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Please help me evaluate this seemingly simple integral

  1. Jan 5, 2012 #1
    1. The problem statement, all variables and given/known data

    Let x(t) = e-100tu(t)

    u(t) = 0 for t < 0
    u(t) = 1 for t > 0

    Evaluate the following integral (from -∞ to ∞):

    X(ω) = ∫ x(t)e-iωtdt

    2. Relevant equations

    See below.

    3. The attempt at a solution

    I tried to evaluate the integral by splitting it in two parts, since x(t) takes two different values. Keep in mind I'm replacing ∞ with a to evaluate the limit later.

    from -∞ to 0:
    X1(ω) = ∫ e-iωtdt

    X1(ω) = (1/-iω) e-iωt (from -∞ to 0)

    X1(ω) = (1/-iω)(1 - eiωa)

    from 0 to ∞:
    X2(ω) = e-100∫ e-iωtdt

    X2(ω) = e-100 (1/-iω) e-iωt (from 0 to ∞)

    X2(ω) = e-100 (1/-iω)(e-iωa - 1)

    X(ω) = X1(ω) + X2(ω) = (1/-iω)(1 - eiωa) + e-100 (1/-iω)(e-iωa - 1)

    = (1/-iω) (1 - eiωa + e-100e-iωa - e-100)

    This is where I'm stuck. The answer is supposed to be X(ω) = 1 / (100 + iω), and I have absolutely no idea how I'm supposed to get there. Would someone mind helping me out?
    Last edited: Jan 5, 2012
  2. jcsd
  3. Jan 5, 2012 #2
    Doh, it looks like I failed hard, I accidently put x(t) = e-100 instead of x(t) = e-100t. However when I tried to solve the problem earlier I got that right and I still didn't end up with the right answer (I re-did the problem from scratch when posting here). Is there a "trick" to solve this?
  4. Jan 5, 2012 #3


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    Homework Helper

    shouldn't the negative half of the integral be zero?

    also don't forget to take the limit for a as well
    Last edited: Jan 5, 2012
  5. Jan 5, 2012 #4
    edit: sorry of some of my vocabulary is hard to understand, english isn't my primary language. I got the positive part of the integral right (it gives me the correct answer), I fixed the mistake I did in my first post. However I can't see how the negative part is supposed to be zero.

    Well I see that it's supposed to be zero given the form of the answer, but I don't see how lim a -> ∞(1 - eiωa) can give zero, unless it's a special property of the exponent with the imaginary number i. Note that I'm taking the limit of a to infinity since I replaced the -∞ with -a earlier, thus cancelling the negative sign of the exponent.
  6. Jan 5, 2012 #5


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    Homework Helper

    based on x(t)=0 for t<0, it should just be the integral of zero
  7. Jan 5, 2012 #6
    You have defined [itex]u(t)=0[/itex] when [itex]t<0[/itex] and [itex]x(t)=e^{-100t}\cdot u(t).[/itex]
  8. Jan 5, 2012 #7
    Yes, but I'm not evaluating the integral of x(t), I'm evaluating the integral of X(ω), which is different from x(t). Sorry if the notation is confusing, I used the variables given in the problem.
  9. Jan 5, 2012 #8
    Did you forget x(t) from the integrand?
  10. Jan 5, 2012 #9
    Wow I'm a dumbass, somehow I had x(t)=1 when t **< 0. Thanks everyone.
  11. Jan 5, 2012 #10
    No problem. I suspect that we have all made these kinds of mistakes, it often happens to me when the problem statement is complicated. :)
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