Fourier transform of $1/x$, $x>0$

Question

I'm new to Fourier transform. From Does the Fourier Transform exist for f(t) = 1/t? ,I saw that the Fourier transform of $1/x$ is $\text{sgn}(x)$. Define $$f(x)=\begin{cases} \frac{1}{x} &\text{if}\,\,x>0\\ 0 &\text{otherwise} \end{cases}$$ It seems to me that the Fourier transform of $f$ is still $1/x$ (up to a multiple of constant). Am I correct? I just "followed" the same method as in that post.