Another Comparison of Integrals

In summary, the box provided predicts that the integral \int^\infty_1 \frac{x^2+1}{x^3 + 3x + 2} dx diverges based on the behavior of rational and exponential functions as x \rightarrow \infty. The integral can be simplified to \int^\infty_1 \frac{1}{x} dx, which also diverges. Using an estimate, it can be shown that the original integral is greater than 1/(7x), which also diverges. Therefore, it can be concluded that the original integral diverges.
  • #1
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Homework Statement


Use the box and the behavior of rational and exponential functions as [tex] x \rightarrow \infty [/tex] to predict whether the integrals converge or diverge.

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

[tex] \int^1_0 \frac{1}{x^p} dx [/tex] converges for p < 1 and diverges for p > 1.

[tex] \int^\infty_0 e^{-ax} dx [/tex] converges for a > 0.

Here is the problem I need help with:
[tex] \int^\infty_1 \frac{x^2+1}{x^3 + 3x + 2} dx [/tex]

Homework Equations


The box above.

The Attempt at a Solution


I know that this integral is less than [tex] \int^\infty_1 \frac{1}{x} dx [/tex]. I also know that [tex] \int^\infty_1 \frac{1}{x} dx [/tex] 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.
 
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  • #2
Is it greater than say, 1/(100*x)? Does the integral of 1/(100*x) converge or diverge?
 
  • #3
1 / (100x) diverges.

How do I tell which integral is greater?
 
  • #4
Estimate ruthlessly. Eg. would you believe (x^2+1)/(x^3+3*x+3)>x^2/(x^3+3*x^3+3*x^3).

You should. I've made the numerator less and the denominator bigger. (x>1). Now simplify the RHS.
 
  • #5
So it's more of guessing to pick something that's bigger?

Your RHS simplifies to 1/(7x).. which diverges..

So, we can also say that the original integral diverges..

So, for your RHS, did you just pull random numbers out of the sky until you figured it would be smaller than the original integral.. so that you could prove divergence?
 
  • #6
Sure. I just made the numerator smaller and the denominator bigger in such a way that I could easily simplify and still have a divergence. Not that much 'guesswork' involved.
 
  • #7
All right, makes a lot of sense.

Thanks a lot.
 

1. What is "Another Comparison of Integrals"?

"Another Comparison of Integrals" is a mathematical method used to evaluate integrals by comparing them to known integrals that are easier to solve. It is often used when the given integral is difficult to solve using traditional methods.

2. How does "Another Comparison of Integrals" work?

This method works by comparing the given integral to a known integral that has a similar form. By using properties of integrals such as linearity and substitution, we can manipulate the given integral to match the known one, making it easier to solve.

3. What are the benefits of using "Another Comparison of Integrals"?

"Another Comparison of Integrals" can be a useful tool in solving difficult integrals that may not have a straightforward solution. It also allows us to use known integrals and their properties to simplify the problem and make it more manageable.

4. When should "Another Comparison of Integrals" be used?

This method is best used when the given integral is difficult to solve using traditional methods such as substitution, integration by parts, or trigonometric substitution. It is also useful when the integrand has a complicated form or involves special functions.

5. What are some common examples of "Another Comparison of Integrals"?

Some common examples of "Another Comparison of Integrals" include using known integrals such as the power rule, exponential functions, and trigonometric functions to solve more complex integrals. It can also be used to compare integrals involving rational functions or logarithmic functions to known integrals with similar forms.

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