Find the sup of an integral succession












0














Let
$$
a_n=int^pi_0e^{-nx}sin(n^2x)dx
$$



find $sup{a_n:ninmathbb N}$




I found myself the solution.



Integrating twice by parts we have
$$
a_n=nint^pi_0e^{-nx}cos(n^2x)dx=left[-e^{-nx}cos(n^2x)right]^pi_{x=0}-n^2int^pi_0e^{-nx}sin(n^2x)dx=1-e^{-npi}cos(n^2pi)-n^2a_n\
a_n=frac{1-e^{-npi}cos(n^2pi)}{1+n^2}=frac{1-(-1)^ne^{-npi}}{1+n^2}\
a_{2k}=frac{1-e^{-2kpi}}{1+(2k)^2}\
a_{2k+1}=frac{1+e^{-(2k+1)pi}}{1+(2k+1)^2}
$$
then
$$
a_{2k}<a_{2k-1}<a_1=frac{1+e^{-pi}}{2}
$$






calculus integration supremum-and-infimum







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  • 1




    Did you compute $a_n$ ? If you did, what I hope and wish, include your result in the post.
    – Claude Leibovici
    Nov 22 '18 at 11:23


















0














Let
$$
a_n=int^pi_0e^{-nx}sin(n^2x)dx
$$



find $sup{a_n:ninmathbb N}$




I found myself the solution.



Integrating twice by parts we have
$$
a_n=nint^pi_0e^{-nx}cos(n^2x)dx=left[-e^{-nx}cos(n^2x)right]^pi_{x=0}-n^2int^pi_0e^{-nx}sin(n^2x)dx=1-e^{-npi}cos(n^2pi)-n^2a_n\
a_n=frac{1-e^{-npi}cos(n^2pi)}{1+n^2}=frac{1-(-1)^ne^{-npi}}{1+n^2}\
a_{2k}=frac{1-e^{-2kpi}}{1+(2k)^2}\
a_{2k+1}=frac{1+e^{-(2k+1)pi}}{1+(2k+1)^2}
$$
then
$$
a_{2k}<a_{2k-1}<a_1=frac{1+e^{-pi}}{2}
$$






calculus integration supremum-and-infimum







share|cite|improve this question




















  • 1




    Did you compute $a_n$ ? If you did, what I hope and wish, include your result in the post.
    – Claude Leibovici
    Nov 22 '18 at 11:23
















0












0








0







Let
$$
a_n=int^pi_0e^{-nx}sin(n^2x)dx
$$



find $sup{a_n:ninmathbb N}$




I found myself the solution.



Integrating twice by parts we have
$$
a_n=nint^pi_0e^{-nx}cos(n^2x)dx=left[-e^{-nx}cos(n^2x)right]^pi_{x=0}-n^2int^pi_0e^{-nx}sin(n^2x)dx=1-e^{-npi}cos(n^2pi)-n^2a_n\
a_n=frac{1-e^{-npi}cos(n^2pi)}{1+n^2}=frac{1-(-1)^ne^{-npi}}{1+n^2}\
a_{2k}=frac{1-e^{-2kpi}}{1+(2k)^2}\
a_{2k+1}=frac{1+e^{-(2k+1)pi}}{1+(2k+1)^2}
$$
then
$$
a_{2k}<a_{2k-1}<a_1=frac{1+e^{-pi}}{2}
$$






calculus integration supremum-and-infimum







share|cite|improve this question















Let
$$
a_n=int^pi_0e^{-nx}sin(n^2x)dx
$$



find $sup{a_n:ninmathbb N}$




I found myself the solution.



Integrating twice by parts we have
$$
a_n=nint^pi_0e^{-nx}cos(n^2x)dx=left[-e^{-nx}cos(n^2x)right]^pi_{x=0}-n^2int^pi_0e^{-nx}sin(n^2x)dx=1-e^{-npi}cos(n^2pi)-n^2a_n\
a_n=frac{1-e^{-npi}cos(n^2pi)}{1+n^2}=frac{1-(-1)^ne^{-npi}}{1+n^2}\
a_{2k}=frac{1-e^{-2kpi}}{1+(2k)^2}\
a_{2k+1}=frac{1+e^{-(2k+1)pi}}{1+(2k+1)^2}
$$
then
$$
a_{2k}<a_{2k-1}<a_1=frac{1+e^{-pi}}{2}
$$






calculus integration supremum-and-infimum




calculus integration supremum-and-infimum






share|cite|improve this question















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edited Nov 23 '18 at 3:13









Saad

19.7k92352




19.7k92352










asked Nov 22 '18 at 11:07









P De Donato

4147




4147








  • 1




    Did you compute $a_n$ ? If you did, what I hope and wish, include your result in the post.
    – Claude Leibovici
    Nov 22 '18 at 11:23
















  • 1




    Did you compute $a_n$ ? If you did, what I hope and wish, include your result in the post.
    – Claude Leibovici
    Nov 22 '18 at 11:23










1




1




Did you compute $a_n$ ? If you did, what I hope and wish, include your result in the post.
– Claude Leibovici
Nov 22 '18 at 11:23






Did you compute $a_n$ ? If you did, what I hope and wish, include your result in the post.
– Claude Leibovici
Nov 22 '18 at 11:23












1 Answer
1






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oldest

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0














I found myself the solution.



Integrating twice by parts we have
$$
a_n=nint^pi_0e^{-nx}cos(n^2x)dx=left[-e^{-nx}cos(n^2x)right]^pi_{x=0}-n^2int^pi_0e^{-nx}sin(n^2x)dx=1-e^{-npi}cos(n^2pi)-n^2a_n\
a_n=frac{1-e^{-npi}cos(n^2pi)}{1+n^2}=frac{1-(-1)^ne^{-npi}}{1+n^2}\
a_{2k}=frac{1-e^{-2kpi}}{1+(2k)^2}\
a_{2k+1}=frac{1+e^{-(2k+1)pi}}{1+(2k+1)^2}
$$
then
$$
a_{2k}<a_{2k-1}<a_1=frac{1+e^{-pi}}{2}
$$






share|cite|improve this answer





















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    I found myself the solution.



    Integrating twice by parts we have
    $$
    a_n=nint^pi_0e^{-nx}cos(n^2x)dx=left[-e^{-nx}cos(n^2x)right]^pi_{x=0}-n^2int^pi_0e^{-nx}sin(n^2x)dx=1-e^{-npi}cos(n^2pi)-n^2a_n\
    a_n=frac{1-e^{-npi}cos(n^2pi)}{1+n^2}=frac{1-(-1)^ne^{-npi}}{1+n^2}\
    a_{2k}=frac{1-e^{-2kpi}}{1+(2k)^2}\
    a_{2k+1}=frac{1+e^{-(2k+1)pi}}{1+(2k+1)^2}
    $$
    then
    $$
    a_{2k}<a_{2k-1}<a_1=frac{1+e^{-pi}}{2}
    $$






    share|cite|improve this answer


























      0














      I found myself the solution.



      Integrating twice by parts we have
      $$
      a_n=nint^pi_0e^{-nx}cos(n^2x)dx=left[-e^{-nx}cos(n^2x)right]^pi_{x=0}-n^2int^pi_0e^{-nx}sin(n^2x)dx=1-e^{-npi}cos(n^2pi)-n^2a_n\
      a_n=frac{1-e^{-npi}cos(n^2pi)}{1+n^2}=frac{1-(-1)^ne^{-npi}}{1+n^2}\
      a_{2k}=frac{1-e^{-2kpi}}{1+(2k)^2}\
      a_{2k+1}=frac{1+e^{-(2k+1)pi}}{1+(2k+1)^2}
      $$
      then
      $$
      a_{2k}<a_{2k-1}<a_1=frac{1+e^{-pi}}{2}
      $$






      share|cite|improve this answer
























        0












        0








        0






        I found myself the solution.



        Integrating twice by parts we have
        $$
        a_n=nint^pi_0e^{-nx}cos(n^2x)dx=left[-e^{-nx}cos(n^2x)right]^pi_{x=0}-n^2int^pi_0e^{-nx}sin(n^2x)dx=1-e^{-npi}cos(n^2pi)-n^2a_n\
        a_n=frac{1-e^{-npi}cos(n^2pi)}{1+n^2}=frac{1-(-1)^ne^{-npi}}{1+n^2}\
        a_{2k}=frac{1-e^{-2kpi}}{1+(2k)^2}\
        a_{2k+1}=frac{1+e^{-(2k+1)pi}}{1+(2k+1)^2}
        $$
        then
        $$
        a_{2k}<a_{2k-1}<a_1=frac{1+e^{-pi}}{2}
        $$






        share|cite|improve this answer












        I found myself the solution.



        Integrating twice by parts we have
        $$
        a_n=nint^pi_0e^{-nx}cos(n^2x)dx=left[-e^{-nx}cos(n^2x)right]^pi_{x=0}-n^2int^pi_0e^{-nx}sin(n^2x)dx=1-e^{-npi}cos(n^2pi)-n^2a_n\
        a_n=frac{1-e^{-npi}cos(n^2pi)}{1+n^2}=frac{1-(-1)^ne^{-npi}}{1+n^2}\
        a_{2k}=frac{1-e^{-2kpi}}{1+(2k)^2}\
        a_{2k+1}=frac{1+e^{-(2k+1)pi}}{1+(2k+1)^2}
        $$
        then
        $$
        a_{2k}<a_{2k-1}<a_1=frac{1+e^{-pi}}{2}
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 22 '18 at 21:29









        P De Donato

        4147




        4147






























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