
2






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7












$begingroup$

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









$endgroup$





















    2












    $begingroup$

    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






      active

      oldest

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      active

      oldest

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      active

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      7












      $begingroup$

      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









      $endgroup$


















        7












        $begingroup$

        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









        $endgroup$
















          7












          7








          7





          $begingroup$

          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









          $endgroup$



          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 Jan 5 at 14:09









          YankoYanko

          6,3981528




          6,3981528























              2












              $begingroup$

              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











              $endgroup$


















                2












                $begingroup$

                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











                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  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











                  $endgroup$



                  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



                  share|cite|improve this answer








                  edited Jan 5 at 15:40

























                  answered Jan 5 at 14:32









                  Robert ZRobert Z

                  95.2k1063134




                  95.2k1063134






























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