# Integral to Infinity of Function over Argument

## 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:

$\ds \int_0^\infty {\dfrac {\map f t} t} = \int_0^{\to \infty} \map F u \rd u$

provided the integrals converge.

## Proof

 $\ds \laptrans {\dfrac {\map f t} t}$ $=$ $\ds \int_s^{\to \infty} \map F u \rd u$ Integral of Laplace Transform $\ds \leadsto \ \$ $\ds \int_0^\infty e^{-s t} {\dfrac {\map f t} t} \rd t$ $=$ $\ds \int_s^{\to \infty} \map F u \rd u$ Definition of Laplace Transform $\ds \leadsto \ \$ $\ds \lim_{s \mathop \to 0} \int_0^\infty e^{-s t} \dfrac {\map f t} t \rd t$ $=$ $\ds \lim_{s \mathop \to 0} \int_s^{\to \infty} \map F u \rd u$ $\ds \leadsto \ \$ $\ds \int_0^\infty \dfrac {\map f t} t \rd t$ $=$ $\ds \int_0^{\to \infty} \map F u \rd u$

$\blacksquare$