# Improper Multiple Integrals

1. Jan 27, 2008

### kingwinner

I am having some trouble with the following 2 questions on improper multiple integrals. I hope that someone can help me out!

1) Determine whether
I=∫∫ cos(sqrt(x2+y2)) / (x2+y2) converges or diverges.
x,y>1

Solution:
Let R=[0,1]x[0,1]
B(0,1)=ball of radius 1 centered at origin
R+xR+=positive xy-plane
f(x,y)=the integrand

The solution says that
I=∫∫ f(x,y) dA - ∫∫ f(x,y) dA
R+xR+ \B(0,1) R\B(0,1)

And then showed that both integrals converge, so the given improper integral I converges.
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Now, I am having a lot of trouble understanding the red part, WHY is it true?

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2) Determine whether the following converges or diverges.
Let S=[-1,1]x[-1,1]
2a)
∫∫ x2 / (x2+y2) dA
S
2b)
∫∫ sqrt|x| / (x2+y2) dA
S

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In the solutions manual, seemingly, they know the answer at the beginning; they inscribed a circle within the rectangle S for 2a) and inscribed the rectangle S in a circle for 2b), said that the integrand >0 except the origin, and used the comparsion test to conclude the first one diverges and the second converges.
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Now, I just want to know HOW I can get a first feeling about whether the above improper integrals will converge or diverge before going into the details. It's nice to know the answer ahead of time, so that I can know which direction to push forward the proof. Otherwise, I will just be doubling my amount of time and effort to finish.

Thank you for explaining!

Last edited: Jan 27, 2008
2. Jan 27, 2008

### foxjwill

I'm not sure exactly what the red part for 1) is saying. Try converting it to cylindrical coordinates and set $$R = \{\theta \in [0,2\pi], r \in (0,1] \}$$. You'll basically end up needing to show that

$$\int^1_0 \frac{\cos r}{r}dr$$

converges.

3. Jan 28, 2008

### kingwinner

The red part:
I=∫∫ f(x,y) dA - ∫∫ f(x,y) dA
R+xR+ \B(0,1) R\B(0,1)

R+xR+ \B(0,1), (xy-plane take away ball centered at origin of radius 1), is region of integration of the first double integral

R\B(0,1), [0,1] x [0,1] take away ball centered at origin of radius 1, is region of integration of the second double integral

But I don't understand the equality in the red part.

In cylindrical coordinates, the limits of integration are not going to be "nice" since the region is x>1 and y>1. How can we describe this in cylindrical coordinates?? Besides, it's a double integral, how can cylindrical coordinates work in R^2?

Last edited: Jan 29, 2008
4. Jan 28, 2008

### Vid

r goes from 1 to infinity so it's then just a single variable convergence problem.

5. Jan 29, 2008

### kingwinner

I don't think r goes from 1 to infinity since the region x,y>1 is not a ring...

Can someone please at least help me with one of the 2 questions? Any help of any kind would be appreciated!

6. Jan 29, 2008

### kingwinner

1) I think the solutions have mistaken the interpretation of x,y>1, they think that x,y>1 is the region in the positive xy-plane take away [0,1]x[0,1], but this is clearly not the case, so the solution is wrong.

How can we actually solve this problem?

7. Jan 29, 2008

### Troels

Last edited by a moderator: Apr 23, 2017
8. Jan 30, 2008

### kingwinner

1) But in our case, since it's a double integral (2-D), it's impossible to use a 3-D cylindrical coordinates

Also, how can we possibly know whether the integral converges or diverges? If we don't know what to prove, how can we solve it?

Last edited by a moderator: Apr 23, 2017
9. Jan 30, 2008

### kingwinner

2) Any help with this problem? I have the answers to these, but I don't understand how they figured out convergence/divergence