Are there functions that are not of exponential order for which you can define a Laplace transform?

Question

I'am in a course of Introduction to Linear Differential Equations and teacher made us this question in class.

we work in $\mathbb{R}$, and any help to answer this is welcome

Answer 1

$\newcommand{\dd}{\mathrm{d}}$A prime example that you find in many textbooks is

$$ f(t) = 2te^{t^2}\cos(e^{t^2}) $$

We may observe that $f(t)=\frac{\dd}{\dd t}\sin(e^{t^2})$, therefore,

\begin{align} (\mathscr{L}f)(s) {}={}& \int_0^\infty e^{-s\tau} \frac{\dd}{\dd \tau}\sin(e^{\tau^2}) \dd \tau \\ {}={}& \left.e^{-s\tau}\sin(e^{\tau^2})\right|_0^\infty - \int_0^\infty (-s)e^{-s\tau}\sin(e^{\tau^2})\dd\tau \\ {}={}& -\sin 1 - \int_0^\infty (-s)e^{-s\tau}\sin(e^{\tau^2})\dd\tau \end{align}

The last integral converges because

\begin{align} \int_0^\infty (-s)e^{-s\tau}\sin(e^{\tau^2})\dd\tau {}\leq{}& \left|\int_0^\infty (-s)e^{-s\tau}\sin(e^{\tau^2})\dd\tau\right| \\ {}={}&\int_0^\infty s e^{-s\tau}|\sin(e^{\tau^2})|\dd\tau \\ {}\leq{}& \int_0^\infty s e^{-s\tau}\dd\tau {}={} \frac{1}{s}, \end{align}

so the integral is defined for $s\in \mathbb{C}$ with $\Re(s) > 0$.

There are several other examples such as the one mentioned by @SimonS in a comment (see here).

Yet another example is the following function

$$ f(t) = \begin{cases} e^{n^2}, &\text{if } t\in [n, n+e^{-n^2}) \\ 0,&\text{otherwise} \end{cases} $$

It is not difficult to show that the Laplace transform of $f$ exists.

Answer 2

A good example is $\ln x$ .

Its laplace transform is $-\dfrac{\gamma+\ln s}{s}$ .