Fourier Transform of a function not in Schwartz Space I want to calculate the Fourier transform of $$f(x)=\frac{x}{1+x^2}.$$
It is an exercise in my textbook.
I know that $f(x) \notin \mathcal{S}$, in which $\mathcal{S}$ is the Schwartz Space, and that $\int_{-\infty}^\infty xf(x) dx$ doesn't converge (even if $g(x)=\frac{1}{1+x^2} \in L^1$ and its Transform is known). 
Can I use the formula $$\mathcal{F}(f(x))(w)=\mathcal{F}(xg(x))(w)= i \frac{d \widehat{g}}{dw}(w)?$$ What are the conditions to guarantee that this property is valid?
My doubt is because in the books of Fourier Analysis (and even in questions here in the fórum), the previous result is applied only in $\mathcal{S}$ (that has good properties) and I'm not not sure if only the fact of $g \in \mathcal{L}^1$ is enough to apply it.
Thanks a lot
 A: Notice that $\frac{1}{\pi}\int \frac{1}{1+x^2}e^{-2\pi i xt}\,dx= e^{-2\pi|t|}$. 
This can be seen in terms of probability: The characteristic function of the double exponential distribution $\phi(y)=\frac12 e^{-|y|}$ is $g(t)=\frac{1}{1+t^2}$. That later is integrable and so one can use the Fourier inverse theorem.
Recall that for any tempere distribution  $u$,  $\mathcal{F}(u)$ is define as
$\widehat{u}(\phi)=u\big(\mathcal{F}(\phi)\big)\,dx$. In your case, you want to see what the Fourier transform of the distribution $\phi\mapsto \int\frac{x}{1+x^2}\phi(x)\,dx$ is .
Recall that for any $\phi\in\mathcal{S}$, $\mathcal{F}(\phi)\in\mathcal{S}$, and $x\phi(x)\in\mathcal{S}$. You may try to the use
$$\int \frac{1}{\pi}\frac{1}{1+x^2}x\phi(x)\,dx= \langle \mathcal{F}(e^{-2\pi|x|}),x\mathcal{F}(\phi(x))\rangle$$
where $\langle \cdot,\cdot\rangle$ is the inner product on $L1$, that is $\langle f,g\rangle =\int f\overline{g}$.
The following identities will come handy: 
$$\langle \mathcal{F}f,\phi\rangle = \langle f,\widetilde{\mathcal{F}\phi}\rangle=\langle f,\mathcal{F}(\widetilde{\phi})\rangle$$
for all $f\in L_1$ nd $\phi\in\mathcal{S}$, where $\widetilde{g}(x)=g(-x)$, and also
\begin{aligned}
D(\mathcal{F}\phi)(t)&=-2\pi i\,\mathcal{F}(x\phi(x))(t)\\
\mathcal{F}(\mathcal{F}\phi)(x) &= \widetilde{\phi}(x)
\end{aligned}
for all $\phi\in\mathcal{S}$.
