Why is the Laplace transform unchanged when t is replaced with -t?

[SamRoss]

TL;DR

In the book I'm reading, it says "...since we integrate from 0 to infinity, [the Laplace transform] is the same no matter how [the original function] is defined for negative t." Why is this so?

In Mathematical Methods in the Physical Sciences by Mary Boas, the author defines the Laplace transform as...

$${L(f)=}\int_0^\infty{f(t)}e^{-pt}{dt=F(p)}$$

The author then states that "...since we integrate from 0 to $\infty$, ${L(f)}$ is the same no matter how ${f(t)}$ is defined for negative t." Why is this so?

 

[Orodruin]

Because the integral is only over positive $t$, the value of $f(t)$ for $t < 0$ is irrelevant.

Edit: Note that this is not the same thing as you are stating in the title.

 

[SamRoss]

Because the integral is only over positive $t$, the value of $f(t)$ for $t < 0$ is irrelevant.

Edit: Note that this is not the same thing as you are stating in the title.

Thanks very much for the quick response, and especially the edit. I thought I might have been reading that the wrong way.