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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.
    [itex]\int_{0}^{1}\frac{e^{-x}}{\sqrt{x}}[/itex]

    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

    Mark44

    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:
     
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