# Fourier transform using residues

1. Jan 18, 2010

### libelec

1. The problem statement, all variables and given/known data
Find the Fourier transform of f(t) = e-at2, with a > 0 using the residue theorem.

3. 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?

2. Jan 18, 2010

### 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.

Last edited: Jan 18, 2010
3. Jan 20, 2010

### 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?

4. Jan 20, 2010

### 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.

5. Jan 20, 2010

### 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)?

6. Jan 21, 2010

### libelec

This is what I don't understand.

7. Jan 21, 2010

### Count Iblis

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

8. Jan 21, 2010

### Dick

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.

9. Jan 23, 2010

### libelec

sin[at]*exp(-iwt).

10. Jan 23, 2010

### libelec

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

11. Jan 23, 2010

### Count Iblis

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).