In QM and QFT course, we need to use the Residue Theorem to get do a integral in the calculation in propagator

\[\int_{-\infty}^{+\infty} \frac{\mathrm{e}^{\mathrm{i}\omega t}}{\omega - \omega_0 + \mathrm{i}\varepsilon}\mathrm{d}\omega\]

To calculate this integral, we need to use the Residue theorem. And another important thing is the sign of $t$.

Choose $\omega$ as the point in the complex plane. The integral is from $-\infty \in \mathbb{R}$ to $+\infty \in \mathbb{R}$. There are two ways to construct the closed curve (the red curve and the blue curve in the figure below). And the function to be integrated is

\[\frac{\mathrm{e}^{\mathrm{i}\omega t}}{\omega - \omega_0 + \mathrm{i}\varepsilon}\]

This function have only one singularity which is $\omega=\omega_0-\mathrm{i}\varepsilon$.

Note that when the close curve is clockwise, there should be a negative value sign ‘’$-$’’ before residue in Residue Formula.

If $t>0$

We will choose the blue way

\[\begin{align} \int_{-\infty}^{+\infty} \frac{\mathrm{e}^{\mathrm{i}\omega t}}{\omega - \omega_0 + \mathrm{i}\varepsilon}\mathrm{d}\omega &=-\lim_{R\to\infty}\int_0^\pi \frac{\mathrm{e}^{\mathrm{i}R\mathrm{e}^{\mathrm{i}\theta} t}}{R\mathrm{e}^{\mathrm{i}\theta} - \omega_0 + \mathrm{i}\varepsilon}\mathrm{d}(R\mathrm{e}^{\mathrm{i}\theta}) \\ &=-\lim_{R\to\infty}\int_0^\pi \frac{\mathrm{e}^{\mathrm{i}Rt(\cos \theta+\mathrm{i\sin\theta})}(R\mathrm{i\mathrm{e}^{\mathrm{i}\theta}})}{R\mathrm{e}^{\mathrm{i}\theta} - \omega_0 + \mathrm{i}\varepsilon}\mathrm{d}\theta \\ &=-\lim_{R\to\infty}\int_0^\pi \mathrm{e}^{-Rt\sin\theta}\frac{\mathrm{i}\mathrm{e}^{\mathrm{i}(Rt\cos\theta+\theta)}}{\mathrm{e}^{\mathrm{i}\theta} -\frac{\omega_0 - \mathrm{i}\varepsilon}{R}}\mathrm{d}\theta \end{align}\]

The integrant

\[\left|\mathrm{e}^{-Rt\sin\theta}\frac{\mathrm{i}\mathrm{e}^{\mathrm{i}(Rt\cos\theta+\theta)}}{\mathrm{e}^{\mathrm{i}\theta} -\frac{\omega_0 - \mathrm{i}\varepsilon}{R}}\right|\leq 2\mathrm{e}^{-Rt\sin\theta}\]

For a great enough $R$.

Thus

\[\left|\int_{-\infty}^{+\infty} \frac{\mathrm{e}^{\mathrm{i}\omega t}}{\omega - \omega_0 + \mathrm{i}\varepsilon}\mathrm{d}\omega \right|\leq 2\lim_{R\to\infty}\int_0^\pi \mathrm{e}^{-Rt\sin\theta}\mathrm{d}\theta\]

For any $\delta>0$ which is small enough(no need to be infinitesimal)

\[\begin{align} \int_0^\pi \mathrm{e}^{-Rt\sin\theta}\mathrm{d}\theta &\leq\left(\int_0^{\delta}+\int_{\pi-\delta}^\pi \right)\mathrm{e}^{-Rt\sin\theta}\mathrm{d}\theta+\int_\delta^{\pi-\delta}\mathrm{e}^{-Rt\sin\theta}\mathrm{d}\theta \\ &\leq 2\delta+\pi \cdot \mathrm{e}^{-Rt\sin\delta} \end{align}\]

If take $\lim_{R\to\infty}$ on both side

\[\lim_{R\to\infty}\int_0^\pi \mathrm{e}^{-Rt\sin\theta}\mathrm{d}\theta \leq 2\delta\]

This means that $\lim_{R\to\infty}\int_0^\pi \mathrm{e}^{-Rt\sin\theta}\mathrm{d}\theta$ can be infinitesimal(smaller than any positive number)

\[\lim_{R\to\infty}\int_0^\pi \mathrm{e}^{-Rt\sin\theta}\mathrm{d}\theta=0\]

So when $t>0$

\[\int_{-\infty}^{+\infty} \frac{\mathrm{e}^{\mathrm{i}\omega t}}{\omega - \omega_0 + \mathrm{i}\varepsilon}\mathrm{d}\omega=0\]

If $t<0$

We will choose the red way

\[\begin{align} \int_{-\infty}^{+\infty} \frac{\mathrm{e}^{\mathrm{i}\omega t}}{\omega - \omega_0 + \mathrm{i}\varepsilon}\mathrm{d}\omega &=-2\pi\mathrm{i}\,\mathrm{Res}(\omega_0-\mathrm{i\varepsilon})-\\ &\lim_{R\to\infty}\int_0^{-\pi} \frac{\mathrm{e}^{\mathrm{i}R\mathrm{e}^{\mathrm{i}\theta} t}}{R\mathrm{e}^{\mathrm{i}\theta} - \omega_0 + \mathrm{i}\varepsilon}\mathrm{d}(R\mathrm{e}^{\mathrm{i}\theta}) \\ &=-2\pi\mathrm{i}\mathrm{e}^{\mathrm{i}\omega_0 t}-0 \end{align}\]

The first term $\mathrm{e}^{\mathrm{i}\omega_0 t}$ is the residue of $\frac{\mathrm{e}^{\mathrm{i}\omega t}}{\omega - \omega_0 + \mathrm{i}\varepsilon}$ at $\omega_0-\mathrm{i}\varepsilon$.

\[\omega':=\omega_0-\mathrm{i}\varepsilon \\ \frac{\mathrm{e}^{\mathrm{i}\omega t}}{\omega - \omega_0 + \mathrm{i}\varepsilon} =\frac{\mathrm{e}^{\mathrm{i}(\omega-\omega') t}}{\omega -\omega'}\cdot \mathrm{e}^{\mathrm{i}\omega't} =\frac{\mathrm{e}^{\mathrm{i}\omega't}}{\omega-\omega'} + \cdots\]

Thus the residue is $\mathrm{e}^{\mathrm{i}\omega’t}=\mathrm{e}^{\mathrm{i}\omega_0t}$.

The second term is zero due to the same reason before. The only difference is this time $t<0$ and $\theta$ is form $0$ to $-\pi$.

Total result

\[\int_{-\infty}^{+\infty} \frac{\mathrm{e}^{\mathrm{i}\omega t}}{\omega - \omega_0 + \mathrm{i}\varepsilon}\mathrm{d}\omega= -2\pi\mathrm{i}\mathrm{e}^{\mathrm{i}\omega_0t}\Theta(-t) \tag{1}\]

If $\omega$ is shifted to $-\omega$ on the numerator. We can just shift $t$ to $-t$

\[\int_{-\infty}^{+\infty} \frac{\mathrm{e}^{-\mathrm{i}\omega t}}{\omega - \omega_0 + \mathrm{i}\varepsilon}\mathrm{d}\omega =-2\pi\mathrm{i}\mathrm{e}^{-\mathrm{i}\omega_0t}\Theta(t) \tag{2}\]

If I want to shift $\varepsilon$ to $-\varepsilon$, keeping $\varepsilon>0$. Of cause you can just repeat the above process. But you can play a trick by applying complex conjugate on both side of equation (1).

\[\int_{-\infty}^{+\infty} \frac{\mathrm{e}^{-\mathrm{i}\omega t}}{\omega - \omega_0 - \mathrm{i}\varepsilon}\mathrm{d}\omega =2\pi\mathrm{i}\mathrm{e}^{-\mathrm{i}\omega_0t}\Theta(-t) \tag{3}\]

Or we can rewrite them as the Inverse Fourier Transform

\[\begin{align} \int_{-\infty}^{+\infty}\frac{\mathrm{d}\omega}{2\pi}\mathrm{e}^{-\mathrm{i}\omega t} \frac{\mathrm{i}}{\omega - \omega_0 + \mathrm{i}\varepsilon}= \mathrm{e}^{-\mathrm{i}\omega_0t}\Theta(t) \tag{2'} \\ \int_{-\infty}^{+\infty} \frac{\mathrm{d}\omega}{2\pi}\mathrm{e}^{-\mathrm{i}\omega t}\frac{\mathrm{i}}{\omega - \omega_0 - \mathrm{i}\varepsilon} =-\mathrm{e}^{-\mathrm{i}\omega_0t}\Theta(-t) \tag{3'} \end{align}\]