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Set $a_n=1-1/(2n)$, $b_n=1+1/(2n)$. The finite pulse has Laplace transform ${\bar{S}}_n(p)=n(e^{-p{a}_n}-e^{-p{b}_n})/p$. Since $f(0)=0$, transforming the equation givesThe partial fraction decomposition $1/[p(p-1)]=1/(p-1)-1/p$ has inverse $e^t-1$. By the Laplace transform time-shift rule,This finite-pulse response of a first-order linear equation gives the requested outer regions:For completeness, while the pulse is active the solution is $n(e^{t-a_n}-1)$. It is continuous at both switching times. After the forcing stops it grows as $e^t$, because the homogeneous equation is $f^{\prime }=f$. The apparent pole at $p=0$ in the full transform is removable; the Bromwich contour may be taken with $c>1$.