# Improper Integral in 3-Space

1. Jul 31, 2007

### engin

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. Jul 31, 2007

### dhris

To make the substitution that you made, you also need to change the limits of integration. Does that help?

3. Jul 31, 2007

### engin

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. Jul 31, 2007

### dhris

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. Jul 31, 2007

### Dick

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. Jul 31, 2007

### dhris

I think that was just a typo, as his substitution appears to correctly apply to the original statement.