Laplace transform of a function

Question

How do I find the Laplace transform of the function $U_n(t) = t^n$?

I normally don't post a question without at least some attempt but I have no idea how to start this. Any suggestions are greatly appreciated.

Answer 1

Hint:

Start by noting that \begin{align} F(s) &= \lim_{A \rightarrow \infty}\int_0^A e^{-st}t^{n}dt \\ &= \lim_{A \rightarrow \infty} \Big \{t^n \frac{e^{-st}}{-s}\Bigg |_0^A - \int_0^A \frac{nt^{n-1}e^{-st}}{-s}dt\Big \} \\ &= 0 + \frac{n}{s}\lim_{A \rightarrow \infty}\int_0^Ae^{-st}t^{n-1}dt \\ &= \frac{n}{s} \mathcal{L}\{t^{n-1}\} \end{align} So a recurrence relation exists. Now use this method again to find a relation to $\mathcal{L}\{t^{n-2}\}$ and then in general.

Answer 2

$$F(s) = \int\limits_0^\infty f (t){e^{ - st}}{\mkern 1mu} dt$$ do it for constant number first $$f(t)=1 \to \\F(s)= \int\limits_0^\infty 1{e^{ - st}}{\mkern 1mu} dt=\frac{1}{s}$$ then for $$f(t)=t \to \\F(s)=\int\limits_0^\infty t{e^{ - st}}{\mkern 1mu} dt= ?$$integration by part show you The idea to induction $$F(s)=\frac{te^{-st}}{-s}|_{0}^{\infty}-\int_{0}^{\infty}\frac{1e^{-st}}{-s}ds=\\0+\frac{1}{s}\int_{0}^{\infty}e^{-st}ds=\\\dfrac{1}{s}Laplace(f(t)=1)=\dfrac{1}{s}dfrac{1}{s}=\dfrac{1}{s^2}$$ $$f(t)=t^2 \to \\ F(s)=\int\limits_0^\infty t^2{e^{ - st}}{\mkern 1mu} dt=\\ \dfrac{t^2e^{-st}}{-s}| -\int\limits_0^\infty \dfrac{(2t)e^{-st}}{-s}{\mkern 1mu} dt=\\0+\dfrac{2}{s}\int_{0}^{\infty} te^{-st}ds=\\\dfrac2s \times laplace(f(t)=t) =\\ \dfrac{2}{s}\dfrac{1}{s^2}=\dfrac{2!}{s^3}$$ continue this method will give you $$f(t)=t^3 \to F(s)=\dfrac{3!}{s^4}\\ f(t)=t^4 \to F(s)=\dfrac{4!}{s^5}\\\vdots \\ f(t)=t^n \to F(s)=\dfrac{n!}{s^{n+1}}\\$$