Does the improper integral $int_{R^2} frac{ln(x^2+y^2)}{x^2+y^2}$ converge?












1












$begingroup$


Does the improper integral $int_{R^2} frac{ln(x^2+y^2)}{x^2+y^2}$ converge?



So I know to solve such integrals when the function is all positive, but here it can be also negative.



I tried using the definition $int f =int f^+ - int f^-$ and calculate for each part.



When I try for $f^-$, that means for { $x^2+y^2 <1 $ } I get to the integral (after using polar coordinates) $$int_{0}^{1} frac{ln(r^2)}{r}dr$$ which goes to $- infty$.



While the same integral for $f^+$ goes to $infty$.



Does that mean the the original integral goes to $infty$ ? From what I remember either the integral for $f^-$ or $f^+$ has to converge for the integral to be defined.



Another direction which I thought would be calculating the integral for $$int_{R^2} frac{|ln(x^2+y^2)|}{x^2+y^2}$$ and since it is less than $$int_{x^2+y^2 > 1} frac{|ln(x^2+y^2)|}{x^2+y^2}$$ then it must not converge.
Am I right here?



Any help would be appreciated.






integration multivariable-calculus improper-integrals







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    The last part I am not sure on, as the last integral you've written does converge... the problematic point is at $(0,0)$, where, indeed, the integral fails to converge.
    $endgroup$
    – Simply Beautiful Art
    Jan 10 at 1:14










  • $begingroup$
    Are you sure it does not converge? I can use polar coordinates and the integral would be $ln^2(infty) - ln^2(1)$
    $endgroup$
    – Gabi G
    Jan 10 at 1:17












  • $begingroup$
    @GabiG: maybe think back to $1$ dimensional improper integrals. Those are defined by taking the integral on finite pieces and then taking the limit. You need to take separate limits for each singularity. You can't just cancel $infty$'s willy-nilly.
    $endgroup$
    – Cheerful Parsnip
    Jan 10 at 1:21












  • $begingroup$
    But since the function (with the absolute value) is positive if I can show that at one singularity it doesn't converge that it won't converge generally, or am I mistaken here?
    $endgroup$
    – Gabi G
    Jan 10 at 1:25












  • $begingroup$
    Since both $int f^+$ and $int f^-$ are infinite, the integral $int f$ does not exist as in Lebesgue-integral sense.
    $endgroup$
    – Sangchul Lee
    Jan 10 at 1:30
















1












$begingroup$


Does the improper integral $int_{R^2} frac{ln(x^2+y^2)}{x^2+y^2}$ converge?



So I know to solve such integrals when the function is all positive, but here it can be also negative.



I tried using the definition $int f =int f^+ - int f^-$ and calculate for each part.



When I try for $f^-$, that means for { $x^2+y^2 <1 $ } I get to the integral (after using polar coordinates) $$int_{0}^{1} frac{ln(r^2)}{r}dr$$ which goes to $- infty$.



While the same integral for $f^+$ goes to $infty$.



Does that mean the the original integral goes to $infty$ ? From what I remember either the integral for $f^-$ or $f^+$ has to converge for the integral to be defined.



Another direction which I thought would be calculating the integral for $$int_{R^2} frac{|ln(x^2+y^2)|}{x^2+y^2}$$ and since it is less than $$int_{x^2+y^2 > 1} frac{|ln(x^2+y^2)|}{x^2+y^2}$$ then it must not converge.
Am I right here?



Any help would be appreciated.






integration multivariable-calculus improper-integrals







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    The last part I am not sure on, as the last integral you've written does converge... the problematic point is at $(0,0)$, where, indeed, the integral fails to converge.
    $endgroup$
    – Simply Beautiful Art
    Jan 10 at 1:14










  • $begingroup$
    Are you sure it does not converge? I can use polar coordinates and the integral would be $ln^2(infty) - ln^2(1)$
    $endgroup$
    – Gabi G
    Jan 10 at 1:17












  • $begingroup$
    @GabiG: maybe think back to $1$ dimensional improper integrals. Those are defined by taking the integral on finite pieces and then taking the limit. You need to take separate limits for each singularity. You can't just cancel $infty$'s willy-nilly.
    $endgroup$
    – Cheerful Parsnip
    Jan 10 at 1:21












  • $begingroup$
    But since the function (with the absolute value) is positive if I can show that at one singularity it doesn't converge that it won't converge generally, or am I mistaken here?
    $endgroup$
    – Gabi G
    Jan 10 at 1:25












  • $begingroup$
    Since both $int f^+$ and $int f^-$ are infinite, the integral $int f$ does not exist as in Lebesgue-integral sense.
    $endgroup$
    – Sangchul Lee
    Jan 10 at 1:30














1












1








1





$begingroup$


Does the improper integral $int_{R^2} frac{ln(x^2+y^2)}{x^2+y^2}$ converge?



So I know to solve such integrals when the function is all positive, but here it can be also negative.



I tried using the definition $int f =int f^+ - int f^-$ and calculate for each part.



When I try for $f^-$, that means for { $x^2+y^2 <1 $ } I get to the integral (after using polar coordinates) $$int_{0}^{1} frac{ln(r^2)}{r}dr$$ which goes to $- infty$.



While the same integral for $f^+$ goes to $infty$.



Does that mean the the original integral goes to $infty$ ? From what I remember either the integral for $f^-$ or $f^+$ has to converge for the integral to be defined.



Another direction which I thought would be calculating the integral for $$int_{R^2} frac{|ln(x^2+y^2)|}{x^2+y^2}$$ and since it is less than $$int_{x^2+y^2 > 1} frac{|ln(x^2+y^2)|}{x^2+y^2}$$ then it must not converge.
Am I right here?



Any help would be appreciated.






integration multivariable-calculus improper-integrals







share|cite|improve this question











$endgroup$




Does the improper integral $int_{R^2} frac{ln(x^2+y^2)}{x^2+y^2}$ converge?



So I know to solve such integrals when the function is all positive, but here it can be also negative.



I tried using the definition $int f =int f^+ - int f^-$ and calculate for each part.



When I try for $f^-$, that means for { $x^2+y^2 <1 $ } I get to the integral (after using polar coordinates) $$int_{0}^{1} frac{ln(r^2)}{r}dr$$ which goes to $- infty$.



While the same integral for $f^+$ goes to $infty$.



Does that mean the the original integral goes to $infty$ ? From what I remember either the integral for $f^-$ or $f^+$ has to converge for the integral to be defined.



Another direction which I thought would be calculating the integral for $$int_{R^2} frac{|ln(x^2+y^2)|}{x^2+y^2}$$ and since it is less than $$int_{x^2+y^2 > 1} frac{|ln(x^2+y^2)|}{x^2+y^2}$$ then it must not converge.
Am I right here?



Any help would be appreciated.






integration multivariable-calculus improper-integrals




integration multivariable-calculus improper-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 10 at 1:11









Zacky

7,2701961




7,2701961










asked Jan 10 at 0:23









Gabi GGabi G

408110




408110








  • 1




    $begingroup$
    The last part I am not sure on, as the last integral you've written does converge... the problematic point is at $(0,0)$, where, indeed, the integral fails to converge.
    $endgroup$
    – Simply Beautiful Art
    Jan 10 at 1:14










  • $begingroup$
    Are you sure it does not converge? I can use polar coordinates and the integral would be $ln^2(infty) - ln^2(1)$
    $endgroup$
    – Gabi G
    Jan 10 at 1:17












  • $begingroup$
    @GabiG: maybe think back to $1$ dimensional improper integrals. Those are defined by taking the integral on finite pieces and then taking the limit. You need to take separate limits for each singularity. You can't just cancel $infty$'s willy-nilly.
    $endgroup$
    – Cheerful Parsnip
    Jan 10 at 1:21












  • $begingroup$
    But since the function (with the absolute value) is positive if I can show that at one singularity it doesn't converge that it won't converge generally, or am I mistaken here?
    $endgroup$
    – Gabi G
    Jan 10 at 1:25












  • $begingroup$
    Since both $int f^+$ and $int f^-$ are infinite, the integral $int f$ does not exist as in Lebesgue-integral sense.
    $endgroup$
    – Sangchul Lee
    Jan 10 at 1:30














  • 1




    $begingroup$
    The last part I am not sure on, as the last integral you've written does converge... the problematic point is at $(0,0)$, where, indeed, the integral fails to converge.
    $endgroup$
    – Simply Beautiful Art
    Jan 10 at 1:14










  • $begingroup$
    Are you sure it does not converge? I can use polar coordinates and the integral would be $ln^2(infty) - ln^2(1)$
    $endgroup$
    – Gabi G
    Jan 10 at 1:17












  • $begingroup$
    @GabiG: maybe think back to $1$ dimensional improper integrals. Those are defined by taking the integral on finite pieces and then taking the limit. You need to take separate limits for each singularity. You can't just cancel $infty$'s willy-nilly.
    $endgroup$
    – Cheerful Parsnip
    Jan 10 at 1:21












  • $begingroup$
    But since the function (with the absolute value) is positive if I can show that at one singularity it doesn't converge that it won't converge generally, or am I mistaken here?
    $endgroup$
    – Gabi G
    Jan 10 at 1:25












  • $begingroup$
    Since both $int f^+$ and $int f^-$ are infinite, the integral $int f$ does not exist as in Lebesgue-integral sense.
    $endgroup$
    – Sangchul Lee
    Jan 10 at 1:30








1




1




$begingroup$
The last part I am not sure on, as the last integral you've written does converge... the problematic point is at $(0,0)$, where, indeed, the integral fails to converge.
$endgroup$
– Simply Beautiful Art
Jan 10 at 1:14




$begingroup$
The last part I am not sure on, as the last integral you've written does converge... the problematic point is at $(0,0)$, where, indeed, the integral fails to converge.
$endgroup$
– Simply Beautiful Art
Jan 10 at 1:14












$begingroup$
Are you sure it does not converge? I can use polar coordinates and the integral would be $ln^2(infty) - ln^2(1)$
$endgroup$
– Gabi G
Jan 10 at 1:17






$begingroup$
Are you sure it does not converge? I can use polar coordinates and the integral would be $ln^2(infty) - ln^2(1)$
$endgroup$
– Gabi G
Jan 10 at 1:17














$begingroup$
@GabiG: maybe think back to $1$ dimensional improper integrals. Those are defined by taking the integral on finite pieces and then taking the limit. You need to take separate limits for each singularity. You can't just cancel $infty$'s willy-nilly.
$endgroup$
– Cheerful Parsnip
Jan 10 at 1:21






$begingroup$
@GabiG: maybe think back to $1$ dimensional improper integrals. Those are defined by taking the integral on finite pieces and then taking the limit. You need to take separate limits for each singularity. You can't just cancel $infty$'s willy-nilly.
$endgroup$
– Cheerful Parsnip
Jan 10 at 1:21














$begingroup$
But since the function (with the absolute value) is positive if I can show that at one singularity it doesn't converge that it won't converge generally, or am I mistaken here?
$endgroup$
– Gabi G
Jan 10 at 1:25






$begingroup$
But since the function (with the absolute value) is positive if I can show that at one singularity it doesn't converge that it won't converge generally, or am I mistaken here?
$endgroup$
– Gabi G
Jan 10 at 1:25














$begingroup$
Since both $int f^+$ and $int f^-$ are infinite, the integral $int f$ does not exist as in Lebesgue-integral sense.
$endgroup$
– Sangchul Lee
Jan 10 at 1:30




$begingroup$
Since both $int f^+$ and $int f^-$ are infinite, the integral $int f$ does not exist as in Lebesgue-integral sense.
$endgroup$
– Sangchul Lee
Jan 10 at 1:30










1 Answer
1






active

oldest

votes


















1












$begingroup$

You are almost right. Let us do the calculation in a bit detailed way



$int_{-infty}^{infty}int_{-infty}^{infty}frac{ln(x^2+y^2)}{x^2 + y^2}dxdy = 2piint_{0}^{infty}frac{ln r^2}{r}dr = 4piint_{0}^{infty}frac{ln r}{r}dr$



Now, using variable transformation of $ln r = u$, we obtain
$$int_{-infty}^{infty}int_{-infty}^{infty}frac{ln(x^2+y^2)}{x^2 + y^2}dxdy = 4pi int_{-infty}^{infty}u du$$



If you are strictly speaking in the context of elementary calculus, then yes the integral does not converge. However, the Cauchy Principle Value of the integral exists and it tends to zero. You might want to look for Cauchy Principle Value.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Well, in the sense of what I learned, even if $int_{-infty}^{infty}u du$ exists but $int_{-infty}^{infty}|u| du$ doesn't, then we say it doesn't converge
    $endgroup$
    – Gabi G
    Jan 10 at 1:30













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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

You are almost right. Let us do the calculation in a bit detailed way



$int_{-infty}^{infty}int_{-infty}^{infty}frac{ln(x^2+y^2)}{x^2 + y^2}dxdy = 2piint_{0}^{infty}frac{ln r^2}{r}dr = 4piint_{0}^{infty}frac{ln r}{r}dr$



Now, using variable transformation of $ln r = u$, we obtain
$$int_{-infty}^{infty}int_{-infty}^{infty}frac{ln(x^2+y^2)}{x^2 + y^2}dxdy = 4pi int_{-infty}^{infty}u du$$



If you are strictly speaking in the context of elementary calculus, then yes the integral does not converge. However, the Cauchy Principle Value of the integral exists and it tends to zero. You might want to look for Cauchy Principle Value.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Well, in the sense of what I learned, even if $int_{-infty}^{infty}u du$ exists but $int_{-infty}^{infty}|u| du$ doesn't, then we say it doesn't converge
    $endgroup$
    – Gabi G
    Jan 10 at 1:30


















1












$begingroup$

You are almost right. Let us do the calculation in a bit detailed way



$int_{-infty}^{infty}int_{-infty}^{infty}frac{ln(x^2+y^2)}{x^2 + y^2}dxdy = 2piint_{0}^{infty}frac{ln r^2}{r}dr = 4piint_{0}^{infty}frac{ln r}{r}dr$



Now, using variable transformation of $ln r = u$, we obtain
$$int_{-infty}^{infty}int_{-infty}^{infty}frac{ln(x^2+y^2)}{x^2 + y^2}dxdy = 4pi int_{-infty}^{infty}u du$$



If you are strictly speaking in the context of elementary calculus, then yes the integral does not converge. However, the Cauchy Principle Value of the integral exists and it tends to zero. You might want to look for Cauchy Principle Value.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Well, in the sense of what I learned, even if $int_{-infty}^{infty}u du$ exists but $int_{-infty}^{infty}|u| du$ doesn't, then we say it doesn't converge
    $endgroup$
    – Gabi G
    Jan 10 at 1:30
















1












1








1





$begingroup$

You are almost right. Let us do the calculation in a bit detailed way



$int_{-infty}^{infty}int_{-infty}^{infty}frac{ln(x^2+y^2)}{x^2 + y^2}dxdy = 2piint_{0}^{infty}frac{ln r^2}{r}dr = 4piint_{0}^{infty}frac{ln r}{r}dr$



Now, using variable transformation of $ln r = u$, we obtain
$$int_{-infty}^{infty}int_{-infty}^{infty}frac{ln(x^2+y^2)}{x^2 + y^2}dxdy = 4pi int_{-infty}^{infty}u du$$



If you are strictly speaking in the context of elementary calculus, then yes the integral does not converge. However, the Cauchy Principle Value of the integral exists and it tends to zero. You might want to look for Cauchy Principle Value.






share|cite|improve this answer









$endgroup$



You are almost right. Let us do the calculation in a bit detailed way



$int_{-infty}^{infty}int_{-infty}^{infty}frac{ln(x^2+y^2)}{x^2 + y^2}dxdy = 2piint_{0}^{infty}frac{ln r^2}{r}dr = 4piint_{0}^{infty}frac{ln r}{r}dr$



Now, using variable transformation of $ln r = u$, we obtain
$$int_{-infty}^{infty}int_{-infty}^{infty}frac{ln(x^2+y^2)}{x^2 + y^2}dxdy = 4pi int_{-infty}^{infty}u du$$



If you are strictly speaking in the context of elementary calculus, then yes the integral does not converge. However, the Cauchy Principle Value of the integral exists and it tends to zero. You might want to look for Cauchy Principle Value.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 10 at 1:25









Evan William ChandraEvan William Chandra

626313




626313












  • $begingroup$
    Well, in the sense of what I learned, even if $int_{-infty}^{infty}u du$ exists but $int_{-infty}^{infty}|u| du$ doesn't, then we say it doesn't converge
    $endgroup$
    – Gabi G
    Jan 10 at 1:30




















  • $begingroup$
    Well, in the sense of what I learned, even if $int_{-infty}^{infty}u du$ exists but $int_{-infty}^{infty}|u| du$ doesn't, then we say it doesn't converge
    $endgroup$
    – Gabi G
    Jan 10 at 1:30


















$begingroup$
Well, in the sense of what I learned, even if $int_{-infty}^{infty}u du$ exists but $int_{-infty}^{infty}|u| du$ doesn't, then we say it doesn't converge
$endgroup$
– Gabi G
Jan 10 at 1:30






$begingroup$
Well, in the sense of what I learned, even if $int_{-infty}^{infty}u du$ exists but $int_{-infty}^{infty}|u| du$ doesn't, then we say it doesn't converge
$endgroup$
– Gabi G
Jan 10 at 1:30




















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