# Laplace Transform of Positive Integer Power/Proof 2

## Theorem

Let $\laptrans f$ denote the Laplace transform of a function $f$.

Let $t^n: \R \to \R$ be $t$ to the $n$th power for some $n \in \N_{\ge 0}$.

Then:

$\laptrans {t^n} = \dfrac {n!} { s^{n + 1} }$

for $\map \Re s > 0$.

## Proof

The proof proceeds by induction on $n$ for $t^n$.

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

$\laptrans {t^n} = \dfrac {n!} { s^{n + 1} }$

### Basis for the Induction

$\map P 0$ is the case:

 $\ds \laptrans {t^0}$ $=$ $\ds \laptrans 1$ $\ds$ $=$ $\ds \frac 1 s$ Laplace Transform of 1 $\ds$ $=$ $\ds \frac {0!} {s^{0 + 1} }$ Definition of Factorial: $0! = 1$

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:

$\laptrans {t^k} = \dfrac {k!} {s^{k + 1} }$

from which it is to be shown that:

$\laptrans {t^{k + 1} } = \dfrac {\paren {k + 1}!} {s^{k + 2} }$

### Induction Step

This is our induction step:

 $\ds \laptrans {t^{k + 1} }$ $=$ $\ds \int_0^{\to +\infty} t^{k + 1} e^{-s t} \rd t$ Definition of Laplace Transform

From Integration by Parts:

$\displaystyle \int f g' \rd t = f g - \int f' g \rd t$

Here:

 $\ds f$ $=$ $\ds t^{k + 1}$ $\ds \leadsto \ \$ $\ds f'$ $=$ $\ds \paren {k + 1} t^n$ Power Rule for Derivatives $\ds g'$ $=$ $\ds e^{-s t}$ $\ds \leadsto \ \$ $\ds g$ $=$ $\ds -\frac 1 s e^{-s t}$ Primitive of Exponential Function

So:

 $\ds \int t^{k + 1} e^{-s t} \rd t$ $=$ $\ds -\frac {t^{k + 1} } s e^{-s t} + \frac {k + 1} s \int t^k e^{-s t} \rd t$

Evaluating at $t = 0$ and $t \to +\infty$:

 $\ds \laptrans {t^{k + 1} }$ $=$ $\ds -\intlimits {\frac 1 s t^{k + 1} e^{-s t} } {t \mathop = 0} {t \mathop \to +\infty} + \frac {k + 1} s \laptrans {t^n}$ $\ds$ $=$ $\ds -\intlimits {\frac {s^{-1} t^{k + 1} } {e^{s t} } } {t \mathop = 0} {t \mathop \to +\infty} + \frac {k + 1} s \laptrans {t^n}$ Exponent Combination Laws $\ds$ $=$ $\ds 0 - 0 + \frac {k + 1} s \laptrans {t^n}$ Limit at Infinity of Polynomial over Complex Exponential $\ds$ $=$ $\ds \frac {k + 1} s \times \frac {k!} {s^{k + 1} }$ Induction Hypothesis $\ds$ $=$ $\ds \frac {\paren {k + 1}!} {s^{k + 1 + 1} }$ Exponent Combination Laws, Definition of Factorial

The result follows by the Principle of Mathematical Induction.

$\blacksquare$