# Understanding the Initial-value Theorem in the Laplace Transform Theory

I am an Electrical Engineering student and I study from the book "Engineering Circuit Analysis 8th Ed." authored by Hayt, Kemmerly and Durbin. On Chapter 14, page 562 dealing with an COMPLEX FREQUENCY AND THE LAPLACE TRANSFORM, there is a subsection on the Initial-Value Theorem.

Now, what don't understand is the part where the initial integral is split into two:

$$\lim_{s\to\infty}\left(\int_{0^{-}}^{\infty} e^{-st}\frac{df}{dt}dt\right) = \lim_{s\to\infty}\left(\int_{0^{-}}^{0^+} e^{0}\frac{df}{dt}dt + \int_{0^{+}}^{\infty} e^{-st}\frac{df}{dt}dt\right)$$

Why did $$e^{-st}$$ become $$e^{0}$$ in the first part of the split integral when taking the limit to infinity? I have taken Calculus course before but I am just lost in here ($$s$$ is a complex number in this context).

I would be really grateful if someone answers my question.

In a very hand wavy way, what is the only number between $$0^+$$ and $$0^-$$? $$0$$, of course. So the only number that gets plugged in for $$t$$ is $$0$$. However, this is a terrible way to prove the Initial Value Theorem. Here is a better way. We already have that
$$\int_0^\infty f'(t) e^{-st}dt = sF(s) -f(0)$$
Then taking $$s\to\infty$$, and assuming $$f'(t)$$ was bounded, we get that
$$\int_0^\infty f'(t)\cdot 0 dt = 0 = \lim_{s\to\infty} sF(s) -f(0)$$