Puzzled with a double integration












2












$begingroup$


Double integration of $f(x,y) = frac{(x-y)}{(x+y)^3}$ for which $x$ ranges from $0$ to $1$ and
$y$ ranges from $0$ to $1$. It comes out to be $frac{1}{2}$.
But suppose that for $(a,b)$ the value of $f$ is $frac{(a-b)}{(a+b)^3}$ and there will be a pair $(b,a)$ where $f$ will be $frac{(b-a)}{(a+b)^3}$. both opposite in sign with equal value. for each $(a,b)$ pair we have $(b,a)$ hence overall integration should be $0$.
Please tell where am i wrong?






calculus integration







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    2












    $begingroup$


    Double integration of $f(x,y) = frac{(x-y)}{(x+y)^3}$ for which $x$ ranges from $0$ to $1$ and
    $y$ ranges from $0$ to $1$. It comes out to be $frac{1}{2}$.
    But suppose that for $(a,b)$ the value of $f$ is $frac{(a-b)}{(a+b)^3}$ and there will be a pair $(b,a)$ where $f$ will be $frac{(b-a)}{(a+b)^3}$. both opposite in sign with equal value. for each $(a,b)$ pair we have $(b,a)$ hence overall integration should be $0$.
    Please tell where am i wrong?






    calculus integration







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      Double integration of $f(x,y) = frac{(x-y)}{(x+y)^3}$ for which $x$ ranges from $0$ to $1$ and
      $y$ ranges from $0$ to $1$. It comes out to be $frac{1}{2}$.
      But suppose that for $(a,b)$ the value of $f$ is $frac{(a-b)}{(a+b)^3}$ and there will be a pair $(b,a)$ where $f$ will be $frac{(b-a)}{(a+b)^3}$. both opposite in sign with equal value. for each $(a,b)$ pair we have $(b,a)$ hence overall integration should be $0$.
      Please tell where am i wrong?






      calculus integration







      share|cite|improve this question











      $endgroup$




      Double integration of $f(x,y) = frac{(x-y)}{(x+y)^3}$ for which $x$ ranges from $0$ to $1$ and
      $y$ ranges from $0$ to $1$. It comes out to be $frac{1}{2}$.
      But suppose that for $(a,b)$ the value of $f$ is $frac{(a-b)}{(a+b)^3}$ and there will be a pair $(b,a)$ where $f$ will be $frac{(b-a)}{(a+b)^3}$. both opposite in sign with equal value. for each $(a,b)$ pair we have $(b,a)$ hence overall integration should be $0$.
      Please tell where am i wrong?






      calculus integration




      calculus integration






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      edited Jan 30 at 21:03









      GSofer

      8831313




      8831313










      asked Jan 30 at 20:49









      Jaswant ShekhawatJaswant Shekhawat

      133




      133






















          2 Answers
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          $begingroup$

          This integral is not absolutely convergent due to the singularity at $(x,y) = (0,0)$. Hence, a lot of weird things can happen if you try to evaluate the integral.



          For example, $$int_{0}^{1}int_{0}^{1}dfrac{x-y}{(x+y)^3},dx,dy = -dfrac{1}{2}$$ but if we reverse the order of integration, we get $$int_{0}^{1}int_{0}^{1}dfrac{x-y}{(x+y)^3},dy,dx = dfrac{1}{2}.$$






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          • 1




            $begingroup$
            Yes Fubini’s theorem doesn’t apply here!
            $endgroup$
            – mathcounterexamples.net
            Jan 30 at 20:56



















          0












          $begingroup$

          If you write the integral as $lim_{epsilon to 0}int_epsilon^1int_epsilon^1 frac{x-y}{(x+y)^3}dxdy$ you will get $0$.






          share|cite|improve this answer









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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3












            $begingroup$

            This integral is not absolutely convergent due to the singularity at $(x,y) = (0,0)$. Hence, a lot of weird things can happen if you try to evaluate the integral.



            For example, $$int_{0}^{1}int_{0}^{1}dfrac{x-y}{(x+y)^3},dx,dy = -dfrac{1}{2}$$ but if we reverse the order of integration, we get $$int_{0}^{1}int_{0}^{1}dfrac{x-y}{(x+y)^3},dy,dx = dfrac{1}{2}.$$






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              Yes Fubini’s theorem doesn’t apply here!
              $endgroup$
              – mathcounterexamples.net
              Jan 30 at 20:56
















            3












            $begingroup$

            This integral is not absolutely convergent due to the singularity at $(x,y) = (0,0)$. Hence, a lot of weird things can happen if you try to evaluate the integral.



            For example, $$int_{0}^{1}int_{0}^{1}dfrac{x-y}{(x+y)^3},dx,dy = -dfrac{1}{2}$$ but if we reverse the order of integration, we get $$int_{0}^{1}int_{0}^{1}dfrac{x-y}{(x+y)^3},dy,dx = dfrac{1}{2}.$$






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              Yes Fubini’s theorem doesn’t apply here!
              $endgroup$
              – mathcounterexamples.net
              Jan 30 at 20:56














            3












            3








            3





            $begingroup$

            This integral is not absolutely convergent due to the singularity at $(x,y) = (0,0)$. Hence, a lot of weird things can happen if you try to evaluate the integral.



            For example, $$int_{0}^{1}int_{0}^{1}dfrac{x-y}{(x+y)^3},dx,dy = -dfrac{1}{2}$$ but if we reverse the order of integration, we get $$int_{0}^{1}int_{0}^{1}dfrac{x-y}{(x+y)^3},dy,dx = dfrac{1}{2}.$$






            share|cite|improve this answer









            $endgroup$



            This integral is not absolutely convergent due to the singularity at $(x,y) = (0,0)$. Hence, a lot of weird things can happen if you try to evaluate the integral.



            For example, $$int_{0}^{1}int_{0}^{1}dfrac{x-y}{(x+y)^3},dx,dy = -dfrac{1}{2}$$ but if we reverse the order of integration, we get $$int_{0}^{1}int_{0}^{1}dfrac{x-y}{(x+y)^3},dy,dx = dfrac{1}{2}.$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 30 at 20:55









            JimmyK4542JimmyK4542

            41.3k245107




            41.3k245107








            • 1




              $begingroup$
              Yes Fubini’s theorem doesn’t apply here!
              $endgroup$
              – mathcounterexamples.net
              Jan 30 at 20:56














            • 1




              $begingroup$
              Yes Fubini’s theorem doesn’t apply here!
              $endgroup$
              – mathcounterexamples.net
              Jan 30 at 20:56








            1




            1




            $begingroup$
            Yes Fubini’s theorem doesn’t apply here!
            $endgroup$
            – mathcounterexamples.net
            Jan 30 at 20:56




            $begingroup$
            Yes Fubini’s theorem doesn’t apply here!
            $endgroup$
            – mathcounterexamples.net
            Jan 30 at 20:56











            0












            $begingroup$

            If you write the integral as $lim_{epsilon to 0}int_epsilon^1int_epsilon^1 frac{x-y}{(x+y)^3}dxdy$ you will get $0$.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              If you write the integral as $lim_{epsilon to 0}int_epsilon^1int_epsilon^1 frac{x-y}{(x+y)^3}dxdy$ you will get $0$.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                If you write the integral as $lim_{epsilon to 0}int_epsilon^1int_epsilon^1 frac{x-y}{(x+y)^3}dxdy$ you will get $0$.






                share|cite|improve this answer









                $endgroup$



                If you write the integral as $lim_{epsilon to 0}int_epsilon^1int_epsilon^1 frac{x-y}{(x+y)^3}dxdy$ you will get $0$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 30 at 21:28









                herb steinbergherb steinberg

                3,1132311




                3,1132311






























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