Cauchy integral formula and contour I was doing fourier transform of lorentzian function
$$
B(k)=\frac{1}{1+\left(\frac{k-k0}{\Delta k}\right)^2}
=\frac{\Delta k^2}{\Delta k^2+\left(k-k_0\right)^2}
$$
$B(k)$ is neither even nor odd function, so Fourier transform of $B(k)$
$$
\mathscr{F}[B(k)]=\Delta k^2\int_{\infty}^{\infty}
\frac{e^{-ik\delta}}{\Delta k^2+\left(k-k_0\right)^2}dk
$$
substitution
$
k-k_0=\kappa
$
, $\qquad
k=k_0+\kappa
$
$$
\mathscr{F}[B(k)]=\Delta k^2e^{-ik_0\delta}\int_{\infty}^{\infty}
\frac{e^{-i\kappa\delta}}{\Delta k^2+\kappa^2}d\kappa
=
\Delta k^2e^{-ik_0\delta}\int_{\infty}^{\infty}
\frac{e^{-i\kappa\delta}}{(\kappa+i\Delta k)(\kappa-i\Delta k)}
d\kappa
$$
Cauchy's integral formula, I used the formula once for each pole.
When $\delta<0$, the contour is on the upper plane, and it is counter clockwise
$$
\int_{\infty}^{\infty}f(\kappa)e^{-i\kappa\delta}d\kappa=2\pi i \sum Res[f(\kappa)e^{-i\kappa\delta}]
$$
$$
\mathscr{F}[B(k)]
=
2\pi i \Delta k^2e^{-ik_0\delta} \frac{e^{-\Delta k\delta}}{2i\Delta k}
=
\pi \Delta k e^{-ik_0\delta}e^{-\Delta k\delta}
$$
When $\delta>0$, the contour is on the lower plane, and it is clockwise (which adds minus(-) sign to the integral)
$$
\mathscr{F}[B(k)]
=
-2\pi i \Delta k^2e^{-ik_0\delta} \frac{e^{\Delta k\delta}}{-2i\Delta k}
=
\pi \Delta k e^{-ik_0\delta}e^{\Delta k\delta}
$$
Together they comprise
$$
\pi \Delta k e^{-\Delta k |\delta|-ik_0\delta}
$$
This is some other's work, and there is something I quite don't understand.
How does $\delta>0$ corresponds to upper plane and $\delta<0$ to lower plane?
(besides, did I wrote the correct answer for FT?)
 A: Cauchy's formula requires a closed contour in the complex plane.  We typically close the contour with a semicircle in the upper or lower half plane.  When $\delta \lt 0$, the integral over the semicircle in the upper half plane vanishes (and that over the lower half plane diverges), so we use the upper half plane.  Vice-versa for $\delta \lt 0$.
A: Note that for $\delta<0$ the integration over a semi-circular contour $C_R$, with radius $R$ and centered at the origin in the upper-half plane becomes
$$\int_{C_R}\frac{e^{-i\delta z}}{z^2-\Delta k^2}\,dz = \int_0^\pi \frac{e^{-i\delta R e^{i\phi}}}{R^2e^{i2\phi}-\Delta k^2}(iRe^{i\phi})\,d\phi$$
Note that the term $e^{-i\delta R e^{i\phi}}=e^{-i\delta R\cos(\phi)}e^{\delta R\sin(\phi)}$ decays exponentially for $0<\phi <\pi$ as $R\to \infty$ when $\delta<0$.  Therefore, as $R\to \infty$ the integral over $C_R$ vanishes.

It is important to point out that this last statement is not obvious since for $\phi \approx 0$ and $\phi \approx \pi$ the exponential is close to zero for fixed $R$.  To show that the integral vanishes, we write the following estimates for $R>\Delta k$:
$$\begin{align}\left|\int_0^\pi \frac{e^{-i\delta R e^{i\phi}}}{R^2e^{i2\phi}-\Delta k^2}(iRe^{i\phi})\,d\phi\right|&\le \int_0^\pi \frac{Re^{\delta R\sin(\phi)}}{|R^2e^{i2\phi}-\Delta k^2|}\,d\phi\\\\&\le \frac{2R}{R^2-\Delta k^2}\int_0^{\pi/2} e^{2 \delta R\,\phi/\pi }\,d\phi\\\\&=\frac{\pi(1-e^{-|\delta| R})}{|\delta| (R^2-\Delta k^2)}\end{align}$$  So, it is evident that as $R\to \infty$, the integral over $C_R$ tends to $0$.  Note that in the development, we used $\sin(\phi)\ge 2\phi/\pi$ for $0\le \phi\le \pi/2$.

However, when $\delta>0$, the term $e^{-i\delta R e^{i\phi}}=e^{-i\delta R\cos(\phi)}e^{\delta R\sin(\phi)}$ increases exponentially for $0<\phi <\pi$ as $R\to \infty$.  Therefore, when $\delta >0$, we close the contour with the semi-circular arc $C'_{R}$, with radius $R$ and centered at the origin in the lower-half plane.  On $C'_{R}$, the exponential term decays as $R\to \infty$ and the integration over $C'_{R}$ approaches zero.
