# College Math Teaching

## October 25, 2013

### A Laplace Transform of a function of non-exponential order

Many differential equations textbooks (“First course” books) limit themselves to taking Laplace transforms of functions of exponential order. That is a reasonable thing to do. However I’ll present an example of a function NOT of exponential order that has a valid (if not very useful) Laplace transform.

Consider the following function: $n \in \{1, 2, 3,...\}$

$g(t)= \begin{cases} 1,& \text{if } 0 \leq t \leq 1\\ 10^n, & \text{if } n \leq t \leq n+\frac{1}{100^n} \\ 0, & \text{otherwise} \end{cases}$

Now note the following: $g$ is unbounded on $[0, \infty)$, $lim_{t \rightarrow \infty} g(t)$ does not exist and
$\int^{\infty}_0 g(t)dt = 1 + \frac{1}{10} + \frac{1}{100^2} + .... = \frac{1}{1 - \frac{1}{10}} = \frac{10}{9}$

One can think of the graph of $g$ as a series of disjoint “rectangles”, each of width $\frac{1}{100^n}$ and height $10^n$ The rectangles get skinnier and taller as $n$ goes to infinity and there is a LOT of zero height in between the rectangles.

Needless to say, the “boxes” would be taller and skinnier.

Note: this is an example can be easily modified to provide an example of a function which is $l^2$ (square integrable) which is unbounded on $[0, \infty)$. Hat tip to Ariel who caught the error.

It is easy to compute the Laplace transform of $g$:

$G(s) = \int^{\infty}_0 g(t)e^{-st} dt$. The transform exists if, say, $s \geq 0$ by routine comparison test as $|e^{-st}| \leq 1$ for that range of $s$ and the calculation is easy:

$G(s) = \int^{\infty}_0 g(t)e^{-st} dt = \frac{1}{s} (1-e^{-s}) + \frac{1}{s} \sum^{\infty}_{n=1} (\frac{10}{e^s})^n(1-e^{\frac{-s}{100^n}})$

Note: if one wants to, one can see that the given series representation converges for $s \geq 0$ by using the ratio test and L’Hoptial’s rule.