Laplace Transform of Positive Integer Power

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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 1

\(\displaystyle \laptrans {t^n}\) \(=\) \(\displaystyle \int_0^{\to +\infty} t^n e^{-s t} \rd t\) $\quad$ Definition of Laplace Transform $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \intlimits {\frac {e^{-s t} } {-s} \sum_{k \mathop = 0}^n \paren {\paren {-1}^k \frac {n^{\underline k} t^{n - k} } {\paren {-s}^k} } } {t \mathop = 0} {t \mathop \to +\infty}\) $\quad$ Primitive of $x^n e^{a x}$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{k \mathop = 0}^n \paren {0 - \frac 1 {-s} \paren {-1}^k \frac {n^{\underline k} 0^{n - k} } {\paren {-s}^k} }\) $\quad$ Exponential Tends to Zero and Infinity, Exponent Combination Laws: Negative Power $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{k \mathop = 0}^n \paren {\dfrac 1 {s^{k + 1} } n^{\underline k} 0^{n - k} }\) $\quad$ simplifying $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \dfrac 1 {s^{k + 1} } n^{\underline n}\) $\quad$ $0^{n - k} = 0$ for all terms except $n = k$, when $0^{n - n} = 1$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {n!} {s^{k + 1} }\) $\quad$ Integer to Power of Itself Falling is Factorial $\quad$

$\blacksquare$


Proof 2

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:

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


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:

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

From Integration by Parts:

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

Here:

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

So:

\(\displaystyle \int t^{k + 1} e^{-s t} \rd t\) \(=\) \(\displaystyle -\frac {t^{k + 1} } s e^{-s t} + \frac {k + 1} s \int t^k e^{-s t} \rd t\) $\quad$ $\quad$


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

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


The result follows by the Principle of Mathematical Induction.

$\blacksquare$


Also see


Sources