Initial Value Theorem of Laplace Transform

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

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

if those limits exist.

Proof
We have that $\map {f'} t$ is piecewise continuous with one-sided limits and of exponential order.

Hence:
 * $\displaystyle \lim_{s \mathop \to \infty} \int_0^\infty e^{-s t} \map {f'} t \rd t = 0$

Suppose that $f$ is continuous at $t = 0$.

From Laplace Transform of Derivative:


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

Taking the limit as $s \to \infty$ in $(1)$, where it is assumed that $\map f t$ is continuous at $t = 0$:


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

or:


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

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:


 * $(2): \quad \laptrans {\map {f'} t} = s \, \map F s - \displaystyle \lim_{u \mathop \to 0} \map f u$

Similarly taking the limit as $s \to \infty$ in $(2)$, where it is assumed that $\map f t$ is continuous at $t = 0$:


 * $0 = \displaystyle \lim_{s \mathop \to \infty} s \, \map F s - \lim_{u \mathop \to 0} \map f u$

and so:


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