Is $f\left(t\right)=\frac{1}{t^2+1}$ of exponential order? I'm learning Laplace Transforms and one of the questions I'm working on is the following:
$$\text{Is}\:\:f\left(t\right)=\frac{1}{t^2+1}\:\:\:\text{of exponential order?}$$
If so or if not, how do I show that? Obviously (or at least my guess) is that it would perhaps involve solving the Laplace Transform, resulting in
$$F\left(s\right)= \lim_{N\rightarrow\infty}\int_0^N \frac{e^{-st}}{t^2+1}\: dt=$$
but I don't know how to evaluate the integral...
 A: Usually the definition is that $f$ is of exponential order if there exist constants $C, a > 0$ such that
$$|f(t)| \leq Ce^{at} \ \ \ \ \hbox{ for all } \ t > 0$$
I suggest it's not hard to find such constants $C$ and $a$ here, as in fact $f$ is bounded: $$ \left|\frac{1}{t^2 +1}\right| \leq 1 \ \ \ \ \hbox{ for all } t$$
A: The easiest way to show it, is using the following definition:

A function $f(t)$ is of exponential order as $t \to \infty$ if:
  $$\lim_{t \to \infty}\frac{|f(t)|}{e^{bt}}<\infty$$That is, if the limit exists and is finite.

Now, $f(t)=\frac{1}{t^2+1}$
Thus
\begin{align*}\lim_{t \to \infty}\frac{|\frac{1}{t^2+1}|}{e^{bt}} &= \lim_{t \to \infty} \frac{\frac{1}{t^2+1}}{e^{bt}} =0\end{align*} 
(I will leave it to you to verify the above limit).
Hence, we know $f(t)=\frac{1}{t^2 +1}$ is of exponential order , as $t\to \infty$, provided that $b>0$.

Note:
This fits in with the "traditional" definition

A function $f$ is said to be or exponential order as $t \to \infty$ if there are constants $M$, $b$ and $t_1$ existing such that $$|f(t)|<Me^{bt} \ \forall \ t \geq t_1$$

For example:
$$\text{If} \displaystyle \lim_{t \to \infty} \frac{|f(t)|}{e^{bt}}=N, \text{then there exists a } t_1 \ \text{such that} \frac{|f(t)|}{e^{bt}}<N+1, \  \forall t > t_1, $$
that is $$f(t)< (N+1)e^{bt} = Me^{bt}, \ \forall t \geq t_1$$

