Integral to Infinity of Function over Argument

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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:

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

provided the integrals converge.


Proof

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

$\blacksquare$


Also see


Sources