orthogonality of Laguerre polynomials

We use the definition of Laguerre polynomials $L_n(x)$ via their Rodrigues formula (http://planetmath.org/RodriguesFormula)

$L_n(x):=e^x{\displaystyle \frac{d^n}{d{x}^n}}(x^n{e}^{-x}).$(1)

The polynomials (1) themselves are not orthogonal to each other, but the expressions $e^{-\frac{x}{2}}{L}_n(x)$  ($n=0,\mathrm{\hspace{0.17em}1},\mathrm{\hspace{0.17em}2},\mathrm{\dots }$) are orthogonal (http://planetmath.org/OrthogonalPolynomials) on the interval from 0 to $\mathrm{\infty }$, i.e. the polynomials are orthogonal with respect to the weighting function $e^{-x}$ on that interval, as is seen in the following.

Let $m$ be another nonnegative integer.  We integrate by parts (http://planetmath.org/IntegrationByParts) $m$ times in

$${\int }_0^{\mathrm{\infty }}{e}^{-x}{x}^m{L}_n(x)𝑑x={\int }_0^{\mathrm{\infty }}{x}^m\frac{d^n}{d{x}^n}(x^n{e}^{-x})𝑑x={(-1)}^{m}m!{\int }_0^{\mathrm{\infty }}\frac{d^{n-m}}{d{x}^{n-m}}(x^m{e}^{-x})𝑑x.$$

When  ,  this yields

${\displaystyle {\int }_0^{\mathrm{\infty }}}{e}^{-x}{x}^m{L}_n(x)𝑑x={(-1)}^{m}m!\underset{x=0}{\overset{\mathrm{\infty }}{/}}{\displaystyle \frac{d^{n-m-1}}{d{x}^{n-m-1}}}(x^m{e}^{-x})=\mathrm{\hspace{0.33em}0}.$(2)

and for  $m=n$  it gives

${\displaystyle {\int }_0^{\mathrm{\infty }}}{e}^{-x}{x}^m{L}_n(x)𝑑x={(-1)}^{n}n!{\displaystyle {\int }_0^{\mathrm{\infty }}}{x}^n{e}^{-x}𝑑x={(-1)}^n{(n!)}^2.$(3)

The result (2) implies, because $L_m(x)$ is a polynomial of degree $m$, that

whence also

${\displaystyle {\int }_0^{\mathrm{\infty }}}{e}^{-x}{L}_m(x)L_n(x)dx=\mathrm{\hspace{0.33em}0}\mathit{\hspace{1em}\hspace{1em}}(m\ne n).$(4)

Thus the orthogonality has been shown.  Therefore, since the leading term of $L_n(x)$ is ${(-1)}^n{x}^n$, we infer by (3) and (4) that

$${\int }_0^{\mathrm{\infty }}{e}^{-x}{[L_n(x)]}^2𝑑x={(-1)}^n{\int }_0^{\mathrm{\infty }}{e}^{-x}{x}^n{L}_n(x)𝑑s={(n!)}^2,$$

so that the expressions $\frac{L_n(x)}{n!}$ form a system of orthonormal polynomials.