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Convergence/Divergence of an Integral

  1. Nov 5, 2011 #1
    Determine whether the integral converges or diverges:

    [itex]\int_0^{\infty}(t^{-2}e^t) dt[/itex]

    What would be the first step here in determining convergence or divergence?
  2. jcsd
  3. Nov 6, 2011 #2
    You can apply several tests. If you take integral calculus..... you use the comparsion test.. testing the bounds or evaluating a similar function via comparsion, test. You should find whether or not these converges.
  4. Nov 6, 2011 #3


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    Try determining convergence or divergence of
    [itex]\int_0^{1}t^{-2} dt[/itex]
    [itex]\int_1^{\infty}e^t dt[/itex]
    then deduce the convergence by comparison or otherwise of
    [itex]\int_0^{\infty}t^{-2}e^t dt[/itex]
  5. Nov 6, 2011 #4
    My method is that I draw a rough sketch of function.If f(t)→ 0 while t→∞,then we can say that integral is convergent as integral is area under the curve.
  6. Nov 6, 2011 #5


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    ^That does not work. Convergence can depend upon behavior at points other than infinity and even when behavior at infinity is conclusive the function going to zero is not strong enough consider f(t)=1/t.
  7. Nov 6, 2011 #6
    So if both have no limit and diverge, then by comparison test the whole integrand diverges?

    Edit: Just noticed you have the limits a little different. How did you determine the limits to be those when evaluating them separately?

    I get [itex]\int_0^1 t^{-2} dt = -1[/itex]

    And [itex]\int_1^{\infty} e^t dt = \infty[/itex]

    What would I determine from these exactly?
    Last edited: Nov 6, 2011
  8. Nov 6, 2011 #7


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    The limits can be broken up to study the convergence of each region. So if any part does not converge the integral cannot. Show et>t2 for large t to conclude et/t2>1 and thus diverges.
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