a basic definite integration and its result used in evaluating limit












2












$begingroup$


Question :



$mathbfOmega(n)=displaystyleint _{0}^{2pi}log(n^2-2ncos t+1)dt , ngeq1 $



then , find :



$displaystyle lim_{n to infty} left(1+dfrac{mathbf Omega(n)}{4pi}right)^{log(n+1)}$



my attempt:



$mathbfOmega(n)=displaystyleint _{0}^{2pi}log(n-e^{it})dt+displaystyleint _{0}^{2pi}log(n-e^{-it})dt $



$mathbfOmega(n)=2displaystyleint _{0}^{2pi} log n dt+displaystyle int_{0}^{2pi}logleft(1-dfrac{e^{it}}{n}right)dt+displaystyle int_{0}^{2pi}logleft(1-dfrac{e^{-it}}{n}right)dt$



$mathbfOmega(n)=2pi log n-displaystyle int_{0}^{2pi}displaystyle sum_{k=1}^{infty} dfrac{e^{ikt}}{k n^k}dt-displaystyle int_{0}^{2pi}displaystyle sum_{k=1}^{infty} dfrac{e^{-ikt}}{k n^k}dt$



changing order of summation and integration



$mathbfOmega(n)=2pi log n -displaystyle sum_{k=1}^{infty}dfrac{1}{kn^k}displaystyle int_{0}^{2pi}e^{ikt}dt - displaystyle sum_{k=1}^{infty}dfrac{1}{kn^k}displaystyle int_{0}^{2pi}e^{-ikt}dt$



both integrals becomes zero becuase they are integration over a period of sinusoid



$mathbfOmega(n)=2pi log n $



therefore,



$displaystyle lim_{n to infty} left(1+dfrac{mathbf Omega(n)}{4pi}right)^{log(n+1)}=displaystyle lim_{n to infty} left(1+dfrac{log n}{2}right)^{log(n+1)}$



after this step, i don't know how to proceed to find final answer. my integral might be wrong so, please help me re-evaluate if possible and also help me find limit



thanks in advance






integration complex-analysis definite-integrals contest-math complex-integration







share|cite|improve this question











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    2












    $begingroup$


    Question :



    $mathbfOmega(n)=displaystyleint _{0}^{2pi}log(n^2-2ncos t+1)dt , ngeq1 $



    then , find :



    $displaystyle lim_{n to infty} left(1+dfrac{mathbf Omega(n)}{4pi}right)^{log(n+1)}$



    my attempt:



    $mathbfOmega(n)=displaystyleint _{0}^{2pi}log(n-e^{it})dt+displaystyleint _{0}^{2pi}log(n-e^{-it})dt $



    $mathbfOmega(n)=2displaystyleint _{0}^{2pi} log n dt+displaystyle int_{0}^{2pi}logleft(1-dfrac{e^{it}}{n}right)dt+displaystyle int_{0}^{2pi}logleft(1-dfrac{e^{-it}}{n}right)dt$



    $mathbfOmega(n)=2pi log n-displaystyle int_{0}^{2pi}displaystyle sum_{k=1}^{infty} dfrac{e^{ikt}}{k n^k}dt-displaystyle int_{0}^{2pi}displaystyle sum_{k=1}^{infty} dfrac{e^{-ikt}}{k n^k}dt$



    changing order of summation and integration



    $mathbfOmega(n)=2pi log n -displaystyle sum_{k=1}^{infty}dfrac{1}{kn^k}displaystyle int_{0}^{2pi}e^{ikt}dt - displaystyle sum_{k=1}^{infty}dfrac{1}{kn^k}displaystyle int_{0}^{2pi}e^{-ikt}dt$



    both integrals becomes zero becuase they are integration over a period of sinusoid



    $mathbfOmega(n)=2pi log n $



    therefore,



    $displaystyle lim_{n to infty} left(1+dfrac{mathbf Omega(n)}{4pi}right)^{log(n+1)}=displaystyle lim_{n to infty} left(1+dfrac{log n}{2}right)^{log(n+1)}$



    after this step, i don't know how to proceed to find final answer. my integral might be wrong so, please help me re-evaluate if possible and also help me find limit



    thanks in advance






    integration complex-analysis definite-integrals contest-math complex-integration







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      Question :



      $mathbfOmega(n)=displaystyleint _{0}^{2pi}log(n^2-2ncos t+1)dt , ngeq1 $



      then , find :



      $displaystyle lim_{n to infty} left(1+dfrac{mathbf Omega(n)}{4pi}right)^{log(n+1)}$



      my attempt:



      $mathbfOmega(n)=displaystyleint _{0}^{2pi}log(n-e^{it})dt+displaystyleint _{0}^{2pi}log(n-e^{-it})dt $



      $mathbfOmega(n)=2displaystyleint _{0}^{2pi} log n dt+displaystyle int_{0}^{2pi}logleft(1-dfrac{e^{it}}{n}right)dt+displaystyle int_{0}^{2pi}logleft(1-dfrac{e^{-it}}{n}right)dt$



      $mathbfOmega(n)=2pi log n-displaystyle int_{0}^{2pi}displaystyle sum_{k=1}^{infty} dfrac{e^{ikt}}{k n^k}dt-displaystyle int_{0}^{2pi}displaystyle sum_{k=1}^{infty} dfrac{e^{-ikt}}{k n^k}dt$



      changing order of summation and integration



      $mathbfOmega(n)=2pi log n -displaystyle sum_{k=1}^{infty}dfrac{1}{kn^k}displaystyle int_{0}^{2pi}e^{ikt}dt - displaystyle sum_{k=1}^{infty}dfrac{1}{kn^k}displaystyle int_{0}^{2pi}e^{-ikt}dt$



      both integrals becomes zero becuase they are integration over a period of sinusoid



      $mathbfOmega(n)=2pi log n $



      therefore,



      $displaystyle lim_{n to infty} left(1+dfrac{mathbf Omega(n)}{4pi}right)^{log(n+1)}=displaystyle lim_{n to infty} left(1+dfrac{log n}{2}right)^{log(n+1)}$



      after this step, i don't know how to proceed to find final answer. my integral might be wrong so, please help me re-evaluate if possible and also help me find limit



      thanks in advance






      integration complex-analysis definite-integrals contest-math complex-integration







      share|cite|improve this question











      $endgroup$




      Question :



      $mathbfOmega(n)=displaystyleint _{0}^{2pi}log(n^2-2ncos t+1)dt , ngeq1 $



      then , find :



      $displaystyle lim_{n to infty} left(1+dfrac{mathbf Omega(n)}{4pi}right)^{log(n+1)}$



      my attempt:



      $mathbfOmega(n)=displaystyleint _{0}^{2pi}log(n-e^{it})dt+displaystyleint _{0}^{2pi}log(n-e^{-it})dt $



      $mathbfOmega(n)=2displaystyleint _{0}^{2pi} log n dt+displaystyle int_{0}^{2pi}logleft(1-dfrac{e^{it}}{n}right)dt+displaystyle int_{0}^{2pi}logleft(1-dfrac{e^{-it}}{n}right)dt$



      $mathbfOmega(n)=2pi log n-displaystyle int_{0}^{2pi}displaystyle sum_{k=1}^{infty} dfrac{e^{ikt}}{k n^k}dt-displaystyle int_{0}^{2pi}displaystyle sum_{k=1}^{infty} dfrac{e^{-ikt}}{k n^k}dt$



      changing order of summation and integration



      $mathbfOmega(n)=2pi log n -displaystyle sum_{k=1}^{infty}dfrac{1}{kn^k}displaystyle int_{0}^{2pi}e^{ikt}dt - displaystyle sum_{k=1}^{infty}dfrac{1}{kn^k}displaystyle int_{0}^{2pi}e^{-ikt}dt$



      both integrals becomes zero becuase they are integration over a period of sinusoid



      $mathbfOmega(n)=2pi log n $



      therefore,



      $displaystyle lim_{n to infty} left(1+dfrac{mathbf Omega(n)}{4pi}right)^{log(n+1)}=displaystyle lim_{n to infty} left(1+dfrac{log n}{2}right)^{log(n+1)}$



      after this step, i don't know how to proceed to find final answer. my integral might be wrong so, please help me re-evaluate if possible and also help me find limit



      thanks in advance






      integration complex-analysis definite-integrals contest-math complex-integration




      integration complex-analysis definite-integrals contest-math complex-integration






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      share|cite|improve this question













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      edited Dec 6 '18 at 17:12









      Arturo Magidin

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      asked Dec 6 '18 at 15:17









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          $begingroup$

          You don't really need to calculate $Omega(n)$.



          Note that $Omega(n)geq clog n$ for some $c>0$ and all sufficiently large $n$ (e.g., from bounding the integrand $log(n^2-2nlog t+1)$ by $2log(n-1)$), so $(1+Omega(n)/(4pi))^{log(n+1)}$ is of the form $infty^infty$ so must $toinfty$ as $ntoinfty$.






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            $begingroup$

            You don't really need to calculate $Omega(n)$.



            Note that $Omega(n)geq clog n$ for some $c>0$ and all sufficiently large $n$ (e.g., from bounding the integrand $log(n^2-2nlog t+1)$ by $2log(n-1)$), so $(1+Omega(n)/(4pi))^{log(n+1)}$ is of the form $infty^infty$ so must $toinfty$ as $ntoinfty$.






            share|cite|improve this answer









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              2












              $begingroup$

              You don't really need to calculate $Omega(n)$.



              Note that $Omega(n)geq clog n$ for some $c>0$ and all sufficiently large $n$ (e.g., from bounding the integrand $log(n^2-2nlog t+1)$ by $2log(n-1)$), so $(1+Omega(n)/(4pi))^{log(n+1)}$ is of the form $infty^infty$ so must $toinfty$ as $ntoinfty$.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                You don't really need to calculate $Omega(n)$.



                Note that $Omega(n)geq clog n$ for some $c>0$ and all sufficiently large $n$ (e.g., from bounding the integrand $log(n^2-2nlog t+1)$ by $2log(n-1)$), so $(1+Omega(n)/(4pi))^{log(n+1)}$ is of the form $infty^infty$ so must $toinfty$ as $ntoinfty$.






                share|cite|improve this answer









                $endgroup$



                You don't really need to calculate $Omega(n)$.



                Note that $Omega(n)geq clog n$ for some $c>0$ and all sufficiently large $n$ (e.g., from bounding the integrand $log(n^2-2nlog t+1)$ by $2log(n-1)$), so $(1+Omega(n)/(4pi))^{log(n+1)}$ is of the form $infty^infty$ so must $toinfty$ as $ntoinfty$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 6 '18 at 17:59









                user10354138user10354138

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