Integrable on square [closed]
0
For which $s>0$ the functions $f(x,y)= frac{1}{|x-y|^s}$ and $f(x,y)=|y-sin(frac{1}{x})|^{-s}$ are integrable on the square $[-1,1]^2$?
Help me please. We have to look at what values $s>0$ the integral $intlimits_{[-1,1]^2} f(x)d mu $ diverges converges, but i have not idea what to do.
integration measure-theory
asked Dec 9 '18 at 21:21
GThompson
266
closed as off-topic by GNUSupporter 8964民主女神 地下教會, RRL, Saad, José Carlos Santos, user10354138 Dec 23 '18 at 19:40
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." – GNUSupporter 8964民主女神 地下教會, RRL, Saad, José Carlos Santos, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
0
For which $s>0$ the functions $f(x,y)= frac{1}{|x-y|^s}$ and $f(x,y)=|y-sin(frac{1}{x})|^{-s}$ are integrable on the square $[-1,1]^2$?
Help me please. We have to look at what values $s>0$ the integral $intlimits_{[-1,1]^2} f(x)d mu $ diverges converges, but i have not idea what to do.
integration measure-theory
asked Dec 9 '18 at 21:21
GThompson
266
closed as off-topic by GNUSupporter 8964民主女神 地下教會, RRL, Saad, José Carlos Santos, user10354138 Dec 23 '18 at 19:40
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." – GNUSupporter 8964民主女神 地下教會, RRL, Saad, José Carlos Santos, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
1
As a start, we notice the first function is well-defined except on $x = y$ and the second function is well-defined except on $x = 0$ and $y = sin frac{1}{x}$. Now lines are of measure $0$, and the topologist's sine curve is also of measure $0$. Besides, both functions are Borel measurable (being composition of Borel measurable functions). So both functions are well-defined and non-negative measurable a.e. on $[-1,1]^2$. Hence both integrals exists in $[0, +infty]$ (being defined as supremum).
– Alex Vong
Dec 10 '18 at 19:32
1
The only way for the above functions to be non-integrable is when their integrals turn out to be $+infty$.
– Alex Vong
Dec 10 '18 at 19:34
1
For the first integral, I think you can partition the domain into two triangles. Then ,approximate the triangle using smaller triangles converging to it. By monotone convergence theorem, it suffices to compute the integral on the smaller triangles, which can be computed by using Fubini's theorem. But I don't think the same idea works for the second integral since it is not obvious how to partition the domain in this case.
– Alex Vong
Dec 12 '18 at 21:30
1
Actually, I figure out how to partition the domain for the second integral. One simply partition the domain using the curve $y = sin frac{1}{x}$. Then we can apply MCT and Fubini as before. The remaining problem is that at some point, one has to integrate $(sin frac{1}{x} + 1)^{-s + 1}$, which doesn't seem to have a closed form solution. Perhaps some estimation is required, but I don't know how to proceed.
– Alex Vong
Dec 13 '18 at 14:44
add a comment |
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0
For which $s>0$ the functions $f(x,y)= frac{1}{|x-y|^s}$ and $f(x,y)=|y-sin(frac{1}{x})|^{-s}$ are integrable on the square $[-1,1]^2$?
Help me please. We have to look at what values $s>0$ the integral $intlimits_{[-1,1]^2} f(x)d mu $ diverges converges, but i have not idea what to do.
integration measure-theory
asked Dec 9 '18 at 21:21
GThompson
266
For which $s>0$ the functions $f(x,y)= frac{1}{|x-y|^s}$ and $f(x,y)=|y-sin(frac{1}{x})|^{-s}$ are integrable on the square $[-1,1]^2$?
Help me please. We have to look at what values $s>0$ the integral $intlimits_{[-1,1]^2} f(x)d mu $ diverges converges, but i have not idea what to do.
integration measure-theory
integration measure-theory
asked Dec 9 '18 at 21:21
GThompson
266
asked Dec 9 '18 at 21:21
GThompson
266
asked Dec 9 '18 at 21:21
GThompson
266
asked Dec 9 '18 at 21:21
GThompson
266
asked Dec 9 '18 at 21:21
GThompson
266
266
closed as off-topic by GNUSupporter 8964民主女神 地下教會, RRL, Saad, José Carlos Santos, user10354138 Dec 23 '18 at 19:40
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." – GNUSupporter 8964民主女神 地下教會, RRL, Saad, José Carlos Santos, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by GNUSupporter 8964民主女神 地下教會, RRL, Saad, José Carlos Santos, user10354138 Dec 23 '18 at 19:40
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." – GNUSupporter 8964民主女神 地下教會, RRL, Saad, José Carlos Santos, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
1
As a start, we notice the first function is well-defined except on $x = y$ and the second function is well-defined except on $x = 0$ and $y = sin frac{1}{x}$. Now lines are of measure $0$, and the topologist's sine curve is also of measure $0$. Besides, both functions are Borel measurable (being composition of Borel measurable functions). So both functions are well-defined and non-negative measurable a.e. on $[-1,1]^2$. Hence both integrals exists in $[0, +infty]$ (being defined as supremum).
– Alex Vong
Dec 10 '18 at 19:32
1
The only way for the above functions to be non-integrable is when their integrals turn out to be $+infty$.
– Alex Vong
Dec 10 '18 at 19:34
1
For the first integral, I think you can partition the domain into two triangles. Then ,approximate the triangle using smaller triangles converging to it. By monotone convergence theorem, it suffices to compute the integral on the smaller triangles, which can be computed by using Fubini's theorem. But I don't think the same idea works for the second integral since it is not obvious how to partition the domain in this case.
– Alex Vong
Dec 12 '18 at 21:30
1
Actually, I figure out how to partition the domain for the second integral. One simply partition the domain using the curve $y = sin frac{1}{x}$. Then we can apply MCT and Fubini as before. The remaining problem is that at some point, one has to integrate $(sin frac{1}{x} + 1)^{-s + 1}$, which doesn't seem to have a closed form solution. Perhaps some estimation is required, but I don't know how to proceed.
– Alex Vong
Dec 13 '18 at 14:44
add a comment |
1
As a start, we notice the first function is well-defined except on $x = y$ and the second function is well-defined except on $x = 0$ and $y = sin frac{1}{x}$. Now lines are of measure $0$, and the topologist's sine curve is also of measure $0$. Besides, both functions are Borel measurable (being composition of Borel measurable functions). So both functions are well-defined and non-negative measurable a.e. on $[-1,1]^2$. Hence both integrals exists in $[0, +infty]$ (being defined as supremum).
– Alex Vong
Dec 10 '18 at 19:32
1
The only way for the above functions to be non-integrable is when their integrals turn out to be $+infty$.
– Alex Vong
Dec 10 '18 at 19:34
1
For the first integral, I think you can partition the domain into two triangles. Then ,approximate the triangle using smaller triangles converging to it. By monotone convergence theorem, it suffices to compute the integral on the smaller triangles, which can be computed by using Fubini's theorem. But I don't think the same idea works for the second integral since it is not obvious how to partition the domain in this case.
– Alex Vong
Dec 12 '18 at 21:30
1
Actually, I figure out how to partition the domain for the second integral. One simply partition the domain using the curve $y = sin frac{1}{x}$. Then we can apply MCT and Fubini as before. The remaining problem is that at some point, one has to integrate $(sin frac{1}{x} + 1)^{-s + 1}$, which doesn't seem to have a closed form solution. Perhaps some estimation is required, but I don't know how to proceed.
– Alex Vong
Dec 13 '18 at 14:44
1
1
As a start, we notice the first function is well-defined except on $x = y$ and the second function is well-defined except on $x = 0$ and $y = sin frac{1}{x}$. Now lines are of measure $0$, and the topologist's sine curve is also of measure $0$. Besides, both functions are Borel measurable (being composition of Borel measurable functions). So both functions are well-defined and non-negative measurable a.e. on $[-1,1]^2$. Hence both integrals exists in $[0, +infty]$ (being defined as supremum).
– Alex Vong
Dec 10 '18 at 19:32
As a start, we notice the first function is well-defined except on $x = y$ and the second function is well-defined except on $x = 0$ and $y = sin frac{1}{x}$. Now lines are of measure $0$, and the topologist's sine curve is also of measure $0$. Besides, both functions are Borel measurable (being composition of Borel measurable functions). So both functions are well-defined and non-negative measurable a.e. on $[-1,1]^2$. Hence both integrals exists in $[0, +infty]$ (being defined as supremum).
– Alex Vong
Dec 10 '18 at 19:32
1
1
The only way for the above functions to be non-integrable is when their integrals turn out to be $+infty$.
– Alex Vong
Dec 10 '18 at 19:34
The only way for the above functions to be non-integrable is when their integrals turn out to be $+infty$.
– Alex Vong
Dec 10 '18 at 19:34
1
1
For the first integral, I think you can partition the domain into two triangles. Then ,approximate the triangle using smaller triangles converging to it. By monotone convergence theorem, it suffices to compute the integral on the smaller triangles, which can be computed by using Fubini's theorem. But I don't think the same idea works for the second integral since it is not obvious how to partition the domain in this case.
– Alex Vong
Dec 12 '18 at 21:30
For the first integral, I think you can partition the domain into two triangles. Then ,approximate the triangle using smaller triangles converging to it. By monotone convergence theorem, it suffices to compute the integral on the smaller triangles, which can be computed by using Fubini's theorem. But I don't think the same idea works for the second integral since it is not obvious how to partition the domain in this case.
– Alex Vong
Dec 12 '18 at 21:30
1
1
Actually, I figure out how to partition the domain for the second integral. One simply partition the domain using the curve $y = sin frac{1}{x}$. Then we can apply MCT and Fubini as before. The remaining problem is that at some point, one has to integrate $(sin frac{1}{x} + 1)^{-s + 1}$, which doesn't seem to have a closed form solution. Perhaps some estimation is required, but I don't know how to proceed.
– Alex Vong
Dec 13 '18 at 14:44
Actually, I figure out how to partition the domain for the second integral. One simply partition the domain using the curve $y = sin frac{1}{x}$. Then we can apply MCT and Fubini as before. The remaining problem is that at some point, one has to integrate $(sin frac{1}{x} + 1)^{-s + 1}$, which doesn't seem to have a closed form solution. Perhaps some estimation is required, but I don't know how to proceed.
– Alex Vong
Dec 13 '18 at 14:44
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1
As a start, we notice the first function is well-defined except on $x = y$ and the second function is well-defined except on $x = 0$ and $y = sin frac{1}{x}$. Now lines are of measure $0$, and the topologist's sine curve is also of measure $0$. Besides, both functions are Borel measurable (being composition of Borel measurable functions). So both functions are well-defined and non-negative measurable a.e. on $[-1,1]^2$. Hence both integrals exists in $[0, +infty]$ (being defined as supremum).
– Alex Vong
Dec 10 '18 at 19:32
1
The only way for the above functions to be non-integrable is when their integrals turn out to be $+infty$.
– Alex Vong
Dec 10 '18 at 19:34
1
For the first integral, I think you can partition the domain into two triangles. Then ,approximate the triangle using smaller triangles converging to it. By monotone convergence theorem, it suffices to compute the integral on the smaller triangles, which can be computed by using Fubini's theorem. But I don't think the same idea works for the second integral since it is not obvious how to partition the domain in this case.
– Alex Vong
Dec 12 '18 at 21:30
1
Actually, I figure out how to partition the domain for the second integral. One simply partition the domain using the curve $y = sin frac{1}{x}$. Then we can apply MCT and Fubini as before. The remaining problem is that at some point, one has to integrate $(sin frac{1}{x} + 1)^{-s + 1}$, which doesn't seem to have a closed form solution. Perhaps some estimation is required, but I don't know how to proceed.
– Alex Vong
Dec 13 '18 at 14:44