---
tag: 484GKH3ZS8
kind: definition
name: Abscissa of absolute convergence
msc: [44A10]
defines: YCSH2WK69M
uses: [05XT5AVJBK, FTHFSJ30WJ, XVEWR93EMV]
sources:
  - url: https://en.wikipedia.org/wiki/Laplace_transform
---

## Definition

Let $f \colon [0, \infty) \to \C$ be integrable on each interval $[0, T]$ with $T > 0$.
The *abscissa of absolute convergence* of the [[Laplace transform|05XT5AVJBK]] of $f$ is
$$\sigma_a = \inf \set{ \sigma \in \R : \int_0^\infty \abs{f(t)} e^{-\sigma t}\, dt < \infty } \in [-\infty, +\infty],$$
with $\inf \emptyset = +\infty$.

## Remarks

For $\operatorname{Re} s > \sigma_a$ the Laplace integral converges as a [[Lebesgue integral|XVEWR93EMV]], and for $\operatorname{Re} s < \sigma_a$ it does not converge absolutely.
Always $\sigma_c \le \sigma_a$, where $\sigma_c$ is the [[abscissa of convergence|FTHFSJ30WJ]].
Example: for $f(t) = e^{at}$ with $a \in \R$, $\abs{f} = f$, so $\sigma_a = \sigma_c = a$.
Example with $\sigma_c < \sigma_a$: let $f(t) = e^t \sin(e^t)$.
The substitution $u = e^t$ turns $\int_0^\infty f(t) e^{-\sigma t}\, dt$ into $\int_1^\infty \sin(u)\, u^{-\sigma}\, du$.
This converges for $\sigma > 0$ by the Dirichlet test and diverges for $\sigma \le 0$, so $\sigma_c = 0$.
It converges absolutely exactly for $\sigma > 1$, since $\int_1^\infty \abs{\sin u}\, u^{-\sigma}\, du < \infty$ if and only if $\sigma > 1$; so $\sigma_a = 1$.
For a function of exponential order $a$ that is integrable on bounded intervals, $\sigma_a \le a$.