Prove $\int_0^\infty e^{-x} x^k [L^k_n (x)]^2 \, dx=\frac{(n+k)!}{n!}$ How can I prove the normalization ratio of associated Laguerre polynomials: $$\int_0^\infty e^{-x} x^k [L^k_n (x)]^2 \, dx=\frac{(n+k)!}{n!}$$ using the generator function of Laguerre polynomial, $$\sum_{n=0}^\infty L_n^k (x) t^n=\frac{1}{(1-t)^{k+1}}\exp\left(\frac{-xt}{1-t}\right)?$$
 A: I think that the easiest (and well-known) way to do it is via Rodrigues' formula $$L_n^k(x)=\frac{x^{-k}e^x}{n!}\frac{d^n}{dx^n}(x^{n+k}e^{-x}).\label{rodrigues}\tag{*}$$
To obtain it from the generating function, one may use the Cauchy integral $$L_n^k(x)=\frac{1}{2\pi i}\oint_{(t=0)^+}\frac{1}{(1-t)^{k+1}}\exp\left(\frac{-xt}{1-t}\right)\frac{dt}{t^{n+1}}$$ (with the contour, say, a small circle around $t=0$), and substitute $t=1-x/z$ to get $$L_n^k(x)=\frac{x^{-k}e^x}{2\pi i}\oint_{(z=x)^+}\frac{z^{n+k}e^{-z}}{(z-x)^{n+1}}\,dz,$$ exactly the Cauchy integral for \eqref{rodrigues}. Now, returning to the question, we get
$$\int_0^\infty x^k e^{-x}\left(L_n^k(x)\right)^2 dx
=\frac{1}{n!}\int_0^\infty L_n^k(x)\frac{d^n}{dx^n}(x^{n+k}e^{-x})\,dx
\\=\frac{(-1)^m}{n!}\int_0^\infty\left(\frac{d^m}{dx^m}L_n^k(x)\right)\left(\frac{d^{n-m}}{dx^{n-m}}(x^{n+k}e^{-x})\right)dx$$
for $m\leqslant n$ (using induction on $m$ and integration by parts). Taking $m=n$ (recall that $L_n^k(x)$ is a polynomial of degree $n$ with the leading coefficient equal to $(-1)^n/n!$), we get the needed $$\int_0^\infty x^k e^{-x}\left(L_n^k(x)\right)^2 dx=\frac{1}{n!}\int_0^\infty x^{n+k}e^{-x}\,dx=\frac{(n+k)!}{n!}.$$ (The orthogonality $\int_0^\infty x^k e^{-x} L_m^k(x) L_n^k(x)\,dx=0$ for $m\neq n$ is shown the same way.)
