Laplace Transform of Positive Integer Power/Proof 2

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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.

$\blacksquare$


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