Construct an example of a continuous function $y=f(x)$ defined on $[0,\infty)$, such that it is absolutely integrable, i.e.,
$\int^\infty_0 |f(x)|dx<\infty$,
but not of exponential order.
What about a function of exponential order?
For reference "A real function $y=f(x)$ on $[0,\infty)$ or $(0,\infty)$ is defined to be of exponential order $u $ if there are constants $u\geq 0$, $M\geq 0$, and $A\geq 0$ , such that $|f(x)|\leq Me^{ux},$ for all $x\in[A,\infty)$
It seems to me that any old function that is continuous ought to be of exponential order so I'm not sure how to proceed.
We can build such a function as follows: As we want $f$ not to be of exponential order, we will choose $f$ such that $$ f(n) = n\exp(nx), \qquad n \in \mathbb N $$ Then $f$ is not of exponential order. To make $f$ absolutely integrable, define $a_n := (1+n)^{-2}\exp(-2nx)$, then $f(n)a_n \to 0$ and define $f$ by $$ f(x) := \begin{cases} n\exp(nx)\frac 1{a_n}(x-n+a_n) & x \in [n - a_n, n]\\ -n\exp(nx)\frac 1{a_n}(x-n-a_n) & x \in [n, n+a_n]\\ 0 & \text{otherwise} \end{cases} $$ Then $f$ is continous, and we have $$ \int_0^K f(x)\, dx \le \sum_{k=0}^{\lceil K\rceil} k\exp(kx) \cdot \frac 12 \cdot 2a_k \le \sum_{k=0}^{\lceil K \rceil} (1+k)^{-1}\exp(-kx) \le \sum_{k=0}^\infty (1+k)^{-1}\exp(-kx) < \infty. $$