---
tag: 6X05N89XDP
kind: statement
role: theorem
name: Analyticity of the Laplace transform in its half-plane
msc: [44A10]
uses: [05XT5AVJBK, FTHFSJ30WJ, 0DG0Z72CQD]
about: [0V49JEZW06]
sources:
  - url: https://en.wikipedia.org/wiki/Laplace_transform
---

## Statement

Let $f \colon [0, \infty) \to \C$ be integrable on each interval $[0, T]$ with $T > 0$.
Assume that the [[Laplace integral|05XT5AVJBK]] $\int_0^\infty f(t) e^{-s_0 t}\, dt$ converges, as an improper integral, at some $s_0 \in \C$.
Let $\sigma_c$ be the [[abscissa of convergence|FTHFSJ30WJ]] of $f$.
Then:

1. The integral $F(s) = \int_0^\infty f(t) e^{-st}\, dt$ converges for every $s \in \C$ with $\operatorname{Re} s > \operatorname{Re} s_0$.
2. The function $F$ is [[holomorphic|0DG0Z72CQD]] on the half-plane $\set{s \in \C : \operatorname{Re} s > \sigma_c}$.
3. For every $n \in \N$ and every $s$ with $\operatorname{Re} s > \sigma_c$,
$$F^{(n)}(s) = \int_0^\infty (-t)^n f(t) e^{-st}\, dt,$$
and this integral converges.

## 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 $\operatorname{Re} s \le 0$.

## Explanations

### L1 human

The Laplace transform of a function is a very smooth object: wherever the integral converges (off the boundary line), the transform is a holomorphic function of the complex variable $s$.
The region of convergence is always a right half-plane.
Derivatives of the transform are transforms of $f$ multiplied by powers of $-t$.

### L2 human

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.