Solve $int_{[0,infty[}(int_{[0,infty[} e^{-xy}sin{x} sin{y} dlambda(x)) dlambda(y)$












0












$begingroup$


Hint Required for :



$int_{[0,infty[}(int_{[0,infty[} e^{-xy}sin{x} sin{y} dlambda(x)) dlambda(y)$



My ideas:



Looking at the first inetgral $int_{[0,infty[} e^{-xy}sin{x} sin{y} dlambda(x)$



I would say $int_{[0,infty[} e^{-xy}sin{x} sin{y} dlambda(x)=sin{y}int_{[0,infty[}e^{-xy}sin{x}dlambda(x)$



And we know $sin(x)=frac{e^{ix}-e^{-ix}}{2i}$



So $sin{y}int_{[0,infty[}e^{-xy}sin{x}dlambda(x)=sin{y}int_{[0,infty[}e^{-xy}frac{e^{ix}-e^{-ix}}{2i}dlambda(x)$



and then I do not know what to do. Should I use this route or is partial integration better suited?



I've just started with multivariable integration, so any pointers would be of great assistance.






real-analysis integration measure-theory multivariable-calculus







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    0












    $begingroup$


    Hint Required for :



    $int_{[0,infty[}(int_{[0,infty[} e^{-xy}sin{x} sin{y} dlambda(x)) dlambda(y)$



    My ideas:



    Looking at the first inetgral $int_{[0,infty[} e^{-xy}sin{x} sin{y} dlambda(x)$



    I would say $int_{[0,infty[} e^{-xy}sin{x} sin{y} dlambda(x)=sin{y}int_{[0,infty[}e^{-xy}sin{x}dlambda(x)$



    And we know $sin(x)=frac{e^{ix}-e^{-ix}}{2i}$



    So $sin{y}int_{[0,infty[}e^{-xy}sin{x}dlambda(x)=sin{y}int_{[0,infty[}e^{-xy}frac{e^{ix}-e^{-ix}}{2i}dlambda(x)$



    and then I do not know what to do. Should I use this route or is partial integration better suited?



    I've just started with multivariable integration, so any pointers would be of great assistance.






    real-analysis integration measure-theory multivariable-calculus







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Hint Required for :



      $int_{[0,infty[}(int_{[0,infty[} e^{-xy}sin{x} sin{y} dlambda(x)) dlambda(y)$



      My ideas:



      Looking at the first inetgral $int_{[0,infty[} e^{-xy}sin{x} sin{y} dlambda(x)$



      I would say $int_{[0,infty[} e^{-xy}sin{x} sin{y} dlambda(x)=sin{y}int_{[0,infty[}e^{-xy}sin{x}dlambda(x)$



      And we know $sin(x)=frac{e^{ix}-e^{-ix}}{2i}$



      So $sin{y}int_{[0,infty[}e^{-xy}sin{x}dlambda(x)=sin{y}int_{[0,infty[}e^{-xy}frac{e^{ix}-e^{-ix}}{2i}dlambda(x)$



      and then I do not know what to do. Should I use this route or is partial integration better suited?



      I've just started with multivariable integration, so any pointers would be of great assistance.






      real-analysis integration measure-theory multivariable-calculus







      share|cite|improve this question









      $endgroup$




      Hint Required for :



      $int_{[0,infty[}(int_{[0,infty[} e^{-xy}sin{x} sin{y} dlambda(x)) dlambda(y)$



      My ideas:



      Looking at the first inetgral $int_{[0,infty[} e^{-xy}sin{x} sin{y} dlambda(x)$



      I would say $int_{[0,infty[} e^{-xy}sin{x} sin{y} dlambda(x)=sin{y}int_{[0,infty[}e^{-xy}sin{x}dlambda(x)$



      And we know $sin(x)=frac{e^{ix}-e^{-ix}}{2i}$



      So $sin{y}int_{[0,infty[}e^{-xy}sin{x}dlambda(x)=sin{y}int_{[0,infty[}e^{-xy}frac{e^{ix}-e^{-ix}}{2i}dlambda(x)$



      and then I do not know what to do. Should I use this route or is partial integration better suited?



      I've just started with multivariable integration, so any pointers would be of great assistance.






      real-analysis integration measure-theory multivariable-calculus




      real-analysis integration measure-theory multivariable-calculus






      share|cite|improve this question













      share|cite|improve this question











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      asked Dec 17 '18 at 12:22









      SABOYSABOY

      649311




      649311






















          2 Answers
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          1












          $begingroup$

          HINT: note that the inner integral is equivalent to the imaginary part of the integral



          $$int_{[0,infty)}e^{-xy}(cos x+isin x), dx=int_{[0,infty)} e^{-x(y-i)}, dx=-frac{e^{-x(y-i)}}{y-i}bigg|_0^infty=frac1{y-i}$$



          where the result is taken assuming that $y> 0$ (the integral doesn't exists for $y=0$, however we can ignore this point because the set ${0}$ have measure zero). In any case it is easy to check that the above integral is absolutely Lebesgue integrable for $y>0$, hence Lebesgue integrable.



          Of course you can also use the identity $sin x=frac{e^{ix}-e^{-ix}}{2i}$ as you was doing, and you will find the same answer.




          However the outer integral doesn't have a closed form solution.






          share|cite|improve this answer











          $endgroup$





















            0












            $begingroup$

            If you did not check any sort of absolute convergence for the multivariate integral, I strongly suggest integrating wrt $x$ first, then $y$ second. Note that $sin(x)$ is the imaginary part of $e^{ix}$, that should help you compute your integral better (you know the integral from $0$ to $infty$ of $t longmapsto e^{-pt}$ for reasonable complex numbers $p$, right?)






            share|cite|improve this answer









            $endgroup$













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              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              1












              $begingroup$

              HINT: note that the inner integral is equivalent to the imaginary part of the integral



              $$int_{[0,infty)}e^{-xy}(cos x+isin x), dx=int_{[0,infty)} e^{-x(y-i)}, dx=-frac{e^{-x(y-i)}}{y-i}bigg|_0^infty=frac1{y-i}$$



              where the result is taken assuming that $y> 0$ (the integral doesn't exists for $y=0$, however we can ignore this point because the set ${0}$ have measure zero). In any case it is easy to check that the above integral is absolutely Lebesgue integrable for $y>0$, hence Lebesgue integrable.



              Of course you can also use the identity $sin x=frac{e^{ix}-e^{-ix}}{2i}$ as you was doing, and you will find the same answer.




              However the outer integral doesn't have a closed form solution.






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                HINT: note that the inner integral is equivalent to the imaginary part of the integral



                $$int_{[0,infty)}e^{-xy}(cos x+isin x), dx=int_{[0,infty)} e^{-x(y-i)}, dx=-frac{e^{-x(y-i)}}{y-i}bigg|_0^infty=frac1{y-i}$$



                where the result is taken assuming that $y> 0$ (the integral doesn't exists for $y=0$, however we can ignore this point because the set ${0}$ have measure zero). In any case it is easy to check that the above integral is absolutely Lebesgue integrable for $y>0$, hence Lebesgue integrable.



                Of course you can also use the identity $sin x=frac{e^{ix}-e^{-ix}}{2i}$ as you was doing, and you will find the same answer.




                However the outer integral doesn't have a closed form solution.






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  HINT: note that the inner integral is equivalent to the imaginary part of the integral



                  $$int_{[0,infty)}e^{-xy}(cos x+isin x), dx=int_{[0,infty)} e^{-x(y-i)}, dx=-frac{e^{-x(y-i)}}{y-i}bigg|_0^infty=frac1{y-i}$$



                  where the result is taken assuming that $y> 0$ (the integral doesn't exists for $y=0$, however we can ignore this point because the set ${0}$ have measure zero). In any case it is easy to check that the above integral is absolutely Lebesgue integrable for $y>0$, hence Lebesgue integrable.



                  Of course you can also use the identity $sin x=frac{e^{ix}-e^{-ix}}{2i}$ as you was doing, and you will find the same answer.




                  However the outer integral doesn't have a closed form solution.






                  share|cite|improve this answer











                  $endgroup$



                  HINT: note that the inner integral is equivalent to the imaginary part of the integral



                  $$int_{[0,infty)}e^{-xy}(cos x+isin x), dx=int_{[0,infty)} e^{-x(y-i)}, dx=-frac{e^{-x(y-i)}}{y-i}bigg|_0^infty=frac1{y-i}$$



                  where the result is taken assuming that $y> 0$ (the integral doesn't exists for $y=0$, however we can ignore this point because the set ${0}$ have measure zero). In any case it is easy to check that the above integral is absolutely Lebesgue integrable for $y>0$, hence Lebesgue integrable.



                  Of course you can also use the identity $sin x=frac{e^{ix}-e^{-ix}}{2i}$ as you was doing, and you will find the same answer.




                  However the outer integral doesn't have a closed form solution.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 17 '18 at 13:46

























                  answered Dec 17 '18 at 12:29









                  MasacrosoMasacroso

                  13.1k41747




                  13.1k41747























                      0












                      $begingroup$

                      If you did not check any sort of absolute convergence for the multivariate integral, I strongly suggest integrating wrt $x$ first, then $y$ second. Note that $sin(x)$ is the imaginary part of $e^{ix}$, that should help you compute your integral better (you know the integral from $0$ to $infty$ of $t longmapsto e^{-pt}$ for reasonable complex numbers $p$, right?)






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        If you did not check any sort of absolute convergence for the multivariate integral, I strongly suggest integrating wrt $x$ first, then $y$ second. Note that $sin(x)$ is the imaginary part of $e^{ix}$, that should help you compute your integral better (you know the integral from $0$ to $infty$ of $t longmapsto e^{-pt}$ for reasonable complex numbers $p$, right?)






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          If you did not check any sort of absolute convergence for the multivariate integral, I strongly suggest integrating wrt $x$ first, then $y$ second. Note that $sin(x)$ is the imaginary part of $e^{ix}$, that should help you compute your integral better (you know the integral from $0$ to $infty$ of $t longmapsto e^{-pt}$ for reasonable complex numbers $p$, right?)






                          share|cite|improve this answer









                          $endgroup$



                          If you did not check any sort of absolute convergence for the multivariate integral, I strongly suggest integrating wrt $x$ first, then $y$ second. Note that $sin(x)$ is the imaginary part of $e^{ix}$, that should help you compute your integral better (you know the integral from $0$ to $infty$ of $t longmapsto e^{-pt}$ for reasonable complex numbers $p$, right?)







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 17 '18 at 12:32









                          MindlackMindlack

                          4,740210




                          4,740210






























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