Use Fubini's Theorem to calculate the integral $int_{0}^{pi/2}int_{y}^{pi/2}yfrac{sin x}{x}dx dy$












0












$begingroup$


Here is my attempt:
As $[0,pi/2] times [0,pi/2]$ is a finite measure space(hence $sigma$ finite) and also its $x times y$ integrable therefore hypothesis of the fubini's theorem are satisfied and



By change of limit we get:



$int_{0}^{pi/2}int_{x}^{pi/2}yfrac{sin x}{x}dx dy$



$implies$ $frac{pi^2}{8}int_{0}^{pi/2}frac{sin x}{x}-int_{0}^{pi/2}frac{xsin x}{2}dx$



and then I don't know how to integrate the first integral $int_0^{pi/2}frac{sin x}{x}dx$.



Thanks for any help!!






real-analysis measure-theory lebesgue-integral







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    0












    $begingroup$


    Here is my attempt:
    As $[0,pi/2] times [0,pi/2]$ is a finite measure space(hence $sigma$ finite) and also its $x times y$ integrable therefore hypothesis of the fubini's theorem are satisfied and



    By change of limit we get:



    $int_{0}^{pi/2}int_{x}^{pi/2}yfrac{sin x}{x}dx dy$



    $implies$ $frac{pi^2}{8}int_{0}^{pi/2}frac{sin x}{x}-int_{0}^{pi/2}frac{xsin x}{2}dx$



    and then I don't know how to integrate the first integral $int_0^{pi/2}frac{sin x}{x}dx$.



    Thanks for any help!!






    real-analysis measure-theory lebesgue-integral







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Here is my attempt:
      As $[0,pi/2] times [0,pi/2]$ is a finite measure space(hence $sigma$ finite) and also its $x times y$ integrable therefore hypothesis of the fubini's theorem are satisfied and



      By change of limit we get:



      $int_{0}^{pi/2}int_{x}^{pi/2}yfrac{sin x}{x}dx dy$



      $implies$ $frac{pi^2}{8}int_{0}^{pi/2}frac{sin x}{x}-int_{0}^{pi/2}frac{xsin x}{2}dx$



      and then I don't know how to integrate the first integral $int_0^{pi/2}frac{sin x}{x}dx$.



      Thanks for any help!!






      real-analysis measure-theory lebesgue-integral







      share|cite|improve this question











      $endgroup$




      Here is my attempt:
      As $[0,pi/2] times [0,pi/2]$ is a finite measure space(hence $sigma$ finite) and also its $x times y$ integrable therefore hypothesis of the fubini's theorem are satisfied and



      By change of limit we get:



      $int_{0}^{pi/2}int_{x}^{pi/2}yfrac{sin x}{x}dx dy$



      $implies$ $frac{pi^2}{8}int_{0}^{pi/2}frac{sin x}{x}-int_{0}^{pi/2}frac{xsin x}{2}dx$



      and then I don't know how to integrate the first integral $int_0^{pi/2}frac{sin x}{x}dx$.



      Thanks for any help!!






      real-analysis measure-theory lebesgue-integral




      real-analysis measure-theory lebesgue-integral






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      share|cite|improve this question













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      edited Dec 31 '18 at 10:57









      Davide Giraudo

      127k17154268




      127k17154268










      asked Dec 29 '18 at 1:04









      InfinityInfinity

      352113




      352113






















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          $begingroup$

          I think you messed up the change of order. In the first integral, we have $0leq y leq frac pi 2$. Then, in the second integral, we have $y leq x leq fracpi 2$. In total, this gives us:



          $$0 leq y leq x leq frac pi 2$$



          Now, if we want to switch the order, we need to remember abide by this. First, $0leq xleq fracpi 2$, which determines the limits of the new outer integral. Then, we have $0leq y leq x$, which determines the limits of the new inner integral. Thus, the new integral becomes:



          $$int_0^{frac pi 2}int_0^x yfrac{sin x}{x}dydx$$



          Now, we can take $frac{sin x}{x}$ out of the inner integral since it has nothing to do with $y$, so we get:



          $$int_0^{frac pi 2}frac{sin x}{x}int_0^x ydydx$$



          $int_0^x ydy=frac{x^2}{2}$, so when we substitute back in and multiply it with $frac{sin x}{x}$, we get:



          $$int_0^{frac pi 2}frac{xsin x}{2}dx$$



          I will leave the final integration by parts for you to do.






          share|cite|improve this answer











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            1 Answer
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            $begingroup$

            I think you messed up the change of order. In the first integral, we have $0leq y leq frac pi 2$. Then, in the second integral, we have $y leq x leq fracpi 2$. In total, this gives us:



            $$0 leq y leq x leq frac pi 2$$



            Now, if we want to switch the order, we need to remember abide by this. First, $0leq xleq fracpi 2$, which determines the limits of the new outer integral. Then, we have $0leq y leq x$, which determines the limits of the new inner integral. Thus, the new integral becomes:



            $$int_0^{frac pi 2}int_0^x yfrac{sin x}{x}dydx$$



            Now, we can take $frac{sin x}{x}$ out of the inner integral since it has nothing to do with $y$, so we get:



            $$int_0^{frac pi 2}frac{sin x}{x}int_0^x ydydx$$



            $int_0^x ydy=frac{x^2}{2}$, so when we substitute back in and multiply it with $frac{sin x}{x}$, we get:



            $$int_0^{frac pi 2}frac{xsin x}{2}dx$$



            I will leave the final integration by parts for you to do.






            share|cite|improve this answer











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              2












              $begingroup$

              I think you messed up the change of order. In the first integral, we have $0leq y leq frac pi 2$. Then, in the second integral, we have $y leq x leq fracpi 2$. In total, this gives us:



              $$0 leq y leq x leq frac pi 2$$



              Now, if we want to switch the order, we need to remember abide by this. First, $0leq xleq fracpi 2$, which determines the limits of the new outer integral. Then, we have $0leq y leq x$, which determines the limits of the new inner integral. Thus, the new integral becomes:



              $$int_0^{frac pi 2}int_0^x yfrac{sin x}{x}dydx$$



              Now, we can take $frac{sin x}{x}$ out of the inner integral since it has nothing to do with $y$, so we get:



              $$int_0^{frac pi 2}frac{sin x}{x}int_0^x ydydx$$



              $int_0^x ydy=frac{x^2}{2}$, so when we substitute back in and multiply it with $frac{sin x}{x}$, we get:



              $$int_0^{frac pi 2}frac{xsin x}{2}dx$$



              I will leave the final integration by parts for you to do.






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                I think you messed up the change of order. In the first integral, we have $0leq y leq frac pi 2$. Then, in the second integral, we have $y leq x leq fracpi 2$. In total, this gives us:



                $$0 leq y leq x leq frac pi 2$$



                Now, if we want to switch the order, we need to remember abide by this. First, $0leq xleq fracpi 2$, which determines the limits of the new outer integral. Then, we have $0leq y leq x$, which determines the limits of the new inner integral. Thus, the new integral becomes:



                $$int_0^{frac pi 2}int_0^x yfrac{sin x}{x}dydx$$



                Now, we can take $frac{sin x}{x}$ out of the inner integral since it has nothing to do with $y$, so we get:



                $$int_0^{frac pi 2}frac{sin x}{x}int_0^x ydydx$$



                $int_0^x ydy=frac{x^2}{2}$, so when we substitute back in and multiply it with $frac{sin x}{x}$, we get:



                $$int_0^{frac pi 2}frac{xsin x}{2}dx$$



                I will leave the final integration by parts for you to do.






                share|cite|improve this answer











                $endgroup$



                I think you messed up the change of order. In the first integral, we have $0leq y leq frac pi 2$. Then, in the second integral, we have $y leq x leq fracpi 2$. In total, this gives us:



                $$0 leq y leq x leq frac pi 2$$



                Now, if we want to switch the order, we need to remember abide by this. First, $0leq xleq fracpi 2$, which determines the limits of the new outer integral. Then, we have $0leq y leq x$, which determines the limits of the new inner integral. Thus, the new integral becomes:



                $$int_0^{frac pi 2}int_0^x yfrac{sin x}{x}dydx$$



                Now, we can take $frac{sin x}{x}$ out of the inner integral since it has nothing to do with $y$, so we get:



                $$int_0^{frac pi 2}frac{sin x}{x}int_0^x ydydx$$



                $int_0^x ydy=frac{x^2}{2}$, so when we substitute back in and multiply it with $frac{sin x}{x}$, we get:



                $$int_0^{frac pi 2}frac{xsin x}{2}dx$$



                I will leave the final integration by parts for you to do.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 29 '18 at 14:51

























                answered Dec 29 '18 at 1:12









                Noble MushtakNoble Mushtak

                15.3k1835




                15.3k1835






























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