Analyticity of the Laplace transform in its half-plane

Remarks

For part 1, let $\varphi(t) = \int_0^t f(u) e^{-s_0 u}\, du$; it is continuous and bounded on $[0, \infty)$ because it has a finite limit. Integration by parts gives $\int_0^T f(t) e^{-st}\, dt = \varphi(T) e^{-(s - s_0)T} + (s - s_0) \int_0^T \varphi(t) e^{-(s - s_0)t}\, dt$, and both terms converge when $\operatorname{Re}(s - s_0) > 0$. The same formula shows that the truncated integrals converge locally uniformly on the half-plane $\operatorname{Re} s > \operatorname{Re} s_0$. By part 1 the integral converges at real points $\sigma$ arbitrarily close to $\sigma_c$ from the right, so these half-planes cover $\operatorname{Re} s > \sigma_c$. A locally uniform limit of holomorphic functions is holomorphic, which gives part 2. Part 3 follows by differentiating under the integral sign on $[0, T]$ and passing to the limit. Example: for $f = 1$, $F(s) = 1/s$ is holomorphic on $\operatorname{Re} s > 0$, and $F'(s) = -1/s^2 = \int_0^\infty (-t) e^{-st}\, dt$. The half-plane is in general not the largest domain of holomorphy: $1/s$ extends to $\C \setminus \set{0}$, but the integral diverges for $\operatorname{Re} s \le 0$.

Going deeper

For a student

This theorem is why complex analysis applies to Laplace transforms. Residues, contour shifts and analytic continuation are used to invert transforms, as in the Bromwich integral, and to read off the asymptotic behaviour of $f$ from the singularities of $F$. Part 3 is the rule "multiplication by $-t$ corresponds to differentiation in $s$". It is the analogue of the fact that a power series converges on a disc and is holomorphic inside it; Dirichlet series obey the same pattern.