# Deriving Laplace Transform of Laguerre polynomial

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!

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.

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 $$\int_0^\infty x^k e^{-sx}dx= k! s^{-k-1}.$$