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:
- $\ds \int f g' \rd t = f g - \int f' g \rd t$
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.
$\blacksquare$
Sources
- For a video presentation of the contents of this page, visit the Khan Academy.