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Comparison Test for improper integral

  1. Nov 16, 2015 #1
    1. The problem statement, all variables and given/known data
    use the comparison theorem to determine whether ∫ 0→1 (e^-x/√x) dx converges.

    2. Relevant equations
    I used ∫ 0 → 1 (1/√x) dx to compare with the integral above

    3. The attempt at a solution
    i found that ∫ 0 → 1 (1/√x) dx = 2 ( by substituting 0 for t and take the limit of the defenite integral as t → 0^+) thus it is convergent. and (1/√x) > (e^-x/√x) so ∫ 0→1 (e^-x/√x) dx also converges.

    But in my text book they only use the comparison theorem for the limits of integration from 0 to ∞. would it still be acceptable to use the comparison theorem for this problem for limits of integration from 0 to 1.
    Thanks for reading :).
     
  2. jcsd
  3. Nov 16, 2015 #2

    Samy_A

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    I don't see any particular reason why the comparison theorem for integrals wouldn't be valid with other limits of integration than ##0## to ##\infty##.

    See for example here.
     
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