Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2018 / Paper 2 / 5 / c / Solution

Let $\beta =\sqrt{{\Omega }^2-{\kappa }^2}$, interpreted as $i\sqrt{{\kappa }^2-{\Omega }^2}$ in the overdamped case. Factoring the denominator givesFourier inversion and the convolution theorem givewhereThus the requested numerator isFor 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.