Laplace transform of $t^n$ using integration by parts

Question

I am trying to solve the following question:

Let n be a positive integer. Show that $\mathcal{L(t^n)(s)=\frac{n}{s}} \mathcal{L(t^{n-1})}$ and use mathematical induction to prove $\mathcal{L(t^n)(s)=\frac{n!}{s^{n+1}}}$ for every positive integer $n$.

How would I go about doing this? I am not sure how to manipulate the indefinite integral or use the Laplace transform to get the result. I tried to begin by using the derivative of a Laplace transform, but ended up in a loop.

Any guidance is greatly appreciated!

Answer 1

Your question title says it: Integrate by parts. $$\int_{a}^{b}u dv = uv|_{a}^{b}- \int_{a}^{b} vdu$$ Let $u = t^n$, so $dv = e^{-st}dt$, and so $du = nt^{n-1}dt$ while $$v = \frac{e^{-st}}{-s}$$ The limit of the $uv$ term as $t \rightarrow 0$ and as $t \rightarrow \infty$ are both $0$. Now look at what you have left: $$\int_{0}^{\infty}\frac{1}{s}nt^{n-1}e^{-st} dt$$ Pull out constants and you have it.

Answer 2

$$\mathcal{L}(t\mapsto t^n)(s)=\int_0^\infty \underbrace{t^n}_{u}\underbrace{e^{-st}\mathrm{d}t}_{\mathrm{d}v}=(uv)|^\infty_0-\int_0^\infty v~\mathrm{d}u$$ $$=\left(-\frac{1}{s}t^ne^{-st}\right)\Bigg|^\infty_0+\dots$$ Can you take it from here?

Answer 3

$$\mathcal{L}(t^n)(s)=\frac{n}{s}\mathcal{L(t^{n-1})}$$ You can integrate by part or you can use the Laplace transform of a derivative: $$s\mathcal{L}(t^n)(s)= \mathcal{L}{(nt^{n-1})}$$ $$s\mathcal{L}(t^n)(s)= \mathcal{L}\{(t^{n})'\}$$ $$s\mathcal{L}(t^n)(s)-0= \mathcal{L}\{(t^{n})'\}$$ This is just: $$ \mathcal{L}\{y'\}=s\mathcal{L}(y)-y(0)$$ Where $y(t)=t^n$