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Another Comparison of Integrals

  1. Feb 20, 2007 #1
    1. The problem statement, all variables and given/known data
    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]

    2. Relevant equations
    The box above.

    3. 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.
     
  2. jcsd
  3. Feb 20, 2007 #2

    Dick

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    Is it greater than say, 1/(100*x)? Does the integral of 1/(100*x) converge or diverge?
     
  4. Feb 20, 2007 #3
    1 / (100x) diverges.

    How do I tell which integral is greater?
     
  5. Feb 20, 2007 #4

    Dick

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    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.
     
  6. Feb 20, 2007 #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?
     
  7. Feb 20, 2007 #6

    Dick

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    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.
     
  8. Feb 20, 2007 #7
    All right, makes a lot of sense.

    Thanks a lot.
     
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