Laplace Transform of Multiple Integral

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:

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}$