About a Fourier transform of a non- integrable function. I'm trying to obtain the Fourier transform of the following function:
$$F(x)=\frac{x}{1+x^2}$$
I have tried using Residue Theorem, but i think it can't be applied because the difference between the  polynomial grade of numerator and denominator is not 2 or more...
Also, $F$ is not an integrable function...
It must be some "distributional argument" to solve this. Any ideas?
Thanks a lot for any help!
 A: Well, apparently, $F$ is not a Schwartz function, but it's in $L^2$. In addition, one can show that this function is (clearly) $L^1_{loc}$ and is, in fact, a tempered distribution (if you have problems with proving it, ask in comments). Thus, we can indeed, take a Fourier transform of it in the sense of distributions and in the sense of $L^2$.
How we can find it? As @mrf showed, one can still use residues. Another approach is the fact that you can reverse the Fourier transfrom by applying it again to the "mirrored" function. Or, in other words (up to a certain multiplicative constant that depends only on the exact definition of Fourier transform), 
$$f=\mathcal F \left[\mathcal F[f](-\xi)\right],$$
where $\mathcal F$ stands for Fourier transform.
Now, we can represent your function $F$ as $\frac {ix}{i}\frac{1}{1+x^2}$, which looks very similar to the Fourier transform of the derivative. One can find (via residues, for example), that $$\mathcal F\left[\frac{1}{1+x^2}\right](\xi) = e^{-|\xi|}$$
(up to a certain multiplicative constant that depends only on the exact definition of Fourier transform (again, elas)). The derivative is equal to $$-e^{-\xi}H(\xi)+e^{\xi}H(-\xi),$$
where $H$ is the Heavyside function.
Finally, when we combine everythig back, we obtain the final result: $$Const\cdot \frac 1i\left(-e^{-\xi}H(\xi)+e^{\xi}H(-\xi)\right).$$
You can check it on wolframalpha, the input would be FourierTransform[x/(1+x^2),x].
A: Even though the function is not $L^1$, residue calculus still helps.
Put
$$ f(z) = \frac{z}{1+z^2}. $$
If $\xi < 0$, then $f(z)e^{-ix\xi} \to 0$ on a semi-circle in the upper half-plane. Jordans's lemma (details omitted) shows that
$$
\int_{-\infty}^\infty \frac{xe^{-ix\xi}}{1+x^2}\,dx = 
2\pi i \operatorname{Res}\limits_{z=i} f(z)e^{-iz\xi} = i\pi e^{\xi}.
$$
Similarly, if $\xi > 0$, then $f(z)e^{-ix\xi} \to 0$ on a semi-circle in the lower half-plane. Again, Jordan's lemma gives
$$
\int_{-\infty}^\infty \frac{xe^{-ix\xi}}{1+x^2}\,dx = 
-2\pi i \operatorname{Res}\limits_{z=-i} f(z)e^{-iz\xi} = -i\pi e^{-\xi}.
$$
The extra minus sign comes from the negative orientation of the semi-circle. (Your normalization of the Fourier transform may be different.)
