I was looking for the FT of the symmetric functions:
where $n>0$. Here $-\infty \leq x \leq \infty$. I expect the FT to be symmetric and exponentially deceasing. Could you please guide me to the answer or provide a proof? Thanks in advance.
Similarly as in my asnwer to the other question here, these functions can be bounded above by $C/|x|$ and do not have an exponential decay (except if $n$ is an even integer and $f$ is of the form 1.). Perhaps what you did not remark is just the fact that $(1+|x|^2)^{-n}$ is infinitely smooth, while $(1+|x|)^{-n}$ is as smooth as $|x|$ at $x=0$, and this is what prevents the exponential decay of their Fourier transform.
I do not know any simple expression for their Fourier transform in general, however, one can still obtain the decay of their Fourier transform by comparing to a function with a similar local behavior. For example, if $n=1$, then $f(x) = (1+|x|)^{-1}$ has the same behavior as $g(x) = e^{-|x|}$. More precisely, $$ h(x) = f(x) - g(x) $$ has its (weak) third derivative in $L^\infty$ and so $|h(x)| \leq C/|x|^3$ for large $x$. Now, the explicit Fourier transform of $e^{-|x|}$ is (see here) $$ \widehat{g}(y) = \frac{2}{1+|y|^2} $$ and so one deduces that $$ \mathcal F((1+|x|)^{-1}) = \frac{2}{|x|^2} + O\!\left(\frac{1}{|x|^3}\right) $$ when $|x|\to\infty$.
The same is true for any $n>0$, $\mathcal F((1+|x|)^{-n}) \sim \frac{C}{|x|^2}$ when $|x|\to\infty$.
