Applying Green's function for one dimensional wave equation The Green's function of the one dimensional wave equation
$$
(\partial_t^2-\partial_z^2)\phi=0
$$
fulfills
$$
(\partial_t^2-\partial_z^2)G(z,t)=\delta(z)\delta(t)
$$
I calculated that its retarded part is given by:
$$
G_+(z,t)=\Theta(t)\Theta(t-|z|).
$$
In Wikipedia I find a very similar expression without the first $\Theta(t)$. I think this has to do with the fact that in Wikipedia the full Green's function is given and not the retarded part, right?
I now want to apply the retarded Green's function to solve the wave equation with the source
$$
\begin{cases}
(\partial_t^2-\partial_z^2)\phi=\kappa(z,t)\\
\kappa(z,t)=e^{-i\omega t+i \omega z}\Theta_L(z)
\end{cases}
$$
where
$$
\Theta_L(z)=
\begin{cases}
1&0<z<L\\
0 &\text{otherwise}
\end{cases}
$$
The formal solution is now given by:
$$
\begin{split}
\phi(z,t)&=\int\limits_{-\infty}^{\infty}dt'\int\limits_{-\infty}^{\infty}dz'\Theta(t-t')\Theta\big(t-t'-|z-z'|\big)\kappa(z',t')\\
&=\int\limits_{-\infty}^{t}dt'\int\limits_{-\infty}^{\infty}dz'\Theta\big(t-t'-|z-z'|\big)e^{-i\omega t'+i \omega z'}\Theta_L(z')\\
&=\int\limits_{-\infty}^{t}dt'\int\limits_{0}^{L}dz'\Theta\big(t-t'-|z-z'|\big)e^{-i\omega t'+i \omega z'}\\
&=\int\limits_{0}^{L}dz'e^{i \omega z'}\int\limits_{-\infty}^{t-|z-z'|}dt'e^{-i\omega t'}
\end{split}
$$
I have problems with evaluating the $t'$ integral. I get
$$
\int\limits_{-\infty}^{t-|z-z'|}dt'e^{-i\omega t'}=\frac{1}{-i\omega}\left(e^{-i\omega(t-|z-z'|)}-e^{-i\omega(-\infty)}\right)
$$
where the last term is clearly not defined!
What am I doing wrong? Is it something about the boundary conditions which I should impose?
Many thanks in advance!
 A: You haven't done anything wrong, rather your question is actually ill-posed. Notice that your problem does not have any boundary conditions so it can't have an unambiguous answer (what happens if you add a constant to $\phi$). If your source term cut off at some point in the past, say $\kappa(z, t) = e^{- i \omega t + i \omega z} \Theta_L(z) \Theta(t - t_0)$ then your integral would look like,
$$ \int_{t_0}^{t - |z - z'|} \mathrm{d}t' e^{- i \omega t'} = \frac{i}{\omega} \left( e^{- i \omega (t - |z - z'|)} - e^{- i \omega t_0} \right) $$
which is perfectly well behaved. This corresponds to implicitly imposing boundary conditions $\phi(z, t) = 0$ in the past for $t 
\le t_0$ (I encourage you to think about how boundary contitions are incorporated into the general solution throught the Green's function). However, in your case you are taking $t_0 \to - \infty$ but we can't impose this sort of boundary condition at $- \infty$ since $\phi(z, -\infty) = 0$ isn't meaningful! This correspond to the fact that, for the solutions $\phi_{t_0}$ computed for a cutoff set at $t_0$, the limit $\lim\limits_{t_0 \to - \infty} \phi_{t_0}(z, t)$ does not exist. This is exactly the same ill-defined limit that you noticed when you couldn't compute the limit in the improper integral,
$$ \int_{-\infty}^{t - |z - z'|} \mathrm{d}t' e^{- i \omega t'} = \lim_{t_0 \to - \infty}  \int_{t_0}^{t - |z - z'|} \mathrm{d}t' e^{- i \omega t'} = \lim_{t_0 \to - \infty} \frac{i}{\omega} \left( e^{- i \omega (t - |z - z'|)} - e^{- i \omega t_0} \right) $$
Now what would make your problem well-posed while retaininig the same source term $\kappa(z, t)$. Well, suppose at some $t_0$ we know the value of $\phi(z, t_0)$ and $\partial_t \phi(z, t) |_{t_0}$. Then we need to modify our Green's function $G(z,z',t,t')$ to take this information into account. Where $G$ satisfies,
$$ (\partial_t^2 - \partial_x^2) G(z, z', t, t') = \delta(t - t') \delta(x - x') $$
and the Green's function must be a function of $t$ and $t'$ (not of the form $G(z - z', t - t')$ as you had before since we require that $G(z, z', t_0, t')$ satisfy the boundary contition for all $t'$). Suppose we know $\phi(z, t_0) = 0$ and $\partial_t \phi(z, t)|_{t_0} = 0$. You will find something piecewise like,
$$ G(z, z', t, t') = 
\begin{cases}
\Theta(t - t')\Theta(t - t' - |z - z'|)  & t' > t_0
\\
\Theta(t' - t)\Theta(t' - t - |z - z'|) & t' < t_0
\end{cases}  $$
Notice the Green's function is advanced before $t_0$ and retarded afterwards conforming to the sort of causality we expect for the propagation of information about the solution at $t_0$ to information about the solution at all $t$. Now our solution takes the form,
$$ \phi(z, t) = \int_{-\infty}^{\infty} \mathrm{d}{z'} \int_{-\infty}^{\infty} \mathrm{d}{t'} G(z, z', t, t') \kappa(z', t') $$
which becomes
$$ \phi(z, t) = \int_{0}^{L} \mathrm{d}{z'}
\begin{cases}
\int_{t_0}^{t - |z - z'|} \mathrm{d}{t'} e^{- i \omega t' + i \omega z'} & t > t_0
\\
\int_{t + |z - z'|}^{t_0} \mathrm{d}{t'} e^{- i \omega t' + i \omega z'} & t < t_0
\end{cases}
\quad =
\frac{i}{\omega} \int_{0}^{L} \mathrm{d}{z'} e^{i \omega z'}
\begin{cases}
\Theta(t - t_0 - |z - z'|)\left( e^{- i \omega (t - |z - z'|)} - e^{- i \omega t_0} \right) & t > t_0
\\
\Theta(t_0 - t - |z - z'|)\left(-e^{- i \omega (t + |z - z'|)} + e^{- i \omega t_0} \right)  & t < t_0
\end{cases} $$
and thus,
$$ \phi(z, t) = \frac{i}{\omega} \mathrm{sign}(t - t_0) \left( \int_{0}^{L} \mathrm{d}{z'} \Theta(|t - t_0| - |z - z'|) \left( e^{i \omega z'} e^{\mathrm{sign}(t - t_0) i \omega | z - z'|} e^{- i \omega t} - e^{- i \omega t_0} \right)  \right) 
$$
If we replace our source with a delta function at the origin $\kappa(z, t) = e^{- i \omega t} \delta(z)$ it is easier to see what is going on. In this case we get,
$$ \phi(z, t) = \frac{i}{\omega} \mathrm{sign}(t - t_0) \Theta(|t - t_0| - |z|)  \left( e^{\mathrm{sign}(t - t_0) i \omega | z |} e^{- i \omega t} - e^{- i \omega t_0} \right)   $$
which after $t_0$ is a solution with outgoing waves from the origin and before $t_0$ is a solution with incoming waves exactly absorbed at the origin such that at $t = t_0$ all the waves cancel.
