The <Bromwich inversion formula> gives
$$
f(t)=\frac1{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}\frac{e^{st}}{s^2}\,ds,
\qquad \gamma>0.
$$
For $t>0$, close the <Bromwich contour> in the left half-plane. The only enclosed singularity is the <double pole> at $s=0$, whose <residue> is
$$
\operatorname*{Res}_{s=0}\frac{e^{st}}{s^2}
=\left.\frac d{ds}e^{st}\right|_{s=0}=t.
$$
The <residue theorem> therefore yields
$$
\boxed{\mathcal L^{-1}\{s^{-2}\}(t)=t}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / ib / Paper 4 / 14D / a / Solution

For $f(t)=\sin\omega t$,
$$
\widehat f(p)=\frac{\omega}{p^2+\omega^2},
\qquad
\widehat y(x,p)=\frac{\omega e^{-px/c}}{p^2+\omega^2}.
$$
The <Bromwich inversion formula> gives
$$
y(x,t)=\frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}
\frac{\omega e^{p(t-x/c)}}{p^2+\omega^2}\,dp.
$$
When $t>x/c$, close the contour to the left. The residues at $p=\pm i\omega$ sum to $\sin(\omega(t-x/c))$. When $t<x/c$, close it to the right, where there are no poles, and the integral vanishes. Hence
$$
\boxed{y(x,t)=
\begin{cases}
0,&t<x/c,\\
\sin\!\bigl(\omega(t-x/c)\bigr),&t\ge x/c.
\end{cases}}
$$
Equivalently, $y(x,t)=\Theta(t-x/c)\sin(\omega(t-x/c))$ in terms of the <Heaviside step function>. The boundary oscillation travels to the right at <wave speed> $c$ without reflection or dispersion; a point at $x$ begins moving after the propagation delay $x/c$.

Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 2 / 13E / ii / Solution

{c}

Under suitable regularity and growth conditions, a function with <Laplace transform> $F$ is recovered by the inverse contour integral
$$
f(t)=\frac1{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}e^{st}F(s)\,ds,
$$
where the vertical line lies to the right of every singularity of $F$.

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 Incoming links: Bromwich inversion formula