I'm messing around with some manual computation of Fourier transforms, and I'm getting conflicting results. First of all though, just to make sure I'm seeing things the right way: the Heaviside function is used to limit the domain of a function in order to keep it in the $L^1$ space - thus making it transformable - right? For example, $e^{-x}$ isn't transformable in $R^n$ because it diverges for $x\rightarrow-\infty$, but $e^{-x}u(x)$ (with $u(x)$ being the Heaviside function) is lower bounded to $0$ which makes it F-transformable. Am I making sense, or am I being blasphemous?
That being said, I'm trying to find the transform of: $$x(t)=u(t-1)e^{-2t}+u(t)-u(t+2)$$
Assuming what I said earlier is correct, I did this: $$\hat x(\xi)=\int_{1}^{\infty}e^{-t(2+i\xi)}dt+\int_{0}^{\infty}e^{-i\xi t}dt-\int_{-2}^{\infty}e^{-i\xi t}dt$$ with the integration intervals shifted according to the 3 Heavisde functions, which returns $$\hat x(\xi)={e^{-2-i\xi} \over 2+i\xi}+{1 \over i\xi}-{e^{2i\xi}\over i\xi}$$ Now, my problem is that the result I am given is $$\hat x(\xi)={e^{-2-i\xi} \over 2+i\xi}-{2e^{i\xi}sin\xi \over \xi}$$
I see the similarities, but I think I might've missed something along the path. Thanks