How does the Fourier transform of f(t)=exp(-ltl) behave at infinity?

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SUMMARY

The Fourier transform of the function f(t) = exp(-|t|) is analyzed, revealing that as t approaches ±∞, the limit of the expression exp(-|t|)exp(-iωt) converges to zero. The discussion clarifies that while exp(-iωt) oscillates indefinitely, the factor exp(-|t|) ensures the overall expression approaches the origin in the complex plane. The final result of the Fourier transform is confirmed to be 1/√2π * (1/(1+ω²)).

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zezima1
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Homework Statement


Find the Fourier transform of f(t)=exp(-ltl)

Homework Equations


The expression for the Fourier transform.

The Attempt at a Solution


Applying the Fourier transform I get an expression, where I have to take the limit of t->-∞ of exp(-i\omegat) - how do I do that?
 
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zezima1 said:

Homework Statement


Find the Fourier transform of f(t)=exp(-ltl)


Homework Equations


The expression for the Fourier transform.

The Attempt at a Solution


Applying the Fourier transform I get an expression, where I have to take the limit of t->-∞ of exp(-i\omegat) - how do I do that?

What is the expression you obtained?
 
Yes I better post it - just tried to avoid it since I'm not good at latex. I get:
1/√2π(∫-∞0exp(t-iωt)dt + ∫0exp(-t-iωt)dt)
=
1/√2π([1/(1-iω)exp(t-iωt)]-∞0+[-1/(1+iω)exp(-t-iωt)]0
 
Last edited:
zezima1 said:
Yes I better post it - just tried to avoid it since I'm not good at latex. I get:
1/√2π(∫-∞0exp(t-iωt)dt + ∫0exp(-t-iωt)dt)
=
1/√2π([1/(1-iω)exp(t-iωt)]-∞0+[-1/(1+iω)exp(-t-iωt)]0

OK, that looks right. But look more carefully: the limit you need is not

\lim_{t \rightarrow \infty} \exp(-i\omega t)
but rather
\lim_{t \rightarrow \infty} \exp(-|t|)\exp(-i\omega t)
 
oh god yes :)

But still: What if exp(-iwt) tends to infinity? Okay I don't think it does since its a sum of a real cos and an imaginary sin but still I want to know how to calculate the limit of the term with the complex exponential :)
 
Either way I get:

1/√2π *(1/(1+ω2)

Do you also get that? :)
 
nvm that was almost correct but got the right one now :)
 
\exp(-i\omega t) has no limit as t \rightarrow \infty. It just spins endlessly around the unit circle in the complex plane.

\exp(-|t|)\exp(-i\omega t) spirals endlessly around the origin, but the \exp(-|t|) factor pushes it closer and closer to the origin as t increases toward \infty (or decreases toward -\infty). This is why it has a limit even though one of its factors does not.

You can see this more formally by looking at the magnitude:

|\exp(-|t|)\exp(-i\omega t)| = |\exp(-|t|)|\cdot |\exp(-i\omega t)| = |\exp(-|t|)| \cdot 1 = |\exp(-|t|)| = \exp(-|t|)

which goes to zero as t goes to either +\infty or -\infty. And the magnitude of a function goes to zero if and only if the function itself goes to zero.
 

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