Solving for C when we have $Cint_0^infty int_0^infty frac{e^frac{-(x_1+x_2)}{2}}{x_1+x_2} ,dx_1 ,dx_2=1$












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Solving for C when we have $Cint_0^infty int_0^infty frac{e^frac{-(x_1+x_2)}{2}}{x_1+x_2} ,dx_1 ,dx_2=1$



$$int_0^infty int_0^infty frac{e^frac{-(x_1+x_2)}{2}}{x_1+x_2} ,dx_1 ,dx_2$$



Let $u=x_1+x_2$ and $,du=,dx_1$.



$$int_0^infty int_{x_2}^infty frac{e^frac{-u}{2}}{u} ,du ,dx_2 $$



How do I compute the inner integral?






calculus integration improper-integrals







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    $begingroup$


    Solving for C when we have $Cint_0^infty int_0^infty frac{e^frac{-(x_1+x_2)}{2}}{x_1+x_2} ,dx_1 ,dx_2=1$



    $$int_0^infty int_0^infty frac{e^frac{-(x_1+x_2)}{2}}{x_1+x_2} ,dx_1 ,dx_2$$



    Let $u=x_1+x_2$ and $,du=,dx_1$.



    $$int_0^infty int_{x_2}^infty frac{e^frac{-u}{2}}{u} ,du ,dx_2 $$



    How do I compute the inner integral?






    calculus integration improper-integrals







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Solving for C when we have $Cint_0^infty int_0^infty frac{e^frac{-(x_1+x_2)}{2}}{x_1+x_2} ,dx_1 ,dx_2=1$



      $$int_0^infty int_0^infty frac{e^frac{-(x_1+x_2)}{2}}{x_1+x_2} ,dx_1 ,dx_2$$



      Let $u=x_1+x_2$ and $,du=,dx_1$.



      $$int_0^infty int_{x_2}^infty frac{e^frac{-u}{2}}{u} ,du ,dx_2 $$



      How do I compute the inner integral?






      calculus integration improper-integrals







      share|cite|improve this question











      $endgroup$




      Solving for C when we have $Cint_0^infty int_0^infty frac{e^frac{-(x_1+x_2)}{2}}{x_1+x_2} ,dx_1 ,dx_2=1$



      $$int_0^infty int_0^infty frac{e^frac{-(x_1+x_2)}{2}}{x_1+x_2} ,dx_1 ,dx_2$$



      Let $u=x_1+x_2$ and $,du=,dx_1$.



      $$int_0^infty int_{x_2}^infty frac{e^frac{-u}{2}}{u} ,du ,dx_2 $$



      How do I compute the inner integral?






      calculus integration improper-integrals




      calculus integration improper-integrals






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 21 '14 at 3:34









      dustin

      6,72892969




      6,72892969










      asked Nov 21 '14 at 3:00









      Username UnknownUsername Unknown

      1,25242058




      1,25242058






















          2 Answers
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          $begingroup$

          From my initial assessment, the substitution $x_1 = x^2$ and $x_2 = y^2$ should work. After that let $x^2 + y^2 = r^2$ where the Jacobian is then $rdrdtheta$. Then you can make a u sub where $u = r^2/2$. At this point, you can integrate your u integral and you left with a $theta$ integral of $sin(theta)cos(theta)$
          begin{align}
          4iintfrac{exp(-r^2/2)}{r^2}r^3sin(theta)cos(theta)drdtheta
          &= 4iintexp(-u)sin(theta)cos(theta)dudtheta\
          &= -4intexp(-u)Bigl|_0^{infty}sin(theta)cos(theta)dtheta\
          &=cdots
          end{align}




          If you obtained $frac{1}{16pi^2}$, I would suggest you try again. If you still cant figure it out, show what you have done.



          When we convert to $r$ and $theta$, the bounds become $0<r<infty$ and $0<theta<pi/2$



          Mouse over for solution:




          begin{align}&= 4int_0^{pi/2}sin(theta)cos(theta)dtheta\&=2int_0^{pi/2}sin(2theta)dtheta\&= -cos(2theta)Bigr|_0^{pi/2}\&=1 + 1\&=2end{align}







          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Perfect so $c=frac{1}{16pi^2}$
            $endgroup$
            – Username Unknown
            Nov 21 '14 at 3:37










          • $begingroup$
            @UsernameUnknown You should get $C=1/2$
            $endgroup$
            – dustin
            Nov 21 '14 at 3:37





















          -1












          $begingroup$

          Integration by parts:))
          $$
          int_0^{infty}int_{x_2}^{infty}frac{e^{-frac u2}}udu, dx_2=int_0^{infty}x_2frac{e^{-frac {x_2}2}}{x_2}dx_2=2.
          $$






          share|cite|improve this answer











          $endgroup$













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            2 Answers
            2






            active

            oldest

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            2 Answers
            2






            active

            oldest

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            active

            oldest

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            active

            oldest

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            2












            $begingroup$

            From my initial assessment, the substitution $x_1 = x^2$ and $x_2 = y^2$ should work. After that let $x^2 + y^2 = r^2$ where the Jacobian is then $rdrdtheta$. Then you can make a u sub where $u = r^2/2$. At this point, you can integrate your u integral and you left with a $theta$ integral of $sin(theta)cos(theta)$
            begin{align}
            4iintfrac{exp(-r^2/2)}{r^2}r^3sin(theta)cos(theta)drdtheta
            &= 4iintexp(-u)sin(theta)cos(theta)dudtheta\
            &= -4intexp(-u)Bigl|_0^{infty}sin(theta)cos(theta)dtheta\
            &=cdots
            end{align}




            If you obtained $frac{1}{16pi^2}$, I would suggest you try again. If you still cant figure it out, show what you have done.



            When we convert to $r$ and $theta$, the bounds become $0<r<infty$ and $0<theta<pi/2$



            Mouse over for solution:




            begin{align}&= 4int_0^{pi/2}sin(theta)cos(theta)dtheta\&=2int_0^{pi/2}sin(2theta)dtheta\&= -cos(2theta)Bigr|_0^{pi/2}\&=1 + 1\&=2end{align}







            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Perfect so $c=frac{1}{16pi^2}$
              $endgroup$
              – Username Unknown
              Nov 21 '14 at 3:37










            • $begingroup$
              @UsernameUnknown You should get $C=1/2$
              $endgroup$
              – dustin
              Nov 21 '14 at 3:37


















            2












            $begingroup$

            From my initial assessment, the substitution $x_1 = x^2$ and $x_2 = y^2$ should work. After that let $x^2 + y^2 = r^2$ where the Jacobian is then $rdrdtheta$. Then you can make a u sub where $u = r^2/2$. At this point, you can integrate your u integral and you left with a $theta$ integral of $sin(theta)cos(theta)$
            begin{align}
            4iintfrac{exp(-r^2/2)}{r^2}r^3sin(theta)cos(theta)drdtheta
            &= 4iintexp(-u)sin(theta)cos(theta)dudtheta\
            &= -4intexp(-u)Bigl|_0^{infty}sin(theta)cos(theta)dtheta\
            &=cdots
            end{align}




            If you obtained $frac{1}{16pi^2}$, I would suggest you try again. If you still cant figure it out, show what you have done.



            When we convert to $r$ and $theta$, the bounds become $0<r<infty$ and $0<theta<pi/2$



            Mouse over for solution:




            begin{align}&= 4int_0^{pi/2}sin(theta)cos(theta)dtheta\&=2int_0^{pi/2}sin(2theta)dtheta\&= -cos(2theta)Bigr|_0^{pi/2}\&=1 + 1\&=2end{align}







            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Perfect so $c=frac{1}{16pi^2}$
              $endgroup$
              – Username Unknown
              Nov 21 '14 at 3:37










            • $begingroup$
              @UsernameUnknown You should get $C=1/2$
              $endgroup$
              – dustin
              Nov 21 '14 at 3:37
















            2












            2








            2





            $begingroup$

            From my initial assessment, the substitution $x_1 = x^2$ and $x_2 = y^2$ should work. After that let $x^2 + y^2 = r^2$ where the Jacobian is then $rdrdtheta$. Then you can make a u sub where $u = r^2/2$. At this point, you can integrate your u integral and you left with a $theta$ integral of $sin(theta)cos(theta)$
            begin{align}
            4iintfrac{exp(-r^2/2)}{r^2}r^3sin(theta)cos(theta)drdtheta
            &= 4iintexp(-u)sin(theta)cos(theta)dudtheta\
            &= -4intexp(-u)Bigl|_0^{infty}sin(theta)cos(theta)dtheta\
            &=cdots
            end{align}




            If you obtained $frac{1}{16pi^2}$, I would suggest you try again. If you still cant figure it out, show what you have done.



            When we convert to $r$ and $theta$, the bounds become $0<r<infty$ and $0<theta<pi/2$



            Mouse over for solution:




            begin{align}&= 4int_0^{pi/2}sin(theta)cos(theta)dtheta\&=2int_0^{pi/2}sin(2theta)dtheta\&= -cos(2theta)Bigr|_0^{pi/2}\&=1 + 1\&=2end{align}







            share|cite|improve this answer











            $endgroup$



            From my initial assessment, the substitution $x_1 = x^2$ and $x_2 = y^2$ should work. After that let $x^2 + y^2 = r^2$ where the Jacobian is then $rdrdtheta$. Then you can make a u sub where $u = r^2/2$. At this point, you can integrate your u integral and you left with a $theta$ integral of $sin(theta)cos(theta)$
            begin{align}
            4iintfrac{exp(-r^2/2)}{r^2}r^3sin(theta)cos(theta)drdtheta
            &= 4iintexp(-u)sin(theta)cos(theta)dudtheta\
            &= -4intexp(-u)Bigl|_0^{infty}sin(theta)cos(theta)dtheta\
            &=cdots
            end{align}




            If you obtained $frac{1}{16pi^2}$, I would suggest you try again. If you still cant figure it out, show what you have done.



            When we convert to $r$ and $theta$, the bounds become $0<r<infty$ and $0<theta<pi/2$



            Mouse over for solution:




            begin{align}&= 4int_0^{pi/2}sin(theta)cos(theta)dtheta\&=2int_0^{pi/2}sin(2theta)dtheta\&= -cos(2theta)Bigr|_0^{pi/2}\&=1 + 1\&=2end{align}








            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 21 '14 at 16:25

























            answered Nov 21 '14 at 3:17









            dustindustin

            6,72892969




            6,72892969












            • $begingroup$
              Perfect so $c=frac{1}{16pi^2}$
              $endgroup$
              – Username Unknown
              Nov 21 '14 at 3:37










            • $begingroup$
              @UsernameUnknown You should get $C=1/2$
              $endgroup$
              – dustin
              Nov 21 '14 at 3:37




















            • $begingroup$
              Perfect so $c=frac{1}{16pi^2}$
              $endgroup$
              – Username Unknown
              Nov 21 '14 at 3:37










            • $begingroup$
              @UsernameUnknown You should get $C=1/2$
              $endgroup$
              – dustin
              Nov 21 '14 at 3:37


















            $begingroup$
            Perfect so $c=frac{1}{16pi^2}$
            $endgroup$
            – Username Unknown
            Nov 21 '14 at 3:37




            $begingroup$
            Perfect so $c=frac{1}{16pi^2}$
            $endgroup$
            – Username Unknown
            Nov 21 '14 at 3:37












            $begingroup$
            @UsernameUnknown You should get $C=1/2$
            $endgroup$
            – dustin
            Nov 21 '14 at 3:37






            $begingroup$
            @UsernameUnknown You should get $C=1/2$
            $endgroup$
            – dustin
            Nov 21 '14 at 3:37













            -1












            $begingroup$

            Integration by parts:))
            $$
            int_0^{infty}int_{x_2}^{infty}frac{e^{-frac u2}}udu, dx_2=int_0^{infty}x_2frac{e^{-frac {x_2}2}}{x_2}dx_2=2.
            $$






            share|cite|improve this answer











            $endgroup$


















              -1












              $begingroup$

              Integration by parts:))
              $$
              int_0^{infty}int_{x_2}^{infty}frac{e^{-frac u2}}udu, dx_2=int_0^{infty}x_2frac{e^{-frac {x_2}2}}{x_2}dx_2=2.
              $$






              share|cite|improve this answer











              $endgroup$
















                -1












                -1








                -1





                $begingroup$

                Integration by parts:))
                $$
                int_0^{infty}int_{x_2}^{infty}frac{e^{-frac u2}}udu, dx_2=int_0^{infty}x_2frac{e^{-frac {x_2}2}}{x_2}dx_2=2.
                $$






                share|cite|improve this answer











                $endgroup$



                Integration by parts:))
                $$
                int_0^{infty}int_{x_2}^{infty}frac{e^{-frac u2}}udu, dx_2=int_0^{infty}x_2frac{e^{-frac {x_2}2}}{x_2}dx_2=2.
                $$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 8 '18 at 7:16









                Brahadeesh

                6,32942363




                6,32942363










                answered Dec 7 '18 at 23:20









                AAKAAK

                365




                365






























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