If $\mathcal{L}\{f(t)\} = F(s)$, then $\lim_{s \to 0}sF(s) = \lim_{t\to +\infty } f(t)$. provided the limits exist.

Question

$f$ is sectionally continuous and of exponential order. $F(s)$ denotes it's Laplace transform.

this is a stated theorem with no proof in 'Laplace Transforms - Murray R. Spiegel'.

my attempt at proving this :

$\int_{0}^{+\infty} f'(t)e^{-st}dt = sF(s) - \lim_{t\to0}f(t)$

taking the limit as $s \to 0$ on both sides we get :

$\lim_{s \to 0}\int_{0}^{+\infty} f'(t)e^{-st}dt = \lim_{s \to 0}sF(s) - \lim_{t\to0}f(t)$

now assuming I could move the limit inside the integral then I'm basically done. however I can't think of any justification. how do you justify my last step ?

Answer 1

As already mentioned in the comments, since $s$ is a fixed constant independent of $t$, shifting the limit inside the integral is perfectly valid.

In fact, this is known as the final value theorem and you can see the proof (given in the link) also does exactly this: writes the limit of integral as integral of limit.