# Comparison of Integrals

• Feb 19th 2007, 07:19 PM
Jacobpm64
Comparison of Integrals
Use the box and the behavior of rational and exponential functions as $\displaystyle x \rightarrow \infty$ to predict whether the integrals converge or diverge.

Here is the box:
$\displaystyle \int^\infty_1 \frac{1}{x^p} dx$ converges for p > 1 and diverges for p < 1.

$\displaystyle \int^1_0 \frac{1}{x^p} dx$ converges for p < 1 and diverges for p > 1.

$\displaystyle \int^\infty_0 e^{-ax} dx$ converges for a > 0.

Here is the problem I need help with .. along with my work:
$\displaystyle \int^\infty_1 \frac{x^2+1}{x^3 + 3x + 2} dx$
I know that this integral is less than $\displaystyle \int^\infty_1 \frac{1}{x} dx$. I also know that $\displaystyle \int^\infty_1 \frac{1}{x} dx$ diverges. This does not help me though because I can not use a diverging integral to say that a smaller integral is also diverging. This is where I'm confused.
• Feb 20th 2007, 01:28 PM
CaptainBlack
Quote:

Originally Posted by Jacobpm64
Use the box and the behavior of rational and exponential functions as $\displaystyle x \rightarrow \infty$ to predict whether the integrals converge or diverge.

Here is the box:
$\displaystyle \int^\infty_1 \frac{1}{x^p} dx$ converges for p > 1 and diverges for p < 1.

$\displaystyle \int^1_0 \frac{1}{x^p} dx$ converges for p < 1 and diverges for p > 1.

$\displaystyle \int^\infty_0 e^{-ax} dx$ converges for a > 0.

Here is the problem I need help with .. along with my work:
$\displaystyle \int^\infty_1 \frac{x^2+1}{x^3 + 3x + 2} dx$
I know that this integral is less than $\displaystyle \int^\infty_1 \frac{1}{x} dx$. I also know that $\displaystyle \int^\infty_1 \frac{1}{x} dx$ diverges. This does not help me though because I can not use a diverging integral to say that a smaller integral is also diverging. This is where I'm confused.

The trick here is to trap the integrand between two function the integrals
of which both diverge.

The integrand may be written:

f(x) = (1+1/x)/(x+3/x+2/x^2)

Then for x>1:

1 < 1+1/x <2,

and for x>5 (say, this can be made tighter):

x < x + 3/x +2/x^2 < 2x.

So if x>5, we have:

1/(2x) < f(x) < 2/x

but both of the integrals of 1/(2x) and 2/x diverge, so the integral f(x)
from 5 to infinity diverges, and so the integral from 1 to infinity also
diverges.

RonL