Integral of Laplace Transform

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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:

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


Also see


Sources

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