# Decompose integral of derivative and $e^{st}$ (laplace transform)

I'm reading on Laplace transform and stumbled upon the transform of a derived function. Could someone explain me this? $$$$\int_{0^{-}}^\infty \frac{d}{dt}f(t)e^{-st} dt = e^{-st}f(t)|^{\infty}_{0} + \int^{\infty}_{0^{-}}sf(t)e^{-st}dt = -f(0) + sF(s)$$$$

I'm just curious how you pass from the first statement, to the second, to the third, and especially, from the first to the second. They use the 0+ and 0- notation to respectively design 'just after 0' and 'just before 0'.

Could someone explain?

Thanks

First in the entire process, you assume that $f(t)$ grows very slowly than exponential for the integral to make sense, i.e., $\lim_{t\to \infty}e^{-st} f(t) = 0$. Assuming this, by integration by parts, we have $$\int_0^{\infty} e^{-st} f'(t)dt = \int_0^{\infty} e^{-st}d\left(f(t)\right) = \left. e^{-st}f(t)\right \vert_{t=0}^{\infty} - \int_0^{\infty}f(t) d(e^{-st}) = -f(0) + s\int_0^{\infty}f(t)e^{-st}dt$$ where we made use of the fact that $\lim_{t\to \infty}e^{-st} f(t) = 0$, while evaluating the upper limit.
The first is simply integration by parts, http://en.wikipedia.org/wiki/Integration_by_parts, in the third, since $s$ is simply a constant $\int_0^\infty fe^{-st} dt=F(s)$ as pr the definition of the laplace transform, with the first term simply coming from inserting the limits, ie. $\lim_{a\to\infty}f(a)e^{-sa}-f(0)e^{-s0}=-f(0)$.