I have the definition of the unilateral $\mathcal{L}$-transform, valid for causal signals, to be:
$$\mathcal{L}\left[f(t)\right](s)=\int_{0^-}^{+\infty}f(t)e^{-st}dt$$
My question is regarding the lower integration bound, $t=0^-$. Is this so (as opposed to say, $0^+$, or simply $0$) as to ensure that initial conditions are still considered?
For example, say $f=H$ the Heaviside step function, and there's some non-zero initial condition $f(0)$. From a bound $t=0^+$ we'd surely conclude the value of $f(0)$ played no role; from $t=0$, I suppose it would be more ambiguous, and would - I believe - depend on your definition of $H$: potentially forcing $f(0)=H(0)=0.5$ in the three-valued definition.
Is this correct, or is there some other reason?
$0^- $ means you would include a delta function placed at $t=0$. You still would have to deal with initial conditions whether you had $0^+$ or $0^-$
One of the most common reasons a delta function may appear at $t=0$ is if you differentiate a function multiplied by the step function.