How can I determine if an improper multiple integral converges or diverges?

In summary, the conversation discusses two questions on improper multiple integrals, specifically the convergence or divergence of the integrals. The first question involves determining whether an integral converges or diverges using cylindrical coordinates. The second question involves using the comparison test to determine the convergence or divergence of two integrals. The conversation also mentions difficulty understanding certain parts of the solution and asks for help with the problems.
  • #1
kingwinner
1,270
0
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.
============================
Now, I am having a lot of trouble understanding the red part, WHY is it true?



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

[tex]\int^1_0 \frac{\cos r}{r}dr[/tex]

converges.
 
  • #3
foxjwill said:
I'm not sure exactly what the red part for 1) is saying. Try converting it to cylindrical coordinates and set [tex]R = \{\theta \in [0,2\pi], r \in (0,1] \}[/tex]. You'll basically end up needing to show that

[tex]\int^1_0 \frac{\cos r}{r}dr[/tex]

converges.
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.




About your method:

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?
 
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  • #4
r goes from 1 to infinity so it's then just a single variable convergence problem.
 
  • #5
Vid said:
r goes from 1 to infinity so it's then just a single variable convergence problem.
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
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
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  • #8
Troels said:
Ask https://www.physicsforums.com/showpost.php?p=1586354&postcount=15". Slightly different boundaries, but the same idea.
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?
 
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  • #9
2) Any help with this problem? I have the answers to these, but I don't understand how they figured out convergence/divergence
 

FAQ: How can I determine if an improper multiple integral converges or diverges?

1. What is an improper multiple integral?

An improper multiple integral is an integral where one or both of the limits of integration are infinite or the integrand is unbounded on the interval of integration. This can occur when integrating over an unbounded region or when the integrand has a vertical asymptote within the interval of integration.

2. How do you evaluate an improper multiple integral?

To evaluate an improper multiple integral, you must first identify the type of improper integral (Type 1 or Type 2) and then use the appropriate method to solve it. For Type 1 improper integrals, you can use the limit comparison test, comparison test, or direct evaluation. For Type 2 improper integrals, you can use partial fractions decomposition, trigonometric substitution, or integration by parts.

3. What is the difference between a Type 1 and Type 2 improper multiple integral?

A Type 1 improper multiple integral has one or both limits of integration as infinite, while a Type 2 improper multiple integral has an unbounded integrand within the interval of integration. Type 1 improper integrals can be solved using the limit comparison test or direct evaluation, while Type 2 improper integrals require more advanced methods such as partial fractions decomposition or trigonometric substitution.

4. Can improper multiple integrals diverge?

Yes, improper multiple integrals can diverge, meaning they do not have a finite value. This can occur when the integrand approaches infinity or oscillates infinitely within the interval of integration. In these cases, the integral is said to be divergent and cannot be evaluated.

5. How are improper multiple integrals used in real-world applications?

Improper multiple integrals are used in various fields of science, such as physics, engineering, and economics, to model real-world phenomena. For example, they can be used to calculate the work done by a varying force or the net profit of a company over a certain time period. Improper multiple integrals also play a crucial role in solving differential equations and finding the area under a curve with infinite bounds.

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