# 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?)

• You have a sign error in the exponent, it's $e^{-\Delta k \lvert\delta\rvert - i k_0\delta}$. Mar 11, 2016 at 16:25
• Thank you for your keen observation Mar 11, 2016 at 16:31
• uh.. I checked again, i think it is still $e^{\Delta k}$ Mar 11, 2016 at 17:04
• No, for $\delta < 0$, when we use the semicircle in the upper half-plane, we must plug $i\Delta k$ into $e^{-i\kappa\delta}$, and $-i\cdot (i\Delta k) = \Delta k$, so we get $e^{\Delta k\delta} = e^{-\Delta k\lvert\delta\rvert}$ then. Mar 11, 2016 at 17:11
• oh. you're right. sorry for the confusion Mar 11, 2016 at 20:07

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 $\delta \lt 0$.

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.