Cauchy's integral formula
Cauchys integral formula
Let $f$ be holomorphic and $\gamma$ a closed curve in the complex plane $$\int_\gamma \frac{f(z)}{z-z_0} dz = 2\pi if(z_0)$$ and $$\int_\gamma \frac{f(z)}{(z-z_0)^n} dz = \frac{2\pi i}{(n-1)!}f^{(n-1)}(z_0)$$
Example: Time ordered exponentials
One example for using Cauchys integral formula, are time ordered exponentials as they appear in the Feynman propagator $$e^{-iw_k\tau}\Theta(\tau) + e^{iw_k\tau}\Theta(-\tau) = \lim_{\epsilon\to 0}\frac{-2w_k}{2\pi i}\int^\infty_{-\infty}\frac{dw}{w^2-w^2_k + i\epsilon}e^{iw\tau}.$$
To proof this identity, we calculate the right side with Cauchys integral formla, by calculating the integral.
To do so we write the fraction $$\frac{1}{w^2-w^2_k + i\epsilon}=\frac{1}{[w-(w_k-i\epsilon)][w-(-w_k+i\epsilon)]}=\frac{1}{2w_k}\left[\frac{1}{w-(w_k-i\epsilon)} - \frac{1}{w-(-w_k+i\epsilon)}\right]$$ where we rewrote the terms $\epsilon^2,2w_k\epsilon \to \epsilon$ since inside the limes it is the same.
To calculate it we notice that $$\int_\gamma dw \frac{1}{w-(w_k-i\epsilon)} = \int^\infty_{-\infty} dw \frac{1}{w-(w_k-i\epsilon)} + \int_\Gamma dw \frac{1}{w-(w_k-i\epsilon)},$$ with $\Gamma$ the countour which closes together with the real axis to a half circle in the complex plane. We know that $$\int_\Gamma dw \frac{1}{w-(w_k-i\epsilon)}e^{iw\tau}=0,$$ because the points on $\Gamma$ have infinite imaginary value (since they have $|w|=\infty$ and the polar angle is not finite since the values dont lie on the real axis.) leading to $e^{iw\tau}$ becoming zero. However this means, we have to close the contour on the upper side (Im$w>0$) when $\tau >0$, since the exponent will be negative and on the bottom side (Im$w<0$) when $\tau <0$. Therefore $$\lim_{\epsilon\to 0}\frac{2w_k}{2\pi i}\int^\infty_{-\infty}\frac{dw}{w^2-w^2_k + i\epsilon}e^{iw\tau}=\lim_{\epsilon\to 0} \frac{1}{2\pi i} \int_\gamma\left[\frac{1}{w-(w_k-i\epsilon)} - \frac{1}{w-(-w_k+i\epsilon)}\right]e^{iw\tau},$$ where we close the contour of $\gamma$ depending on the sign of $\tau$.
So if $\tau>0$ we close the contour upwards and therefore, only the second term has a pole inside $\gamma$ (having positive imaginary value) which gives $$\int^\infty_{-\infty} \frac{d\omega}{\omega-(-\omega_k+i\epsilon)}e^{iw\tau} = 2\pi i e^{-iw_k\tau}\Theta(\tau)+O(\epsilon)$$
If $\tau<0$ we close the contour downwards where only the first term has a pole inside $\gamma$ (having neagtive imaginary value) which gives $$\int^\infty_{-\infty} \frac{d\omega}{\omega-(\omega_k-i\epsilon)}e^{iw\tau} = -2\pi i e^{iw_k\tau}\Theta(-\tau)+O(\epsilon),$$ where the minus sign comes from having a clockwise contour (which is enforce by integration from minus infinity to plus infinity).
Putting both together we get $$\lim_{\epsilon\to 0}\frac{-2w_k}{2\pi i}\int^\infty_{-\infty}\frac{dw}{w^2-w^2_k + i\epsilon}e^{iw\tau}=e^{i-w_k\tau}\Theta(\tau) + e^{iw_k\tau}\Theta(-\tau),$$ after setting $\epsilon\to 0$.