Final Value Theorem of Laplace Transform

Theorem
Let $\laptrans {\map f t} = \map F s$ denote the Laplace transform of the real function $f$.

Then:
 * $\displaystyle \lim_{t \mathop \to \infty} \map f t = \lim_{s \mathop \to 0} s \, \map F s$

if those limits exist.

Proof
From Laplace Transform of Derivative:


 * $(1): \quad \laptrans {\map {f'} t} = s \, \map F s - \map f 0$

We have that:

Hence:

Suppose that $f$ is not continuous at $t = 0$.

From Laplace Transform of Derivative with Discontinuity at Zero:


 * $\laptrans {\map {f'} t} = s \, \map F s - \map f {0^+}$

which means:


 * $(3): \quad \laptrans {\map {f'} t} = s \, \map F s - \displaystyle \lim_{u \mathop \to 0} \map f u$

We have that:

Hence: