Inversion of rational transforms by partial fractions

Remarks

The identity follows from $\mathcal{L}\{t^{k-1}\}(s) = (k-1)!/s^k$ and the frequency shift theorem. Example: $F(s) = \frac{s + 3}{(s + 1)(s + 2)} = \frac{2}{s + 1} - \frac{1}{s + 2}$, so $f(t) = 2 e^{-t} - e^{-2t}$; check $f(0) = 1 = \lim_{s \to \infty} s F(s)$. Non-example: $F(s) = s/(s + 1)$ has $\deg P = \deg Q$, so the method does not apply as stated. Indeed $F(s) \to 1$ as $s \to \infty$, while the Laplace transform of a function tends to $0$ at infinity; here $F = 1 - 1/(s+1)$ is the transform of the distribution $\delta - e^{-t}$. For real $P$ and $Q$, the complex roots come in conjugate pairs, and the two terms of a simple pair combine into $e^{\alpha t}(A \cos \beta t + B \sin \beta t)$. The coefficients of simple roots are quickly found with the Heaviside cover-up method.