Laplace Transform of Integral

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Theorem

Let $f: \R \to \R$ or $\R \to \C$ be a function.

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


Then:

$\ds \laptrans {\int_0^t \map f u \rd u} = \dfrac {\map F s} s$

wherever $\laptrans f$ exists.


Proof

Let $\map g t = \ds \int_0^t \map f u \rd u$.

Then:

$\map {g'} t = \map f t$

and:

$\map g 0 = 0$


Thus:

\(\ds \laptrans {\map {g'} t}\) \(=\) \(\ds s \laptrans {\map g t} - \map g 0\) Laplace Transform of Derivative
\(\ds \) \(=\) \(\ds s \laptrans {\map g t}\)
\(\ds \) \(=\) \(\ds \map F s\) as $\map F s = \laptrans {\map f t}$
\(\ds \leadsto \ \ \) \(\ds \laptrans {\map g t}\) \(=\) \(\ds \dfrac {\map F s} s\)
\(\ds \leadsto \ \ \) \(\ds \laptrans {\int_0^t \map f u \rd u}\) \(=\) \(\ds \dfrac {\map F s} s\)

$\blacksquare$


Examples

Example $1$

$\ds \laptrans {\int_0^1 \sin 2 u \rd u} = \dfrac 2 {s \paren {s^2 + 4} }$


Sources

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