$int_{[-1,1]}int_{[-1,1]}frac{xy}{(x^2+y^2)^2}dlambda(x) dlambda(y)=...

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I have calculated



$int_{[-1,1]}int_{[-1,1]}frac{xy}{(x^2+y^2)^2}dlambda(x) dlambda(y)= int_{[-1,1]}int_{[-1,1]}frac{xy}{(x^2+y^2)^2}dlambda(y) dlambda(x)=0$



Why would the double integral w.r.t. $lambda ^2$ not exist?










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closed as off-topic by Saad, Carl Mummert, Adrian Keister, jgon, Lord_Farin Dec 19 '18 at 18:24


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Carl Mummert, Adrian Keister, jgon, Lord_Farin

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    0












    $begingroup$


    I have calculated



    $int_{[-1,1]}int_{[-1,1]}frac{xy}{(x^2+y^2)^2}dlambda(x) dlambda(y)= int_{[-1,1]}int_{[-1,1]}frac{xy}{(x^2+y^2)^2}dlambda(y) dlambda(x)=0$



    Why would the double integral w.r.t. $lambda ^2$ not exist?










    share|cite|improve this question









    $endgroup$



    closed as off-topic by Saad, Carl Mummert, Adrian Keister, jgon, Lord_Farin Dec 19 '18 at 18:24


    This question appears to be off-topic. The users who voted to close gave this specific reason:


    • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Carl Mummert, Adrian Keister, jgon, Lord_Farin

    If this question can be reworded to fit the rules in the help center, please edit the question.



















      0












      0








      0





      $begingroup$


      I have calculated



      $int_{[-1,1]}int_{[-1,1]}frac{xy}{(x^2+y^2)^2}dlambda(x) dlambda(y)= int_{[-1,1]}int_{[-1,1]}frac{xy}{(x^2+y^2)^2}dlambda(y) dlambda(x)=0$



      Why would the double integral w.r.t. $lambda ^2$ not exist?










      share|cite|improve this question









      $endgroup$




      I have calculated



      $int_{[-1,1]}int_{[-1,1]}frac{xy}{(x^2+y^2)^2}dlambda(x) dlambda(y)= int_{[-1,1]}int_{[-1,1]}frac{xy}{(x^2+y^2)^2}dlambda(y) dlambda(x)=0$



      Why would the double integral w.r.t. $lambda ^2$ not exist?







      integration measure-theory multivariable-calculus lebesgue-integral






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      asked Dec 19 '18 at 9:32









      SABOYSABOY

      656311




      656311




      closed as off-topic by Saad, Carl Mummert, Adrian Keister, jgon, Lord_Farin Dec 19 '18 at 18:24


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Carl Mummert, Adrian Keister, jgon, Lord_Farin

      If this question can be reworded to fit the rules in the help center, please edit the question.







      closed as off-topic by Saad, Carl Mummert, Adrian Keister, jgon, Lord_Farin Dec 19 '18 at 18:24


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Carl Mummert, Adrian Keister, jgon, Lord_Farin

      If this question can be reworded to fit the rules in the help center, please edit the question.






















          1 Answer
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          The rectangle $[-1,1] times [-1,1]$ contains the unit disk. Consider $int _{{x^{2}+y^{2} leq 1}} frac {|xy|} {(x^{2}+y^{2})^{2}} dxdy$. If you use polar coordinates you will get a constant times $int_0^{1} frac 1 r , dr$ which is $infty$. [The constant is $int_0^{2pi} |sin, theta cos , theta|, dtheta$].






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            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            The rectangle $[-1,1] times [-1,1]$ contains the unit disk. Consider $int _{{x^{2}+y^{2} leq 1}} frac {|xy|} {(x^{2}+y^{2})^{2}} dxdy$. If you use polar coordinates you will get a constant times $int_0^{1} frac 1 r , dr$ which is $infty$. [The constant is $int_0^{2pi} |sin, theta cos , theta|, dtheta$].






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              The rectangle $[-1,1] times [-1,1]$ contains the unit disk. Consider $int _{{x^{2}+y^{2} leq 1}} frac {|xy|} {(x^{2}+y^{2})^{2}} dxdy$. If you use polar coordinates you will get a constant times $int_0^{1} frac 1 r , dr$ which is $infty$. [The constant is $int_0^{2pi} |sin, theta cos , theta|, dtheta$].






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                The rectangle $[-1,1] times [-1,1]$ contains the unit disk. Consider $int _{{x^{2}+y^{2} leq 1}} frac {|xy|} {(x^{2}+y^{2})^{2}} dxdy$. If you use polar coordinates you will get a constant times $int_0^{1} frac 1 r , dr$ which is $infty$. [The constant is $int_0^{2pi} |sin, theta cos , theta|, dtheta$].






                share|cite|improve this answer









                $endgroup$



                The rectangle $[-1,1] times [-1,1]$ contains the unit disk. Consider $int _{{x^{2}+y^{2} leq 1}} frac {|xy|} {(x^{2}+y^{2})^{2}} dxdy$. If you use polar coordinates you will get a constant times $int_0^{1} frac 1 r , dr$ which is $infty$. [The constant is $int_0^{2pi} |sin, theta cos , theta|, dtheta$].







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 19 '18 at 9:47









                Kavi Rama MurthyKavi Rama Murthy

                63.7k42463




                63.7k42463















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