Inverse Fourier Transform of $\omega ^2$ in $[-\omega _0,\omega _0]$

Question

I've been learning about signals for a while now, and I'm just starting to learn about Continuous time Fourier transforms. In this particular case, we were asked to get the inverse Fourier Transform of

$$F(\omega)=\begin{cases}\omega^2\,&\ -\omega_0\leq \omega \leq \omega_0\\0,&\textrm{otherwise}\end{cases}$$

and I just wanted to make sure that the correct way to go about this was to solve for the integral

$$f(t)=\int_{-\omega_0}^{\omega_0}\omega^2e^{j\omega t}\frac{d\omega}{2\pi}$$ since the Fourier Transform itself is 0 for any values less than $-\omega_0$ or more than $\omega_0$

then the final answer is something ugly like $$f(t)=\frac{1}{2\pi} \left[\frac{\omega_0^2}{jt}(e^{j\omega t}-e^{-j\omega t})+\frac{2\omega_0}{t^2}(e^{j\omega t}+e^{-j\omega t})+\frac{2}{jt^3}(-e^{j\omega t}+e^{-j\omega t})\right]$$

from doing integration by parts twice. And sometimes this kind of answers end up ringing a bell that something feels wrong.

Answer 1

Your result looks correct. I just want to show you another way that's probably faster. I'm sure you know the inverse Fourier transform of a function that equals $1$ for $-\omega_0<\omega<\omega_0$ and vanishes elsewhere. Let's call this inverse Fourier transform $g(t)$. If you don't know it, you can compute it very easily. Now note that a multiplication with $j\omega$ in the frequency domain corresponds to differentiation in the time domain. Consequently, a multiplication with $\omega^2=-(j\omega)^2$ corresponds to the second derivative and a sign inversion. So in order to compute the inverse Fourier transform of the function in your example you can differentiate $g(t)$ twice and change its sign:

$$f(t)=-g''(t)$$