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Can someone explain the direct comparison test to me in detail?

  1. Feb 10, 2013 #1
    1. The problem statement, all variables and given/known data
    A few examples done in class that I didn't understand are as follows:

    (Find out if convergent or divergent)

    integral 0-->infinity ( x / (x^3 + 1) )

    integral 1-->infinity ( (2+e^(-x) ) / x )

    2. Relevant equations

    None

    3. The attempt at a solution

    Basically, I have NO idea where to begin. My professor seriously glossed over this concept in class; he basically just stated the theorem and then said "just compare it to another function". What functions? How? How do we know if we need a converging or diverging function? Can I get a careful walkthrough of some example problems so that this concept sinks in?

    Thank you.
     
  2. jcsd
  3. Feb 10, 2013 #2

    Ray Vickson

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    What functions? Well, to prove divergence of ∫ f(x) dx for f ≥ 0 you need to find a g ≥ 0 with f(x) ≥ g(x) for all x, and for which you already know that ∫ g(x) dx diverges.

    Testing for convergence is similar: find g such that 0 ≤ f(x) ≤ g(x) and for which you know already that ∫ g(x) dx converges.

    So, you need to look at your f(x) and try to come up with functions g(x) that 'work'.
     
  4. Feb 10, 2013 #3

    Zondrina

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    So you're asked to determine if those integrals converge by using the comparison test.

    Basically what the comparison test states is that you either have to find a bigger integral that you KNOW will converge or a smaller integral that you KNOW will diverge.

    Why you ask?

    I'm sure you're familiar with the concept that the integral represents the area under your curve right?

    If you can find something BIGGER to compare your integral to (i.e a bigger function) which converges to a finite area, then certainly the thing which is smaller must also converge right?

    If you can find something SMALLER to compare your integral to (i.e a smaller function) which diverges to infinity, then certainly, if you can't find the area under the smaller curve, what makes you think you can find the area of the bigger one?
     
  5. Feb 10, 2013 #4

    Mute

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    Say the function you are integrating is ##f(x) \geq 0##, and you are integrating from x=a to b. If you can find a function ##g(x)## such that ##f(x) \leq g(x)## for ##a \leq x \leq b##, then the inequality holds even when you integrate over the functions. That is, if ##0 \leq f(x) \leq g(x)##, then

    $$0 \leq \int_a^b dx~f(x) \leq \int_a^b dx~g(x).$$

    Say you know f(x), but you don't know whether or not the integral over f(x) converges. But, suppose you suspect if converges. If you can find a function g(x) that is greatly than f(x) on the integration region, and you can show that the integral over g(x) converges, then you know that the integral over f(x) converges. However, not that if the g(x) you choose diverges when integrated, it doesn't tell you anything about whether or not the integral over f(x) diverges, because the integral over f(x) could be a finite number while still satisfying the inequality ##\int dx~f(x) < \int dx~g(x) = \infty##. So, you either have to choose a better comparison function that is 'closer' to your actual integrand, or you may have to consider that the integral diverges and you should check that with the next test.

    Similarly, say you know g(x), but you don't know whether or not the integral over g(x) converges. But, suppose you suspect if diverges. If you can find a function f(x) that is less than g(x), and you can show that the integral over f(x) diverges, then you know the integral over g(x) diverges. However, if the integral over f(x) converages, it doesn't tell you if the integral over g(x) diverges or not.

    Choosing an appropriate function to compare your integrand to is a matter of experience, in some sense. Typical things to try are cutting out factors in the denominator (e.g., if you integrand were x/(x+1), then you know that making the denominator smaller results in a larger overall number, so if we cut out the 1, we know that x/(x+1) < x/x = 1. Note that in this example the constant 1 would be a bad comparison function because if we integrate out to infinity we know the integral diverges and it wouldn't tell us anything about x/(x+1).)

    In particular, it helps if you can find a comparison function for which you know the antiderivative. For a simple example problem, consider your first integral,

    $$\int_0^\infty dx~\frac{x}{x^3+1}.$$

    Can you find a function which is greater than your integrand, but whose antiderivative you know, using the hints I mentioned above?
     
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