Integration in the transform variable

Remarks

Write $\int_s^\infty F(\sigma) \, d\sigma = \int_s^\infty \int_0^\infty f(t) e^{-\sigma t} \, dt \, d\sigma$ and swap the integrals with Fubini's theorem. The swap is allowed because $\int_0^\infty \abs{f(t)} e^{-s t} / t \, dt < \infty$ under the hypotheses. The inner integral is $\int_s^\infty e^{-\sigma t} \, d\sigma = e^{-s t} / t$. Example: with $f(t) = \sin t$, one gets $\mathcal{L}\{\sin t / t\}(s) = \int_s^\infty d\sigma / (\sigma^2 + 1) = \pi/2 - \arctan s$. The hypothesis at $0$ is needed: for $f = 1$, the function $1/t$ has no Laplace transform, and $\int_s^\infty d\sigma / \sigma$ diverges.

Going deeper

For a student

The rule gives transforms of functions such as $\sin t / t$ and $(1 - e^{-t})/t$ that are hard to compute directly. Letting $s \to 0^+$ in the example, with a limit argument, recovers the Dirichlet integral $\int_0^\infty \sin t / t \, dt = \pi / 2$. Compare differentiation in the transform variable.