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







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
    2 days ago










  • (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
    2 days ago










  • (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 2 days ago









md2perpe

7,72111028




7,72111028










asked 2 days ago









Math LadyMath Lady

1196




1196








  • 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
    2 days ago










  • (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
    2 days ago










  • (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
2 days ago




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
2 days ago












(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






active

oldest

votes


















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























      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3062748%2fmeasure-theory-integral%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      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 2 days ago









          YankoYanko

          6,469727




          6,469727























              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



                  share|cite|improve this answer








                  edited 2 days ago

























                  answered 2 days ago









                  Robert ZRobert Z

                  93.8k1061132




                  93.8k1061132






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.





                      Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                      Please pay close attention to the following guidance:


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3062748%2fmeasure-theory-integral%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      -

                      Popular posts from this blog

                      Mario Kart Wii

                      Image
                      Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro...
                      Read more

                      The Binding of Isaac: Rebirth/Afterbirth

                      Image
                      Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me...
                      Read more

                      What does “Dominus providebit” mean?

                      Image
                      5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1 ...
                      Read more