Fourier Transform of 1/x

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I am trying to calculate the Fourier Transform of $h(x)=1/x$

So I was thinking:

Since the FT of $g(x)=sign(x)$ is $G(f)=1/(i\pi f)$ ,

then using property of duality gives that $G(x)=1/(i\pi x)$ transforms to $g(-f)=-sign(f)$

Next I used scaling property, and it turns out that -going this way- $h(x)=1/x$ transforms to $H(f)=-\pi sign(i\pi f)=-i\pi^2 sign(f)$, which is different from MATLAB output $-i\pi sign(f)$

I've set MATLAB to be consistent with this definition:

$$F\{f(x)\}=\int_{-\infty}^{\infty}{f(x)exp(-i2\pi fx)dx}$$

Here I am using f as frequency $f=\omega/(2\pi)$ , and the properties used are shown here..

Any idea of what's happening?

Thanks in advance.-