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}$
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 32$: Table of General Properties of Laplace Transforms: $32.14$