# 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:

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

$\blacksquare$