Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2007 / Paper 2 / 5

The unilateral Laplace transform is $\bar{f}(p)={\int }_0^{\infty }{e}^{-pt}f(t)dt$ in its region of absolute convergence of a Laplace transform. Under the usual inversion hypotheses, the Bromwich inversion formula isThe vertical Bromwich contour is directed upward, with $c$ inside the transform's convergence half-plane and to the right of all its singularities. At discontinuities the standard symmetric inversion gives the mean of the one-sided values.

For a piecewise continuously differentiable $f$ with an integrable transformed derivative, integration by parts gives the Laplace transform of a derivative:Here the upper endpoint vanishes by the given decay condition, and right continuity at zero supplies the lower endpoint.

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 $c>1$.