Computing the normalisation constant of the Laguerre polynomials How does one compute the normalisation constant for the Laguerre polynomials from the Rodrigues formula, i.e. $\tfrac{\Gamma(n+\alpha+1)}{n!}\delta_{n,m}$? I tried:
$$
\int_0^\infty w(x)L_n^{(\alpha)}L_n^{(\alpha)}\mathrm{d}x
=\int_0^\infty e^{-x}x^{\alpha}
 \left(\frac{e^x x^{-\alpha}}{n!}
 \frac{\mathrm{d}^n}{\mathrm{d}x^n}(e^{-x}x^{n+\alpha}) 
 \right)^2\mathrm{d}x
$$
$$=\int_0^\infty
 \frac{e^x x^{-\alpha}}{(n!)^2}
    \left(
 \frac{\mathrm{d}^n}{\mathrm{d}x^n}(e^{-x}x^{n+\alpha}) 
 \right)^2\mathrm{d}x
$$
But here I am not seeing what to do next. I tried using the general Leibniz rule but I don't think this will lead me somewhere
$$\frac{\mathrm{d}^n}{\mathrm{d}x^n}(e^{-x}x^{n+\alpha})=\sum\limits_{k=0}^n\binom{n}{k}(e^{-x})^{(n-k)}(x^{n+\alpha})^{(k)}=e^{-x}(-1)^{n-k}\sum\limits_{k=0}^n\binom{n}{k}(x^{n+\alpha})^{(k)}$$
Edit: maybe it is helpful to rewrite the gamma function
$$\frac{\Gamma(n+\alpha+1)}{n!}=\frac{1}{n!}\int_0^{\infty} x^{n+\alpha}e^{-x}\mathrm{d}x=\cdots =\frac{1}{n!}\int_0^\infty 
 \frac{e^x x^{-\alpha}}{n!}\left(
 \frac{\mathrm{d}^n}{\mathrm{d}x^n}(e^{-x}x^{n+\alpha}) 
 \right)^2\mathrm{d}x$$
 A: Too long for a comment
Suppose that $m>n$. Using integration by part
$$I_{mn}=\int_0^\infty w(x)L_n^{(\alpha)}L_m^{(\alpha)}\mathrm{d}x$$ $$=\int_0^\infty e^{-x}x^{\alpha}\left(\frac{e^x x^{-\alpha}}{n!}\frac{\mathrm{d}^n}{\mathrm{d}x^n}(e^{-x}x^{n+\alpha}) \right)\left(\frac{e^x x^{-\alpha}}{m!}\frac{\mathrm{d}^m}{\mathrm{d}x^m}(e^{-x}x^{m+\alpha}) \right)\mathrm{d}x$$
$$=(-1)^m\int_0^\infty e^{-x}x^{m+\alpha}\frac{\mathrm{d}^m}{\mathrm{d}x^m}\Bigg(\frac{e^x x^{-\alpha}}{m!n!}\frac{\mathrm{d}^n}{\mathrm{d}x^n}(e^{-x}x^{n+\alpha})\Bigg)\mathrm{d}x$$
In the big parentheses, after differentiation and multiplication, there is a polynomial of degree $n$ and, therefore, the integral is zero for $m>n$.
For $m=n$
$$I_{nn}=\frac{(-1)^n}{(n!)^2}\int_0^\infty e^{-x}x^{n+\alpha}\frac{\mathrm{d}^n}{\mathrm{d}x^n}\Bigg(e^x x^{-\alpha}\frac{\mathrm{d}^n}{\mathrm{d}x^n}(e^{-x}x^{n+\alpha})\Bigg)\mathrm{d}x$$
We get a polynomial of degree $n$ in the parentheses, and only one term with the highest power (with the power $n$ - when we differentiated only $e^{-x}\, \,n $ times) of this polynomial survives the second $n$-times differentiation. All other terms vanish.
$$I_{nn}=\frac{(-1)^n}{(n!)^2}\int_0^\infty e^{-x}x^{n+\alpha}\frac{\mathrm{d}^n}{\mathrm{d}x^n}\Big((-1)^n\,x^n\Big)\mathrm{d}x=\frac{1}{n!}\int_0^\infty e^{-x}x^{n+\alpha}\mathrm{d}x=\frac{\Gamma(n+\alpha+1)}{n!}$$
