Convergence of Improper Integral in 3-Space

In summary: However, in order to be sure, you could try solving for rho^2 instead of rho. Doing so would give you the correct answer.In summary, the integral is convergent if and only if rho^2 = 1.
  • #1
engin
8
0

Homework Statement



Discuss the convergence of the integral
1/[x^2 + y^2 + z^2 + 1]^2 dxdydz in the whole space.


Homework Equations





The Attempt at a Solution



Since the space is unbounded, the integral is an improper integral so we can consider a sphere with radius N and take the limit as N goes to infinity. I have used spherical coordinates. Theta is between 0 and 2Pi, Phi is between 0 and Pi, and rho is between 0 and N and the integrand becomes
(rho^2)sin(Phi)/[1 + (rho^2)] d(rho) d(phi) d(theta) .
Here again we use substitution : rho = tan x and the integrand becomes
((sin x)^2)d(x). But i can't figure out how to go on then? Is this integral convergent?
 
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  • #2
To make the substitution that you made, you also need to change the limits of integration. Does that help?
 
  • #3
Yes, i know that the limits of integration change. For rho = 0, tan(theta) = 0 but for rho = N, tan(theta) = ? I am a little confused there and passing to the limit. Can you help me with this?
 
  • #4
Sure, since tan x=sin x/cos x, and cos x goes to 0 at x=pi/2, tan x blows up at x=pi/2. So as rho goes to infinity, x goes to pi/2.
 
  • #5
engin said:

Homework Statement



Discuss the convergence of the integral
1/[x^2 + y^2 + z^2 + 1]^2 dxdydz in the whole space.


Homework Equations





The Attempt at a Solution



Since the space is unbounded, the integral is an improper integral so we can consider a sphere with radius N and take the limit as N goes to infinity. I have used spherical coordinates. Theta is between 0 and 2Pi, Phi is between 0 and Pi, and rho is between 0 and N and the integrand becomes
(rho^2)sin(Phi)/[1 + (rho^2)] d(rho) d(phi) d(theta) .
Here again we use substitution : rho = tan x and the integrand becomes
((sin x)^2)d(x). But i can't figure out how to go on then? Is this integral convergent?

If the original problem statement is correct, then the denominator should be (1+rho^2)^2. Note the extra square. You may be solving the wrong problem.
 
  • #6
I think that was just a typo, as his substitution appears to correctly apply to the original statement.
 

1. What is an improper integral in 3-space?

An improper integral in 3-space is a type of integral where the limits of integration involve points at infinity, making the integral undefined or infinite. This type of integral is often used in mathematics and physics to solve problems involving infinite or unbounded regions in 3-dimensional space.

2. How is an improper integral in 3-space different from a regular integral?

The main difference between an improper integral in 3-space and a regular integral is that the limits of integration for an improper integral involve points at infinity, while the limits for a regular integral are finite. This means that an improper integral may not converge to a finite value, while a regular integral will always have a finite solution.

3. What are some examples of problems that can be solved using improper integrals in 3-space?

Improper integrals in 3-space are commonly used to solve problems involving infinite or unbounded regions, such as calculating the volume of a cone or the surface area of a sphere. They are also used in physics to solve problems related to electric and magnetic fields in 3-dimensional space.

4. How do you evaluate an improper integral in 3-space?

Evaluating an improper integral in 3-space involves breaking down the integral into smaller, finite integrals and then taking the limit as the boundaries of those integrals approach infinity. This is done using various techniques, such as substitution, partial fractions, or integration by parts.

5. What are some common challenges in solving improper integrals in 3-space?

One common challenge in solving improper integrals in 3-space is determining the convergence or divergence of the integral. This can be difficult because the limits of integration involve points at infinity, making it harder to visualize and understand the behavior of the integral. Another challenge is finding the appropriate techniques to evaluate the integral, as different techniques may be necessary for different types of integrals.

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