# 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:

$\ds \lim_{t \mathop \to \infty} \map f t = \lim_{s \mathop \to 0} s \, \map F s$

if those limits exist.

### General Result

Let $\ds \lim_{t \mathop \to \infty} \dfrac {\map f t} {\map g t} = 1$.

Then:

$\ds \lim_{s \mathop \to 0} \dfrac {\map F s} {\map G s} = 1$

if those limits exist.

## Proof

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

We have that:

 $\ds \lim_{s \mathop \to 0} \laptrans {\map {f'} t}$ $=$ $\ds \lim_{s \mathop \to 0} \int_0^\infty e^{-s t} \map {f'} t \rd t$ Definition of Laplace Transform $\ds$ $=$ $\ds \int_0^\infty \map {f'} t \rd t$ $\ds$ $=$ $\ds \lim_{L \mathop \to \infty} \int_0^L \map {f'} t \rd t$ $\ds$ $=$ $\ds \lim_{L \mathop \to \infty} \paren {\map f L - \map f 0}$ Fundamental Theorem of Calculus $\text {(2)}: \quad$ $\ds$ $=$ $\ds \lim_{t \mathop \to \infty} \map f t - \map f 0$

Hence:

 $\ds \lim_{s \mathop \to 0} \laptrans {\map {f'} t}$ $=$ $\ds \lim_{s \mathop \to 0} s \, \map F s - \map f 0$ from $(1)$ $\ds \leadsto \ \$ $\ds \lim_{t \mathop \to \infty} \map f t - \map f 0$ $=$ $\ds \lim_{s \mathop \to 0} s \, \map F s - \map f 0$ from $(2)$ $\ds \leadsto \ \$ $\ds \lim_{t \mathop \to \infty} \map f t$ $=$ $\ds \lim_{s \mathop \to 0} s \, \map F s$

$\Box$

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

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

which means:

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

We have that:

 $\ds \lim_{s \mathop \to 0} \laptrans {\map {f'} t}$ $=$ $\ds \lim_{u \mathop \to 0} \paren {\lim_{s \mathop \to 0} \int_u^\infty e^{-s t} \map {f'} t \rd t}$ Definition of Laplace Transform $\ds$ $=$ $\ds \lim_{u \mathop \to 0} \paren {\int_u^\infty \map {f'} t \rd t}$ $\ds$ $=$ $\ds \lim_{u \mathop \to 0} \paren {\lim_{L \mathop \to \infty} \int_u^L \map {f'} t \rd t}$ $\ds$ $=$ $\ds \lim_{u \mathop \to 0} \paren {\lim_{L \mathop \to \infty} \paren {\map f L - \map f u} }$ Fundamental Theorem of Calculus $\ds$ $=$ $\ds \lim_{u \mathop \to 0} \paren {\lim_{t \mathop \to \infty} \map f t - \map f u}$ $\text {(4)}: \quad$ $\ds$ $=$ $\ds \lim_{t \mathop \to \infty} \map f t - \lim_{u \mathop \to 0} \map f u$

Hence:

 $\ds \lim_{s \mathop \to 0} \laptrans {\map {f'} t}$ $=$ $\ds \lim_{s \mathop \to 0} s \, \map F s - \lim_{u \mathop \to 0} \map f u$ from $(3)$ $\ds \leadsto \ \$ $\ds \lim_{t \mathop \to \infty} \map f t - \lim_{u \mathop \to 0} \map f u$ $=$ $\ds \lim_{s \mathop \to 0} s \, \map F s - \lim_{u \mathop \to 0} \map f u$ from $(4)$ $\ds \leadsto \ \$ $\ds \lim_{t \mathop \to \infty} \map f t$ $=$ $\ds \lim_{s \mathop \to 0} s \, \map F s$

$\blacksquare$

## Examples

### Example 1

Consider the real function $f: \R \to \R$ defined as:

$\map f t = 3 e^{-2 t}$
$\laptrans {\map f t} = \dfrac 3 {s + 2}$

Then by the Initial Value Theorem of Laplace Transform:

 $\ds \lim_{t \mathop \to 0} 3 e^{-2 t}$ $=$ $\ds \lim_{s \mathop \to \infty} \dfrac 3 {s + 2}$ $\ds \leadsto \ \$ $\ds 3$ $=$ $\ds 3$