# Integral of Laplace Transform

## Theorem

Let $f: \R \to \R$ or $\R \to \C$ be a continuous function on any interval of the form $0 \le t \le A$.

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

Then:

$\ds \laptrans {\dfrac {\map f t} t} = \int_s^{\to \infty} \map F u \rd u$

wherever $\ds \lim_{t \mathop \to 0} \dfrac {\map f t} t$ and $\laptrans f$ exist.

## Proof

Let $\map g t := \dfrac {\map f t} t$.

Then:

 $\ds \map f t$ $=$ $\ds t \map g t$ $\ds \leadsto \ \$ $\ds \laptrans {\map f t}$ $=$ $\ds \laptrans {t \map g t}$ $\ds \leadsto \ \$ $\ds \laptrans {\map f t}$ $=$ $\ds -\dfrac \d {\d s} \laptrans {\map g t}$ Derivative of Laplace Transform $\ds \leadsto \ \$ $\ds \map F s$ $=$ $\ds -\dfrac {\d G} {\d s}$ $\map F s := \laptrans {\map f t}$, and so on $\ds \leadsto \ \$ $\ds \map G s$ $=$ $\ds -\int_{-\infty}^s \map f u \rd u$ $\ds$ $=$ $\ds \int_s^\infty \map f u \rd u$

The result follows.

$\blacksquare$

## Examples

### Example $1$

Let $\sin$ denote the real sine function.

Let $\laptrans f$ denote the Laplace transform of a real function $f$.

Then:

$\laptrans {\dfrac {\sin t} t} = \arctan \dfrac 1 s$