Problem with (classical) Fourier transform 
Problem Find the Fourier transformation of $u(x) = \frac{1}{1+x^2}$

I want $\int_\mathbb R e^{-itx} \frac{1}{1+x^2} dx$.
Let $f(z) = e^{-itz} \frac{1}{1+z^2}$, $z \in \mathbb C$, let's integrate this over the semi-circunference of radius $R$ and the line $[-R, R]$.
First, the integral on the circumference vanishes (with the change $z = Re^{i\theta}$)
$$\left|\int_{\gamma_R} \frac{e^{-izt}}{z^2+1}dz\right| = \left|\int_0^\pi \frac{e^{-itRe^{i\theta}}}{R^2e^{2i\theta}+1}Rie^{i\theta}d\theta\right| \le \int_0^\pi \frac{R}{|R^2e^{2i\theta}+1|}d\theta \le \int_0^\pi\frac1R d\theta= \frac\pi R$$
That goes to $0$ as $R \to \infty$.
The integral on the whole path is $2\pi i \text{ Res }(f, i)$, and since the integral on the circumference vanishes we have
$$\int_\mathbb R \frac{e^{-ixt}}{1+x^2} = 2 \pi i \text{ Res }(f, i)$$
I know that 

If $z_0 \neq \infty$ is a pole of order $m$ for $f$, then set $g(z) =
 (z-z_0)^mf(z)$ and $\displaystyle \text{ Res }(f, z_0) =
 \frac{g^{(m-1)}(z_0)}{(m-1)!}$

Since $z_0 = i$ is a pole of order $1$, I set $\displaystyle g(z) = \frac{e^{-izt}}{z+i}$ and the residue should be $\displaystyle g(i) = \frac{e^t}{2i}$
And the integral then is $$\int_\mathbb R \frac{e^{-ixt}}{1+x^2} = \pi e^t$$
I know this is wrong (the fourier transform isn't even bounded) but I don't know where the error is. 
I am pretty sure the residue is wrong, but why?
 A: The mistake is your answer to the question

What is $$\bigl\lvert e^{-izt}\bigr\rvert\,?$$

Answer that correctly, and you will see what goes on.
We have $\lvert e^{-izt}\rvert = e^{t\operatorname{Im} z}$, so the exponential factor is bounded only if $t\operatorname{Im} z \leqslant 0$.
Thus we can only use the semicircle of radius $R$ in the upper half-plane to close the contour and evaluate the integral over the real line by the residue theorem if $t \leqslant 0$. In that case, it indeed evaluates to $\pi e^t$.
For $t > 0$, we can argue with the parity, $\frac{1}{1+x^2}$ is an even function, so
$$\int_{-\infty}^\infty \frac{\sin (xt)}{1+x^2}\,dx = 0$$
for all $t\in \mathbb{R}$ and therefore
$$\int_{-\infty}^\infty \frac{e^{-ixt}}{1+x^2}\,dx = \int_{-\infty}^\infty \frac{e^{ixt}}{1+x^2}\,dx,$$
so the integral is $\pi e^{-t}$ for $t > 0$, altogether $\pi e^{-\lvert t\rvert}$. Alternatively, one can close the contour with a semicircle in the lower half-plane for $t > 0$, and the residue theorem then yields
$$\int_{-\infty}^\infty \frac{e^{-ixt}}{1+x^2}\,dx = -2\pi i \operatorname{Res}\left(\frac{e^{-izt}}{1+z^2}; -i\right) = -2\pi i \frac{e^{-t}}{-2i} = \pi e^{-t}.$$
