$int_{[0,1]} int_{[0,1]} frac {x^2-y^2}{(x^2+y^2)^2}dxdy$ and $int_{[0,1]} int_{[0,1]} frac...












3












$begingroup$


I am facing problem in calculating $$I_1=int_{[0,1]} int_{[0,1]} frac {x^2-y^2}{(x^2+y^2)^2} dxdy$$ and $$I_2=int_{[0,1]} int_{[0,1]} frac {x^2-y^2}{(x^2+y^2)^2} dydx$$ after substituting $x=rcos theta$ and $y=rsin theta$ we have $I_1=int_{[-pi,pi]} int_{[0,1]}frac{cos(2theta)}{r^2}rdr dtheta$ so I am getting $infty times 0=0$ So I am not getting $pi/4$ and $-pi/4$ resp.



Maybe there is a silly point I am missing. Please help.






calculus integration multivariable-calculus definite-integrals







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    How about just doing the integrals, as asked, and not "substituting"?
    $endgroup$
    – Lord Shark the Unknown
    Dec 16 '18 at 7:45












  • $begingroup$
    Yeah I am confused in not getting. May be a result of doing a lot of pure math :( .. sorry
    $endgroup$
    – Gimgim
    Dec 16 '18 at 7:54










  • $begingroup$
    Yeah I might get downvote but as I asked this question what I didn't get. So I am not going to erase the question. Thanks all of you for helping
    $endgroup$
    – Gimgim
    Dec 16 '18 at 8:00






  • 1




    $begingroup$
    Its a good question and you showed some effort. Clearly a "trick" was needed.
    $endgroup$
    – RRL
    Dec 16 '18 at 8:01






  • 1




    $begingroup$
    @Gimgim: The anti-derivatives don't exactly jump out at you so we can all move on.
    $endgroup$
    – RRL
    Dec 16 '18 at 8:16
















3












$begingroup$


I am facing problem in calculating $$I_1=int_{[0,1]} int_{[0,1]} frac {x^2-y^2}{(x^2+y^2)^2} dxdy$$ and $$I_2=int_{[0,1]} int_{[0,1]} frac {x^2-y^2}{(x^2+y^2)^2} dydx$$ after substituting $x=rcos theta$ and $y=rsin theta$ we have $I_1=int_{[-pi,pi]} int_{[0,1]}frac{cos(2theta)}{r^2}rdr dtheta$ so I am getting $infty times 0=0$ So I am not getting $pi/4$ and $-pi/4$ resp.



Maybe there is a silly point I am missing. Please help.






calculus integration multivariable-calculus definite-integrals







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    How about just doing the integrals, as asked, and not "substituting"?
    $endgroup$
    – Lord Shark the Unknown
    Dec 16 '18 at 7:45












  • $begingroup$
    Yeah I am confused in not getting. May be a result of doing a lot of pure math :( .. sorry
    $endgroup$
    – Gimgim
    Dec 16 '18 at 7:54










  • $begingroup$
    Yeah I might get downvote but as I asked this question what I didn't get. So I am not going to erase the question. Thanks all of you for helping
    $endgroup$
    – Gimgim
    Dec 16 '18 at 8:00






  • 1




    $begingroup$
    Its a good question and you showed some effort. Clearly a "trick" was needed.
    $endgroup$
    – RRL
    Dec 16 '18 at 8:01






  • 1




    $begingroup$
    @Gimgim: The anti-derivatives don't exactly jump out at you so we can all move on.
    $endgroup$
    – RRL
    Dec 16 '18 at 8:16














3












3








3


1



$begingroup$


I am facing problem in calculating $$I_1=int_{[0,1]} int_{[0,1]} frac {x^2-y^2}{(x^2+y^2)^2} dxdy$$ and $$I_2=int_{[0,1]} int_{[0,1]} frac {x^2-y^2}{(x^2+y^2)^2} dydx$$ after substituting $x=rcos theta$ and $y=rsin theta$ we have $I_1=int_{[-pi,pi]} int_{[0,1]}frac{cos(2theta)}{r^2}rdr dtheta$ so I am getting $infty times 0=0$ So I am not getting $pi/4$ and $-pi/4$ resp.



Maybe there is a silly point I am missing. Please help.






calculus integration multivariable-calculus definite-integrals







share|cite|improve this question









$endgroup$




I am facing problem in calculating $$I_1=int_{[0,1]} int_{[0,1]} frac {x^2-y^2}{(x^2+y^2)^2} dxdy$$ and $$I_2=int_{[0,1]} int_{[0,1]} frac {x^2-y^2}{(x^2+y^2)^2} dydx$$ after substituting $x=rcos theta$ and $y=rsin theta$ we have $I_1=int_{[-pi,pi]} int_{[0,1]}frac{cos(2theta)}{r^2}rdr dtheta$ so I am getting $infty times 0=0$ So I am not getting $pi/4$ and $-pi/4$ resp.



Maybe there is a silly point I am missing. Please help.






calculus integration multivariable-calculus definite-integrals




calculus integration multivariable-calculus definite-integrals






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 16 '18 at 7:44









GimgimGimgim

25713




25713








  • 1




    $begingroup$
    How about just doing the integrals, as asked, and not "substituting"?
    $endgroup$
    – Lord Shark the Unknown
    Dec 16 '18 at 7:45












  • $begingroup$
    Yeah I am confused in not getting. May be a result of doing a lot of pure math :( .. sorry
    $endgroup$
    – Gimgim
    Dec 16 '18 at 7:54










  • $begingroup$
    Yeah I might get downvote but as I asked this question what I didn't get. So I am not going to erase the question. Thanks all of you for helping
    $endgroup$
    – Gimgim
    Dec 16 '18 at 8:00






  • 1




    $begingroup$
    Its a good question and you showed some effort. Clearly a "trick" was needed.
    $endgroup$
    – RRL
    Dec 16 '18 at 8:01






  • 1




    $begingroup$
    @Gimgim: The anti-derivatives don't exactly jump out at you so we can all move on.
    $endgroup$
    – RRL
    Dec 16 '18 at 8:16














  • 1




    $begingroup$
    How about just doing the integrals, as asked, and not "substituting"?
    $endgroup$
    – Lord Shark the Unknown
    Dec 16 '18 at 7:45












  • $begingroup$
    Yeah I am confused in not getting. May be a result of doing a lot of pure math :( .. sorry
    $endgroup$
    – Gimgim
    Dec 16 '18 at 7:54










  • $begingroup$
    Yeah I might get downvote but as I asked this question what I didn't get. So I am not going to erase the question. Thanks all of you for helping
    $endgroup$
    – Gimgim
    Dec 16 '18 at 8:00






  • 1




    $begingroup$
    Its a good question and you showed some effort. Clearly a "trick" was needed.
    $endgroup$
    – RRL
    Dec 16 '18 at 8:01






  • 1




    $begingroup$
    @Gimgim: The anti-derivatives don't exactly jump out at you so we can all move on.
    $endgroup$
    – RRL
    Dec 16 '18 at 8:16








1




1




$begingroup$
How about just doing the integrals, as asked, and not "substituting"?
$endgroup$
– Lord Shark the Unknown
Dec 16 '18 at 7:45






$begingroup$
How about just doing the integrals, as asked, and not "substituting"?
$endgroup$
– Lord Shark the Unknown
Dec 16 '18 at 7:45














$begingroup$
Yeah I am confused in not getting. May be a result of doing a lot of pure math :( .. sorry
$endgroup$
– Gimgim
Dec 16 '18 at 7:54




$begingroup$
Yeah I am confused in not getting. May be a result of doing a lot of pure math :( .. sorry
$endgroup$
– Gimgim
Dec 16 '18 at 7:54












$begingroup$
Yeah I might get downvote but as I asked this question what I didn't get. So I am not going to erase the question. Thanks all of you for helping
$endgroup$
– Gimgim
Dec 16 '18 at 8:00




$begingroup$
Yeah I might get downvote but as I asked this question what I didn't get. So I am not going to erase the question. Thanks all of you for helping
$endgroup$
– Gimgim
Dec 16 '18 at 8:00




1




1




$begingroup$
Its a good question and you showed some effort. Clearly a "trick" was needed.
$endgroup$
– RRL
Dec 16 '18 at 8:01




$begingroup$
Its a good question and you showed some effort. Clearly a "trick" was needed.
$endgroup$
– RRL
Dec 16 '18 at 8:01




1




1




$begingroup$
@Gimgim: The anti-derivatives don't exactly jump out at you so we can all move on.
$endgroup$
– RRL
Dec 16 '18 at 8:16




$begingroup$
@Gimgim: The anti-derivatives don't exactly jump out at you so we can all move on.
$endgroup$
– RRL
Dec 16 '18 at 8:16










1 Answer
1






active

oldest

votes


















3












$begingroup$

Hint:



$$frac {x^2-y^2}{(x^2+y^2)^2} = frac{partial}{partial y} left(frac{y}{x^2 + y^2}right) = -frac{partial}{partial x} left(frac{x}{x^2 + y^2}right) $$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks a lot....
    $endgroup$
    – Gimgim
    Dec 16 '18 at 7:58






  • 1




    $begingroup$
    You're welcome. Perhaps you didn't need the second clue.
    $endgroup$
    – RRL
    Dec 16 '18 at 7:59










  • $begingroup$
    Yeah I might get downvote but as I asked this question what I didn't get. So I am not going to erase the question. Thanks all of you for helping
    $endgroup$
    – Gimgim
    Dec 16 '18 at 8:00











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









3












$begingroup$

Hint:



$$frac {x^2-y^2}{(x^2+y^2)^2} = frac{partial}{partial y} left(frac{y}{x^2 + y^2}right) = -frac{partial}{partial x} left(frac{x}{x^2 + y^2}right) $$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks a lot....
    $endgroup$
    – Gimgim
    Dec 16 '18 at 7:58






  • 1




    $begingroup$
    You're welcome. Perhaps you didn't need the second clue.
    $endgroup$
    – RRL
    Dec 16 '18 at 7:59










  • $begingroup$
    Yeah I might get downvote but as I asked this question what I didn't get. So I am not going to erase the question. Thanks all of you for helping
    $endgroup$
    – Gimgim
    Dec 16 '18 at 8:00
















3












$begingroup$

Hint:



$$frac {x^2-y^2}{(x^2+y^2)^2} = frac{partial}{partial y} left(frac{y}{x^2 + y^2}right) = -frac{partial}{partial x} left(frac{x}{x^2 + y^2}right) $$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks a lot....
    $endgroup$
    – Gimgim
    Dec 16 '18 at 7:58






  • 1




    $begingroup$
    You're welcome. Perhaps you didn't need the second clue.
    $endgroup$
    – RRL
    Dec 16 '18 at 7:59










  • $begingroup$
    Yeah I might get downvote but as I asked this question what I didn't get. So I am not going to erase the question. Thanks all of you for helping
    $endgroup$
    – Gimgim
    Dec 16 '18 at 8:00














3












3








3





$begingroup$

Hint:



$$frac {x^2-y^2}{(x^2+y^2)^2} = frac{partial}{partial y} left(frac{y}{x^2 + y^2}right) = -frac{partial}{partial x} left(frac{x}{x^2 + y^2}right) $$






share|cite|improve this answer











$endgroup$



Hint:



$$frac {x^2-y^2}{(x^2+y^2)^2} = frac{partial}{partial y} left(frac{y}{x^2 + y^2}right) = -frac{partial}{partial x} left(frac{x}{x^2 + y^2}right) $$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 16 '18 at 7:58

























answered Dec 16 '18 at 7:55









RRLRRL

51k42573




51k42573












  • $begingroup$
    Thanks a lot....
    $endgroup$
    – Gimgim
    Dec 16 '18 at 7:58






  • 1




    $begingroup$
    You're welcome. Perhaps you didn't need the second clue.
    $endgroup$
    – RRL
    Dec 16 '18 at 7:59










  • $begingroup$
    Yeah I might get downvote but as I asked this question what I didn't get. So I am not going to erase the question. Thanks all of you for helping
    $endgroup$
    – Gimgim
    Dec 16 '18 at 8:00


















  • $begingroup$
    Thanks a lot....
    $endgroup$
    – Gimgim
    Dec 16 '18 at 7:58






  • 1




    $begingroup$
    You're welcome. Perhaps you didn't need the second clue.
    $endgroup$
    – RRL
    Dec 16 '18 at 7:59










  • $begingroup$
    Yeah I might get downvote but as I asked this question what I didn't get. So I am not going to erase the question. Thanks all of you for helping
    $endgroup$
    – Gimgim
    Dec 16 '18 at 8:00
















$begingroup$
Thanks a lot....
$endgroup$
– Gimgim
Dec 16 '18 at 7:58




$begingroup$
Thanks a lot....
$endgroup$
– Gimgim
Dec 16 '18 at 7:58




1




1




$begingroup$
You're welcome. Perhaps you didn't need the second clue.
$endgroup$
– RRL
Dec 16 '18 at 7:59




$begingroup$
You're welcome. Perhaps you didn't need the second clue.
$endgroup$
– RRL
Dec 16 '18 at 7:59












$begingroup$
Yeah I might get downvote but as I asked this question what I didn't get. So I am not going to erase the question. Thanks all of you for helping
$endgroup$
– Gimgim
Dec 16 '18 at 8:00




$begingroup$
Yeah I might get downvote but as I asked this question what I didn't get. So I am not going to erase the question. Thanks all of you for helping
$endgroup$
– Gimgim
Dec 16 '18 at 8:00


















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