Computing iterated integrals over positive integers (question about stack exchange solution)
$begingroup$
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
asked Dec 4 '18 at 19:27
Jane DoeJane Doe
19112
$endgroup$
add a comment |
$begingroup$
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
asked Dec 4 '18 at 19:27
Jane DoeJane Doe
19112
$endgroup$
add a comment |
$begingroup$
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
asked Dec 4 '18 at 19:27
Jane DoeJane Doe
19112
$endgroup$
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
asked Dec 4 '18 at 19:27
Jane DoeJane Doe
19112
asked Dec 4 '18 at 19:27
Jane DoeJane Doe
19112
edited Dec 6 '18 at 2:10
Jane Doe
asked Dec 4 '18 at 19:27
Jane DoeJane Doe
19112
asked Dec 4 '18 at 19:27
Jane DoeJane Doe
19112
asked Dec 4 '18 at 19:27
Jane DoeJane Doe
19112
19112
add a comment |
add a comment |
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