Wrong result in Fourier transform I am trying to calculate the Fourier transform of
$$f(x)=\frac1{1+x^2}$$
with the definition
$$\hat f(\xi)=\int_\mathbb R e^{-ix\xi}\frac{1}{1+x^2}\mathrm dx$$
I tried using the residue theorem. As a consequence of Jordan's Lemma, one should have
$$\int_\mathbb Re^{-ix\xi} \frac{1}{1+x^2}\mathrm d x=2\pi i \text{Res}\left ( e^{-i\xi \cdot}\frac{1}{1+\cdot^2}; i\right )$$
We can compute the Laurent series of the two terms. We obtain
$$\begin{align} \frac{1}{1+x^2}
&=\frac i 2\left( \frac{1}{x+i}-\frac{1}{x-i}\right )
\\
&=-\frac i 2 \frac 1{x-i} +\frac i 2 \sum_{n=0}^{\infty}\left( -\frac 1{2i} \right)^n (x-i)^n\end{align} $$
and 
$$ e^{-ix\xi}=e^\xi e^{-i\xi(x-i)}=e^{\xi}\sum_{n=0}^{\infty}\frac{(-i\xi)^n}{n!} (x-i)^n$$
Taking the product of the two terms, the coefficient of the $(x-i)^{-1}$ term is $-e^\xi \frac i2$, so the integral should evaluate to 
$$\hat f(\xi)= 2\pi i \cdot -e^\xi \frac i2=\pi e^\xi$$
This is clearly wrong, since by the Riemann-Lebesgue Lemma
$$\lim_{|\xi|\to \infty}\hat f(\xi)=0$$
Through another computation, I came to the result that
$$\hat f=\pi e^{-|\xi|}$$
What is wrong in my reasoning through the use of the Residue Theorem?
 A: Observations on the attempt
Let's take a look at the statement of Jordan's Lemma:

Let $f$ be a continuous function on $\mathbb C$ and $\gamma_R$ an arc of a circumference centered in the origin with radius $R$ with an angle span $[\theta_1,\theta_2]$ with $0\le \theta_1<\theta_2\le \pi$. If
  $$\lim_{R\to \infty} \max_{[\theta_1,\theta_2]}|f(Re^{i\theta})|=0 $$
  Then for every $\omega >0$
  $$\lim_{R\to \infty}\int_{\gamma_R}f(z)e^{i\omega z}\mathrm d z=0 \tag {$\dagger$}$$

The essential error in my previous computation regarded the fact that if $\xi>0$ we cannot apply Jordan's Lemma in the upper plane, because $(\dagger)$ holds only for positive $\omega$.
Correct resolution
First of all, we can use Jordan's lemma on the upper plane to compute integrals of the kind
$$\int_\mathbb R e^{ix}R(x)\mathrm d x \tag{$\star$}$$
where $R(x)$ is a rational function of $x$ such that $\lim_{|z|\to \infty}zR(z)=0$. In this case in particular, $R(x)=\frac{1}{1+x^2}$. 
To reconduce the integral
$$\hat f(\xi)=\int_\mathbb Re^{-i\xi x}\frac{1}{1+x^2}\mathrm d x$$
 to $(\star)$ we perform the change of variables $y=-\xi x$. By change of variables in a Lebesgue integral,
$$\begin{align}\int_\mathbb Re^{-i\xi x}\frac{1}{1+x^2}\mathrm d x 
&=\int_\mathbb Re^{iy}\frac{1}{1+\frac{y^2}{\xi^2}}\frac{\mathrm dy}{|\xi|}
\\
&=|\xi|\int_\mathbb Re^{iy}\frac{1}{\xi^2+y^2}\mathrm dy
\end{align}$$
In this form, it is sufficient to apply the residue theorem on the upper plane. There are two cases:


*

*$\xi>0$: in this case, the residue in the upper plane is located in $i\xi$. This yields
$$\begin{align}\hat f(\xi) &
=\xi\int_\mathbb Re^{iy}\frac{1}{(y+i\xi)(y-i\xi)}\mathrm dy
\\
&=\xi \cdot 2\pi i  \text{Res}\left (e^{iy}\frac{1}{(y+i\xi)(y-i\xi)},y=i\xi\right)
\\
&=2\pi i \xi \frac{e^{-\xi}}{2i\xi}=\pi e^{-\xi}
 \end{align}$$

*$\xi<0$: in this case, the residue in the upper plane is located in $-i\xi$. This yields
$$\begin{align}\hat f(\xi) &
=-\xi\int_\mathbb Re^{iy}\frac{1}{(y+i\xi)(y-i\xi)}\mathrm dy
\\
&=-\xi \cdot 2\pi i  \text{Res}\left (e^{iy}\frac{1}{(y+i\xi)(y-i\xi)},y=-i\xi\right)
\\
&=-2\pi i \xi \frac{e^{\xi}}{-2i\xi}=\pi e^{\xi}
 \end{align}$$
The final result can be written as
$$\hat f(\xi)=\pi e^{-|\xi|}$$
