Mark is completely correct; Chaojie needs to explain what is $\frac{1}{k}$ in the sense of distributions. If $\theta_-(x):=\theta (-x)=1-\theta(x)$ then
$\newcommand{\sgn}{\operatorname{sgn}} \sgn(x)=\theta(x)-\theta(-x)$
\begin{align}
&\hat{\theta}(k)= \frac{1}{2i\pi}(k-i0)^{-1}=\lim_{\varepsilon\to +0} \frac{1}{2i\pi}(k-i\varepsilon)^{-1},\\
&\hat{\theta_-}(k)= -\frac{1}{2i\pi}(k+i0)^{-1}=\lim_{\varepsilon\to +0} -\frac{1}{2i\pi}(k+i\varepsilon)^{-1},\\
&\hat{1}=\frac{1}{2i\pi} \bigl[ (k-i0)^{-1}-(k+i0)^{-1}\bigr]= \delta(k),\\
&\widehat{\sgn}(k)=\frac{1}{2i\pi} \bigl[ (k-i0)^{-1}+(k+i0)^{-1}\bigr] = \frac{1}{i\pi}pv (k^{-1})
\end{align}
with
$$pv(k^{-1}) (\varphi)=pv \int_{-\infty}^\infty k^{-1}\varphi(k)\,dk.$$
