How to construct a function $f(x)$ such that $f(x)e^{-px}$ wouldn't tend to $0$ as $x$ tends to infinity How to construct a function $f(x)$ such that $f(x)e^{-px}$ wouldn't tend to $0$ as $x$ tends to infinity?
This question is motivated when studying Laplace Transform when I encountered the following result

Suppose $f$ and $f'$ both have Laplace Transform on some half plane $\Re(p)>p_0$, provided that $f(x)e^{-px}\to 0$ as $x\to \infty$ for $p$ such that $\Re(p)>p_0$. Then we have $\hat {f'}(p)=p\hat {f}(p)-f(0)$.

Where the hat notation is meant to be the Laplace Transform of function.
Now, I am just quite curious if there exists a function $f$ such that the condition of "provided that $f(x)e^{-px}\to 0$ as $x\to \infty$ for $p$ such that $\Re(p)>p_0$" would fail. Surely we can just choose $p_0$ to be very large in its real part unless we can construct something that can grow more rapidly than exponentials?
Many thanks in advance!
 A: How about $f(x) = \exp(x^2)$.
EDIT: Note that
$$ \dfrac{d}{dx} \left(f(x) e^{-px}\right) = f'(x) e^{-px} - p f(x) e^{-px} $$
so $$\int_0^b f'(x) e^{-px}\; dx - p \int_0^b f(x) e^{-px}\; dx = f(b) e^{-pb} - f(0)$$
If $f(x)$ and $f'(x)$ both have Laplace transforms at $p$, the left side must go to a finite limit as $b \to \infty$, so the right side must also, and in particular $f(x) e^{-px}$ must be bounded, and for $\text{Re}(c) > \text{Re}(p)$ we must have $f(x) e^{-cx} \to 0$ as $x \to \infty$.  So it is not possible for both $f$ and $f'$ to have a Laplace transform for $\text{Re}(p) > p_0$ without
$|f(x) e^{-px}| \to 0$ as $x \to \infty$ for $\text{Re}(p) > p_0$.
A: $f(x)=\Gamma(x)$ works as well.
A: A way to go is to find a function that is guaranteed to grow faster than $-px$ descends, regardless of $p$.  For instance, $x^2$, $x^3$, ..., $\mathrm{e}^x$, $x \ln x$, and so on.  So each of the following is a potential $f(x)$:  \begin{align*}
&\mathrm{e}^{x^2}  \text{,}  \\
&\mathrm{e}^{x^3}  \text{,}  \\
&\vdots  \text{,}  \\
&\mathrm{e}^{\mathrm{e}^x}  \text{, and }  \\
&\mathrm{e}^{x \ln x} = x^x  \text{.}
\end{align*}
And of course, once one sees these, ideas for even more rapidly growing functions should spring to mind.
