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Limit of e^(i*x) at infinity?

  1. Nov 30, 2013 #1
    1. The problem statement, all variables and given/known data
    delta dirac function(x) * e^(-i*x) at ∞ and -∞

    2. Relevant equations
    delta dirac(x)*e^(i*x)


    3. The attempt at a solution
    I'm wondering how e^(i*x) looks like at infinity/-infinity. I know it has some sort of oscillating property i.e e^(pi*i)=e^(3pi*i). The problem is I'm trying to find the delta dirac function * e^(-i*x) at infinity and negative infinity but I don't know how to evaluate the exponential. I know the delta function should be 0 in both cases but I don't know if the exponential will turn out to be indeterminate or something else. I should get 0 from the whole equation.
     
  2. jcsd
  3. Nov 30, 2013 #2

    vela

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  4. Nov 30, 2013 #3
    Ok. Using Euler's if I'm not mistaken just tells me the limit doesn't exist. What do I do then to evaluate this problem?
     
  5. Nov 30, 2013 #4
    How are you defining the Dirac delta? What does that imply about the product?
     
  6. Nov 30, 2013 #5
    I'm not sure what you mean by how I'm defining the dirac delta.
     
  7. Nov 30, 2013 #6
    Well, what does the Dirac delta evaluate to at any nonzero ##x\in\mathbb{R}##?

    I'm assuming your "*" means multiplication, correct?
     
  8. Nov 30, 2013 #7
    Yes "*" means multiplication. Anywhere except the origin is 0.
     
  9. Nov 30, 2013 #8

    vela

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    If you're referring to the limit ##\displaystyle \lim_{x \to \infty} e^{ix}##, then you're correct. That limit doesn't exist because ##e^{ix}## doesn't converge to a single value. But that's not the limit you were asking about. You should cogitate on the implication of your (correct) answer to Mandelbroth's question.
     
  10. Nov 30, 2013 #9
    So I have the delta function at infinity which is 0 * a non converging exponential=...0?
     
  11. Nov 30, 2013 #10

    vela

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    Yes, that's the right answer, but you're not thinking about this correctly. Since you're interested in the limit as x goes to +∞, you can assume x>0. For any such value of x, ##\delta(x)e^{ix} = 0\times e^{ix} = 0##. Now you take the limit of this as x goes to +∞.
    $$\lim_{x \to \infty} \delta(x)e^{ix} = \lim_{x \to \infty} 0 = 0.$$ The behavior of ##e^{ix}## doesn't matter.
     
  12. Nov 30, 2013 #11
    Oh, ok. Thank you!
     
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