Fourier transform of propagator in Lehmann spectral representation I need help showing that:
$$ iG(k,t)=\int_{\mu/\hbar}^{+\infty}  e^{-i\omega t}A(k,\omega)d\omega $$
knowing that
$$ G(k,\omega) = \int_{-\infty}^{+\infty} \frac{A(k,\omega')}{\omega -\omega'+i\eta\ sgn(\omega'-\frac{\mu}{\hbar})}d\omega' $$
where $\eta$ is a positive infinitesimal and $A(k,\omega)$ is the spectral function in Lehmann representation. (See here eq. 5 for a reference)
So I did the Fourier transform
$$iG(k,t) =\frac{i}{2\pi}\int_{-\infty}^{+\infty}  e^{-i\omega t}d\omega\int_{-\infty}^{+\infty} \frac{A(k,\omega')}{\omega -\omega'+i\eta\ sgn(\omega'-\frac{\mu}{\hbar})}d\omega'$$
and I think that I probably need to split the integral into two, one going from $-\infty $ to $\mu/\hbar$ and the other one from $\mu/\hbar$ to $+\infty$ , and then use residue method but can't manage to get to the final result.
 A: Rewriting the labels in your last equation, we have
$$iG(k,t)={i \over 2\pi}\int_{-\infty}^{+\infty}{d\omega A(k,\omega)\int_{-\infty}^{+\infty}{d\omega' {e^{-i\omega't} \over \omega' - (\omega - i\eta \text{sgn}(\omega-{\mu \over \hbar}))}}}$$
Let us recall the Cauchy's integral formula
$$f(a)={1 \over 2\pi i}\int_{C}{{f(z) \over z-a}dz}$$
where $C$ is some closed non-intersecting curve, $a$ is enclosed by $C$, and $f$ is holomorphic. Therefore, to compute the integral over $\omega'$, we have to choose some well-designed curve. Let $C=C_1+C_2$ where $C_1$ is a line starting from $-R$ to $+R$ with $R>0$, and $C_2=\{Re^{-i\theta} | \theta \in [0,\pi]\}$. This curve is a semicircle on the complex plane for visualization. Given $\omega > {\mu \over \hbar}$, the point $\omega-i\eta\text{sgn}(\omega-{\mu \over \hbar})$ is enclosed by $C$. And the integral (there is an additional minus sign on the left hand side since the path along $C$ is clockwise) is
$${-2 \pi i}e^{-i (\omega - i \eta \text{sgn}(\omega - {\mu \over \hbar}))t}=\int_{C_1}{d\omega' {e^{-i\omega't} \over \omega' - (\omega - i\eta \text{sgn}(\omega-{\mu \over \hbar}))}} + \int_{C_2}{d\omega' {e^{-i\omega't} \over \omega' - (\omega - i\eta \text{sgn}(\omega-{\mu \over \hbar}))}}$$
Notice if we make $R \rightarrow +\infty$,
$$\lim_{R \rightarrow +\infty}{\int_{C_1}{d\omega' {e^{-i\omega't} \over \omega' - (\omega - i\eta \text{sgn}(\omega-{\mu \over \hbar}))}}} = \int_{-\infty}^{+\infty}{d\omega' {e^{-i\omega't} \over \omega' - (\omega - i\eta \text{sgn}(\omega-{\mu \over \hbar}))}}$$
In addition, for large enough $R$ and $t>0$,
\begin{align}
\bigg|\int_{C_2}{d\omega' {e^{-i\omega't} \over \omega' - (\omega - i\eta \text{sgn}(\omega-{\mu \over \hbar}))}}\bigg| & \leq 
\int_{0}^{\pi}{{|e^{-i Re^{-i\theta} \ \ t}| \over {1 \over 2}|Re^{-i\theta}t|} |Re^{-i\theta}(-it)|d\theta} \\
& = \int_{0}^{\pi}{2te^{-R\sin{\theta}t} d\theta}
\end{align}
When $R \rightarrow +\infty$, $\int_{0}^{\pi}{2te^{-R\sin{\theta }t} d\theta} \rightarrow 0$. Therefore,
$$\int_{-\infty}^{+\infty}{d\omega' {e^{-i\omega't} \over \omega' - (\omega - i\eta \text{sgn}(\omega-{\mu \over \hbar}))}} = - 2\pi ie^{-i (\omega - i \eta \text{sgn}(\omega - {\mu \over \hbar}))t} = - 2\pi ie^{-i (\omega - i \eta)}$$
What if $\omega < {\mu \over \hbar}$? The process works similarly except the function
$$f(\omega')={e^{-i\omega't} \over \omega' - (\omega - i\eta \text{sgn}(\omega-{\mu \over \hbar}))}$$
is holomorphic on the region enclosed by $C$ (the only pole at $\omega+i\eta$ is in the upper side of the complex plane). Therefore, we have $0$ on the left hand side instead of $-2\pi ie^{-i (\omega - i \eta \text{sgn}(\omega - {\mu \over \hbar}))t}$ at the start of computing the path integral along $C$. Therefore, if $\omega < {\mu \over \hbar}$,
$$\int_{-\infty}^{+\infty}{d\omega' {e^{-i\omega't} \over \omega' - (\omega - i\eta \text{sgn}(\omega-{\mu \over \hbar}))}} = 0$$
This implies
$$iG(k,\omega) = \int_{\mu \over \hbar}^{+\infty}{e^{-i\omega t}A(k,\omega)d\omega}$$
