What happens to the constant term when finding integral using Laplace transform

Question

Suppose I have a function $g(t)=\frac{df(t)}{dt}$. Then $f(t) = f(0)+\int_0^tg(\tau)d\tau$.

Now if I apply the Laplace transform to the first equation I get $g(s) = sf(s)$. Hence, $f(s)=\frac{g(s)}{s}$. But then what happened to the $f(0)$?

Answer 1

We don't have $g(s)=sf(s)$, because $g,f$ are the original functions, not their transforms.

Under the assumption that $\lim_{t\to\infty} f(t)e^{-st}=0$ for all $s>0$, we get \begin{align*} G(s) & = \mathcal{L}\{f'\}(s) = \int_0^\infty f'(t)e^{-st}dt \\ & = f(t)e^{-st}\Bigr]_{t=0}^\infty + s\int_0^\infty f(t)e^{-st}dt \\ & = -f(0)+sF(s).\end{align*}

So the transform of $g$ is $sF(s)-f(0)$, and the constant remains.