= Solution
For $0<k^2<4/\gamma^2$, define
$$
a_k=\frac{\gamma k^2}{2},\qquad \Omega_k=\sqrt{k^2-\frac{\gamma^2k^4}{4}}>0.
$$
The simple poles are $\omega_\pm=\pm\Omega_k-ia_k$. In the inverse transform $\widetilde G(k,t)=\int e^{-i\omega t}\widehat G\,d\omega/(2\pi)$, close the contour below for $t>0$. Its clockwise orientation gives $-i$ times the sum of residues. That sum is
$$
\frac{e^{-i\omega_+t}-e^{-i\omega_-t}}{2\Omega_k}
=-i e^{-a_kt}\frac{\sin(\Omega_kt)}{\Omega_k}.
$$
For $t<0$, close above; there are no poles. The <residue theorem> therefore gives
$$
\boxed{\widetilde G_R(k,t)=-\Theta(t)e^{-\gamma k^2t/2}\frac{\sin(\Omega_kt)}{\Omega_k}.}
$$
Here $\Theta$ is the <Heaviside step function>. At $k=0$ the continuous retarded limit is $-t\Theta(t)$. At critical damping, $\Omega_k\to0$ likewise replaces $\sin(\Omega_kt)/\Omega_k$ by $t$; beyond the stated oscillatory range an imaginary $\Omega_k$ converts it to a hyperbolic sine. These continuations retain decaying temporal exponents.
For each fixed $k$, taking $\gamma\to0^+$ gives
$$
\widetilde G_R(k,t)\longrightarrow-\Theta(t)\frac{\sin(|k|t)}{|k|}.
$$
For $t>0$, invert in space and set $s=kt$:
$$
G_R(x,t)=-\frac{\Theta(t)}{2\pi}\int_{-\infty}^{\infty}\frac{\sin(kt)}k e^{-ikx}\,dk
=-\frac{\Theta(t)}{2\pi}\int_{-\infty}^{\infty}\frac{\sin s}s e^{-isx/t}\,ds.
$$
The correctly normalized <sinc function> transform is
$$
\frac1\pi\int_{-\infty}^{\infty}\frac{\sin s}s e^{-isu}\,ds
=\begin{cases}1,&|u|<1,\\0,&|u|>1.\end{cases}
$$
For instance, $\sin s/s=\frac12\int_{-1}^{1}e^{isy}\,dy$, and <Fourier inversion> immediately gives the factor $\pi$. The PDF's prefactor $\sqrt{2/\pi}$ is inconsistent with its top hat of unit height; it would give height $\sqrt{2\pi}$. Using the normalized identity yields
$$
\boxed{G_R(x,t)=-\frac12\Theta(t)\Theta(t-|x|)
=-\frac12\Theta(t-|x|),}
$$
as a <distribution>. The values assigned precisely at a wave front do not change that distribution. This is the negative of the standard <one-dimensional retarded wave Green function>, because the PDF's differential operator is $-\partial_t^2+\partial_x^2$.
The sign and initial data can be checked without any contour calculation. Integrating the spatial-Fourier equation through the impulse gives
$$
\widetilde G_R(k,0^+)=0,\qquad \partial_t\widetilde G_R(k,0^+)=-1.
$$
In real space these are $G_R(x,0^+)=0$ and $\partial_tG_R(x,0^+)=-\delta(x)$. In particular,
$$
\lim_{t\downarrow0}\int_{\mathbb R}\frac{G_R(x,t)}t f(x)\,dx=-f(0)
$$
for a smooth test function $f$, confirming the negative delta-function initial velocity. Before the impulse the displacement and velocity are zero. The disturbance fills the forward <causal cone for the constant-speed wave equation>, whose fronts move at unit speed, consistent with $\rho=\kappa=1$. The final question's printed $\rho=k$ is interpreted as $\rho=\kappa$, referring to the elastic constant, not to the Fourier wavevector. No initial conditions were otherwise specified, so the causal choice has been made explicitly.
\Image[/past-exam-of-the-theoretical-physics-i-course-of-the-university-of-cambridge/2018-q4-retarded-rod-response.png]
{title=Retarded rod response: damping of a Fourier mode and the undamped causal cone}
{description=Original calculated Fourier-mode responses for $k=1$, and the zero-damping spacetime Green function for the operator $-\partial_t^2+\partial_x^2$. Its value is $-1/2$ inside the forward cone and zero outside; the negative sign enforces the stated positive delta-function source.}
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