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Improper Integral

  1. Jul 11, 2012 #1
    1. The problem statement, all variables and given/known data

    I'm trying to test whether the sequence converges or not:
    [itex]\sum^{∞}_{k = 1}ke^{-2k^2}[/itex]

    2. The attempt at a solution

    I tried to evaluate this in two ways, each of which produced different answers. I was able to eventually discover that this series does converge, but I still don't see what was wrong with the first method I tried (which told me it diverged.)

    Could someone please take a look at my work and tell me what I did wrong?

    [itex]\sum^{∞}_{k = 1}ke^{-2k^2}[/itex]

    [itex]\int{^{∞}_{1}xe^{-2x^2} dx}[/itex]

    Let [itex]u = -2x^2[/itex]
    [itex]du = -4x dx[/itex]

    [itex]\frac{-1}{4}\int{^{∞}_{1}-4xe^{-2x^2} dx}[/itex]

    [itex]\frac{-1}{4}\int{^{∞}_{-2}e^{u} du}[/itex]

    [itex]\frac{-1}{4}{lim}_{b → ∞}[e^u]^{b}_{-2}[/itex]

    [itex]\frac{-1}{4}[{lim}_{b → ∞}(e^b) - \frac{1}{e^{2}}][/itex]

    [itex]\frac{-1}{4}[∞ - \frac{1}{e^{2}}][/itex]

    [itex]= -∞[/itex]

    However, if you instead let [itex]u = 2x^{2}[/itex] it can be shown that the series converges. (Along with the integral)

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  2. jcsd
  3. Jul 11, 2012 #2

    Dick

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    Science Advisor
    Homework Helper

    Your u limits should be -2 to MINUS infinity. Right?
     
  4. Jul 11, 2012 #3
    They sure should. Thanks :)
     
  5. Jul 12, 2012 #4

    Zondrina

    User Avatar
    Homework Helper

    [itex]\int{^{∞}_{1}xe^{-2x^2} dx}[/itex]
    = [itex]\int{^{∞}_{1}x/e^{2x^2} dx}[/itex]

    u=2x^2
    1/4du = xdx

    =1/4[itex]\int{^{∞}_{1}1/e^{u} du}[/itex]
    =1/4[itex]\int{^{∞}_{1}e^{-u} du}[/itex]

    Integrate that, sub back in for u, take the limit, and you should be done.
     
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