Exponential order in Laplace Transform:constructing a function such that it is of exponential order but its derivative is not.

I don't understand how to construct such function.I know the definition of a function being of exponential order which states that:

$f(t)$ is of exponential order if there exist constants $c,M>0,T_0>0$ such that $|f(t)|e^{-ct} \le M$ for all $t>T_0$.

Second Question:Any example which is not of exponential order but its Laplace tranform exists?If I consider $e^{t^2}$,it's not of exponential order.But does its Laplace tranform exist??

Any help is appreciated,as always.Thanks in advance.

Solution 1:

For the first question, take something like $$ f(t) = \sin(e^{t^2}). $$ Then $f$ is bounded (so in particular of exponential growth), but $$ f'(t) = 2t e^{t^2} \cos( e^{t^2} ) $$ is not of exponential growth. (Look at the values for $z=2\pi k$, $k \in \mathbb{Z}$.)

For the second question, see Jyrki's answer.

Solution 2:

Answering the second question with a "standard" example of a function that is not of exponential order, but does have a Laplace transform in the region $\Re s>0$.

Build a function out of spiky triangles $$ \Delta_{H,A}(x)= \begin{cases} H-\frac{H^2}{A}|x|,&\ \text{if $|x|\le A/H$, and}\\ 0,&\ \text{otherwise.} \end{cases} $$ Here $H>0,A>0$ are parameters. You see that $\Delta_{H,A}(x)$ reaches a maximum value $H$ at $x=0$, and $$ \int_{-\infty}^\infty \Delta_{H,A}(x)\,dx=A, $$ because the graph is a triangle of height $H$ and width $2A/H$. Furthermore, $\Delta_{H,A}(x)$ is continuous everywhere.

Because the series $\sum_n\dfrac1{n^2}$ converges, the function $$ f(x):=\sum_{n=1}^\infty \Delta_{e^{n^2},1/n^2}(x-n) $$ will work. We have $f(n)=e^{n^2}$ for all positive integers $n$, so $f(x)$ is not of exponential order. But $$ \int_0^\infty f(x)\,dx=\sum_{n=1}^\infty\dfrac1{n^2}=\frac{\pi^2}6. $$

Consequently, by domination, the integral $$ \int_0^\infty e^{-st}f(t)\,dt $$ converges whenever $\Re s\ge0$.