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Homework Help: Integral help

  1. Jul 20, 2009 #1
    how do i calculate this integral:
    I = \int_{0.5}^{\infty }e^{-jwt}dt

    i though it should be
    I=0 - \frac{e^{-jw/2}}{-jw} = \frac{e^{-jw/2}}{jw}

    but it seems that the first part isn't zero...why ?
  2. jcsd
  3. Jul 20, 2009 #2
    I &= \int_{0.5}^\infty e^{-jwt} \,dt \\
    &= \lim_{a\to\infty} \int_{0.5}^a e^{-jwt} \,dt \\
    &= \lim_{a\to\infty} \frac{e^{-jwt}}{-jw} \,\bigg|_{0.5}^a \\
    &= \lim_{a\to\infty} \frac{e^{-jwa}}{-jw} - \frac{e^{-jw/2}}{-jw} \\
    &= \left( \lim_{a\to\infty} \frac{e^{-jwa}}{-jw} \right) + \frac{e^{-jw/2}}{jw}

    You handle improper integrals by using limits, but the reason why you don't have zero is because the limit on the left doesn't converge for a general w. The complex function e-jwa is a periodic function.
  4. Jul 20, 2009 #3


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    [itex]e^{-i\omega t}= cos(\omega t)- i sin(\omega t)[/itex] for real t. That does NOT go to 0 as t goes to infinity.
  5. Jul 21, 2009 #4
    i see.
    how do i calculate the final result?
  6. Jul 21, 2009 #5
    I think you can use : [tex]\lim_{x \to + \infty} \frac{\cos(x)}{x}= \lim_{x \to + \infty} \frac{\sin(x)}{x}=0[/tex]
  7. Jul 21, 2009 #6
    I don't think that helps in this case, since his limit is with respect to a, and there is no a in the denominator of the limited term
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