Laplace Transform of Multiple 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 for all $n \in \Z_{\ge 0}$:

$\ds \laptrans {\underbrace {\int_0^t \dotsm \int_0^t}_{\text {$n$ times} } \map f u \rd u^n} = \dfrac {\map F s} {s^n}$

wherever $\laptrans f$ exists.


Proof

The proof proceeds by induction.

For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:

$\ds \laptrans {\underbrace {\int_0^t \dotsm \int_0^t}_{\text {$n$ times} } \map f u \rd u^n} = \dfrac {\map F s} {s^n}$


$\map P 0$ is the case:

$\map f u = \map F s$

which is the statement of the Laplace transform.

Thus $\map P 0$ is seen to hold.


Basis for the Induction

$\map P 1$ is the case:

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

which is established in Laplace Transform of Integral

Thus $\map P 1$ is seen to hold.


This is the basis for the induction.


Induction Hypothesis

Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.


So this is the induction hypothesis:

$\ds \laptrans {\underbrace {\int_0^t \dotsm \int_0^t}_{\text {$k$ times} } \map f u \rd u^k} = \dfrac {\map F s} {s^k}$


from which it is to be shown that:

$\ds \laptrans {\underbrace {\int_0^t \dotsm \int_0^t}_{\text {$k + 1$ times} } \map f u \rd u^{k + 1} } = \dfrac {\map F s} {s^{k + 1} }$


Induction Step

This is the induction step:

\(\ds \laptrans {\underbrace {\int_0^t \dotsm \int_0^t}_{\text {$k + 1$ times} } \map f u \rd u^{k + 1} }\) \(=\) \(\ds \laptrans {\int_0^t \paren {\underbrace {\int_0^t \dotsm \int_0^t}_{\text {$k$ times} } \map f u \rd u^k} \rd u}\)
\(\ds \) \(=\) \(\ds \dfrac 1 s \laptrans {\underbrace {\int_0^t \dotsm \int_0^t}_{\text {$k$ times} } \map f u \rd u^k}\) Basis for the Induction
\(\ds \) \(=\) \(\ds \dfrac 1 s \paren {\dfrac {\map F s} {s^k} }\) Induction Hypothesis
\(\ds \) \(=\) \(\ds \dfrac {\map F s} {s^{k + 1} }\) simplification

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.


Therefore:

$\forall n \in \Z_{\ge 0}: \ds \laptrans {\underbrace {\int_0^t \dotsm \int_0^t}_{\text {$n$ times} } \map f u \rd u^n} = \dfrac {\map F s} {s^n}$

$\blacksquare$


Sources

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