How to prove the convergence of a generalized Integral in $mathbb{R}^3$
I know the integral of $displaystyleiiint_{R^3} dfrac{1}{(1+|x-y|)^2|y|^2} dy$ converges for any $xin mathbb{R}^3$. Would someone help me prove the bound of integeral is independent of $x$? i.e. $displaystyleiiint_{R^3} dfrac{1}{(1+|x-y|)^2|y|^2} dyle C$, where $C$ is independent of $x$.
integration
asked Nov 27 '18 at 21:33
user1776247
12
add a comment |
I know the integral of $displaystyleiiint_{R^3} dfrac{1}{(1+|x-y|)^2|y|^2} dy$ converges for any $xin mathbb{R}^3$. Would someone help me prove the bound of integeral is independent of $x$? i.e. $displaystyleiiint_{R^3} dfrac{1}{(1+|x-y|)^2|y|^2} dyle C$, where $C$ is independent of $x$.
integration
asked Nov 27 '18 at 21:33
user1776247
12
1
Do you mean $$displaystyleiiint_{R^3} dfrac{1}{(1+|x-y|)^2|y|^2} dxdydz$$
– gimusi
Nov 27 '18 at 21:36
Fubini + translation invariance of Lebesgue measure.
– Matematleta
Nov 27 '18 at 21:57
No, I mean $x, y in mathbb{R}^3$
– user1776247
Nov 27 '18 at 22:23
Proving a bound does not depend on x does not mean the value of the integral does not depend on x.
– Michael
Nov 28 '18 at 3:36
Thank you very much! I make an error of the expression.
– user1776247
Nov 28 '18 at 14:40
add a comment |
I know the integral of $displaystyleiiint_{R^3} dfrac{1}{(1+|x-y|)^2|y|^2} dy$ converges for any $xin mathbb{R}^3$. Would someone help me prove the bound of integeral is independent of $x$? i.e. $displaystyleiiint_{R^3} dfrac{1}{(1+|x-y|)^2|y|^2} dyle C$, where $C$ is independent of $x$.
integration
asked Nov 27 '18 at 21:33
user1776247
12
I know the integral of $displaystyleiiint_{R^3} dfrac{1}{(1+|x-y|)^2|y|^2} dy$ converges for any $xin mathbb{R}^3$. Would someone help me prove the bound of integeral is independent of $x$? i.e. $displaystyleiiint_{R^3} dfrac{1}{(1+|x-y|)^2|y|^2} dyle C$, where $C$ is independent of $x$.
integration
integration
asked Nov 27 '18 at 21:33
user1776247
12
asked Nov 27 '18 at 21:33
user1776247
12
edited Nov 28 '18 at 14:40
asked Nov 27 '18 at 21:33
user1776247
12
asked Nov 27 '18 at 21:33
user1776247
12
asked Nov 27 '18 at 21:33
user1776247
12
12
1
Do you mean $$displaystyleiiint_{R^3} dfrac{1}{(1+|x-y|)^2|y|^2} dxdydz$$
– gimusi
Nov 27 '18 at 21:36
Fubini + translation invariance of Lebesgue measure.
– Matematleta
Nov 27 '18 at 21:57
No, I mean $x, y in mathbb{R}^3$
– user1776247
Nov 27 '18 at 22:23
Proving a bound does not depend on x does not mean the value of the integral does not depend on x.
– Michael
Nov 28 '18 at 3:36
Thank you very much! I make an error of the expression.
– user1776247
Nov 28 '18 at 14:40
add a comment |
1
Do you mean $$displaystyleiiint_{R^3} dfrac{1}{(1+|x-y|)^2|y|^2} dxdydz$$
– gimusi
Nov 27 '18 at 21:36
Fubini + translation invariance of Lebesgue measure.
– Matematleta
Nov 27 '18 at 21:57
No, I mean $x, y in mathbb{R}^3$
– user1776247
Nov 27 '18 at 22:23
Proving a bound does not depend on x does not mean the value of the integral does not depend on x.
– Michael
Nov 28 '18 at 3:36
Thank you very much! I make an error of the expression.
– user1776247
Nov 28 '18 at 14:40
1
1
Do you mean $$displaystyleiiint_{R^3} dfrac{1}{(1+|x-y|)^2|y|^2} dxdydz$$
– gimusi
Nov 27 '18 at 21:36
Do you mean $$displaystyleiiint_{R^3} dfrac{1}{(1+|x-y|)^2|y|^2} dxdydz$$
– gimusi
Nov 27 '18 at 21:36
Fubini + translation invariance of Lebesgue measure.
– Matematleta
Nov 27 '18 at 21:57
Fubini + translation invariance of Lebesgue measure.
– Matematleta
Nov 27 '18 at 21:57
No, I mean $x, y in mathbb{R}^3$
– user1776247
Nov 27 '18 at 22:23
No, I mean $x, y in mathbb{R}^3$
– user1776247
Nov 27 '18 at 22:23
Proving a bound does not depend on x does not mean the value of the integral does not depend on x.
– Michael
Nov 28 '18 at 3:36
Proving a bound does not depend on x does not mean the value of the integral does not depend on x.
– Michael
Nov 28 '18 at 3:36
Thank you very much! I make an error of the expression.
– user1776247
Nov 28 '18 at 14:40
Thank you very much! I make an error of the expression.
– user1776247
Nov 28 '18 at 14:40
add a comment |
1 Answer
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oldest
votes
We only need to prove when $|x|>1$.
$quaddisplaystyleiiint_{R^3} dfrac{|x|}{(1+|x-y|)^2|y|^2} dy = iiint_{R^3} dfrac{|x|}{(1+|y|)^2|x-y|^2} dy \displaystyle= iiint_{R^3} dfrac{1}{(frac 1{|x|}+|z|)^2|frac x{|x|}-z|^2} dz le iiint_{R^3} dfrac{1}{|z|^2|vec{1}-z|^2} dz$.
answered Nov 28 '18 at 15:15
user1776247
12
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
We only need to prove when $|x|>1$.
$quaddisplaystyleiiint_{R^3} dfrac{|x|}{(1+|x-y|)^2|y|^2} dy = iiint_{R^3} dfrac{|x|}{(1+|y|)^2|x-y|^2} dy \displaystyle= iiint_{R^3} dfrac{1}{(frac 1{|x|}+|z|)^2|frac x{|x|}-z|^2} dz le iiint_{R^3} dfrac{1}{|z|^2|vec{1}-z|^2} dz$.
answered Nov 28 '18 at 15:15
user1776247
12
add a comment |
We only need to prove when $|x|>1$.
$quaddisplaystyleiiint_{R^3} dfrac{|x|}{(1+|x-y|)^2|y|^2} dy = iiint_{R^3} dfrac{|x|}{(1+|y|)^2|x-y|^2} dy \displaystyle= iiint_{R^3} dfrac{1}{(frac 1{|x|}+|z|)^2|frac x{|x|}-z|^2} dz le iiint_{R^3} dfrac{1}{|z|^2|vec{1}-z|^2} dz$.
answered Nov 28 '18 at 15:15
user1776247
12
add a comment |
We only need to prove when $|x|>1$.
$quaddisplaystyleiiint_{R^3} dfrac{|x|}{(1+|x-y|)^2|y|^2} dy = iiint_{R^3} dfrac{|x|}{(1+|y|)^2|x-y|^2} dy \displaystyle= iiint_{R^3} dfrac{1}{(frac 1{|x|}+|z|)^2|frac x{|x|}-z|^2} dz le iiint_{R^3} dfrac{1}{|z|^2|vec{1}-z|^2} dz$.
answered Nov 28 '18 at 15:15
user1776247
12
We only need to prove when $|x|>1$.
$quaddisplaystyleiiint_{R^3} dfrac{|x|}{(1+|x-y|)^2|y|^2} dy = iiint_{R^3} dfrac{|x|}{(1+|y|)^2|x-y|^2} dy \displaystyle= iiint_{R^3} dfrac{1}{(frac 1{|x|}+|z|)^2|frac x{|x|}-z|^2} dz le iiint_{R^3} dfrac{1}{|z|^2|vec{1}-z|^2} dz$.
answered Nov 28 '18 at 15:15
user1776247
12
edited Nov 28 '18 at 15:21
answered Nov 28 '18 at 15:15
user1776247
12
answered Nov 28 '18 at 15:15
user1776247
12
answered Nov 28 '18 at 15:15
user1776247
12
12
add a comment |
add a comment |
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1
Do you mean $$displaystyleiiint_{R^3} dfrac{1}{(1+|x-y|)^2|y|^2} dxdydz$$
– gimusi
Nov 27 '18 at 21:36
Fubini + translation invariance of Lebesgue measure.
– Matematleta
Nov 27 '18 at 21:57
No, I mean $x, y in mathbb{R}^3$
– user1776247
Nov 27 '18 at 22:23
Proving a bound does not depend on x does not mean the value of the integral does not depend on x.
– Michael
Nov 28 '18 at 3:36
Thank you very much! I make an error of the expression.
– user1776247
Nov 28 '18 at 14:40