1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Improper Integral in 3-Space

  1. Jul 31, 2007 #1
    1. The problem statement, all variables and given/known data

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


    2. Relevant equations



    3. 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?
     
  2. jcsd
  3. Jul 31, 2007 #2
    To make the substitution that you made, you also need to change the limits of integration. Does that help?
     
  4. Jul 31, 2007 #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?
     
  5. Jul 31, 2007 #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.
     
  6. Jul 31, 2007 #5

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
  7. Jul 31, 2007 #6
    I think that was just a typo, as his substitution appears to correctly apply to the original statement.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Improper Integral in 3-Space
  1. Improper Integrals (Replies: 3)

  2. Improper integrals (Replies: 5)

  3. Improper integrals (Replies: 2)

  4. Improper Integral (Replies: 7)

Loading...