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

Question

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

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

Answer 1

I believe you're looking for integration by parts to go from the first step to the second.

And to go from the second to the third step, evaluating the first term at t=infinity gives 0 then subtracting it's evaluation at t=0 (f(0)), you get -f(0), then the second integral is simply s*laplace transform of f(t), which is s*F(s)

Answer 2

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.

Answer 3

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. $\lim_{a\to\infty}f(a)e^{-sa}-f(0)e^{-s0}=-f(0)$.