Can the singularity at the center affect the evaluation of this 2D integration?

[shaleba]

1. The problem:
I came up to this integration and noticed that it contains singularity at the center (x,y)=(0,0).
\iint\limits_D \frac{x^2-y^2}{(x^2+y^2)^2} dxdy

Note that the integration domain is a unit circle centered at the origin.

I was thinking that the maybe the integral could be evaluated through the polar coordinate. So I used the follows:

Homework Equations

x^2 +y^2 = r^2
x= r \cdot cos(\theta), y= r \cdot sin(\theta)

The Attempt at a Solution

In that case the integration was transformed to the following integration:
\int_0^1 \frac{1}{r}dr \cdot \int_0^{2\pi}(cos^2(\theta)-sin^2(\theta))d\theta

Evaluation of the first term still includes a singular point with r=0, while the second term gives 0. However, looks like there is a finite value for this integration which is not 0.

Can anybody helps me out on this? Thanks very much.

 

[tiny-tim]

welcome to pf!

hi shaleba! welcome to pf!

by interchanging x and y (or x and -y) in the original integral, the integral (if it is finite) is obviously 0 for any region symmetric about the lines x = y or x = -y

in particular, it's 0 for any ring around the origin

… Evaluation of the first term still includes a singular point with r=0, while the second term gives 0. However, looks like there is a finite value for this integration which is not 0.

why do you say it looks like there is a finite value?

since it's the equivalent of 1 - 1 + 1 + 1 + 1 - … , i suppose it depends exactly how you define the integral

 

[shaleba]

Thank you very much for your reply Tim.
I understand your rationale on the integration. However, I think the key is about the singularity at the center.

For the rationale you mentioned, the 2d delta function looks like satisfy what you mentioned too, but the integration is not 0.

The reason I think it's a finite but not 0, is because this integral appears in one paper and looks like not equal to 0. They didn't show details about evaluating it.

hi shaleba! welcome to pf!

by interchanging x and y (or x and -y) in the original integral, the integral (if it is finite) is obviously 0 for any region symmetric about the lines x = y or x = -y

in particular, it's 0 for any ring around the origin


why do you say it looks like there is a finite value?

since it's the equivalent of 1 - 1 + 1 + 1 + 1 - … , i suppose it depends exactly how you define the integral

 

[tiny-tim]

For the rationale you mentioned, the 2d delta function looks like satisfy what you mentioned too, but the integration is not 0.

what delta function?

The reason I think it's a finite but not 0, is because this integral appears in one paper and looks like not equal to 0. They didn't show details about evaluating it.

which book?

the integral is a two-dimensional limit, and its value depends how you define the approach to that limit

if you define it the obvious way, the integral is 0

but you can define it so as to give*the integral any value

 

[rude man]

1. The problem:
I came up to this integration and noticed that it contains singularity at the center (x,y)=(0,0).
\iint\limits_D \frac{x^2-y^2}{(x^2+y^2)^2} dxdy

Note that the integration domain is a unit circle centered at the origin.

I was thinking that the maybe the integral could be evaluated through the polar coordinate. So I used the follows:

Homework Equations

x^2 +y^2 = r^2
x= r \cdot cos(\theta), y= r \cdot sin(\theta)

The Attempt at a Solution

In that case the integration was transformed to the following integration:
\int_0^1 \frac{1}{r}dr \cdot \int_0^{2\pi}(cos^2(\theta)-sin^2(\theta))d\theta
Evaluation of the first term still includes a singular point with r=0, while the second term gives 0. However, looks like there is a finite value for this integration which is not 0.

Can anybody helps me out on this? Thanks very much.

[/quote]
As I see it:

Your integration of f(r,θ) wrt theta first gave zero and you're worried because the second integration wrt r blows up at r = 0.

But look at what happens to your function f(r,θ) = (cos²θ - sin²θ)/r²

integrated within a small circle about the origin:
lim r → 0 of f(r,θ)(πr²) = πg(θ) where g(θ) = cos²θ - sin²θ and πg(θ) is always between -π and +π so there is no singularity at r = 0.

So I would have to second the motion that the integral of f(r,θ) over the specified area is zero.

 

[Coelum]

Shaleba,
in polar coordinates $(x^2+y^2)^2=r^4$ and the surface element $dx dy$ becomes $r dr dθ$, so the integrand is $(cos^2(θ)-sin^2(θ))/r^3=cos(2θ)/r^3$. Hence, the integral is the product of two terms: one depends on $θ$ only, the other depends on $r$ only. Can you find their values? What definition for an improper integral are you using?

 

[shaleba]

Thanks for your reply. But looks like from the numerator x^2 -y^2, there arises another r^2
The \theta term looks like to be 0 while the other term is singular.

Shaleba,
in polar coordinates $(x^2+y^2)^2=r^4$ and the surface element $dx dy$ becomes $r dr dθ$, so the integrand is $(cos^2(θ)-sin^2(θ))/r^3=cos(2θ)/r^3$. Hence, the integral is the product of two terms: one depends on $θ$ only, the other depends on $r$ only. Can you find their values? What definition for an improper integral are you using?

 

[shaleba]

thanks for reply. the Delta function I mentioned is the Dirac delta function.

what delta function?


which book?

the integral is a two-dimensional limit, and its value depends how you define the approach to that limit

if you define it the obvious way, the integral is 0

but you can define it so as to give*the integral any value

 

[tiny-tim]

ahh!

I understand your rationale on the integration. However, I think the key is about the singularity at the center.

For the rationale you mentioned, the 2d delta function looks like satisfy what you mentioned too, but the integration is not 0.

no, because the dirac delta function (btw, technically it's a distribution, not a function, but that doesn't matter here) is always positive,

and you can always change the order of any positive sum (or integral)

the difficulty with the given function is that it's alternately positive and negative in the four sectors round the origin, so the order in which you take the limits does matter