Ok, this has me a bit stumped.
In my class the teacher "showed" us how to find the bilateral Laplace transform of x(t)=$e^{at}$ where $-\infty<t<\infty$.
Breaking them into the two parts (causal and anticausal) produces:
(H is the Heaviside step function)
x(t)=$e^{at}H(t)+e^{at}H(-t)$
$X(s)=\mathcal{L}[x(t)]=\mathcal{L}[e^{at}H(t)]+\mathcal{L}[e^{at}H(-t)]$
$X(s)=\frac{1}{s-a}-\frac{1}{s-a}=0$
However, note that the region of convergence (ROC) for the left term is Re{s} > a and the ROC for the right term is Re{s} < a so there is no overlap and the Laplace transform does not exist.
The teacher, however, then went on to say that it can easily be shown by "changing the time horizon" that the anticausal term can be ommited leaving the result simply as $X(s)=\frac{1}{s-a}$.
My question is how do you "change the time horizon" and use it to show that the anticausal term is zero when taking the Laplace transform?
I assume it is a trick that you can do because t goes from minus infinity to infinity, but I can't figure out what the trick is.
Additional information: The impulse response of the system that that this signal is passed into is neither causal nor is it guaranteed to be stable.
Any help is appreciated.