Orthogonality of the Laguerre polynomials

The orthogonality of the Laguerre polynomials states that the integral of the product of two distinct Laguerre polynomials over a specified interval is zero.

It is defined mathematically as:

$\int_0^{\infty}L_n(x)L_m(x)e^{-x}dx=0\;\;\;\;n\neq m\;\;\;\;\;\;\;438$

where $e^{-x}$ is known as a weight function.

Question

Why is the weight function included? Can the orthogonality of the Laguerre polynomials be defined as $\int_0^{\infty}L_n(x)L_m(x)dx=0$ , where $n\neq m$ ? Why are the limits of integration from 0 to $\infty$ ?

Answer

The weight function is an integral part of the orthogonality definition of Laguerre polynomials due to its role in ensuring convergence and its practical applications. It is often tied to specific problems, such as those in quantum mechanics. Omitting the weight function would sever this connection and could potentially alter the orthogonality properties of the polynomials. Therefore, defining orthogonality without the weight function would generally be invalid and would not reflect the intended use and properties of the Laguerre polynomials.

The weight function naturally defines the integration range because $e^{-x}\rightarrow 0$ as $x\rightarrow\infty$ , making the integral convergent over this range. This range is also connected to specific problems, such as the radial part of the wave functions in quantum mechanics. In the context of the hydrogen atom, $x$ represents a distance, which is always non-negative.

To prove eq438, we multiply eq430 for $L_n(x)$ and $L_m(x)$ to give

$\frac{e^{-\frac{xt}{1-t}-\frac{xs}{1-s}}}{(1-t)(1-s)}=\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}L_n(x)L_m(x)t^ns^m$

Multiplying through by $e^{-x}$ and integrating with respect to $x$ yields

$\int_0^{\infty}\frac{e^{-(\frac{t}{1-t}+\frac{s}{1-s}+1)x}}{(1-t)(1-s)}dx=\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}\biggr\{\int_0^{\infty}e^{-x}L_n(x)L_m(x)dx\biggr\}t^ns^m$

which simplifies to

$\frac{1}{1-st}=\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}\biggr\{\int_0^{\infty}e^{-x}L_n(x)L_m(x)dx\biggr\}t^ns^m$

Expressing the LHS as a binomial series gives

$\sum_{n=0}^{\infty}(st)^n=\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}\biggr\{\int_0^{\infty}e^{-x}L_n(x)L_m(x)dx\biggr\}t^ns^m$

Equating the coefficients of $t^ns^m$ when $m\neq n$ gives eq438. If we further equate the coefficients of $t^ns^m$ when $m=n$ , we have $\int_0^{\infty}L_n(x)L_m(x)e^{-x}dx=1$ .