Solution
= Solution
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
$$
\boxed{\mathcal L^{-1}\{s^{-2}\}(t)=t}.
$$