= Solution
Let $\beta=\sqrt{\Omega^2-\kappa^2}$, interpreted as $i\sqrt{\kappa^2-\Omega^2}$ in the <overdamped> case. Factoring the denominator gives
$$
\Omega^2-\omega^2+2i\kappa\omega=-(\omega-\omega_-)(\omega-\omega_+),\qquad
\boxed{\omega_\pm=i\kappa\pm\beta}.
$$
<Fourier inversion> and the <convolution theorem> give
$$
\boxed{y(t)=\int_{-\infty}^{\infty}G(t-\xi)f(\xi)\,d\xi},
$$
where
$$
G(t)=-\frac1{2\pi}\int_{-\infty}^{\infty}
\frac{e^{i\omega t}}{(\omega-\omega_-)(\omega-\omega_+)}\,d\omega.
$$
Thus the requested numerator is
$$
\boxed{s(\omega,t)=-\frac1{2\pi}e^{i\omega t}}.
$$
For all positive $\kappa,\Omega$, both <poles> lie in the upper half-plane. In the <overdamped> case their imaginary parts are $\kappa\pm\sqrt{\kappa^2-\Omega^2}$, both positive. At equality they merge into a double <pole>.
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