---
tag: 105G7PWKS1
kind: counterexample
name: Function with no Laplace transform
msc: [44A10]
uses: [05XT5AVJBK, 6X05N89XDP, TKZAV3KR86]
sources:
  - url: https://en.wikipedia.org/wiki/Laplace_transform
---

## Counterexample

The function $f(t) = e^{t^2}$ on $[0, \infty)$ is continuous, but its [[Laplace integral|05XT5AVJBK]] $\int_0^\infty e^{t^2} e^{-st}\, dt$ converges for no $s \in \C$.
Check: suppose it converged at some $s_0 \in \C$.
By [[the half-plane theorem|6X05N89XDP]] it would then converge at every real $s > \operatorname{Re} s_0$.
But for real $s$ and $t \ge \max(s, 0)$, the integrand satisfies $e^{t^2 - st} = e^{t(t - s)} \ge 1$, so $\int_0^T e^{t^2 - st}\, dt \to \infty$ as $T \to \infty$.
This is a contradiction, so the abscissa of convergence of $f$ is $+\infty$.
The function is not of [[exponential order|TKZAV3KR86]]: $e^{t^2} / e^{at} \to \infty$ for every real $a$.

## Remarks

Exponential order is sufficient, not necessary, for a Laplace transform to exist.
For example, $f(t) = 2t e^{t^2} \cos(e^{t^2})$ is not of exponential order, yet $\int_0^T f(t)\, dt = \sin(e^{T^2}) - \sin 1$ stays bounded, and integration by parts shows that its Laplace integral converges for every $s$ with $\operatorname{Re} s > 0$.
A different obstruction is local: $f(t) = 1/t$ has no Laplace transform because $\int_0^1 dt/t$ diverges, whatever $s$ is.

## Explanations

### L1 human

The Laplace transform weighs $f(t)$ by $e^{-st}$, which shrinks at most exponentially fast.
The function $e^{t^2}$ grows faster than any exponential, so no choice of $s$ can tame it and the integral always blows up.
Fast growth is not the only way to fail: a function such as $1/t$ is too large near $0$.