# Comparison of Integrals

1. Feb 19, 2007

### Jacobpm64

1. The problem statement, all variables and given/known data
Use the box and the behavior of rational and exponential functions as $$x \rightarrow \infty$$ to predict whether the integrals converge or diverge.

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

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

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

Problem 1:
$$\int^\infty_1 \frac{x^2}{x^4 + 1} dx$$

Problem 2:
$$\int^\infty_1 \frac{x^2 - 6x + 1}{x^2 + 4} dx$$

2. Relevant equations
The ones in the box above.

3. The attempt at a solution
Problem 1:
I know that this integral is less than $$\int^\infty_1 \frac{1}{x} dx$$. I also know that $$\int^\infty_1 \frac{1}{x} dx$$ 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.

Problem 2:
I know that this integral is less than $$\int^\infty_1 1 dx$$ I also know that $$\int^\infty_1 1 dx$$ diverges. This does not help me though because I can not use a diverging integral to say that a smaller integral is also diverging. So, I'm confused at the same place on this problem.

2. Feb 19, 2007

### eigenglue

For problem 1, isn't (x^2)/((x^4)+1) always less than 1/(x^2) for values of x > 1?

For problem 2, what does the integrand approach as x approaches infinity?

3. Feb 20, 2007

### HallsofIvy

Staff Emeritus
$$\frac{x^2}{x^4+1}< \frac{x^2}{x^4}= \frac{1}{x^2}$$

For the second one, divide first:
[tex]\frac{x^2+ 6x+ 1}{x^2+ 4}= 1+ \frac{6x- 3}{x^2+ 4}[/itex]
What does that first "1" tell you?

4. Feb 20, 2007

### Jacobpm64

Oops, I copied down the wrong first problem.

The second one was correct though.

I understand the first one, thanks for that.

I still don't understand what you mean by what does the first "1" tell me.

Wait, does that mean.. that my integral is bigger than the integral of 1, so if the integral of 1 diverges, I can say that my integral also diverges!

woot!

Thanks!

I'm going to post the correct problem for #1, in a new post. Thanks.