Fourier transform of $f (x)=1 $ on compact carrier $(-1,1)$.

Question

I have a function $f$ defined as: $$ f(x)= \begin{cases} 1 & \text{if $x\in (-1,1)$} \\ 0 & \text{else.} \\ \end{cases} $$ I have calculated that Fourier transform of this is $\sqrt{\frac{2}{\pi}}\frac{\sin\omega}{\omega}$. I calculated this with integration from $-1$ to $1$, because the integral is zero elsewhere. But I am not sure where is my solution defined. On all $\Bbb R$ or just on interval $(-1,1)$, and why? I am wondering this, because I think that I have remembered that for Fourier transforms the domain of original function and its Fourier transform are the same.

Answer 1

the transform you calculated is defined for $\omega \neq 0$. Nevertheless, it is very common to redefine it by saying that its value in $\omega = 0$ is $\sqrt{\frac{2}{\pi}}$ to have a continuous function.