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In order to prove the relation above we perform the inverse Fourier transform:
\[\frac{1}{2\pi}\int\limits _{-\infty}^{\infty}2\pi\bar{\theta}(\omega)e^{-i\omega t}d\omega=\frac{1}{2\pi}\int\limits _{-\infty}^{\infty}\left(\int\limits _{-\infty}^{\infty}\theta(t')e^{i\omega t'}dt'\right)e^{-i\omega t}d\omega=\theta(t).\]
Therefore, on the one hand we must show that:
\[\lim_{\varepsilon\rightarrow0}\frac{1}{2\pi}\int\limits _{-\infty}^{\infty}\frac{i}{\omega+i\varepsilon}e^{-i\omega t}d\omega=\theta(t),\]
where \(\lim_{\varepsilon\rightarrow0}\frac{i}{\omega+i\varepsilon}\) is an expression for \(2\pi\bar{\theta}(\omega)\) and on the other hand we also need to prove:
\[\frac{1}{2\pi}\int\limits _{-\infty}^{\infty}\left(\frac{i}{\omega}+\pi\delta(\omega)\right)e^{-i\omega t}d\omega=\theta(t).\]
1. Proving \(\lim_{\varepsilon\rightarrow0}\frac{1}{2\pi}\int\limits _{-\infty}^{\infty}\frac{i}{\omega+i\varepsilon}e^{-i\omega t}d\omega=\theta(t)\):
As we see the integrand has a pole at \(\omega=-i\varepsilon\), which is in the negative complex plane. Solving the integral we need to consider two cases, where the contour is a closed semicircle in the complex plane:
- Integral in the upper complexe plane \((t<0)\):
Here we obtain upon splitting the integral into its components:
\begin{eqnarray*}\int\limits _{-\infty}^{\infty}\frac{i}{\omega+i\varepsilon}e^{-i\omega t}d\omega&=&\mathcal{P}\int\limits _{-\infty}^{\infty}\frac{i}{\omega+i\varepsilon}e^{-i\omega t}d\omega+\int_{C_{2}}\frac{i}{\omega''+i\varepsilon}e^{\omega''t}d\omega\\&=&\mathcal{P}\int\limits _{-\infty}^{\infty}\frac{i}{\omega+i\varepsilon}e^{i\omega|t|}d\omega+\int_{C_{2}}\frac{i}{\omega''+i\varepsilon}e^{-\omega''|t|}d\omega=0,\end{eqnarray*}
because there are no poles in the upper complex plane and the integral along \(C_{2}\) vanishes for \(\omega^{\prime\prime}\longrightarrow\infty\).
- Integral in the lower complexe plane \((t>0)\):
Using residue theorem the solution is:
\[\lim_{\varepsilon\rightarrow0}\frac{1}{2\pi}\left(\int\limits _{-\infty}^{\infty}\frac{i}{\omega+i\varepsilon}e^{-i\omega t}d\omega\right)=-\lim_{\varepsilon\rightarrow0}\frac{1}{2\pi}\left(2\pi i\,\textrm{Res}\left(\frac{i}{\omega+i\varepsilon}e^{-i\omega t};-i\varepsilon\right)\right)=1.\]
Therefore we get:
\[\Theta\left(t\right)=\lim_{\varepsilon\rightarrow0}\int\limits _{-\infty}^{\infty}\frac{i}{\omega+i\varepsilon}e^{-i\omega t}d\omega=\begin{cases}
1 & t>0\\
0 & else
\end{cases}\]
2. Proving \(\frac{1}{2\pi}\int\limits _{-\infty}^{\infty}\left(\frac{i}{\omega}+\pi\delta(\omega)\right)e^{-i\omega t}d\omega=\theta(t)\):
First step is easy:
\[\frac{1}{2\pi}\int\limits _{-\infty}^{\infty}\left(\frac{i}{\omega}+\pi\delta(\omega)\right)e^{-i\omega t}d\omega=\frac{1}{2}-\frac{1}{2\pi i}\int\limits _{-\infty}^{\infty}\frac{1}{\omega}e^{-i\omega t}d\omega.\]
Now we will rearrange the second integral, obtaining:
\[\int\limits _{-\infty}^{\infty}\frac{1}{\omega}e^{-i\omega t}d\omega=\int\limits _{-\infty}^{0}\frac{1}{\omega}e^{-i\omega t}d\omega+\int\limits _{0}^{\infty}\frac{1}{\omega}e^{-i\omega t}d\omega.\]
In order to solve this integral we will use the trick, that we already used here, deriving the KKR, i.e. we will consider a contour integral in the complex plane with a small semicircle around \(\omega=0\), where we obtain:
\[0=\oint\frac{e^{-i\omega t}}{\omega}d\omega=\int\limits _{-\infty}^{-0}\frac{e^{-i\omega t}}{\omega}d\omega+\int\limits _{+0}^{\infty}\frac{e^{-i\omega t}}{\omega}d\omega+\int\limits _{\pi}^{0}\frac{e^{-i\omega t}}{\omega}d\omega+\int_{C_{2}}\frac{e^{-\omega''|t|}}{\omega''}d\omega''\]
Since we just look at the case \(t<0\) the \(C_{2}\)-integral vanishes, as shown above. Using the parametrization \(\omega=\rho e^{i\phi}\quad\rightarrow\quad d\omega=i\rho e^{i\phi}d\phi\) we obtain:
\[\int\limits _{\pi}^{0}\frac{e^{-i\omega t}}{\omega}d\omega=\int\limits _{\pi}^{0}\frac{i\rho e^{i\phi}}{\rho e^{i\phi}}e^{-i\omega t}d\phi=-i\pi.\]
Also we get depending on the signum of \(t\), i.e. direction of the integration:
\[\int\limits _{-\infty}^{-0}\frac{e^{-i\omega t}}{\omega}d\omega+\int\limits _{+0}^{\infty}\frac{e^{-i\omega t}}{\omega}d\omega=-i\pi\cdot\textrm{sign}(t).\]
This leads to:
\[\frac{1}{2\pi}\int\limits _{-\infty}^{\infty}\left(\frac{i}{\omega}+\pi\delta(\omega)\right)e^{-i\omega t}d\omega=\frac{1}{2}\left[1+\textrm{sign}(t)\right]=\theta(t).\]
Quod erat demonstrandum.
