Cheers, I am trying to find the fourier transform of the signum function, which is
$$ \operatorname{sgn}(t) \triangleq \begin{cases} 1 \qquad & t>0 \\ 0 \qquad & t=0 \\ -1 \qquad & t<0 \\ \end{cases} $$
I rewrite this as:
$$\operatorname{sgn}(t) = 2u(t) -1$$
and find it's first derivative which is:
$$\operatorname{sgn}'(t) = 2 \delta(t)$$
and using the integration rule I know that:
$$\begin{align} \mathscr{F}\{\operatorname{sgn}(t)\} &= \mathscr{F}\left\{\int_{-\infty}^t2\delta(ρ)dρ \right\} \\ &= \frac{1}{j\omega}2 + \pi X(0)\delta(\omega) \\ \end{align}$$
I need to find to prove that $X(0)=0$, as the fourier tranform of the signum function is $\frac{2}{j\omega}$, but I think this transformation always yields 1. Is what I am thinking correct, and if yes how would I go about this? I know that there are alternatives, but I am just checking to see alternative ways to prove things I already know. Thanks =)
Edit: I tried the following thing: I split the Fourier of $\operatorname{sgn}(\cdot)$ to the Fourier of the unit step and the -1 constant. Then, I get that
$$\mathscr{F}\{2u(t)\} = \frac{2}{j\omega} + \pi \delta(\omega) $$
and
$$\mathscr{F}\{1\}= \pi \delta(\omega)$$
so by subtracting, I get the correct thing. Is that the way to do it?