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Convergence of integral

  1. Nov 25, 2009 #1
    1. The problem statement, all variables and given/known data
    what coefficient [tex]\alpha[/tex] this integral is convergent for?:

    [tex]\int_{0}^{\infty}\frac{e^{x \left|\sin x \right|}}{x^{\alpha}} dx[/tex]


    2. Relevant equations



    3. The attempt at a solution
    I've started with this: I = I1 + I2 + I3, where
    [tex]I_1=\int_{0}^{\epsilon}[/tex]

    [tex]I_2=\int_{\epsilon}^{M}[/tex]

    [tex]I_2=\int_{M}^{\infty}[/tex]

    so, I2 is clear, but I have big problem with the third one, it looks like the divergent, but I'm not really sure, can you help me?
    Is it correct that [tex]I_1 \sim \frac{1}{x^{\alpha}} [/tex]?
     
  2. jcsd
  3. Nov 25, 2009 #2

    Mark44

    Staff: Mentor

    You don't need three integrals: two will be enough. For the first one the limits of integration should be epsilon (lower) and b. For the second one, the limits should be b (lower) and M. Both integrals are improper, but for different reasons, and both need to be evaluated using limits. The first integral is evaluated at b and epsilon, and the limit taken as epsilon goes to 0+. The second integral is evaluated at M and b, and the limit taken as M goes to infinity.
     
  4. Nov 26, 2009 #3
    I think both suggestion to solve this looks similar, because in the same way you must solve the limit in the infinity and I consider it quite difficult. Could you suggest me some more precise way of solution. I was trying to make the series from that. I mean: [tex]\sum_{n=M}^{\infty}\int_{n\pi}^{(n+1)\pi}[/tex] ant then I restricted it by high limit and then I was able to integrated when in the end I got some kind of harmonic series. So to sum up [tex]\alpha \in (0;1)[/tex] (according to first integral around zero).
    Thanks for your time.
     
  5. Nov 27, 2009 #4

    Mark44

    Staff: Mentor

    I think you need to work on the integrand.
    1 <= ex|sin(x)| <= ex, so the integrand is bounded by 1/x[itex]\alpha[/itex] and ex/x[itex]\alpha[/itex]. Can you find a value for alpha for which the integrand is small enough that it converges?
     
  6. Nov 27, 2009 #5
    In the first one I´m quite sure that [tex] \alpha [\tex] should be <1.
    In the second one I get into the blind alley, neverthless I think we only one option, so we must restrict for some ending M, then we will get the high bound looks like that: [tex]\leq e^{M}\frac{1}{x^{\alpha}}[\tex], finally [tex] \alpha [\tex] would be >1. But it's very strange resul, isn`t it? Thanks again to everyone
     
    Last edited: Nov 27, 2009
  7. Nov 27, 2009 #6

    Mark44

    Staff: Mentor

    I don't think so. If [itex]\alpha[\itex] < 1, then x[itex]\alpha [\itex] will be small. Since it's in the denominator, the overall expression will be larger than if [itex]\alpha [\itex] were greater than 1.
    My sense is that this integral diverges no matter what [itex]\alpha [\itex] is. The exponential function grows large so fast that division by x[itex]\alpha [\itex] won't make any difference.
     
  8. Nov 27, 2009 #7
    I was optimistic and I believed it could convergent.
    Thanks fot your patience
    See you
     
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