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

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.

• I won't comment about the accuracy of your integration which I haven't checked but I suggest that your answer might be a little more succinct if you apply Euler's formula in reverse to combine your exponential terms into sines and cosines Mar 29 at 16:50

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