Laplace transform of a 100th derivative

Question

$f(t)=e^t\frac{d^{100}}{dt^{100}}(e^{-t}t^{100})$

This is the question. I was able to solve the last part, but how to combine the both part?

Answer 1

Let's call $f(t)=n!L_n(t)$ where $L_n(t)=\frac{\mathrm e^t}{n!}\frac{d^n}{dt^n}\Big(t^ne^{-t}\Big)$ is the Laguerre polynomial of order $n$.

We have $$ L_n(t)=\sum_{k=0}^n {{n} \choose {k}} (-1)^k \frac{t^k}{k!} $$ and then $$\begin{align} \mathcal L\{L_n(t)\}&=\int_0^\infty \mathrm e^{-st}L_n(t)\, \mathrm dt\\ &=\sum_{k=0}^n {{n} \choose {k}} \frac{(-1)^k}{k!} \int_0^\infty t^k\, \mathrm e^{-st}\mathrm dt\\ &=\sum_{k=0}^n {{n} \choose {k}} (-1)^k s^{-k-1}\\ &=s^{-1}\sum_{k=0}^n {{n} \choose {k}} (-s^{-1})^{k} \\ &=\left(\frac{s-1}{s} \right)^n \frac{1}{s} \end{align} $$ and finally $$ F(s)=\mathcal L\{f(t)\}=\left(\frac{s-1}{s} \right)^n \frac{n!}{s} $$

Can you write what it is for $n=100$?

Or you can apply the properties of the Laplace transform to $f(t)=\mathrm e^tg(t)$ and $g(t)=\phi^{(n)}(t)$ with $\phi(t)=t^n\mathrm e^{-t}u(t)$ and then $F(s)=G(s-1)$ and $$G(s)=s^n\Phi(s)-\sum_{i=1}^ns^{i-1}\phi^{(n-i)}(0)$$ with $\Phi(s)=\frac{n!}{(s+1)^n}$ for $\Re(s)>-1$