1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Deciding if Divergent/Convergent via Comparison Theorem

  1. Mar 13, 2012 #1
    1. The problem statement, all variables and given/known data
    Use the Comparison Theorem to determine whether the integral is convergent or divergent.

    2. Relevant equations

    3. The attempt at a solution
    [itex]\int_{0}^{1}\frac{e^{-x}}{\sqrt{x}} \leq \int_{0}^{1}\frac{1}{x^{\frac{1}{2}}}[/itex]

    Because p<1, the integral is divergent. Wolfram, however, is giving me a finite answer, meaning the integral is actually convergent. Where am I going wrong?
    Last edited: Mar 13, 2012
  2. jcsd
  3. Mar 13, 2012 #2
    The improper integral of (1/x.5) is divergent. When you calculate its integral within a closed interval [0,a] you get a finite answer.

    You're actually doing a reasonable comparison because (e-x/x.5) will (for x≥0, real numbers) always be under (1/x.5), therefore, the integral between 0 and a (in this case, a=1) will always be less then the other, since it's the defined area between the graphs and the x axis.

    Hence, you can conclude it's convergent.
  4. Mar 14, 2012 #3
    I'm not sure I quite understand. When the comparison equation is near 0, the graph is up at +∞. So how is it possible that it is convergent?
  5. Mar 14, 2012 #4
    The improper integral (as the endpoint of the interval [0,b] approaches ∞) is divergent. Meaning that if you calculate the area below the graph in this interval [0,∞[ you don't get a finite answer.
    However the improper integral (regarding the startpoint of the interval [0,b]) is convergent, as weird as it may seem, because we're talking about the sqrt(x)-1. If instead of having that you had x-1 (they both have alike graphs), the same improper integral would give ∞ at x→0.

    It's really interesting to see that expressions with basically the same graph have different responses in what concerns limits/integrals.
  6. Mar 14, 2012 #5
    So it's because of the limits that it is convergent. If it were from 1 to 0 it would be divergent?
  7. Mar 14, 2012 #6


    Staff: Mentor

    No. Changing the limits of integration just changes the sign of the value of the integral.

    Assuming that f is integrable on [a, b],

    [tex]\int_a^b f(x)dx = -\int_b^a f(x)dx[/tex]
  8. Mar 14, 2012 #7
    I was mixing up my p-test for 0->∞ and 0->1. :grumpy:
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook