Why is $int_{[0,infty[}e^{-xy}|sin{x}|dlambda(x) leq 2int_{A}e^{-xy}sin{x}dlambda(x)$












1












$begingroup$


Let $A:= bigcup_{k=0}^{infty}[2kpi,(2k+1)pi]$.



I do not understand why the following holds, it was written in our textbook



$int_{[0,infty[}e^{-xy}|sin{x}|dlambda(x) leq 2int_{A}e^{-xy}sin{x}dlambda(x)$



Even reducing it to $int_{[0,2pi[}e^{-xy}|sin{x}|dlambda(x) leq 2int_{[0,2pi[}e^{-xy}sin{x}dlambda(x)$



All I can note is that $e^{-xy}$ is decreasing for a fixed $y$ as $x to infty$



and on $[pi,2pi]$where $sin{x} leq 0$ it is hard to see why $e^{-xy}|sin{x}|leq2e^{-xy}sin{x}$



Surely this cannot be true?






real-analysis integration measure-theory multivariable-calculus







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  • $begingroup$
    Can you prove that $int_0^{2 pi} e^{-xy}|sin y| , dy le 2 int_0^pi e^{-xy} sin y , dy$?
    $endgroup$
    – Hans Engler
    Jan 10 at 21:48












  • $begingroup$
    The problem is unfortunately stated by using $lambda(x)dx$ rather than simply $dx$. In an extreme case $lambda(A)=0$ making the right hand side $=0$, but having a positive value for the left side with $lambda(A')ne 0$.
    $endgroup$
    – herb steinberg
    Jan 10 at 23:36
















1












$begingroup$


Let $A:= bigcup_{k=0}^{infty}[2kpi,(2k+1)pi]$.



I do not understand why the following holds, it was written in our textbook



$int_{[0,infty[}e^{-xy}|sin{x}|dlambda(x) leq 2int_{A}e^{-xy}sin{x}dlambda(x)$



Even reducing it to $int_{[0,2pi[}e^{-xy}|sin{x}|dlambda(x) leq 2int_{[0,2pi[}e^{-xy}sin{x}dlambda(x)$



All I can note is that $e^{-xy}$ is decreasing for a fixed $y$ as $x to infty$



and on $[pi,2pi]$where $sin{x} leq 0$ it is hard to see why $e^{-xy}|sin{x}|leq2e^{-xy}sin{x}$



Surely this cannot be true?






real-analysis integration measure-theory multivariable-calculus







share|cite|improve this question









$endgroup$












  • $begingroup$
    Can you prove that $int_0^{2 pi} e^{-xy}|sin y| , dy le 2 int_0^pi e^{-xy} sin y , dy$?
    $endgroup$
    – Hans Engler
    Jan 10 at 21:48












  • $begingroup$
    The problem is unfortunately stated by using $lambda(x)dx$ rather than simply $dx$. In an extreme case $lambda(A)=0$ making the right hand side $=0$, but having a positive value for the left side with $lambda(A')ne 0$.
    $endgroup$
    – herb steinberg
    Jan 10 at 23:36














1












1








1





$begingroup$


Let $A:= bigcup_{k=0}^{infty}[2kpi,(2k+1)pi]$.



I do not understand why the following holds, it was written in our textbook



$int_{[0,infty[}e^{-xy}|sin{x}|dlambda(x) leq 2int_{A}e^{-xy}sin{x}dlambda(x)$



Even reducing it to $int_{[0,2pi[}e^{-xy}|sin{x}|dlambda(x) leq 2int_{[0,2pi[}e^{-xy}sin{x}dlambda(x)$



All I can note is that $e^{-xy}$ is decreasing for a fixed $y$ as $x to infty$



and on $[pi,2pi]$where $sin{x} leq 0$ it is hard to see why $e^{-xy}|sin{x}|leq2e^{-xy}sin{x}$



Surely this cannot be true?






real-analysis integration measure-theory multivariable-calculus







share|cite|improve this question









$endgroup$




Let $A:= bigcup_{k=0}^{infty}[2kpi,(2k+1)pi]$.



I do not understand why the following holds, it was written in our textbook



$int_{[0,infty[}e^{-xy}|sin{x}|dlambda(x) leq 2int_{A}e^{-xy}sin{x}dlambda(x)$



Even reducing it to $int_{[0,2pi[}e^{-xy}|sin{x}|dlambda(x) leq 2int_{[0,2pi[}e^{-xy}sin{x}dlambda(x)$



All I can note is that $e^{-xy}$ is decreasing for a fixed $y$ as $x to infty$



and on $[pi,2pi]$where $sin{x} leq 0$ it is hard to see why $e^{-xy}|sin{x}|leq2e^{-xy}sin{x}$



Surely this cannot be true?






real-analysis integration measure-theory multivariable-calculus




real-analysis integration measure-theory multivariable-calculus






share|cite|improve this question













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asked Jan 10 at 21:27









MinaThumaMinaThuma

1108




1108












  • $begingroup$
    Can you prove that $int_0^{2 pi} e^{-xy}|sin y| , dy le 2 int_0^pi e^{-xy} sin y , dy$?
    $endgroup$
    – Hans Engler
    Jan 10 at 21:48












  • $begingroup$
    The problem is unfortunately stated by using $lambda(x)dx$ rather than simply $dx$. In an extreme case $lambda(A)=0$ making the right hand side $=0$, but having a positive value for the left side with $lambda(A')ne 0$.
    $endgroup$
    – herb steinberg
    Jan 10 at 23:36


















  • $begingroup$
    Can you prove that $int_0^{2 pi} e^{-xy}|sin y| , dy le 2 int_0^pi e^{-xy} sin y , dy$?
    $endgroup$
    – Hans Engler
    Jan 10 at 21:48












  • $begingroup$
    The problem is unfortunately stated by using $lambda(x)dx$ rather than simply $dx$. In an extreme case $lambda(A)=0$ making the right hand side $=0$, but having a positive value for the left side with $lambda(A')ne 0$.
    $endgroup$
    – herb steinberg
    Jan 10 at 23:36
















$begingroup$
Can you prove that $int_0^{2 pi} e^{-xy}|sin y| , dy le 2 int_0^pi e^{-xy} sin y , dy$?
$endgroup$
– Hans Engler
Jan 10 at 21:48






$begingroup$
Can you prove that $int_0^{2 pi} e^{-xy}|sin y| , dy le 2 int_0^pi e^{-xy} sin y , dy$?
$endgroup$
– Hans Engler
Jan 10 at 21:48














$begingroup$
The problem is unfortunately stated by using $lambda(x)dx$ rather than simply $dx$. In an extreme case $lambda(A)=0$ making the right hand side $=0$, but having a positive value for the left side with $lambda(A')ne 0$.
$endgroup$
– herb steinberg
Jan 10 at 23:36




$begingroup$
The problem is unfortunately stated by using $lambda(x)dx$ rather than simply $dx$. In an extreme case $lambda(A)=0$ making the right hand side $=0$, but having a positive value for the left side with $lambda(A')ne 0$.
$endgroup$
– herb steinberg
Jan 10 at 23:36










2 Answers
2






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1












$begingroup$

Let $I_k =[(2k+1)pi,(2k+2)pi]$ and $J_k =[2kpi,(2k+1)pi]$. Then we have
$$begin{eqnarray}
int_{I_k} e^{-xy}|sin{x}|dlambda(x) &=&int_{J_k} e^{-(x+pi)y}|sin(x+pi)|dlambda(x)\
&=&int_{J_k} e^{-(x+pi)y}sin(x)dlambda(x)\
&le&int_{J_k} e^{-xy}sin(x)dlambda(x).
end{eqnarray}$$ Thus we have
$$begin{eqnarray}
int_{[0,infty[}e^{-xy}|sin{x}|dlambda(x)&=&sum_{kge 0}left[int_{I_k} e^{-xy}|sin{x}|dlambda(x)+int_{J_k} e^{-xy}|sin{x}|dlambda(x)right]\
&le&2sum_{kge 0}int_{J_k} e^{-xy}sin{x}dlambda(x)\
&=&2int_{bigcup_{kge 0} J_k} e^{-xy}sin{x}dlambda(x)=2int_{A} e^{-xy}sin{x}dlambda(x).
end{eqnarray}$$






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    1












    $begingroup$

    You need to compare $I=int_0^{pi}e^{-xy}|sinx|dlambda(x)$ with $J=int_{pi}^{2pi}e^{-xy}|sinx|dlambda(x)$ Because of the exponential term, $Jlt I$. (Note the same comparison holds for each interval of length $2pi$). The right side of your inequality is made up of terms like $2I$, while the left side is made up of terms like $I+J$, so the left side is smaller for all $ygt 0$.



    I suppose if $lambda(x)$ was weird enough, you could get equality.






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

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      active

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      active

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      1












      $begingroup$

      Let $I_k =[(2k+1)pi,(2k+2)pi]$ and $J_k =[2kpi,(2k+1)pi]$. Then we have
      $$begin{eqnarray}
      int_{I_k} e^{-xy}|sin{x}|dlambda(x) &=&int_{J_k} e^{-(x+pi)y}|sin(x+pi)|dlambda(x)\
      &=&int_{J_k} e^{-(x+pi)y}sin(x)dlambda(x)\
      &le&int_{J_k} e^{-xy}sin(x)dlambda(x).
      end{eqnarray}$$       Thus we have
      $$begin{eqnarray}
      int_{[0,infty[}e^{-xy}|sin{x}|dlambda(x)&=&sum_{kge 0}left[int_{I_k} e^{-xy}|sin{x}|dlambda(x)+int_{J_k} e^{-xy}|sin{x}|dlambda(x)right]\
      &le&2sum_{kge 0}int_{J_k} e^{-xy}sin{x}dlambda(x)\
      &=&2int_{bigcup_{kge 0} J_k} e^{-xy}sin{x}dlambda(x)=2int_{A} e^{-xy}sin{x}dlambda(x).
      end{eqnarray}$$






      share|cite|improve this answer









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        1












        $begingroup$

        Let $I_k =[(2k+1)pi,(2k+2)pi]$ and $J_k =[2kpi,(2k+1)pi]$. Then we have
        $$begin{eqnarray}
        int_{I_k} e^{-xy}|sin{x}|dlambda(x) &=&int_{J_k} e^{-(x+pi)y}|sin(x+pi)|dlambda(x)\
        &=&int_{J_k} e^{-(x+pi)y}sin(x)dlambda(x)\
        &le&int_{J_k} e^{-xy}sin(x)dlambda(x).
        end{eqnarray}$$         Thus we have
        $$begin{eqnarray}
        int_{[0,infty[}e^{-xy}|sin{x}|dlambda(x)&=&sum_{kge 0}left[int_{I_k} e^{-xy}|sin{x}|dlambda(x)+int_{J_k} e^{-xy}|sin{x}|dlambda(x)right]\
        &le&2sum_{kge 0}int_{J_k} e^{-xy}sin{x}dlambda(x)\
        &=&2int_{bigcup_{kge 0} J_k} e^{-xy}sin{x}dlambda(x)=2int_{A} e^{-xy}sin{x}dlambda(x).
        end{eqnarray}$$






        share|cite|improve this answer









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          1












          1








          1





          $begingroup$

          Let $I_k =[(2k+1)pi,(2k+2)pi]$ and $J_k =[2kpi,(2k+1)pi]$. Then we have
          $$begin{eqnarray}
          int_{I_k} e^{-xy}|sin{x}|dlambda(x) &=&int_{J_k} e^{-(x+pi)y}|sin(x+pi)|dlambda(x)\
          &=&int_{J_k} e^{-(x+pi)y}sin(x)dlambda(x)\
          &le&int_{J_k} e^{-xy}sin(x)dlambda(x).
          end{eqnarray}$$           Thus we have
          $$begin{eqnarray}
          int_{[0,infty[}e^{-xy}|sin{x}|dlambda(x)&=&sum_{kge 0}left[int_{I_k} e^{-xy}|sin{x}|dlambda(x)+int_{J_k} e^{-xy}|sin{x}|dlambda(x)right]\
          &le&2sum_{kge 0}int_{J_k} e^{-xy}sin{x}dlambda(x)\
          &=&2int_{bigcup_{kge 0} J_k} e^{-xy}sin{x}dlambda(x)=2int_{A} e^{-xy}sin{x}dlambda(x).
          end{eqnarray}$$






          share|cite|improve this answer









          $endgroup$



          Let $I_k =[(2k+1)pi,(2k+2)pi]$ and $J_k =[2kpi,(2k+1)pi]$. Then we have
          $$begin{eqnarray}
          int_{I_k} e^{-xy}|sin{x}|dlambda(x) &=&int_{J_k} e^{-(x+pi)y}|sin(x+pi)|dlambda(x)\
          &=&int_{J_k} e^{-(x+pi)y}sin(x)dlambda(x)\
          &le&int_{J_k} e^{-xy}sin(x)dlambda(x).
          end{eqnarray}$$           Thus we have
          $$begin{eqnarray}
          int_{[0,infty[}e^{-xy}|sin{x}|dlambda(x)&=&sum_{kge 0}left[int_{I_k} e^{-xy}|sin{x}|dlambda(x)+int_{J_k} e^{-xy}|sin{x}|dlambda(x)right]\
          &le&2sum_{kge 0}int_{J_k} e^{-xy}sin{x}dlambda(x)\
          &=&2int_{bigcup_{kge 0} J_k} e^{-xy}sin{x}dlambda(x)=2int_{A} e^{-xy}sin{x}dlambda(x).
          end{eqnarray}$$







          share|cite|improve this answer












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          share|cite|improve this answer










          answered Jan 10 at 21:48









          SongSong

          11.8k628




          11.8k628























              1












              $begingroup$

              You need to compare $I=int_0^{pi}e^{-xy}|sinx|dlambda(x)$ with $J=int_{pi}^{2pi}e^{-xy}|sinx|dlambda(x)$ Because of the exponential term, $Jlt I$. (Note the same comparison holds for each interval of length $2pi$). The right side of your inequality is made up of terms like $2I$, while the left side is made up of terms like $I+J$, so the left side is smaller for all $ygt 0$.



              I suppose if $lambda(x)$ was weird enough, you could get equality.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                You need to compare $I=int_0^{pi}e^{-xy}|sinx|dlambda(x)$ with $J=int_{pi}^{2pi}e^{-xy}|sinx|dlambda(x)$ Because of the exponential term, $Jlt I$. (Note the same comparison holds for each interval of length $2pi$). The right side of your inequality is made up of terms like $2I$, while the left side is made up of terms like $I+J$, so the left side is smaller for all $ygt 0$.



                I suppose if $lambda(x)$ was weird enough, you could get equality.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  You need to compare $I=int_0^{pi}e^{-xy}|sinx|dlambda(x)$ with $J=int_{pi}^{2pi}e^{-xy}|sinx|dlambda(x)$ Because of the exponential term, $Jlt I$. (Note the same comparison holds for each interval of length $2pi$). The right side of your inequality is made up of terms like $2I$, while the left side is made up of terms like $I+J$, so the left side is smaller for all $ygt 0$.



                  I suppose if $lambda(x)$ was weird enough, you could get equality.






                  share|cite|improve this answer









                  $endgroup$



                  You need to compare $I=int_0^{pi}e^{-xy}|sinx|dlambda(x)$ with $J=int_{pi}^{2pi}e^{-xy}|sinx|dlambda(x)$ Because of the exponential term, $Jlt I$. (Note the same comparison holds for each interval of length $2pi$). The right side of your inequality is made up of terms like $2I$, while the left side is made up of terms like $I+J$, so the left side is smaller for all $ygt 0$.



                  I suppose if $lambda(x)$ was weird enough, you could get equality.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 10 at 21:57









                  herb steinbergherb steinberg

                  2,7082310




                  2,7082310






























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