Using this Fourier-transform taken from a table; $$ \mathcal{F}\big\{\frac{1}{1+t^2}\big\} \Longrightarrow \pi e^{-|\omega|} $$
I would like to transform this expression; $$ g(x) = U_0 \frac{L^2}{L^2+x^2}, \quad L,U_0>0 $$
I know that
$$ \mathcal{F}\big\{ U_0f(x) \big\} = U_0\mathcal{F}\big\{ f(x) \big\} $$
Then I thought that maybe I should rewrite it as
$$ g(x) = U_0 \frac{1}{1+(x/L)^2} \Longrightarrow \hat g(\omega)=U_0\pi e^{-|\omega/L|}$$
But the correct answer is
$$ \hat g(\omega)=U_0 \pi L e^{-|\omega|L} $$
Why is my approach wrong, and where should I rethink?
The problem is when you rescale the $x$ variable, there is a factor that comes out of the Fourier transform (and the scaling of frequencies goes the other way around): $$\mathcal F \{f\left(\frac x L\right)\}(\omega)=|L|\mathcal Ff(L\omega)$$