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Homework Help: Comparison Test Question?

  1. Jun 14, 2010 #1
    1. The problem statement, all variables and given/known data

    I'm just curious as to how to think about the following form of equation.

    2. Relevant equations
    [tex] \int_{3}^{\infty } \frac{1}{x + e^x} \,dx[/tex]

    3. The attempt at a solution

    What you're trying to do is to test it;

    [tex] \frac{1}{x \ + \ e^x} \ < \ \frac{1}{x}[/tex]

    [tex] \frac{1}{x} [/tex] diverges

    [tex] \frac{1}{x \ + \ e^x} \ < \ \frac{1}{e^x}[/tex]

    [tex] \lim_{t \to \infty} \int_{3}^{t} e^{-x}\,dx \ = \ \lim_{t \to \infty} - e^{-x} | \ _3 ^t \ = \ \lim_{t \to \infty} - {\frac{1}{e^t} \ + \ \frac{1}{e^3} [/tex]

    so this converges to [tex] \frac{1}{e^3} [/tex].

    I don't get how this means the original eq. will also converge?

    Both [tex]\frac{1}{x}[/tex] and [tex]\frac{1}{e^x}[/tex] are bigger than the original eq. with one converging and the other diverging.

    [tex]\frac{1}{x}[/tex] is bigger than [tex]\frac{1}{e^x}[/tex] but the test of [tex]\frac{1}{x^p}[/tex] is ringing in my ears as a kind of explanation, but I am confused.

    Is there an easy way to link all of this together?
  2. jcsd
  3. Jun 14, 2010 #2
    Indeed, because for [strike]positive x[/strike] x > 1 when p > 1, 1/xp is always less than 1/x. In effect, its graph has less area underneath it for the integration. So there is no contradiction that 1/x leaves just too much area, i.e. it diverges, while 1/xp for p > 1 does not, i.e. it converges. Meanwhile, 1/ex just hugs the x-axis massively more than 1/xp for any constant p, hence it of course converges. Finally, 1/(x+ex) lies even closer to the x-axis and leaves even less area; it converges.

    Knowing 1/x diverges does not mean we know 1/(x+ex) diverges or converges. The graph of the latter has less area, but by how much? Enough to stop the integral from diverging? However, knowing that 1/ex converges does tell us that 1/(x+ex) converges, since it has less area than a finite amount.

    EDIT: It occurred to me that I also need x > 1, not just x positive. Sorry for the mistake.
    Last edited: Jun 14, 2010
  4. Jun 14, 2010 #3
    Ahh Brilliant!

    I think that idea of hugging the x-axis makes everything even clearer, thanks a lot for that :wink:
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