Fourier transform using residues

[libelec]

Homework Statement

Find the Fourier transform of f(t) = e^-at2, with a > 0 using the residue theorem.

The Attempt at a Solution

The problem I have is the function g(t) = f(t).e^-iwt, which is holomorph in all C. Is there another way to do it without residues?

 

[Dick]

f(t)*exp(-iwt)=exp(-a*t^2-iwt). You want to complete the square and displace the integration off the x-axis using that the function is holomorphic. If you want some gory details about a very similar problem check out Count Iblis's advice in https://www.physicsforums.com/showthread.php?t=368440 You can ignore the residues, in the sense that there aren't any because there are no poles. You should still pay some attention the contours, as Count Iblis points out.

 

[libelec]

OK, thanks. I have another question though. I tried this same thing but with the sin(at)/at.

Now, I turned the integral such that $$\[\int\limits_{ - \infty }^\infty {\frac{{Sin(at)}}{{at}}{e^{ - iwt}}dt} = {\mathop{\rm Im}\nolimits} \left( {\int\limits_{ - \infty }^\infty {\frac{{{e^{iat}}{e^{ - iwt}}}}{{at}}dt} } \right) = {\mathop{\rm Im}\nolimits} \left( {\int\limits_{ - \infty }^\infty {\frac{{{e^{i(a - w)t}}}}{{at}}dt} } \right) = {\mathop{\rm Im}\nolimits} \left( {\int\limits_{} {\frac{{{e^{i(a - w)z}}}}{{az}}dz} } \right)\]$$.

Then, evualuating the integral $$\[{\int\limits_\gamma {\frac{{{e^{i(a - w)z}}}}{{az}}dz} }\]$$ I arrive to the known result that it's equal to i$$\[\pi \]$$/a. Then the Fourier transform is $$\[\pi \]$$/a.

But the Fourier transform of sinc is a rectangular function. That means that there are values of w in which the Fourier transform equals the found value, and values of w in which it equals 0. Where do I see that?

 

[Count Iblis]

Taking the imaginary part like you did there doesn't quite work here, because you have to end up with sin(at)exp(-i w t). Also, note that you would also have to take the principal value of the integral when you do that.

Instead of taking the imaginary part, you can substitute:



sin(at) = 1/(2i) [exp(i a t) - exp(-iat)]

in the integrand. Then you replace the integral by the principal value of the integral (i.e. you omit the interval from minus epsilon to epsilon and consider the limit of epsilon to zero). Then you can split up the integral in the two parts with the terms exp(i a t) and exp(- i a t). For each integral you have to consider an appropriate contour integral.

 

[libelec]

So, $$\[\int\limits_{ - \infty }^\infty {\frac{{\sin (at).{e^{ - iwt}}}}{{at}}dt} = \int\limits_{ - \infty }^\infty {\frac{{\left( {{e^{iat}} - {e^{ - iat}}} \right).{e^{ - iwt}}}}{{2i.at}}dt} = \mathop {\lim }\limits_{A \to \infty } \int\limits_{ - A}^A {\frac{{{e^{iat}}.{e^{ - iwt}}}}{{2i.at}}dt} - \int\limits_{ - A}^A {\frac{{{e^{ - iat}}.{e^{ - iwt}}}}{{2i.at}}dt = } \mathop {\lim }\limits_{A \to \infty } \int\limits_{ - A}^A {\frac{{{e^{i(a}}^{ - iw)t}}}{{2i.at}}dt} - \int\limits_{ - A}^A {\frac{{{e^{ - i(a + w)t}}}}{{2i.at}}dt} \]$$

How is that any different from what I did before? Why should I end up with the expression sin(at)exp(-i w t), if that's the imaginary part of exp(iat).exp(-iwt)?

 

[libelec]

Taking the imaginary part like you did there doesn't quite work here, because you have to end up with sin(at)exp(-i w t).

This is what I don't understand.

 

[Count Iblis]

What is Im[exp(i a t) exp(-i w t)] ?

 

[Dick]

OK, thanks. I have another question though. I tried this same thing but with the sin(at)/at.

Now, I turned the integral such that $$\[\int\limits_{ - \infty }^\infty {\frac{{Sin(at)}}{{at}}{e^{ - iwt}}dt} = {\mathop{\rm Im}\nolimits} \left( {\int\limits_{ - \infty }^\infty {\frac{{{e^{iat}}{e^{ - iwt}}}}{{at}}dt} } \right) = {\mathop{\rm Im}\nolimits} \left( {\int\limits_{ - \infty }^\infty {\frac{{{e^{i(a - w)t}}}}{{at}}dt} } \right) = {\mathop{\rm Im}\nolimits} \left( {\int\limits_{} {\frac{{{e^{i(a - w)z}}}}{{az}}dz} } \right)\]$$.

Then, evualuating the integral $$\[{\int\limits_\gamma {\frac{{{e^{i(a - w)z}}}}{{az}}dz} }\]$$ I arrive to the known result that it's equal to i$$\[\pi \]$$/a. Then the Fourier transform is $$\[\pi \]$$/a.

But the Fourier transform of sinc is a rectangular function. That means that there are values of w in which the Fourier transform equals the found value, and values of w in which it equals 0. Where do I see that?

You aren't going to see the step function correctly unless you are a lot more careful with defining the contours. Don't do 'Im' and consider both integrals when you expand (exp(aiz)-exp(-aiz)). Which half plane you need to close each contour in depends on the sign of (a-w) and (a+w). I would also change z into z-i*epsilon with epsilon>0 and then let epsilon -> 0 at the end. That puts the pole definitely in the upper half plane.

 

[libelec]

What is Im[exp(i a t) exp(-i w t)] ?

sin[at]*exp(-iwt).

 

[libelec]

Ah, no, wait. I'm not considering the exp(-iwt) imaginary part, that's what you're saying?

 

[Count Iblis]

Ah, no, wait. I'm not considering the exp(-iwt) imaginary part, that's what you're saying?

That's right, if you take the imaginary part of the whole expression then you get an additional term: the imaginary part of exp(-iwt) times the real part of exp(i a t).