Laguerre polynomials is a kind of orthogonal polynomials whose inner product is zero. (Is this correct?)
To show that two Laguerre polynomials $L_n(x)$ and $L_m(x)$ are orthogonal, they must satisfy the integral $\int\limits_0^\infty e^{-x} L_m (x) L_n (x)dx=0$ with respect to the weight function $e^{-x}$,for $m$ not equal to $n$.
Can you prove this for me?
Recall that
$$
L_n(x) = e^x\frac{d^n}{dx^n}(x^ne^{-x})
$$
Let $m\in \mathbb{N}$, the integration by parts gives
$$
\int_0^{\infty}e^{-x}x^m L_n(x)dx = \int_0^{\infty}x^m\frac{d^n}{dx^n}(x^ne^{-x})dx = (-1)^mm!\int_0^{\infty}\frac{d^{n-m}}{dx^{n-m}}(x^ne^{-x})dx
$$
If $m<n$, we get
$$
\int_0^{\infty}e^{-x}x^m L_n(x)dx = (-1)^mm!\int_0^{\infty}\frac{d^{n-m-1}}{dx^{n-m-1}}(x^ne^{-x})dx = 0
$$
Now $L_m(x)$ is a polynomial of degree $m$, so
$$
\int_0^{\infty}e^{-x}L_m(x)L_n(x)dx = 0 \text{ if } m<n
$$
Similarly, one can also check that
$$
\int_0^{\infty}e^{-x}L_n(x)L_n(x)dx = (n!)^2
$$
And so
$$
\frac{L_n(x)}{n!}
$$
forms an orthonormal system with respect to the weighted measure. $e^{-x}dx$
p.s.I fixed calculation fault in L_n Def.
