$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?
integration measure-theory multivariable-calculus lebesgue-integral
asked Dec 19 '18 at 9:32
SABOYSABOY
656311
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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:
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$begingroup$
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?
integration measure-theory multivariable-calculus lebesgue-integral
asked Dec 19 '18 at 9:32
SABOYSABOY
656311
$endgroup$
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.
add a comment |
$begingroup$
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?
integration measure-theory multivariable-calculus lebesgue-integral
asked Dec 19 '18 at 9:32
SABOYSABOY
656311
$endgroup$
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
integration measure-theory multivariable-calculus lebesgue-integral
asked Dec 19 '18 at 9:32
SABOYSABOY
656311
asked Dec 19 '18 at 9:32
SABOYSABOY
656311
asked Dec 19 '18 at 9:32
SABOYSABOY
656311
asked Dec 19 '18 at 9:32
SABOYSABOY
656311
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.
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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 $int_0^{2pi} |sin, theta cos , theta|, dtheta$].
answered Dec 19 '18 at 9:47
Kavi Rama MurthyKavi Rama Murthy
63.7k42463
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $int_0^{2pi} |sin, theta cos , theta|, dtheta$].
answered Dec 19 '18 at 9:47
Kavi Rama MurthyKavi Rama Murthy
63.7k42463
$endgroup$
add a comment |
$begingroup$
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 $int_0^{2pi} |sin, theta cos , theta|, dtheta$].
answered Dec 19 '18 at 9:47
Kavi Rama MurthyKavi Rama Murthy
63.7k42463
$endgroup$
add a comment |
$begingroup$
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 $int_0^{2pi} |sin, theta cos , theta|, dtheta$].
answered Dec 19 '18 at 9:47
Kavi Rama MurthyKavi Rama Murthy
63.7k42463
$endgroup$
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 $int_0^{2pi} |sin, theta cos , theta|, dtheta$].
answered Dec 19 '18 at 9:47
Kavi Rama MurthyKavi Rama Murthy
63.7k42463
answered Dec 19 '18 at 9:47
Kavi Rama MurthyKavi Rama Murthy
63.7k42463
answered Dec 19 '18 at 9:47
Kavi Rama MurthyKavi Rama Murthy
63.7k42463
answered Dec 19 '18 at 9:47
Kavi Rama MurthyKavi Rama Murthy
63.7k42463
63.7k42463
add a comment |
add a comment |
