The limit of an integral involving Fourier expansion












1












$begingroup$


$F_n(t)=frac{1}{2npi}$ $(frac{sin frac{nt}{2}}{sin frac{t}{2}})^2$



Proof



$int_{deltale |t|le pi} F_n(t) dtrightarrow 0 ,nrightarrow +infty$



My attempt



$frac{1}{2}+cos t+cos 2t+dots +cos nt=frac{sin frac{(2n+1)t}{2}}{2sinfrac{t}{2}}$——(1)



And
$$sum_{m=0}^{n-1}sin (m+frac{1}{2})t=frac{sin ^2frac{nt}{2}}{sin frac{t}{2}}$$



So $int_{deltale |t|le pi} F_n(t) dt= frac{1}{2npi} int_{deltale |t|le pi} frac{sum_{m=0}^{n-1}sin (m+frac{1}{2})t}{sinfrac{t}{2}}dt$
And use the (1)the integral can be change to some integral of cos blabla but i don’t know what should I do next






real-analysis calculus integration limits analysis







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    $F_n(t)=frac{1}{2npi}$ $(frac{sin frac{nt}{2}}{sin frac{t}{2}})^2$



    Proof



    $int_{deltale |t|le pi} F_n(t) dtrightarrow 0 ,nrightarrow +infty$



    My attempt



    $frac{1}{2}+cos t+cos 2t+dots +cos nt=frac{sin frac{(2n+1)t}{2}}{2sinfrac{t}{2}}$——(1)



    And
    $$sum_{m=0}^{n-1}sin (m+frac{1}{2})t=frac{sin ^2frac{nt}{2}}{sin frac{t}{2}}$$



    So $int_{deltale |t|le pi} F_n(t) dt= frac{1}{2npi} int_{deltale |t|le pi} frac{sum_{m=0}^{n-1}sin (m+frac{1}{2})t}{sinfrac{t}{2}}dt$
    And use the (1)the integral can be change to some integral of cos blabla but i don’t know what should I do next






    real-analysis calculus integration limits analysis







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$


      $F_n(t)=frac{1}{2npi}$ $(frac{sin frac{nt}{2}}{sin frac{t}{2}})^2$



      Proof



      $int_{deltale |t|le pi} F_n(t) dtrightarrow 0 ,nrightarrow +infty$



      My attempt



      $frac{1}{2}+cos t+cos 2t+dots +cos nt=frac{sin frac{(2n+1)t}{2}}{2sinfrac{t}{2}}$——(1)



      And
      $$sum_{m=0}^{n-1}sin (m+frac{1}{2})t=frac{sin ^2frac{nt}{2}}{sin frac{t}{2}}$$



      So $int_{deltale |t|le pi} F_n(t) dt= frac{1}{2npi} int_{deltale |t|le pi} frac{sum_{m=0}^{n-1}sin (m+frac{1}{2})t}{sinfrac{t}{2}}dt$
      And use the (1)the integral can be change to some integral of cos blabla but i don’t know what should I do next






      real-analysis calculus integration limits analysis







      share|cite|improve this question











      $endgroup$




      $F_n(t)=frac{1}{2npi}$ $(frac{sin frac{nt}{2}}{sin frac{t}{2}})^2$



      Proof



      $int_{deltale |t|le pi} F_n(t) dtrightarrow 0 ,nrightarrow +infty$



      My attempt



      $frac{1}{2}+cos t+cos 2t+dots +cos nt=frac{sin frac{(2n+1)t}{2}}{2sinfrac{t}{2}}$——(1)



      And
      $$sum_{m=0}^{n-1}sin (m+frac{1}{2})t=frac{sin ^2frac{nt}{2}}{sin frac{t}{2}}$$



      So $int_{deltale |t|le pi} F_n(t) dt= frac{1}{2npi} int_{deltale |t|le pi} frac{sum_{m=0}^{n-1}sin (m+frac{1}{2})t}{sinfrac{t}{2}}dt$
      And use the (1)the integral can be change to some integral of cos blabla but i don’t know what should I do next






      real-analysis calculus integration limits analysis




      real-analysis calculus integration limits analysis






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 2 at 16:35









      Berkheimer

      1,437924




      1,437924










      asked Jan 2 at 15:59









      jacksonjackson

      819




      819






















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          $begingroup$

          You can use the raw form of $F_n(t)$. It's quite a simple estimate. We have
          $$
          frac{1}{2npi}frac{ sin^2 frac{nt}{2}}{sin^2 frac{t}{2}}le frac{1}{2npi}frac{1}{sin^2 frac{delta}{2}},
          $$           for $deltale|t|le pi$ and
          $$
          int_{deltale |t|le pi} F_n(t) dtle int_{deltale |t|le pi} frac{1}{2npi}frac{1}{sin^2 frac{delta}{2}} dtle frac{1}{n}frac{1}{sin^2 frac{delta}{2}}to 0
          $$           as $nto infty$.






          share|cite|improve this answer









          $endgroup$













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            $begingroup$

            You can use the raw form of $F_n(t)$. It's quite a simple estimate. We have
            $$
            frac{1}{2npi}frac{ sin^2 frac{nt}{2}}{sin^2 frac{t}{2}}le frac{1}{2npi}frac{1}{sin^2 frac{delta}{2}},
            $$             for $deltale|t|le pi$ and
            $$
            int_{deltale |t|le pi} F_n(t) dtle int_{deltale |t|le pi} frac{1}{2npi}frac{1}{sin^2 frac{delta}{2}} dtle frac{1}{n}frac{1}{sin^2 frac{delta}{2}}to 0
            $$             as $nto infty$.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              You can use the raw form of $F_n(t)$. It's quite a simple estimate. We have
              $$
              frac{1}{2npi}frac{ sin^2 frac{nt}{2}}{sin^2 frac{t}{2}}le frac{1}{2npi}frac{1}{sin^2 frac{delta}{2}},
              $$               for $deltale|t|le pi$ and
              $$
              int_{deltale |t|le pi} F_n(t) dtle int_{deltale |t|le pi} frac{1}{2npi}frac{1}{sin^2 frac{delta}{2}} dtle frac{1}{n}frac{1}{sin^2 frac{delta}{2}}to 0
              $$               as $nto infty$.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                You can use the raw form of $F_n(t)$. It's quite a simple estimate. We have
                $$
                frac{1}{2npi}frac{ sin^2 frac{nt}{2}}{sin^2 frac{t}{2}}le frac{1}{2npi}frac{1}{sin^2 frac{delta}{2}},
                $$                 for $deltale|t|le pi$ and
                $$
                int_{deltale |t|le pi} F_n(t) dtle int_{deltale |t|le pi} frac{1}{2npi}frac{1}{sin^2 frac{delta}{2}} dtle frac{1}{n}frac{1}{sin^2 frac{delta}{2}}to 0
                $$                 as $nto infty$.






                share|cite|improve this answer









                $endgroup$



                You can use the raw form of $F_n(t)$. It's quite a simple estimate. We have
                $$
                frac{1}{2npi}frac{ sin^2 frac{nt}{2}}{sin^2 frac{t}{2}}le frac{1}{2npi}frac{1}{sin^2 frac{delta}{2}},
                $$                 for $deltale|t|le pi$ and
                $$
                int_{deltale |t|le pi} F_n(t) dtle int_{deltale |t|le pi} frac{1}{2npi}frac{1}{sin^2 frac{delta}{2}} dtle frac{1}{n}frac{1}{sin^2 frac{delta}{2}}to 0
                $$                 as $nto infty$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 2 at 17:10









                SongSong

                7,851423




                7,851423






























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