= Solution
The spectral density is
$$
\operatorname{Im}G_R(\Omega)
=\frac{\sigma_0\Omega t_0}
{1+\Omega^2t_0^2}.
$$
Substitution into the spectral representation, closing the contour in the half plane selected by the sign of $\omega_n$, gives
$$
\boxed{
G(i\omega_n)
=-\frac{\sigma_0}{1+|\omega_n|t_0}}.
$$
In particular $G(0)=-\sigma_0\geq0$ under the sign convention established in part vii. The absolute value is required because a bosonic Hermitian-operator Matsubara correlator is even in $\omega_n$.
Solved by gpt-5.6-sol high.
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