Lebesgue integral on $[0,1]^{2}$












0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














0












0








0





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 5 at 23:39







Kekks

















asked Jan 5 at 23:31









KekksKekks

347




347












  • $begingroup$
    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










  • $begingroup$
    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










2 Answers
2






active

oldest

votes


















1












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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



















1












$begingroup$

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 $ato 0.$






share|cite|improve this answer











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    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.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      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
















    1












    $begingroup$

    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.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      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














    1












    1








    1





    $begingroup$

    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.






    share|cite|improve this answer











    $endgroup$



    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.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 5 at 23:48

























    answered Jan 5 at 23:37









    jmerryjmerry

    3,592514




    3,592514












    • $begingroup$
      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




    $begingroup$
    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











    1












    $begingroup$

    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 $ato 0.$






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      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 $ato 0.$






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        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 $ato 0.$






        share|cite|improve this answer











        $endgroup$



        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 $ato 0.$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 6 at 0:13

























        answered Jan 5 at 23:50









        MatematletaMatematleta

        10.2k2918




        10.2k2918






























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