Solving FT of Decaying Exp: Lorentzian Function

[Amok]

I had to show that:

$$\int_{-\infty}^{\infty} e^{-a \left | t \right |} e^{\omega t} \mathrm{d} t$$

with a positive, is equal to a Lorentzian function.

I simply did this:

$$<br /> \int_{-\infty}^{\infty} e^{-a \left | t \right |} e^{i\omega t} \mathrm{d} t = \int_{0}^{\infty} e^{(-a + i\omega) t} \mathrm{d} t + \int_{-\infty}^{0} e^{(a + i\omega)t } \mathrm{d} t<br /> = \frac{-1}{i\omega\ -a} + \frac{1}{i\omega + a} + \lim_{x \to -\infty} e^{(-a + i\omega )x} - \lim_{y \to +\infty} = \frac{2a}{\omega^2 + a^2} + \lim_{x \to +\infty} e^{-ax} lim_{x \to +\infty} e^{i\omega} - \lim_{y \to -\infty} e^{ax} \lim_{y \to -\infty} e^{i\omega} = \frac{2a}{\omega^2 + a^2}$$

Is this correct? I'm asking because I've seen people do this by using Euler's formula and then calculating an integral with a cosine in it, but I don't really see the point of that.

EDIT: can you guys see the whole thing? I can't, what should I do?

 

[vela]

You're fine up to here:

$$\frac{-1}{i\omega\ -a} + \frac{1}{i\omega + a} + \lim_{t \to \infty} e^{(-a+i\omega)t} - \lim_{t \to -\infty} e^{(a+i\omega)t}$$

Then you just need to say the two limits are equal to 0.

What you can't do is split the limits up like you did:

$$\lim_{t \to \infty} e^{(-a+i\omega)t} \Rightarrow \lim_{t \to \infty} e^{-at}\lim_{t \to \infty} e^{i\omega t}$$

You can only say lim AB = (lim A)(lim B) if both lim A and lim B exist. In this case, the complex exponential oscillates and doesn't converge as t goes to infinity, so the limit doesn't exist.

 

[LCKurtz]

Or, to add a little detail:

|e^(-a+iω)t| = |e^-at||e^iωt| = |e^-at| → 0 as t → ∞

and similarly for the other one.

 

[Amok]

$$\lim_{t \to \infty} e^{(-a+i\omega)t} \Rightarrow \lim_{t \to \infty} e^{-at}\lim_{t \to \infty} e^{i\omega t}$$

You can only say lim AB = (lim A)(lim B) if both lim A and lim B exist. In this case, the complex exponential oscillates and doesn't converge as t goes to infinity, so the limit doesn't exist.

That's what I thought, so how do I show that the limit does exist?

 

[vela]

Do what LCKurtz did.

 

[Amok]

But how did he introduce the modulus in the calculations?

 

[LCKurtz]

If |z| → 0 then doesn't that show z → 0?

 

[Amok]

Of course it does! Thank you guys.

 

[Amok]

If |z| → 0 then doesn't that show z → 0?

I'm sorry to bring thus up again, but I can see that what you're saying is true, simply by picturing the polar form of a complex number on a complex plane, but how do you prove what you're saying? How can you really write it down?

 

[LCKurtz]

Doesn't z → 0 mean |z - 0| gets small?

 

[Amok]

It mosta certainly does. The modulus of my function is

$$\left| e^{-at} \right|$$

which does become arbitrarily small as t goes to infinity. I hadn't given much though to limits of complex functions for a while. This was good practice. Thank you very much Kurtz.