Calculate $lim_{n to infty} int_{mathbb{R_{+}}} exp((cos^n x) -x) dlambda(x)$












3












$begingroup$


the exponential function being increasing we have $| exp((cos^n x) -x)| leq exp(1 -x) in L^1([0,+infty[) $



so $x to exp((cos^n x) -x)$ is Riemann absolutely convergent therefore



$l = lim_{n to infty} int_{mathbb{R_{+}}} exp((cos x^n) -x) dlambda(x) =lim_{n to infty} int_{0}^{+infty} exp((cos^n x) -x) dx $



by the dominated convergence theorem :



$l = int_{0}^{+infty} lim_{n to infty} exp((cos^n x) -x) dx$



I don't know how to deal with this limit, as $x$ is in $mathbb{R_{+}}$ I can't even use a taylor expression around $0$



any hints ?






integration measure-theory improper-integrals lebesgue-integral lebesgue-measure







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    3












    $begingroup$


    the exponential function being increasing we have $| exp((cos^n x) -x)| leq exp(1 -x) in L^1([0,+infty[) $



    so $x to exp((cos^n x) -x)$ is Riemann absolutely convergent therefore



    $l = lim_{n to infty} int_{mathbb{R_{+}}} exp((cos x^n) -x) dlambda(x) =lim_{n to infty} int_{0}^{+infty} exp((cos^n x) -x) dx $



    by the dominated convergence theorem :



    $l = int_{0}^{+infty} lim_{n to infty} exp((cos^n x) -x) dx$



    I don't know how to deal with this limit, as $x$ is in $mathbb{R_{+}}$ I can't even use a taylor expression around $0$



    any hints ?






    integration measure-theory improper-integrals lebesgue-integral lebesgue-measure







    share|cite|improve this question









    $endgroup$















      3












      3








      3





      $begingroup$


      the exponential function being increasing we have $| exp((cos^n x) -x)| leq exp(1 -x) in L^1([0,+infty[) $



      so $x to exp((cos^n x) -x)$ is Riemann absolutely convergent therefore



      $l = lim_{n to infty} int_{mathbb{R_{+}}} exp((cos x^n) -x) dlambda(x) =lim_{n to infty} int_{0}^{+infty} exp((cos^n x) -x) dx $



      by the dominated convergence theorem :



      $l = int_{0}^{+infty} lim_{n to infty} exp((cos^n x) -x) dx$



      I don't know how to deal with this limit, as $x$ is in $mathbb{R_{+}}$ I can't even use a taylor expression around $0$



      any hints ?






      integration measure-theory improper-integrals lebesgue-integral lebesgue-measure







      share|cite|improve this question









      $endgroup$




      the exponential function being increasing we have $| exp((cos^n x) -x)| leq exp(1 -x) in L^1([0,+infty[) $



      so $x to exp((cos^n x) -x)$ is Riemann absolutely convergent therefore



      $l = lim_{n to infty} int_{mathbb{R_{+}}} exp((cos x^n) -x) dlambda(x) =lim_{n to infty} int_{0}^{+infty} exp((cos^n x) -x) dx $



      by the dominated convergence theorem :



      $l = int_{0}^{+infty} lim_{n to infty} exp((cos^n x) -x) dx$



      I don't know how to deal with this limit, as $x$ is in $mathbb{R_{+}}$ I can't even use a taylor expression around $0$



      any hints ?






      integration measure-theory improper-integrals lebesgue-integral lebesgue-measure




      integration measure-theory improper-integrals lebesgue-integral lebesgue-measure






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      asked Jan 9 at 18:26









      rapidracimrapidracim

      1,7191419




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          2 Answers
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          $begingroup$

          For $x in mathbb R_+ setminus{kpi + pi/2 ; k in mathbb N}$ you have:



          $$exp((cos^n x) -x) to e^{-x}$$ as $n to infty$ and



          $$0 le exp((cos^n x) -x) le e cdot e^{-x}$$ for all $x in mathbb R_+$.



          As ${kpi + pi/2 ; k in mathbb N}$ is a null set (for Lebesgue measure) and $int_{mathbb R_+} e^{-x} dx$ converges, you can apply Lebesgue dominated convergence theorem and conclude that



          $$lim_{n to infty} int_{mathbb{R_{+}}} exp((cos^n x) -x) dlambda(x) = int_{mathbb R_+} e^{-x} dx$$






          share|cite|improve this answer











          $endgroup$





















            1












            $begingroup$

            Notice that as $n to infty$ all values of $cos^n(x)$ go to zero except the exact points at which $cos(x)=1$ which have infinitesimally small width so have a value of zero when integrating over them. The integral then becomes simply the integral of $exp(-x)$.






            share|cite|improve this answer









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              2 Answers
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              active

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              2 Answers
              2






              active

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              active

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              active

              oldest

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              1












              $begingroup$

              For $x in mathbb R_+ setminus{kpi + pi/2 ; k in mathbb N}$ you have:



              $$exp((cos^n x) -x) to e^{-x}$$ as $n to infty$ and



              $$0 le exp((cos^n x) -x) le e cdot e^{-x}$$ for all $x in mathbb R_+$.



              As ${kpi + pi/2 ; k in mathbb N}$ is a null set (for Lebesgue measure) and $int_{mathbb R_+} e^{-x} dx$ converges, you can apply Lebesgue dominated convergence theorem and conclude that



              $$lim_{n to infty} int_{mathbb{R_{+}}} exp((cos^n x) -x) dlambda(x) = int_{mathbb R_+} e^{-x} dx$$






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                For $x in mathbb R_+ setminus{kpi + pi/2 ; k in mathbb N}$ you have:



                $$exp((cos^n x) -x) to e^{-x}$$ as $n to infty$ and



                $$0 le exp((cos^n x) -x) le e cdot e^{-x}$$ for all $x in mathbb R_+$.



                As ${kpi + pi/2 ; k in mathbb N}$ is a null set (for Lebesgue measure) and $int_{mathbb R_+} e^{-x} dx$ converges, you can apply Lebesgue dominated convergence theorem and conclude that



                $$lim_{n to infty} int_{mathbb{R_{+}}} exp((cos^n x) -x) dlambda(x) = int_{mathbb R_+} e^{-x} dx$$






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  For $x in mathbb R_+ setminus{kpi + pi/2 ; k in mathbb N}$ you have:



                  $$exp((cos^n x) -x) to e^{-x}$$ as $n to infty$ and



                  $$0 le exp((cos^n x) -x) le e cdot e^{-x}$$ for all $x in mathbb R_+$.



                  As ${kpi + pi/2 ; k in mathbb N}$ is a null set (for Lebesgue measure) and $int_{mathbb R_+} e^{-x} dx$ converges, you can apply Lebesgue dominated convergence theorem and conclude that



                  $$lim_{n to infty} int_{mathbb{R_{+}}} exp((cos^n x) -x) dlambda(x) = int_{mathbb R_+} e^{-x} dx$$






                  share|cite|improve this answer











                  $endgroup$



                  For $x in mathbb R_+ setminus{kpi + pi/2 ; k in mathbb N}$ you have:



                  $$exp((cos^n x) -x) to e^{-x}$$ as $n to infty$ and



                  $$0 le exp((cos^n x) -x) le e cdot e^{-x}$$ for all $x in mathbb R_+$.



                  As ${kpi + pi/2 ; k in mathbb N}$ is a null set (for Lebesgue measure) and $int_{mathbb R_+} e^{-x} dx$ converges, you can apply Lebesgue dominated convergence theorem and conclude that



                  $$lim_{n to infty} int_{mathbb{R_{+}}} exp((cos^n x) -x) dlambda(x) = int_{mathbb R_+} e^{-x} dx$$







                  share|cite|improve this answer














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                  share|cite|improve this answer








                  edited Jan 9 at 21:36

























                  answered Jan 9 at 18:44









                  mathcounterexamples.netmathcounterexamples.net

                  26.9k22157




                  26.9k22157























                      1












                      $begingroup$

                      Notice that as $n to infty$ all values of $cos^n(x)$ go to zero except the exact points at which $cos(x)=1$ which have infinitesimally small width so have a value of zero when integrating over them. The integral then becomes simply the integral of $exp(-x)$.






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        Notice that as $n to infty$ all values of $cos^n(x)$ go to zero except the exact points at which $cos(x)=1$ which have infinitesimally small width so have a value of zero when integrating over them. The integral then becomes simply the integral of $exp(-x)$.






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          Notice that as $n to infty$ all values of $cos^n(x)$ go to zero except the exact points at which $cos(x)=1$ which have infinitesimally small width so have a value of zero when integrating over them. The integral then becomes simply the integral of $exp(-x)$.






                          share|cite|improve this answer









                          $endgroup$



                          Notice that as $n to infty$ all values of $cos^n(x)$ go to zero except the exact points at which $cos(x)=1$ which have infinitesimally small width so have a value of zero when integrating over them. The integral then becomes simply the integral of $exp(-x)$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 9 at 18:36









                          Peter ForemanPeter Foreman

                          2,5441214




                          2,5441214






























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