Propagator in Wave Mechanics Laplace-Fourier transform In my Modern quantum mechanics, J. J. Sakurai p.119-120, when considering the integral of the propagator $K$ in whole space, he gets:
$$G(t)= \int d^3 x' K(\textbf{x'},t;\textbf{x'},0)
= \sum_a \exp \left(\frac{-iE_{a^{'}}t }{\hbar} \right).$$
Then he considers the Laplace-Fourier transform of G(t):
\begin{equation} \label{eq1}
\begin{split}
\tilde{G}(E) = & -i\int^{\infty}_0 dt G(t)\exp(iEt/\hbar)  \\
 & = -i\int^{\infty}_0dt\sum_{a^{'}} \exp(-iE_{a^{'}}t/\hbar) \exp(iEt/\hbar)/ \hbar.
\end{split}
\end{equation}
But why does he say that this integrand oscillates indefinitely, and the result can be changed into a definite one by making the change $E \rightarrow E + i \varepsilon$ to obtain:
$$\tilde{G}(E)=\sum_{a^{'}}\frac{1}{E-E_{a'}}~?$$
 A: This is known as the $\epsilon$ prescription. We have:
\begin{equation}
\begin{aligned}
\tilde{G}(E) =& - \frac{i}{\hbar} \sum_{a^{'}} \int^{\infty}_0 dt \exp\left [ \frac{it (E - E_{a'})}{\hbar} \right]\\
             =& - \frac{i}{\hbar} \sum_{a^{'}} \frac{\hbar}{i} \left( \frac{1}{E - E_{a'}} \right) \exp\left [ \frac{it (E - E_{a'})}{\hbar} \right] \Bigg|_0^{\infty}
\end{aligned}
\end{equation}
Now, the problem is that we cannot evaluate the upper limit of the integral because
$$\lim_{t \to \infty}\exp (itx)$$ 
does not exist for real $x$. The limit does not converge to a definite value; it keeps oscillating instead.
The trick to force convergence upon the integral is to add to $E$ an infinitesimal constant $\epsilon > 0$:
$$ E \rightarrow E + i \epsilon$$
Now, the upper limit of the integral goes to zero because of exponential suppression:
\begin{equation}
\lim_{t \to \infty}- \frac{i}{\hbar} \sum_{a^{'}} \frac{\hbar}{i} \left( \frac{1}{(E + i \epsilon) - E_{a'}} \right) \exp\left [ \frac{it (E - E_{a'})}{\hbar} \right] \exp \left[ \frac{-\epsilon t}{\hbar} \right] = 0
\end{equation}
while the lower limit (with minus sign included) gives:
\begin{equation}
\lim_{t \to 0} \frac{i}{\hbar} \sum_{a^{'}} \frac{\hbar}{i} \left( \frac{1}{(E + i \epsilon) - E_{a'}} \right) \exp\left [ \frac{it (E - E_{a'})}{\hbar} \right] \exp \left[ \frac{-\epsilon t}{\hbar} \right] = \sum_{a^{'}} \left( \frac{1}{(E + i \epsilon) - E_{a'}} \right)
\end{equation}
The $\epsilon$ prescription was used for convergence of the integral. The integral has now been done so we can safely take $\epsilon \to 0$ in the end to get:
$$\tilde{G}(E) =  \sum_{a^{'}} \frac{1}{E - E_{a'}}$$
Analytic continuations such as this $\epsilon$ prescription are used in many places in physics. One of the key motivations is convergence.
