TITLE: Characteristic function of Cauchy distribution.
QUESTION [5 upvotes]: When computing the characteristic function of Cauchy distribution, we applied the Cauchy Integration theorem:
$$

\int_{C_{R}}\frac{e^{i\alpha z}}{z^{2}+1}dz=\int_{-R}^{R}\frac{e^{i\alpha z}}{z^{2}+1}dz+\int_{\Gamma_{R}}\frac{e^{i\alpha z}}{z^{2}+1}dz=I_{R}+J_{R}

$$
We assume $\alpha>0$ and we use the curve from $(-R,0)$ to $(R,0)$ and back to $(-R,0)$ counter clockwise from the positive half plane (imaginary part $>0$) .
$C_{R}$ is the counter-clockwise contour, $\Gamma_{R}$ is the counterclockwise half circle on the upper half plane.
The final result is:
$$

\pi e^{-\alpha}
$$
I am curious where did we use the condition $\alpha>0$ in this derivation? 
I suspect it to be the choice of the integration contour. But how? 
Can we choose $\alpha<0$ while integrate over the same positive contour and get the same result but it is unstable for $\alpha<0$
Thanks in advance.

REPLY [4 votes]: This is to ensure that $J_R\to0$ when $R\to+\infty$. Note that, if $z=R\mathrm e^{\mathrm it}$, $|\mathrm e^{\mathrm i\alpha z}|=\mathrm e^{-\alpha R\sin(t)}$ and that $\sin(t)\gt0$ for every $t$ in $(0,\pi)$, that is, on the path $\Gamma_R$. Hence, when $R\to+\infty$, $|\mathrm e^{\mathrm i\alpha z}|\to0$ if $\alpha\gt0$ but $|\mathrm e^{\mathrm i\alpha z}|\to\infty$ if $\alpha\lt0$.

REPLY [3 votes]: Yes, the requirement $a>0$ determines the contour you choose.  When $\Gamma_R$ is a half-circle in the upper half-plane and $a>0$ it can be shown that
$$\lim_{R \to \infty} \int_{\Gamma_R}\frac{e^{i\alpha z}}{z^{2}+1}dz = 0.$$
(When $a<0$ you choose $\Gamma_R$ to be a circle in the lower half-plane.)  Thus you can conclude that
$$\lim_{R \to \infty} \int_{C_{R}}\frac{e^{i\alpha z}}{z^{2}+1}dz=\int_{-\infty}^{\infty}\frac{e^{i\alpha z}}{z^{2}+1}dz,$$
and the integral on the left can be calculated by the method of residues.
The full derivation can be found in this PDF.