Integral to Infinity of Function over Argument

From ProofWiki
Jump to navigation Jump to search


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 $f$.


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

provided the integrals converge.


\(\ds \laptrans {\dfrac {\map f t} t}\) \(=\) \(\ds \int_s^{\to \infty} \map F u \rd u\) 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
\(\ds \leadsto \ \ \) \(\ds \lim_{s \mathop \to 0} \int_0^\infty e^{-s t} \dfrac {\map f t} t \rd t\) \(=\) \(\ds \lim_{s \mathop \to 0} \int_s^{\to \infty} \map F u \rd u\)
\(\ds \leadsto \ \ \) \(\ds \int_0^\infty \dfrac {\map f t} t \rd t\) \(=\) \(\ds \int_0^{\to \infty} \map F u \rd u\)


Also see