1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Comparison Test Question?
Loading...