Proving inequality $int_a^{pi/2}cos^nxdxle e^{-na^2/2}int_0^{pi/2}cos^nxdx$












5












$begingroup$



How to prove $$int_a^{pi/2}cos^nxdxle e^{-na^2/2}int_0^{pi/2}cos^nxdx,$$
where $ninmathbb N$ and $ain[0,pi/2]$?




I noticed that if we can prove
$$cos^nale nae^{-na^2/2}int_0^{pi/2}cos^nxdx,$$
apply $displaystyleint_a^{pi/2}$ to both side, the conclusion will follow. But unfortunately, this inequality above is not true. When $a=0$, $LHS=1>0=RHS$. Also, Wallis' formula can help us find $displaystyleint_0^{pi/2}cos^nxdx$. I'm not sure if it helps.






calculus integration inequality definite-integrals







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  • 1




    $begingroup$
    It may help to note that $$int_0^{pi/2}sin(x)^acos(x)^bdx=frac12text{B}bigg(frac{a+1}2,frac{b+1}2bigg)$$
    $endgroup$
    – clathratus
    Dec 30 '18 at 9:04


















5












$begingroup$



How to prove $$int_a^{pi/2}cos^nxdxle e^{-na^2/2}int_0^{pi/2}cos^nxdx,$$
where $ninmathbb N$ and $ain[0,pi/2]$?




I noticed that if we can prove
$$cos^nale nae^{-na^2/2}int_0^{pi/2}cos^nxdx,$$
apply $displaystyleint_a^{pi/2}$ to both side, the conclusion will follow. But unfortunately, this inequality above is not true. When $a=0$, $LHS=1>0=RHS$. Also, Wallis' formula can help us find $displaystyleint_0^{pi/2}cos^nxdx$. I'm not sure if it helps.






calculus integration inequality definite-integrals







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    It may help to note that $$int_0^{pi/2}sin(x)^acos(x)^bdx=frac12text{B}bigg(frac{a+1}2,frac{b+1}2bigg)$$
    $endgroup$
    – clathratus
    Dec 30 '18 at 9:04
















5












5








5


1



$begingroup$



How to prove $$int_a^{pi/2}cos^nxdxle e^{-na^2/2}int_0^{pi/2}cos^nxdx,$$
where $ninmathbb N$ and $ain[0,pi/2]$?




I noticed that if we can prove
$$cos^nale nae^{-na^2/2}int_0^{pi/2}cos^nxdx,$$
apply $displaystyleint_a^{pi/2}$ to both side, the conclusion will follow. But unfortunately, this inequality above is not true. When $a=0$, $LHS=1>0=RHS$. Also, Wallis' formula can help us find $displaystyleint_0^{pi/2}cos^nxdx$. I'm not sure if it helps.






calculus integration inequality definite-integrals







share|cite|improve this question









$endgroup$





How to prove $$int_a^{pi/2}cos^nxdxle e^{-na^2/2}int_0^{pi/2}cos^nxdx,$$
where $ninmathbb N$ and $ain[0,pi/2]$?




I noticed that if we can prove
$$cos^nale nae^{-na^2/2}int_0^{pi/2}cos^nxdx,$$
apply $displaystyleint_a^{pi/2}$ to both side, the conclusion will follow. But unfortunately, this inequality above is not true. When $a=0$, $LHS=1>0=RHS$. Also, Wallis' formula can help us find $displaystyleint_0^{pi/2}cos^nxdx$. I'm not sure if it helps.






calculus integration inequality definite-integrals




calculus integration inequality definite-integrals






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asked Dec 30 '18 at 6:38









Kemono ChenKemono Chen

3,1991844




3,1991844








  • 1




    $begingroup$
    It may help to note that $$int_0^{pi/2}sin(x)^acos(x)^bdx=frac12text{B}bigg(frac{a+1}2,frac{b+1}2bigg)$$
    $endgroup$
    – clathratus
    Dec 30 '18 at 9:04
















  • 1




    $begingroup$
    It may help to note that $$int_0^{pi/2}sin(x)^acos(x)^bdx=frac12text{B}bigg(frac{a+1}2,frac{b+1}2bigg)$$
    $endgroup$
    – clathratus
    Dec 30 '18 at 9:04










1




1




$begingroup$
It may help to note that $$int_0^{pi/2}sin(x)^acos(x)^bdx=frac12text{B}bigg(frac{a+1}2,frac{b+1}2bigg)$$
$endgroup$
– clathratus
Dec 30 '18 at 9:04






$begingroup$
It may help to note that $$int_0^{pi/2}sin(x)^acos(x)^bdx=frac12text{B}bigg(frac{a+1}2,frac{b+1}2bigg)$$
$endgroup$
– clathratus
Dec 30 '18 at 9:04












1 Answer
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$begingroup$

Let $0leq xleq frac{pi}{2}$ and $0leq aleq frac{pi}{2}$. Then we can see
$$
cos(x+a) = cos xcos a-sin xsin aleq cos acos x.
$$
Let $$
f(t) = -frac{t^2}{2} -ln (cos t), quad 0leq t<frac{pi}{2}.
$$
Then, $f(0) = 0$ and
$$
f'(t) = -t +tan tgeq 0.
$$ Thus we have
$$
f(a) geq 0,
$$and
$$
cos a leq e^{frac{-a^2}{2}}.
$$ This implies
$$
cos^n(x+a) leq e^{-frac{na^2}{2}}cos^n x
$$ and hence
$$
int_0^{frac{pi}{2}-a}cos^n(x+a)leq e^{-frac{na^2}{2}}int_0^{frac{pi}{2}-a}cos^n xleq e^{-frac{na^2}{2}}int_0^{frac{pi}{2}}cos^n x.
$$






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    6












    $begingroup$

    Let $0leq xleq frac{pi}{2}$ and $0leq aleq frac{pi}{2}$. Then we can see
    $$
    cos(x+a) = cos xcos a-sin xsin aleq cos acos x.
    $$
    Let $$
    f(t) = -frac{t^2}{2} -ln (cos t), quad 0leq t<frac{pi}{2}.
    $$
    Then, $f(0) = 0$ and
    $$
    f'(t) = -t +tan tgeq 0.
    $$     Thus we have
    $$
    f(a) geq 0,
    $$    and
    $$
    cos a leq e^{frac{-a^2}{2}}.
    $$     This implies
    $$
    cos^n(x+a) leq e^{-frac{na^2}{2}}cos^n x
    $$     and hence
    $$
    int_0^{frac{pi}{2}-a}cos^n(x+a)leq e^{-frac{na^2}{2}}int_0^{frac{pi}{2}-a}cos^n xleq e^{-frac{na^2}{2}}int_0^{frac{pi}{2}}cos^n x.
    $$






    share|cite|improve this answer









    $endgroup$


















      6












      $begingroup$

      Let $0leq xleq frac{pi}{2}$ and $0leq aleq frac{pi}{2}$. Then we can see
      $$
      cos(x+a) = cos xcos a-sin xsin aleq cos acos x.
      $$
      Let $$
      f(t) = -frac{t^2}{2} -ln (cos t), quad 0leq t<frac{pi}{2}.
      $$
      Then, $f(0) = 0$ and
      $$
      f'(t) = -t +tan tgeq 0.
      $$       Thus we have
      $$
      f(a) geq 0,
      $$      and
      $$
      cos a leq e^{frac{-a^2}{2}}.
      $$       This implies
      $$
      cos^n(x+a) leq e^{-frac{na^2}{2}}cos^n x
      $$       and hence
      $$
      int_0^{frac{pi}{2}-a}cos^n(x+a)leq e^{-frac{na^2}{2}}int_0^{frac{pi}{2}-a}cos^n xleq e^{-frac{na^2}{2}}int_0^{frac{pi}{2}}cos^n x.
      $$






      share|cite|improve this answer









      $endgroup$
















        6












        6








        6





        $begingroup$

        Let $0leq xleq frac{pi}{2}$ and $0leq aleq frac{pi}{2}$. Then we can see
        $$
        cos(x+a) = cos xcos a-sin xsin aleq cos acos x.
        $$
        Let $$
        f(t) = -frac{t^2}{2} -ln (cos t), quad 0leq t<frac{pi}{2}.
        $$
        Then, $f(0) = 0$ and
        $$
        f'(t) = -t +tan tgeq 0.
        $$         Thus we have
        $$
        f(a) geq 0,
        $$        and
        $$
        cos a leq e^{frac{-a^2}{2}}.
        $$         This implies
        $$
        cos^n(x+a) leq e^{-frac{na^2}{2}}cos^n x
        $$         and hence
        $$
        int_0^{frac{pi}{2}-a}cos^n(x+a)leq e^{-frac{na^2}{2}}int_0^{frac{pi}{2}-a}cos^n xleq e^{-frac{na^2}{2}}int_0^{frac{pi}{2}}cos^n x.
        $$






        share|cite|improve this answer









        $endgroup$



        Let $0leq xleq frac{pi}{2}$ and $0leq aleq frac{pi}{2}$. Then we can see
        $$
        cos(x+a) = cos xcos a-sin xsin aleq cos acos x.
        $$
        Let $$
        f(t) = -frac{t^2}{2} -ln (cos t), quad 0leq t<frac{pi}{2}.
        $$
        Then, $f(0) = 0$ and
        $$
        f'(t) = -t +tan tgeq 0.
        $$         Thus we have
        $$
        f(a) geq 0,
        $$        and
        $$
        cos a leq e^{frac{-a^2}{2}}.
        $$         This implies
        $$
        cos^n(x+a) leq e^{-frac{na^2}{2}}cos^n x
        $$         and hence
        $$
        int_0^{frac{pi}{2}-a}cos^n(x+a)leq e^{-frac{na^2}{2}}int_0^{frac{pi}{2}-a}cos^n xleq e^{-frac{na^2}{2}}int_0^{frac{pi}{2}}cos^n x.
        $$







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        answered Dec 30 '18 at 8:18









        SongSong

        18.5k21550




        18.5k21550






























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