Rectangle paramtrization












0












$begingroup$



Let $Gamma=ABCD$, where:



$$A(0,0),B(a,0),C(a,alpha),D(0,alpha)$$



$$omega=Pdx+Qdy=e^{-x^2+y^2}cos(2xy)dx+e^{-x^2+y^2}sin(2xy)$$




with $alphainmathbb{R}$, and $a$ they don't say anything about it. However I have to use this to calculate:



$$int_0^{infty}e^{-t^2}cos(2alpha t)dt$$



But since $Gamma$ is a closed curve then $int_{Gamma}omega= 0$.
However how do I parametrize this and what is the intuition behind it?



My attempt:



when trying to parametrize I said: $Gamma = ABcup BCcup CDcup DA$



and so:



$AB:$
$$x(t) = t$$
$$y(t) = 0$$
$tin [0,a]$



$BC:$
$$x(t)=a$$
$$y(t) = t$$
$tin [0,alpha ]$



$CD:$
$$x(t)=a-t$$
$$y(t)=alpha$$
$tin [0,a]$



$DA:$
$$x(t)=0$$
$$y(t) = alpha -t$$



$tin [0,alpha]$



Now how do I construct the integral?






integration parametrization







share|cite|improve this question











$endgroup$

















    0












    $begingroup$



    Let $Gamma=ABCD$, where:



    $$A(0,0),B(a,0),C(a,alpha),D(0,alpha)$$



    $$omega=Pdx+Qdy=e^{-x^2+y^2}cos(2xy)dx+e^{-x^2+y^2}sin(2xy)$$




    with $alphainmathbb{R}$, and $a$ they don't say anything about it. However I have to use this to calculate:



    $$int_0^{infty}e^{-t^2}cos(2alpha t)dt$$



    But since $Gamma$ is a closed curve then $int_{Gamma}omega= 0$.
    However how do I parametrize this and what is the intuition behind it?



    My attempt:



    when trying to parametrize I said: $Gamma = ABcup BCcup CDcup DA$



    and so:



    $AB:$
    $$x(t) = t$$
    $$y(t) = 0$$
    $tin [0,a]$



    $BC:$
    $$x(t)=a$$
    $$y(t) = t$$
    $tin [0,alpha ]$



    $CD:$
    $$x(t)=a-t$$
    $$y(t)=alpha$$
    $tin [0,a]$



    $DA:$
    $$x(t)=0$$
    $$y(t) = alpha -t$$



    $tin [0,alpha]$



    Now how do I construct the integral?






    integration parametrization







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$



      Let $Gamma=ABCD$, where:



      $$A(0,0),B(a,0),C(a,alpha),D(0,alpha)$$



      $$omega=Pdx+Qdy=e^{-x^2+y^2}cos(2xy)dx+e^{-x^2+y^2}sin(2xy)$$




      with $alphainmathbb{R}$, and $a$ they don't say anything about it. However I have to use this to calculate:



      $$int_0^{infty}e^{-t^2}cos(2alpha t)dt$$



      But since $Gamma$ is a closed curve then $int_{Gamma}omega= 0$.
      However how do I parametrize this and what is the intuition behind it?



      My attempt:



      when trying to parametrize I said: $Gamma = ABcup BCcup CDcup DA$



      and so:



      $AB:$
      $$x(t) = t$$
      $$y(t) = 0$$
      $tin [0,a]$



      $BC:$
      $$x(t)=a$$
      $$y(t) = t$$
      $tin [0,alpha ]$



      $CD:$
      $$x(t)=a-t$$
      $$y(t)=alpha$$
      $tin [0,a]$



      $DA:$
      $$x(t)=0$$
      $$y(t) = alpha -t$$



      $tin [0,alpha]$



      Now how do I construct the integral?






      integration parametrization







      share|cite|improve this question











      $endgroup$





      Let $Gamma=ABCD$, where:



      $$A(0,0),B(a,0),C(a,alpha),D(0,alpha)$$



      $$omega=Pdx+Qdy=e^{-x^2+y^2}cos(2xy)dx+e^{-x^2+y^2}sin(2xy)$$




      with $alphainmathbb{R}$, and $a$ they don't say anything about it. However I have to use this to calculate:



      $$int_0^{infty}e^{-t^2}cos(2alpha t)dt$$



      But since $Gamma$ is a closed curve then $int_{Gamma}omega= 0$.
      However how do I parametrize this and what is the intuition behind it?



      My attempt:



      when trying to parametrize I said: $Gamma = ABcup BCcup CDcup DA$



      and so:



      $AB:$
      $$x(t) = t$$
      $$y(t) = 0$$
      $tin [0,a]$



      $BC:$
      $$x(t)=a$$
      $$y(t) = t$$
      $tin [0,alpha ]$



      $CD:$
      $$x(t)=a-t$$
      $$y(t)=alpha$$
      $tin [0,a]$



      $DA:$
      $$x(t)=0$$
      $$y(t) = alpha -t$$



      $tin [0,alpha]$



      Now how do I construct the integral?






      integration parametrization




      integration parametrization






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 13 at 7:21







      C. Cristi

















      asked Jan 13 at 7:05









      C. CristiC. Cristi

      1,629218




      1,629218






















          1 Answer
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          0












          $begingroup$

          I solved it myself actually:



          Using that parametrization you get that:



          $$0=int_{Gamma}omega=int_{0}^{a}e^{-t^2}dt+int_0^{alpha}e^{-a^2+t^2}cos(2at)dt-int_0^ae^{-t^2+alpha^2}cos(2alpha t)dt$$



          If we note that desired integral with $I$. We can see that from the last expression we get:



          $$Ie^{alpha^2}=int_{0}^{a}e^{-t^2}dt+int_0^{alpha}e^{-a^2+t^2}cos(2at)dt$$



          Letting $atoinfty$



          We get:
          $$boxed{I=frac {sqrt{pi}}{2}e^{-alpha^2}}$$






          share|cite|improve this answer









          $endgroup$













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            0












            $begingroup$

            I solved it myself actually:



            Using that parametrization you get that:



            $$0=int_{Gamma}omega=int_{0}^{a}e^{-t^2}dt+int_0^{alpha}e^{-a^2+t^2}cos(2at)dt-int_0^ae^{-t^2+alpha^2}cos(2alpha t)dt$$



            If we note that desired integral with $I$. We can see that from the last expression we get:



            $$Ie^{alpha^2}=int_{0}^{a}e^{-t^2}dt+int_0^{alpha}e^{-a^2+t^2}cos(2at)dt$$



            Letting $atoinfty$



            We get:
            $$boxed{I=frac {sqrt{pi}}{2}e^{-alpha^2}}$$






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              I solved it myself actually:



              Using that parametrization you get that:



              $$0=int_{Gamma}omega=int_{0}^{a}e^{-t^2}dt+int_0^{alpha}e^{-a^2+t^2}cos(2at)dt-int_0^ae^{-t^2+alpha^2}cos(2alpha t)dt$$



              If we note that desired integral with $I$. We can see that from the last expression we get:



              $$Ie^{alpha^2}=int_{0}^{a}e^{-t^2}dt+int_0^{alpha}e^{-a^2+t^2}cos(2at)dt$$



              Letting $atoinfty$



              We get:
              $$boxed{I=frac {sqrt{pi}}{2}e^{-alpha^2}}$$






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                I solved it myself actually:



                Using that parametrization you get that:



                $$0=int_{Gamma}omega=int_{0}^{a}e^{-t^2}dt+int_0^{alpha}e^{-a^2+t^2}cos(2at)dt-int_0^ae^{-t^2+alpha^2}cos(2alpha t)dt$$



                If we note that desired integral with $I$. We can see that from the last expression we get:



                $$Ie^{alpha^2}=int_{0}^{a}e^{-t^2}dt+int_0^{alpha}e^{-a^2+t^2}cos(2at)dt$$



                Letting $atoinfty$



                We get:
                $$boxed{I=frac {sqrt{pi}}{2}e^{-alpha^2}}$$






                share|cite|improve this answer









                $endgroup$



                I solved it myself actually:



                Using that parametrization you get that:



                $$0=int_{Gamma}omega=int_{0}^{a}e^{-t^2}dt+int_0^{alpha}e^{-a^2+t^2}cos(2at)dt-int_0^ae^{-t^2+alpha^2}cos(2alpha t)dt$$



                If we note that desired integral with $I$. We can see that from the last expression we get:



                $$Ie^{alpha^2}=int_{0}^{a}e^{-t^2}dt+int_0^{alpha}e^{-a^2+t^2}cos(2at)dt$$



                Letting $atoinfty$



                We get:
                $$boxed{I=frac {sqrt{pi}}{2}e^{-alpha^2}}$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 13 at 7:39









                C. CristiC. Cristi

                1,629218




                1,629218






























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