What is the Fourier Transform of t*f(t)

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SUMMARY

The Fourier Transform of the function f(t) = t * e^(-2t^2) can be derived using integration by parts and completing the square in the exponential argument. The discussion highlights that the Fourier Transform of t * f(t) can be expressed as a convolution integral, specifically \mathcal{F}t*f(t) = \int w*f(w-t)\,\text{d}t. Participants emphasize the importance of breaking the integral into manageable parts, utilizing known transforms, and clarifying that the variable t should be treated as unbounded for the Fourier Transform to be valid.

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Homework Statement


f(t)=t*e^(-2t^2)
Find the Fourier Transform F(w) of f(t).

It is given that when f(t)=e^[(-at^2)/2] F(w)=√(2*pi/a)*e^[(-w^2)/2a]

Homework Equations



The Attempt at a Solution


The transform of e^(-2t^2) is easily obtained from the given information, and I got stuck when I tried to do it with integrate by part.

Is it related to the convolution? something like F(fg)=F(f)F(g) ??
Is there any theorem for the Fourier Transform of t*f(t)?
And also for Laplace Transform too?
 
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Hmm. It is a convolution integral of the transforms, was that what you where looking for so:
\mathcal{F}t*f(t) = \int w*f(w-t)\,\text{d}t
But that makes the problem more complicated.
 
Try completing the square in the argument of the exponential in the transform integral and then doing a change of variables. That should break up the integral into two parts. One of the parts will be easy to integrate, and for the other you can use the information given.
 
dikmikkel said:
Hmm. It is a convolution integral of the transforms, was that what you where looking for so:
\mathcal{F}t*f(t) = \int w*f(w-t)\,\text{d}t
But that makes the problem more complicated.
So there isn't a theorem where I can simply put 2 transforms function together?!


TachyonRunner said:
Try completing the square in the argument of the exponential in the transform integral and then doing a change of variables. That should break up the integral into two parts. One of the parts will be easy to integrate, and for the other you can use the information given.
Hi Tachyon, is this what you mean?

1. ∫t*e^-2t^2 e^-iwt dt
2. -2t^2-iwt = -2(t^2+iwt/2) = -2[(t+iw/4)^2+w^2/16]=-2(t+iw/4)^2 -(w^2)/8
as I let p=t+iw/4 therefore dp=dt
3. ∫t*e^((-w^2)/8)dt +∫e^(-2p^2) dp

Can I used the f(t)=e^[(-at^2)/2] F(w)=√(2*pi/a)*e^[(-w^2)/2a] information here? But isn't the Transform =∫f(p)e-iwp dp? or F(w)=∫f(t)e-iwt dt

And one more stupid question from me...
What happen if we need to do a transform of f(t) to F(w), with no limits of t given??
for example after integration: F(w)= [(t^2)/2] limit ∞ to -∞(?!) + something...

Or there must be a limit??
 
Step 2. looks fine to me, but there is a mistake going into step 3. so take a look at that.

You should have from p=t+i\omega /4 that dp=dt, but also that t=p-i\omega /4. It is the second of these equations that will split your integral into a sum. One of these integrals will be of the form e^{-2p^2} while the other will be of the form pe^{-2p^2}. To evaluate the first of these, you will have to use the information given, and to evaluate the second one you can use basic integration techniques.

I am not sure I understand your second question, but in general to use the Fourier Transform you are using right now, you will want t to be an unbounded variable, that is it should vary from -∞ to +∞. If you were given limits for t, so that for example a<t<b, then you should consider doing a Finite Fourier transform instead.
 

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