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Thread: Comparison of Improper Integrals

  1. #1
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    Comparison of Improper Integrals

    Problem number 1: $\displaystyle \int_1^{\infty} \frac{x^2+1{}}{x^3+3x+2} dx$

    and problem number 2: $\displaystyle \int_1^{\infty} \frac{5x+2{}}{x^4+8x^2+4} dx$
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    Quote Originally Posted by theskyidfalling View Post
    Problem number 1: $\displaystyle \int_1^{\infty} \frac{x^2+1{}}{x^3+3x+2} dx$

    and problem number 2: $\displaystyle \int_1^{\infty} \frac{5x+2{}}{x^4+8x^2+4} dx$
    Problem number 1: $\displaystyle \int_1^{\infty} \frac{x^2+1{}}{x^3+3x+2} dx < \int_1^{\infty} \frac{1}{x} dx$(**)

    Problem number 2: $\displaystyle \int_1^{\infty} \frac{5x+2{}}{x^4+8x^2+4} dx < \int_1^{\infty} \frac{25}{5x^3-2x^2} dx$

    Both the integrals I gave are integrable and converge. (No they don't - not **)
    Last edited by Jester; Mar 15th 2009 at 12:02 PM. Reason: bad mistake- my bad.
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    Thanks! But how does the first one converge? I thought it diverged since $\displaystyle \int_1^{\infty} \frac{1{}}{x} dx$ goes to infinity? And how do you come up with the second integral?
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    Quote Originally Posted by theskyidfalling View Post
    Thanks! But how does the first one converge? I thought it diverged since $\displaystyle \int_1^{\infty} \frac{1{}}{x} dx$ goes to infinity? And how do you come up with the second integral?
    Opps, it doesn't. (Do over)

    $\displaystyle \int_1^{\infty} \frac{1}{x+2} dx < \int_1^{\infty} \frac{x^2+1{}}{x^3+3x+2} dx $
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    On both, look at the long term behaviour of each. For the first

    $\displaystyle \frac{x^2+1}{x^3+3x+2} \approx \frac{1}{x}$ and the second $\displaystyle \frac{5x+2}{x^4+8x^2+4} \approx \frac{5}{x^3}$.

    So we need inequalities that are integrable. For the first

    $\displaystyle \frac{1}{x} \le\frac{x^2+1}{x^3+3x+2} $ isn't true since

    $\displaystyle x^3+3x+2 \le x (x^2+1)$ isn't true because $\displaystyle 3x+2 \le x $ isn't true, so we need little more to have something true for all $\displaystyle x \ge1 $. Try

    $\displaystyle \frac{1}{x+a} \le \frac{x^2+1}{x^3+3x+2} $

    so

    $\displaystyle x^3+3x+2 \le (x+a) (x^2+1)$ so $\displaystyle 3x+2 \le ax^2+x+a$ or $\displaystyle 0 \le ax^2-2 x+a-2$. We see that if $\displaystyle a = 2$ then $\displaystyle 0 \le 2x^2-2 x = 2x(x-1)$ which is true.

    Same for the second problem.

    $\displaystyle \frac{5x+2}{x^4+8x^2+4} \le \frac{5}{x^3}$ isn't true so we try

    $\displaystyle \frac{5x+2}{x^4+8x^2+4} \le \frac{5}{x^3+ax^2}$ and find an a where this is true.
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    Quote Originally Posted by theskyidfalling View Post
    Hi so I was just wondering if anyone could help me figure out how to solve these two particular problems. Thank you!

    Problem number 1: $\displaystyle \int_1^{\infty} \frac{x^2+1{}}{x^3+3x+2} dx$
    $\displaystyle u= x^3+ 3x+ 2$ look like an obvious substitution!

    and problem number 2: $\displaystyle \int_1^{\infty} \frac{5x+2{}}{x^4+8x^2+4} dx$
    $\displaystyle x^4+ 8x^2+ 4= x^4+ 8x^2+ 16- 12= (x^2+ 4)^2- 12$ which factors as $\displaystyle (x^2+4- 2\sqrt{3})(x^2+ 4+ 2\sqrt{3})$. See if that helps.
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  7. #7
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    Quote Originally Posted by theskyidfalling View Post

    Problem number 1: $\displaystyle \int_1^{\infty} \frac{x^2+1{}}{x^3+3x+2} dx$
    $\displaystyle \frac{x^{2}+1}{x^{3}+3x+2}>\frac{x^{2}}{x^{3}+3x^{ 3}+2x^{3}}>\frac{1}{6x},$ thus the integral diverges.

    Quote Originally Posted by theskyidfalling View Post

    and problem number 2: $\displaystyle \int_1^{\infty} \frac{5x+2{}}{x^4+8x^2+4} dx$
    $\displaystyle \frac{5x+2}{x^{4}+8x^{2}+4}\le\frac{5x+2x}{x^{4}}= \frac{7}{x^{3}},$ thus the integral converges.
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