Convergence of Improper Integral in 3-Space

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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?
 
on Phys.org
To make the substitution that you made, you also need to change the limits of integration. Does that help?
 
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?
 
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.
 
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.
 
I think that was just a typo, as his substitution appears to correctly apply to the original statement.