Laplace Transform: Piecewise Function Integrability and Existence of Laplace Transform

Question

I am trying to decide whether the function

$$f(t) = \begin{cases} 1, & \text{$t$ is even} \\ 0, & \text{$t$ is odd} \end{cases}$$

has a Laplace transform, or is even integrable in the first place.

I know that (1) a function is Riemann integrable only if it has a finite number of discontinuities (even if those discontinuities themselves are infinite). I also know that (2) a function has a Laplace transform only if it has a finite number of finite discontinuities.

It seems to me that $f(t)$ satisfies (1) and is therefore Riemann integrable. But then I think about how the graph of such a function would look like, and it doesn't make sense to me how it would be Riemann integrable?

As for (2), although the discontinuities are finite, there is an infinite number of them as we integrate from $0$ to $\infty$, which is what is required for the Fourier transform. Therefore, I'd say that it does not satisfy (2)?

I would greatly appreciate it if people could please take the time to clarify this and explain.

EDIT: What's with all of the down-votes? This is a textbook problem from the end-of-chapter-1 textbook problems of An Introduction to Laplace Transform and Fourier Series by Dyke:

For each of the following functions, determine which has a Laplace transform. If it exists, find it; if it does not, say briefly why.

(f) $$f(t) = \begin{cases} 1, & \text{$t$ is even} \\ 0, & \text{$t$ is odd} \end{cases}$$

I asked this question precisely because I thought it didn't make sense. Not sure why I'm getting down-voted here.

EDIT2: So given the definition of Riemann integrability, what exactly is it about this function that makes it not Riemann integrable? What condition does it fail? $f(t)$ does technically have a finite number of discontinuities, so that condition is satisfied.

Answer 1

I can think of two reasonable explanations.

1. Since the Laplace transfrom is only defined for suitable functions $g:[0,\infty)\to \mathbb R$, and your function is $f:\mathbb N\to \mathbb R$, it does not have a Laplace transform.
2. The author made a mistake, and instead meant to define $f(x)=1$ for even integers and $f(x)=0$ otherwise. In this case, $f$ has infinitely many discontinuities, so (according to the author) it does not have a Laplace transform.