Laplace Transform of Polynomial Functions 📂Odinary Differential Equations

Laplace Transform of Polynomial Functions

Formulas¹

$\mathcal{L} \left\{ t^p \right\} = \dfrac{ \Gamma (p+1) } {s^{p+1}},\quad s>0$

Explanation

The Laplace transform of a polynomial is represented by the Gamma function. If we use $x^p$ instead of $t^p$ , it would be easier to recognize at a glance. Usually, in differential equations, variables represent time, so $x$ is replaced with $t$ .

Derivation

$\begin{align*} \mathcal{L} \left\{ t^p \right\} &= \int_{0}^\infty e^{-st}t^p dt \\ &= \lim \limits_{A \to \infty} \left[ -\dfrac{1}{s} \left[ e^{-st}t^p \right] _{0}^A +\dfrac{p}{s} \int_{0}^A e^{-st}t^{p-1}dt \right] \\ &= \dfrac{p}{s} \mathcal{L} \left\{ t^{p-1} \right\} \\ &= \lim \limits_{A \to \infty} \left[ \dfrac{p}{s} \dfrac{-1}{s}\left[ e^{-st}t^{p-1} \right] _{0}^A + \dfrac{p(p-1)}{s^2}\int _{0}^A e^{-st} t^{p-2} dt\right] \\ &=\dfrac{p(p-1)}{s^2}\mathcal{L} \left\{t^{p-2} \right\} \\ & \vdots \end{align*}$

By continuing in the same manner, after $p$ steps, the following can be obtained.

$\begin{align*} &=\dfrac{p(p-1)(p-2) \cdots 1}{s^p}\mathcal{L} \left\{ t^0=1 \right\} \\ &=\dfrac{p!}{s^{p+1}} \\ &=\dfrac{\Gamma (p+1) }{s^{p+1}} \end{align*}$

However, since $\lim \limits_{A \to \infty} e^{-sA}$ must converge to $0$ , then $s>0$ .

■

After summarizing the Gamma function and substituting $s=1$ , it coincides with the common definition of the Gamma function.

$\Gamma (p+1) = \dfrac{1}{s^{p+1}}\int_{0}^\infty e^{-st}t^p dt = \int_{0}^\infty e^{-t}t^p dt$