# Improper Integral using Comparison to determine Convergence/Divergence

1. Jan 18, 2010

### MiDnite89

1. The problem statement, all variables and given/known data

Use a comparison to determine if the improper integral converges or diverges. If the integral converges, give an upper bound for the value.

Integral of d(theta) / (theta^3 + theta)^1/2 from 1 to infinity

2. Relevant equations

N/A

3. The attempt at a solution

I'm not sure which function would be a good comparison to use to determine convergence or divergence. Earlier in the assignment, I ran across an equation that was dx/(9 - x^2)^1/2 which was arcsin(x/3), and since this is +, it'd be arccos - however that was with x^2 and this is theta^3, I'm not sure if that would be a direct comparison or not.

Any help would be appreciated, thanks.

2. Jan 18, 2010

### Bohrok

∫ dx/√(9 - x2) is arcsin(x/3) + C, but a + between the two terms in the denominator would not involve arccosine in the integral but instead, arctangent. In fact, that integral could also be written as -arccos(x/3) + C. Most integrals involving arcsine or arccosine like this one can differ only by a sign and constant.
For your problem, you don't have to actually find the integral, just find another integral with which to compare it.

Start with the comparison θ3 + θ > θ (for θ > 1) and work with it until you can make the left side look like your integrand 1/√(θ3 + θ)

Last edited: Jan 18, 2010