Why some books define Unilateral Laplace Transform from $0^+$ to $\infty$

Question

I've encountered with three variations of Unilateral Laplace Transform:

1. $$X(s) = \int_{0}^{\infty} x(t) e^{-st} dt$$
2. $$X(s) = \int_{0^-}^{\infty} x(t) e^{-st} dt$$
3. $$X(s) = \int_{0^+}^{\infty} x(t) e^{-st} dt$$

I understand that the 2nd definition (from $0^-$ to $\infty$) uses $0^-$ as the lower limit to include the origin and capture any discontinuity of $x(t)$ at $t = 0$; this is usefull when dealing with singularity functions such as Heaviside unit step $u(t)$ and Dirac delta $\delta(t)$.

What I don't understand is why the 3rd definition uses $0^+$ as lower limit. For example, with this definition the transform is "unable to see" the discontinuity of the function $u(t)$, moreover, is "unable to see" at all the function $\delta(t)$.

Furthermore, in my opinion this 3rd definition creates intrinsic contradictions between some properties of the Unilateral Laplace Transform: For example,

$$x(t)=e^{-3t}u(t) \rightarrow x'(t)=-3e^{-3t}u(t)+e^{-3t}\delta(t) $$

$$ X(s) = \mathcal{L}[x(t)]= \frac{1}{s+3}$$ $$\mathcal{L} [x'(t)]= \mathcal{L} [-3e^{-3t}u(t)+e^{-3t}\delta(t)]=\frac{s}{s+3}$$ $$\mathcal{L} [x'(t)]= sX(s)-x(0^+)=\frac{s}{s+3}-x(0^+)=\frac{s}{s+3}-1$$ So, you get two different results depending on the path you chose to compute!

Please explain the reasons behind this 3rd definition.

Answer 1

To define the unilateral Laplace transform of a function $x(t)$, $$ X(s) := \int_0^\infty x(t) e^{-st} dt $$ you can use any of those definitions and get the same thing. But sometimes strange people want to define the unilateral Laplace transform of something other than a genuine function. Maybe a "fake" function like the delta function $\delta(t)$. Or, more generally, a measure $\mu$. Then the definitions $$ \int_{[0,+\infty)} e^{-st}\;d\mu(t)\qquad\text{and}\qquad \int_{(0,+\infty)} e^{-st}\;d\mu(t) $$ may not be equal. So your definition would have to specify which one you mean.

I guess the point of your "I don't understand" part is: you need to learn about distributions and not treat things like $\delta(t)$ as though it were an ordinary function.