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

1. Nov 30, 2013

Typhon4ever

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. Nov 30, 2013

vela

Staff Emeritus
3. Nov 30, 2013

Typhon4ever

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?

4. Nov 30, 2013

Mandelbroth

How are you defining the Dirac delta? What does that imply about the product?

5. Nov 30, 2013

Typhon4ever

I'm not sure what you mean by how I'm defining the dirac delta.

6. Nov 30, 2013

Mandelbroth

Well, what does the Dirac delta evaluate to at any nonzero $x\in\mathbb{R}$?

I'm assuming your "*" means multiplication, correct?

7. Nov 30, 2013

Typhon4ever

Yes "*" means multiplication. Anywhere except the origin is 0.

8. Nov 30, 2013

vela

Staff Emeritus
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.

9. Nov 30, 2013

Typhon4ever

So I have the delta function at infinity which is 0 * a non converging exponential=...0?

10. Nov 30, 2013

vela

Staff Emeritus
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.

11. Nov 30, 2013

Typhon4ever

Oh, ok. Thank you!