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:

 $\ds \laptrans {\map {g'} t}$ $=$ $\ds s \laptrans {\map g t} - \map g 0$ Laplace Transform of Derivative $\ds$ $=$ $\ds s \laptrans {\map g t}$ $\ds$ $=$ $\ds \map F s$ as $\map F s = \laptrans {\map f t}$ $\ds \leadsto \ \$ $\ds \laptrans {\map g t}$ $=$ $\ds \dfrac {\map F s} s$ $\ds \leadsto \ \$ $\ds \laptrans {\int_0^t \map f u \rd u}$ $=$ $\ds \dfrac {\map F s} s$

$\blacksquare$

Examples

Example $1$

$\ds \laptrans {\int_0^1 \sin 2 u \rd u} = \dfrac 2 {s \paren {s^2 + 4} }$