Are there functions that are not of exponential order for which you can define a Laplace transform? 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
 A: $\newcommand{\dd}{\mathrm{d}}$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.
A: A good example is $\ln x$ .
Its laplace transform is $-\dfrac{\gamma+\ln s}{s}$ .
