Laplace Transform of Integral

Theorem
Let $f: \R \to \R$ or $\R \to \C$ be a function.

Let $\laptrans f = F$ denote the Laplace transform of $f$.

Then:
 * $\ds \laptrans {\int_0^t \map f u \rd u} = \dfrac {\map F s} s$

wherever $\laptrans f$ exists.

Proof
Let $\map g t = \ds \int_0^t \map f u \rd u$.

Then:
 * $\map {g'} t = \map f t$

and:
 * $\map g 0 = 0$

Thus: