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

Question

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.

Answer 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.

Answer 2

EDIT: This proves the wrong thing; ignore this answer.

If $f'$ is of exponential order, so is $f$. Consider the integral $$\int_a^x f'(t) dt$$ From the fundamental theorem of calculus, this differs from $f$ by at most a constant. If $|f(x)| ≤ |g(x)|$ for sufficiently large $x$, $\int_a^x |f(t)| dt ≤ \int_a^x |g(t)| dt$ for sufficiently large $x$ and $a$. $$ |f(x)| ≤ |\int_a^x f'(t) dt| ≤ \int_a^x |f'(t)| dt ≤ \int_a^x |Me^{cx}| dt $$ since $f(x) ≤ Me^{cx}$. $$\int_a^x |Me^{cx}| dt = |M| \int_a^x e^{cx} dt = |M| (\frac{e^{cx}}{c} -\frac{e^a}{c}) ≤ Ke^{cx}$$ for a constant K. Thus $f ≤ K e^{cx}$ and is by definition of exponential order.

Thus, one cannot find $f$ not of exponential so that $f'$ is of exponential order, because if $f'$ is of exponential order then $f$ must be as well, as shown above.

As for your second question, the integral defining a Laplace transform does not converge for a function not of exponential order, so it does not make sense to find a Laplace transform of such a function.

Answer 3

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 $\Re s\ge0$.