Is $f\left(t\right)=\frac{1}{t^2+1}$ of exponential order?

Question

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...

Answer 1

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 $b>0$.

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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 $$f(t)< (N+1)e^{bt} = Me^{bt}, \ \forall t \geq t_1$$

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Answer 2

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: $$ \left|\frac{1}{t^2 +1}\right| \leq 1 \ \ \ \ \hbox{ for all } t$$