Lebesgue integral on $[0,1]^{2}$
$begingroup$
Let $f:[0,1]^2->mathbb{R}$ with
$f(x,y)=frac{x^2-y^2}{(x^2+y^2)^2}$ if $(x,y)neq(0,0)$
$f(x,y)=0$ else
I want to calculate
$lim_{ato 0}int_{[a,1]^2}f(x,y)dmathscr{L}^2$
But how do Ido this? I already calculated
$int_{0}^1int_{0}^1f(x,y)dxdy=-pi/4$
$int_{0}^1int_{0}^1f(x,y)dydx=pi/4$
Doesn't that mean that I cannot use Fubini-Tonelli? But what other choice do I have?
integration measure-theory lebesgue-integral lebesgue-measure
$endgroup$
add a comment |
$begingroup$
Let $f:[0,1]^2->mathbb{R}$ with
$f(x,y)=frac{x^2-y^2}{(x^2+y^2)^2}$ if $(x,y)neq(0,0)$
$f(x,y)=0$ else
I want to calculate
$lim_{ato 0}int_{[a,1]^2}f(x,y)dmathscr{L}^2$
But how do Ido this? I already calculated
$int_{0}^1int_{0}^1f(x,y)dxdy=-pi/4$
$int_{0}^1int_{0}^1f(x,y)dydx=pi/4$
Doesn't that mean that I cannot use Fubini-Tonelli? But what other choice do I have?
integration measure-theory lebesgue-integral lebesgue-measure
$endgroup$
$begingroup$
I think there's a typo - should that be $(x^2+y^2)^2$ in the denominator? I don't believe you want a full line of singularities.
$endgroup$
– jmerry
Jan 5 at 23:38
$begingroup$
Yes, you're right, I'm sorry and edited it.
$endgroup$
– Kekks
Jan 5 at 23:39
add a comment |
$begingroup$
Let $f:[0,1]^2->mathbb{R}$ with
$f(x,y)=frac{x^2-y^2}{(x^2+y^2)^2}$ if $(x,y)neq(0,0)$
$f(x,y)=0$ else
I want to calculate
$lim_{ato 0}int_{[a,1]^2}f(x,y)dmathscr{L}^2$
But how do Ido this? I already calculated
$int_{0}^1int_{0}^1f(x,y)dxdy=-pi/4$
$int_{0}^1int_{0}^1f(x,y)dydx=pi/4$
Doesn't that mean that I cannot use Fubini-Tonelli? But what other choice do I have?
integration measure-theory lebesgue-integral lebesgue-measure
$endgroup$
Let $f:[0,1]^2->mathbb{R}$ with
$f(x,y)=frac{x^2-y^2}{(x^2+y^2)^2}$ if $(x,y)neq(0,0)$
$f(x,y)=0$ else
I want to calculate
$lim_{ato 0}int_{[a,1]^2}f(x,y)dmathscr{L}^2$
But how do Ido this? I already calculated
$int_{0}^1int_{0}^1f(x,y)dxdy=-pi/4$
$int_{0}^1int_{0}^1f(x,y)dydx=pi/4$
Doesn't that mean that I cannot use Fubini-Tonelli? But what other choice do I have?
integration measure-theory lebesgue-integral lebesgue-measure
integration measure-theory lebesgue-integral lebesgue-measure
edited Jan 5 at 23:39
Kekks
asked Jan 5 at 23:31
KekksKekks
518
518
$begingroup$
I think there's a typo - should that be $(x^2+y^2)^2$ in the denominator? I don't believe you want a full line of singularities.
$endgroup$
– jmerry
Jan 5 at 23:38
$begingroup$
Yes, you're right, I'm sorry and edited it.
$endgroup$
– Kekks
Jan 5 at 23:39
add a comment |
$begingroup$
I think there's a typo - should that be $(x^2+y^2)^2$ in the denominator? I don't believe you want a full line of singularities.
$endgroup$
– jmerry
Jan 5 at 23:38
$begingroup$
Yes, you're right, I'm sorry and edited it.
$endgroup$
– Kekks
Jan 5 at 23:39
$begingroup$
I think there's a typo - should that be $(x^2+y^2)^2$ in the denominator? I don't believe you want a full line of singularities.
$endgroup$
– jmerry
Jan 5 at 23:38
$begingroup$
I think there's a typo - should that be $(x^2+y^2)^2$ in the denominator? I don't believe you want a full line of singularities.
$endgroup$
– jmerry
Jan 5 at 23:38
$begingroup$
Yes, you're right, I'm sorry and edited it.
$endgroup$
– Kekks
Jan 5 at 23:39
$begingroup$
Yes, you're right, I'm sorry and edited it.
$endgroup$
– Kekks
Jan 5 at 23:39
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
All right, the function isn't absolutely integrable on the whole square. But that's not the question; we're looking at an improper integral, approaching the full square in a specific way akin to a principal value.
For any $a>0$,
$$I(a) = int_a^1int_a^1 frac{x^2-y^2}{(x^2+y^2)^2},dy,dx = int_a^1int_a^1 frac{x^2-y^2}{(x^2+y^2)^2},dx,dy$$
$$int_a^1int_a^1 frac{x^2-y^2}{(x^2+y^2)^2},dx,dy = int_a^1int_a^1 frac{y^2-x^2}{(x^2+y^2)^2},dy,dx = -I(a)$$
In the first line, we use Fubini's theorem to switch the order (it's a bounded continuous function). In the second line, we just swap the names of the two variables - that isn't really a change of order. Combining the two, we get $I(a)=-I(a)$, so $I(a)=0$. Take the limit as $ato 0$, and we get zero.
$endgroup$
$begingroup$
Thats a really short and nice solution. Looking at the graph of the function I could've come up with this myself. Thanks alot
$endgroup$
– Kekks
Jan 5 at 23:59
add a comment |
$begingroup$
Hint: the integral is $not$ over $[0,1]times [0,1].$ For fixed $a: 0<ale x,yle 1,$ show that $fin L^1([a,1]times [a,1]).$ Then, apply Fubini and finally, take the limit as $ato 0.$
$endgroup$
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
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active
oldest
votes
$begingroup$
All right, the function isn't absolutely integrable on the whole square. But that's not the question; we're looking at an improper integral, approaching the full square in a specific way akin to a principal value.
For any $a>0$,
$$I(a) = int_a^1int_a^1 frac{x^2-y^2}{(x^2+y^2)^2},dy,dx = int_a^1int_a^1 frac{x^2-y^2}{(x^2+y^2)^2},dx,dy$$
$$int_a^1int_a^1 frac{x^2-y^2}{(x^2+y^2)^2},dx,dy = int_a^1int_a^1 frac{y^2-x^2}{(x^2+y^2)^2},dy,dx = -I(a)$$
In the first line, we use Fubini's theorem to switch the order (it's a bounded continuous function). In the second line, we just swap the names of the two variables - that isn't really a change of order. Combining the two, we get $I(a)=-I(a)$, so $I(a)=0$. Take the limit as $ato 0$, and we get zero.
$endgroup$
$begingroup$
Thats a really short and nice solution. Looking at the graph of the function I could've come up with this myself. Thanks alot
$endgroup$
– Kekks
Jan 5 at 23:59
add a comment |
$begingroup$
All right, the function isn't absolutely integrable on the whole square. But that's not the question; we're looking at an improper integral, approaching the full square in a specific way akin to a principal value.
For any $a>0$,
$$I(a) = int_a^1int_a^1 frac{x^2-y^2}{(x^2+y^2)^2},dy,dx = int_a^1int_a^1 frac{x^2-y^2}{(x^2+y^2)^2},dx,dy$$
$$int_a^1int_a^1 frac{x^2-y^2}{(x^2+y^2)^2},dx,dy = int_a^1int_a^1 frac{y^2-x^2}{(x^2+y^2)^2},dy,dx = -I(a)$$
In the first line, we use Fubini's theorem to switch the order (it's a bounded continuous function). In the second line, we just swap the names of the two variables - that isn't really a change of order. Combining the two, we get $I(a)=-I(a)$, so $I(a)=0$. Take the limit as $ato 0$, and we get zero.
$endgroup$
$begingroup$
Thats a really short and nice solution. Looking at the graph of the function I could've come up with this myself. Thanks alot
$endgroup$
– Kekks
Jan 5 at 23:59
add a comment |
$begingroup$
All right, the function isn't absolutely integrable on the whole square. But that's not the question; we're looking at an improper integral, approaching the full square in a specific way akin to a principal value.
For any $a>0$,
$$I(a) = int_a^1int_a^1 frac{x^2-y^2}{(x^2+y^2)^2},dy,dx = int_a^1int_a^1 frac{x^2-y^2}{(x^2+y^2)^2},dx,dy$$
$$int_a^1int_a^1 frac{x^2-y^2}{(x^2+y^2)^2},dx,dy = int_a^1int_a^1 frac{y^2-x^2}{(x^2+y^2)^2},dy,dx = -I(a)$$
In the first line, we use Fubini's theorem to switch the order (it's a bounded continuous function). In the second line, we just swap the names of the two variables - that isn't really a change of order. Combining the two, we get $I(a)=-I(a)$, so $I(a)=0$. Take the limit as $ato 0$, and we get zero.
$endgroup$
All right, the function isn't absolutely integrable on the whole square. But that's not the question; we're looking at an improper integral, approaching the full square in a specific way akin to a principal value.
For any $a>0$,
$$I(a) = int_a^1int_a^1 frac{x^2-y^2}{(x^2+y^2)^2},dy,dx = int_a^1int_a^1 frac{x^2-y^2}{(x^2+y^2)^2},dx,dy$$
$$int_a^1int_a^1 frac{x^2-y^2}{(x^2+y^2)^2},dx,dy = int_a^1int_a^1 frac{y^2-x^2}{(x^2+y^2)^2},dy,dx = -I(a)$$
In the first line, we use Fubini's theorem to switch the order (it's a bounded continuous function). In the second line, we just swap the names of the two variables - that isn't really a change of order. Combining the two, we get $I(a)=-I(a)$, so $I(a)=0$. Take the limit as $ato 0$, and we get zero.
edited Jan 5 at 23:48
answered Jan 5 at 23:37
jmerryjmerry
17k11633
17k11633
$begingroup$
Thats a really short and nice solution. Looking at the graph of the function I could've come up with this myself. Thanks alot
$endgroup$
– Kekks
Jan 5 at 23:59
add a comment |
$begingroup$
Thats a really short and nice solution. Looking at the graph of the function I could've come up with this myself. Thanks alot
$endgroup$
– Kekks
Jan 5 at 23:59
$begingroup$
Thats a really short and nice solution. Looking at the graph of the function I could've come up with this myself. Thanks alot
$endgroup$
– Kekks
Jan 5 at 23:59
$begingroup$
Thats a really short and nice solution. Looking at the graph of the function I could've come up with this myself. Thanks alot
$endgroup$
– Kekks
Jan 5 at 23:59
add a comment |
$begingroup$
Hint: the integral is $not$ over $[0,1]times [0,1].$ For fixed $a: 0<ale x,yle 1,$ show that $fin L^1([a,1]times [a,1]).$ Then, apply Fubini and finally, take the limit as $ato 0.$
$endgroup$
add a comment |
$begingroup$
Hint: the integral is $not$ over $[0,1]times [0,1].$ For fixed $a: 0<ale x,yle 1,$ show that $fin L^1([a,1]times [a,1]).$ Then, apply Fubini and finally, take the limit as $ato 0.$
$endgroup$
add a comment |
$begingroup$
Hint: the integral is $not$ over $[0,1]times [0,1].$ For fixed $a: 0<ale x,yle 1,$ show that $fin L^1([a,1]times [a,1]).$ Then, apply Fubini and finally, take the limit as $ato 0.$
$endgroup$
Hint: the integral is $not$ over $[0,1]times [0,1].$ For fixed $a: 0<ale x,yle 1,$ show that $fin L^1([a,1]times [a,1]).$ Then, apply Fubini and finally, take the limit as $ato 0.$
edited Jan 6 at 0:13
answered Jan 5 at 23:50
MatematletaMatematleta
12.1k21020
12.1k21020
add a comment |
add a comment |
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$begingroup$
I think there's a typo - should that be $(x^2+y^2)^2$ in the denominator? I don't believe you want a full line of singularities.
$endgroup$
– jmerry
Jan 5 at 23:38
$begingroup$
Yes, you're right, I'm sorry and edited it.
$endgroup$
– Kekks
Jan 5 at 23:39