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Integral-Converge or Diverge?

  1. Aug 4, 2013 #1
    1. The problem statement, all variables and given/known data
    Does the following integral converge or diverge?
    [itex] \int_{0}^{∞} √t * e^{-t}\, dt [/itex]
    Exponential function is not under the square root symbol.


    2. Relevant equations



    3. The attempt at a solution
    Using the ratio test, I have:

    sqrt(t+1)* e^(-{t+1})/[sqrt(t)* e^(-t)] = lim t-> ∞ sqrt(t+1)/(e*sqrt(t)) = 0
    0 < 1
    Therefore, I think the series is absolutely convergent.
    Is this correct, and is the logic and method I used correct? Thank you!
     
    Last edited: Aug 4, 2013
  2. jcsd
  3. Aug 4, 2013 #2
    The ratio test is for series, not for integrals. How can you use it here?
     
  4. Aug 4, 2013 #3
    That was a stupid mistake. I shall redo the problem using the integral test:

    v= sqrt(t)
    t= v^2
    dt= 2v dv
    1/2 ∫v*e^(-v^2) dv since u=-v^2 and 0 and ∞ are limits
    1/2 ∫v*e^(-v^2) dv with 0 and z being limits of integration
    lim z-> ∞ \left.\frac{-1}{2}*e^(-v^2)\right|_0^z = lim z-> ∞ (1/2 - 1/2*e^(-z^2))= 1/2

    Therefore, if this integral converges, series also converges by the Integral Test.

    Is this logic accurate? Thanks.
     
    Last edited: Aug 4, 2013
  5. Aug 4, 2013 #4
    If I understand your intent correctly, you want to reduce the analysis of the integral's convergence to a series' convergence, and you do that via the integral test.

    That is an acceptable strategy. But I do not understand at all what you do. I see that you have a substitution, but I do not see how that is related to the integral test. The integral test has a requirement for the integrand, and I would expect that you start by checking that requirement.
     
  6. Aug 4, 2013 #5

    SteamKing

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    e^(-t) is not the natural log function. e^x is the exponential function.
     
  7. Aug 4, 2013 #6

    LCKurtz

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    Hint: On ##[1,\infty)## how do ##\sqrt t## and ##t## compare?
     
  8. Aug 4, 2013 #7
    Thank you SteamKing. I've edited that.

    And I thought the integral test involves picking out a known function, determining convergence for that series, and then comparing the two series?
     
  9. Aug 4, 2013 #8
    The integral test establishes a link between the convergence of an integral and the convergence of a series. There are no "two" series.
     
  10. Aug 4, 2013 #9
    sqrt(t) < t ?
     
  11. Aug 4, 2013 #10

    LCKurtz

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    Right. Do you see how to make a comparison using that fact and making an easier integral to work?
     
  12. Aug 4, 2013 #11
    Yes. So an integral test is essentially a comparison test but for integrals. In this case, I would be able to use the comparison between sqrt(t) and t? I didn't know that-I presumed that I had to use something involving e.
     
  13. Aug 4, 2013 #12
    The integral test is a test for series, involving integrals. It can also be used as a test for integrals involving series. It is not a comparison test.
     
  14. Aug 4, 2013 #13

    LCKurtz

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    Think about what happens if you replace the ##\sqrt t## in the integral with ##t##. You can't ignore the exponential. And you need to say something about the different interval.
     
  15. Aug 4, 2013 #14
    I thought the sqrt(t) just minimizes the function (decreases values)? So the exponential wouldn't matter either way?
     
  16. Aug 4, 2013 #15

    LCKurtz

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    Assuming nonnegative integrand, if you want to show convergence, you have to compare the integral you are given with a new integral that is greater and whose integral converges.
     
  17. Aug 4, 2013 #16
    So could I pick the new integral as t * e^t ? That gives me greater values, I think?
     
  18. Aug 4, 2013 #17
    That integral is wrong (does not converge- increases without bound, I think). Can I try this one?

    (t) e^(-t)
    And the limits of integration should be 1 to infinity, I think? That way, the first integral will be <= the second integral.
     
  19. Aug 4, 2013 #18
    Yes, that will work.
     
  20. Aug 4, 2013 #19

    LCKurtz

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    You have a couple of details left. Your original was$$
    \int_0^\infty \sqrt t e^{-t}\, dt$$You can't compare it directly with $$
    \int_1^\infty te^{-t}\, dt$$because they aren't over the same interval. You have to address that, even if it is easy.
     
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