= Solution
For $t>0$, use the <Bromwich inversion formula> with a vertical contour $\operatorname{Re}s=c>0$:
$$
f(t)=\frac1{2\pi i}\int_{c-i\infty}^{c+i\infty}e^{st}\frac{s}{s^2+a^2}\,ds.
$$
Close the contour to the left, where the exponential decays. The large-arc contribution tends to zero by the usual exponential arc estimate, and the two simple <poles> are at $s=ia$ and $s=-ia$. The <residues> of the complete integrand are $e^{iat}/2$ and $e^{-iat}/2$. The <residue theorem> therefore gives
$$
\boxed{\mathcal L^{-1}\!\left[\frac{s}{s^2+a^2}\right](t)=\cos(at)\quad(t>0).}
$$
The right limit at zero is one. The sign of nonzero real $a$ has no effect on the answer.
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