= Solution
The <convolution theorem> gives the causal response
$$
\phi(t)=\frac{I_0}q\int_0^{\min(t,1)}e^{-p(t-\tau)}\sin\bigl(q(t-\tau)\bigr)\,d\tau\quad(t>0).
$$
Define the integrated <impulse response>, for $s\geq0$, by
$$
F(s)=\frac{1-e^{-ps}\cos(qs)-(p/q)e^{-ps}\sin(qs)}{p^2+q^2}.
$$
Direct differentiation gives $F'(s)=e^{-ps}\sin(qs)/q$ and $F(0)=F'(0)=0$. Thus
$$
\boxed{\phi(t)=\begin{cases}
0,&t\leq0,\\
I_0F(t),&0<t\leq1,\\
I_0\bigl[F(t)-F(t-1)\bigr],&t>1.
\end{cases}}
$$
In particular, after switch-off the answer is
$$
\boxed{\phi(t)=\frac{I_0}{p^2+q^2}\left\{
e^{-p(t-1)}\left[\cos q(t-1)+\frac pq\sin q(t-1)\right]
-e^{-pt}\left[\cos qt+\frac pq\sin qt\right]\right\},\quad t>1.}
$$
The response and its first <derivative> are continuous at both switching times, and the post-pulse motion is a freely decaying <damped harmonic oscillator>. The expression described as valid for $t<1$ in the paper is understood for $0<t<1$; before the pulse the causal output is zero.
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