Computing iterated integrals over positive integers (question about stack exchange solution)












1












$begingroup$


I am referring to this question: here



How does one go about computing the iterated integrals?



I am confused as to how the first one simplifies to $f(1,1)$ and why the second one has:



$sum_{y} 2^{-y-1} - 2^{-y} = - sum_{y} 2^{y-1}$



Any clarification is appreciated.



Edit 1: I assume we are supposed to compute each part by taking the sections of $f$, where we fix one variable.



So, for example, the inside part of the first integral is $sum_{y} f_x(y)$ (where $f_x(y) = f(x,y)$ and $x$ is fixed). For a fixed $x$ this should evaluate to $(2-2^{-x}) + (-2 + 2^{-(x+1)})$ where the first summand is coming from the case where $y=x$ and the second comes from the case where $y=x-1$. In which case, they would evaluate to the same thing.



Edit 2: But obviously this shouldn’t be the case as we violated the non-negativity requirement of Tonelli’s and I think (from what I’ve gathered from the correct solution) we violated integrability of Fubini’s.






real-analysis integration analysis measure-theory product-measure







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$endgroup$

















    1












    $begingroup$


    I am referring to this question: here



    How does one go about computing the iterated integrals?



    I am confused as to how the first one simplifies to $f(1,1)$ and why the second one has:



    $sum_{y} 2^{-y-1} - 2^{-y} = - sum_{y} 2^{y-1}$



    Any clarification is appreciated.



    Edit 1: I assume we are supposed to compute each part by taking the sections of $f$, where we fix one variable.



    So, for example, the inside part of the first integral is $sum_{y} f_x(y)$ (where $f_x(y) = f(x,y)$ and $x$ is fixed). For a fixed $x$ this should evaluate to $(2-2^{-x}) + (-2 + 2^{-(x+1)})$ where the first summand is coming from the case where $y=x$ and the second comes from the case where $y=x-1$. In which case, they would evaluate to the same thing.



    Edit 2: But obviously this shouldn’t be the case as we violated the non-negativity requirement of Tonelli’s and I think (from what I’ve gathered from the correct solution) we violated integrability of Fubini’s.






    real-analysis integration analysis measure-theory product-measure







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I am referring to this question: here



      How does one go about computing the iterated integrals?



      I am confused as to how the first one simplifies to $f(1,1)$ and why the second one has:



      $sum_{y} 2^{-y-1} - 2^{-y} = - sum_{y} 2^{y-1}$



      Any clarification is appreciated.



      Edit 1: I assume we are supposed to compute each part by taking the sections of $f$, where we fix one variable.



      So, for example, the inside part of the first integral is $sum_{y} f_x(y)$ (where $f_x(y) = f(x,y)$ and $x$ is fixed). For a fixed $x$ this should evaluate to $(2-2^{-x}) + (-2 + 2^{-(x+1)})$ where the first summand is coming from the case where $y=x$ and the second comes from the case where $y=x-1$. In which case, they would evaluate to the same thing.



      Edit 2: But obviously this shouldn’t be the case as we violated the non-negativity requirement of Tonelli’s and I think (from what I’ve gathered from the correct solution) we violated integrability of Fubini’s.






      real-analysis integration analysis measure-theory product-measure







      share|cite|improve this question











      $endgroup$




      I am referring to this question: here



      How does one go about computing the iterated integrals?



      I am confused as to how the first one simplifies to $f(1,1)$ and why the second one has:



      $sum_{y} 2^{-y-1} - 2^{-y} = - sum_{y} 2^{y-1}$



      Any clarification is appreciated.



      Edit 1: I assume we are supposed to compute each part by taking the sections of $f$, where we fix one variable.



      So, for example, the inside part of the first integral is $sum_{y} f_x(y)$ (where $f_x(y) = f(x,y)$ and $x$ is fixed). For a fixed $x$ this should evaluate to $(2-2^{-x}) + (-2 + 2^{-(x+1)})$ where the first summand is coming from the case where $y=x$ and the second comes from the case where $y=x-1$. In which case, they would evaluate to the same thing.



      Edit 2: But obviously this shouldn’t be the case as we violated the non-negativity requirement of Tonelli’s and I think (from what I’ve gathered from the correct solution) we violated integrability of Fubini’s.






      real-analysis integration analysis measure-theory product-measure




      real-analysis integration analysis measure-theory product-measure






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 6 '18 at 2:10







      Jane Doe

















      asked Dec 4 '18 at 19:27









      Jane DoeJane Doe

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