Deriving Laplace Transform of Laguerre polynomial

Question

I'm given this definition for the Laguerre polynomials: $$L_n(t)=\frac{e^t}{n!}\frac{d^n}{dt^n}\left[t^ne^{-t}\right],~\text{for }n=0,1,2...$$ and I have to show that the Laplace transform is $$\frac{1}{s}\left(\frac{s-1}{s}\right)^n$$ My first attempt was to find a formula for the $n$th derivative of $t^ne^{-t}$, which I found to be $$\frac{d^n}{dt^n}\left[t^ne^{-t}\right]=e^{-t}\sum_{k=0}^n\frac{n!}{(n-k)!}\binom nk (-t)^{n-k}$$

So from here I tried taking the transform: $$\begin{align*}\mathcal{L}_s\left\{\frac{e^t}{n!}\frac{d^n}{dt^n}\left[t^ne^{-t}\right]\right\}&=\frac{1}{n!}\mathcal{L}_{s-1}\left\{e^{-t}\sum_{k=0}^n\frac{n!}{(n-k)!}\binom nk (-t)^{n-k}\right\}\end{align*}$$ where $\mathcal{L}_s\{f(t)\}=F(s)$ and $\mathcal{L}_s\{e^tf(t)\}=\mathcal{L}_{s-1}\{f(t)\}=F(s-1)$ (in case the notation was unfamiliar). I'm not sure how to continue with this. The factor of $e^{-t}$ seems to undo the shift to $s-1$, and I'm not sure if that's a problem or not.

I don't think I made a mistake with the $n$th derivative formula, but I think there might be a different way to express it that would make computation easier. Any help is appreciated!

Answer 1

Have you considered using the properties of the Laplace transform to simplify calculation?

The one's I'm thinking of are:

$$\mathcal{L}(e^{at}f(t))=F(s-a)$$

and,

$$\mathcal{L}\left(\frac{d^n}{dt^n}f(t)\right)=s^nF(s)-\sum_{i=1}^ns^{i-1}f^{(n-i)}(0)$$

where $F(s)=\mathcal{L}(f(t))$ and $f^{(n)}$ is the n-th derivative of $f$. The $\frac{1}{n!}$ is a constant, so you can just move it out of the integral.

Answer 2

I don't think using the Rodrigues formula is convenient here. If one starts from the standard closed form representation of the Laguerre polynomials $$L_n(x)=\sum_{k=0}^n {{n} \choose {k}} (-1)^k \frac{x^k}{k!}$$ we have the rather straightforward calculation

$$\mathcal L_s(L_n(x)))=\int_0^\infty e^{-sx}L_n(x) dx=\sum_{k=0}^n {{n} \choose {k}} \frac{(-1)^k}{k!} \int_0^\infty x^k e^{-sx}dx=\sum_{k=0}^n {{n} \choose {k}} (-1)^k s^{-k-1}= \\ s^{-1}\sum_{k=0}^n {{n} \choose {k}} (-s^{-1})^{k} =\left(\frac{s-1}{s} \right)^n \frac{1}{s} $$

having used the Gamma function relation $$\int_0^\infty x^k e^{-sx}dx= k! s^{-k-1}.$$