[imath]\displaystyle F(\omega) = -\frac{\pi}{2}e^{-2|\omega|} \text{sgn}(\omega)[/imath]

I did a contour integration, and I got a slightly different result from my first attempt.

Now, I will solve the problem by the Symmetry Property and Differentiation Property.

The main question is to transform the following expression: [imath]\frac{it}{t^2 + 4}[/imath]

1. Using the symmetry property first: [imath]2\pi \frac{i}{4} e^{-2|-\omega|}[/imath]

2. Then applying the differentiation in time property: [imath]i\frac{d}{d\omega} [\frac{i\pi}{2} e^{-2|\omega|}][/imath]

3. Differentiating: [imath]\frac{\pi}{4} e^{-2|\omega|}[/imath]

Is that right, because the solution in the TA's sheet also shows that it's different answer.

Let [imath]\displaystyle g(t) = \frac{i}{t^2 + 4}[/imath] and [imath]\displaystyle f(t) = tg(t)[/imath]

The Symmetry Property will give you:

[imath]\mathcal{F}\{g(t)\} = G(\omega) = 2\pi \frac{i}{4} e^{-2|\omega|} = \pi \frac{i}{2} e^{-2|\omega|}[/imath]

Now we can use the differentiation property:

[imath]\displaystyle \mathcal{F}\{f(t)\} = \mathcal{F}\{tg(t)\} = F(\omega) = i\frac{dG(\omega)}{d\omega} = \frac{\omega}{|\omega|}\pi e^{-2|\omega|}[/imath]

Because [imath]f(t)[/imath] is an odd function, you need to multiply its Fourier transform by [imath]-1[/imath].

[imath]\displaystyle F(\omega) = -\frac{\omega}{|\omega|}\pi e^{-2|\omega|}[/imath]

You can replace the fraction [imath]\displaystyle \frac{\omega}{|\omega|}[/imath] by [imath]\text{sgn}(\omega) \ \text{or} \ u(\omega)[/imath].

[imath]\displaystyle F(\omega) = -\pi e^{-2|\omega|}u(\omega)[/imath]

We know that [imath]\displaystyle -u(\omega) = u(-\omega)[/imath].

[imath]\displaystyle F(\omega) = \pi e^{-2|\omega|}u(-\omega)[/imath]

Which is slightly different than TA's sheet answer! TA's sheet answer has a typo, maybe?!