Laplace transform of $t^n$ from the Laplace transform of $1$

Question

So I have the following question here:

Starting from $\mathcal{L(1)=\frac{1}{s}}$, use the basic properties of Laplace transform to show that $\mathcal{L(t^n)=\frac{n!}{s^{n+1}}}$ for every positive integer $n$.

How would I go about doing this? I am not sure how sure how to manipulate the integral or use the properties of the Laplace transform to get what I want.

I think I have the use the derivative of a laplace transform and go from there but I am not sure.

Answer 1

Apply induction on $n$ and use $\mathcal{L(1)=\dfrac{1}{s}}$ in the base case. We need to show that $$\mathcal{L(t^n)=\dfrac{n!}{s^{n+1}}}$$ for every positive integer $n$.

Base case: When $n = 0$, we have $t^0 = 1$. Therefore

\begin{align*} \mathcal{L} \{ t^0 \} &= \mathcal{L} \{ 1 \} = \dfrac{1}{s} = \dfrac{0!}{s^{0 + 1}} \end{align*}

Induction Hypothesis: Let $n \in \mathbb{N}$ with $n \ge 1$. Then

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

Induction Step: We need to show that

$$\mathcal{L} \{ t^{n+1} \} = \dfrac{(n+1)!}{s^{n + 2}}$$

starting from the LHS

$$\mathcal{L} \{ t^{n + 1} \} = \int_0^\infty t^{n + 1} e^{-st} \ dt$$

we now use integration by parts, which says that

$$\int fg' = fg - \int f'g \ dt$$

let $f = t^{n + 1}$ and $g' = e^{-st}$. Then, $f' = (n + 1)t^n$ and $g = - \dfrac{1}{s} e^{-st}$ so that

$$\int t^{n + 1} e^{-st} \ dt = - \dfrac{t^{n + 1}}{s}e^{-st} + \dfrac{n + 1}{s} \int t^n e^{-st} \ dt$$

evaluating $t = 0$ and $t \to \infty$, we get

\begin{align*} \mathcal{L} \{ t^{n + 1} \} &= \left. - \dfrac{1}{s} t^{n + 1}e^{-st} \right|_0^\infty + \dfrac{n + 1}{s} \mathcal{L} \{ t^n \} \\ &= \left. -\dfrac{s^{-1}t^{n + 1}}{e^{st}} \right|_0^\infty + \dfrac{n + 1}{s} \mathcal{L} \{ t^n \} \\ &= 0 - 0 + \dfrac{n + 1}{s} \mathcal{L} \{ t^n \} \\ &= \dfrac{n + 1}{s} \times \dfrac{n!}{s^{n + 1}} \\ &= \dfrac{(n + 1)!}{s^{n + 2}} \end{align*}

Hence, by the principal of mathematical induction we see that $\mathcal{L(t^n)=\dfrac{n!}{s^{n+1}}}$ is true for every positive integer $n$.

Answer 2

Hint: show that if $$\mathcal{L}(f(t))=F(s)$$ then

$$\mathcal{L}(tf(t))=-F'(s)$$

and then use induction on $n$.

Answer 3

Apply integration by parts.

$$\mathcal{L}(t^n)=\int _0^\infty t^ne^{-st}dt=$$

$$(-1/s)e^{-st}t^n|_0^\infty +(1/s)\int _0^\infty nt^{n-1}e^{-st}dt = $$

$$(n/s)\mathcal {L} (t^{n-1})$$

Thus $$ \mathcal {L} (t)=(1/s)\mathcal {L}( 1) =1/{s^2}$$

and so forth.