Since the Laguerre polynomial is deeply connected to the hydrogen-like electron orbital. The radial solution is

$\displaystyle R_{nl}(r) = A \frac{1}{r} \exp(- \frac{x}{2}) x^{l+1}L_{n-l-1}^{2l+1}(x)$

$\displaystyle x = \frac{2Z}{na_0} r$

From the normalization condition, we can get the coefficient A

$\displaystyle \int_0^{\infty} R_{nl}^2(r) r^2 dr$

Beside of calculating the normalization factor, the general expectation value also need to evaluation of the integration of the Laguerre polynomial.

In here, we will show the calculation for 3 integrals:

$\displaystyle \int_0^{\infty} x^{\alpha+m} e^{-x} L_n^\alpha(x) dx, n,\alpha\geq0$

$\displaystyle \int_0^{\infty} x^{\alpha+k} e^{-x} (L_n^\alpha(x))^2 dx$

$\displaystyle \int_0^{\infty} x^{\alpha +k} e^{-x} L_m^{\beta}(x) L_n^\alpha(x) dx$

Using the Rodrigues formula

$\displaystyle L_n^\alpha(x) = \frac{1}{n!} x^{-\alpha} e^x \frac{d^n}{dx^n}(e^{-x}x^{n+\alpha})$

$\displaystyle \int_0^{\infty} x^{\alpha+m} e^{-x} L_n^\alpha(x) dx = \frac{1}{n!} \int_0^{\infty} x^{m} \frac{d^n}{dx^n}(e^{-x}x^{n+\alpha})dx$

For $m=0$,

$\displaystyle \int_0^{\infty} x^{\alpha+m} e^{-x} L_n^\alpha(x) dx \\ = \frac{1}{n!} \int_0^{\infty} \frac{d^n}{dx^n}(e^{-x}x^{n+\alpha})dx \\ = \frac{1}{n!} (F_{n-1}(\infty)-F_{n-1}(0) )$

$\displaystyle F_{n-1}(x) = \int_0^{\infty} \frac{d^n}{dx^n}(e^{-x}x^{n+\alpha})dx \\ = \sum_{i=0}^{n-1} (-1)^i \begin{pmatrix} n-1 \\ i \end{pmatrix} \frac{(n+\alpha)!}{(\alpha+1+i)!} x^{\alpha+1+i} e^{-x}$

It is obvious that $F(\infty) = 0$ due to the $e^{-x}$. Since $\alpha>0$, then $x^{\alpha+1+i} = 0$ at $x = 0$. Thus,

$\displaystyle \int_0^{\infty} x^{\alpha+m} e^{-x} L_n^\alpha(x) dx = 0 , m=0$

Using integration by path,

$\displaystyle \int_0^{\infty} x^{m} d \left(\frac{d^{n-1}}{dx^{n-1}}(e^{-x}x^{n+\alpha}) \right)dx \\ = [ x^{m} \frac{d^{n-1}}{dx^{n-1}}(e^{-x}x^{n+\alpha}) ]_0^{\infty} - (m) \int_0^{\infty} x^{m-1} \frac{d^{n-1}}{dx^{n-1}}(e^{-x}x^{n+\alpha})dx$

For the same reason, the first term is zero for $\alpha + m \geq 0$. Repeat the integration by path $k$ times,

$\displaystyle \int_0^{\infty} x^{m} d \left(\frac{d^{n-1}}{dx^{n-1}}(e^{-x}x^{n+\alpha}) \right)dx \\ = (-1)^k \frac{(m)!}{(m-k)!} \int_0^{\infty} x^{m-k} \frac{d^{n-k}}{dx^{n-k}}(e^{-x}x^{n+\alpha})dx$

When $m \geq n || 0 > m$, $k$ can go to $n$, then,

$\displaystyle \int_0^{\infty} x^{m} d \left(\frac{d^{n-1}}{dx^{n-1}}(e^{-x}x^{n+\alpha}) \right)dx \\ = (-1)^{n} \frac{(m)!}{(m-n)!} \int_0^{\infty} x^{m-n}(e^{-x}x^{n+\alpha})dx \\ = (-1)^{n} \frac{(m)!}{(m-n)!} (\alpha+m)!$

When $n > m > 0$, $k$ can only go to $m$. then,

$\displaystyle \int_0^{\infty} x^{m} d \left(\frac{d^{n-1}}{dx^{n-1}}(e^{-x}x^{n+\alpha}) \right)dx \\ = (-1)^{m} (m)! \int_0^{\infty} \frac{d^{n-k}}{dx^{n-k}}(e^{-x}x^{n+\alpha})dx \\ = (-1)^{m} (m)! \frac{d^{n-m-1}}{dx^{n-m-1}}(e^{-x}x^{n+\alpha}) = 0$

For the similar result as $F_{n-1}(x)$, the result is zero.

To summarize, the following formula suitable for all $m > n || 0>m$

$\displaystyle \int_0^{\infty} x^{\alpha+m} e^{-x} L_n^\alpha(x) dx = (-1)^n (\alpha+m)! \begin{pmatrix} m \\ n \end{pmatrix}$

To evaluate the integral,

$\displaystyle \int_0^{\infty} x^{\alpha+k} e^{-x} (L_n^\alpha(x))^2 dx$

We need the know that the Laguerre polynomial can be expressed as,

$\displaystyle L_n^\alpha(x) = \sum_{i=0}^{n} (-1)^i \begin{pmatrix} n+\alpha \\ n-i \end{pmatrix} \frac{x^i}{i!}$

Thus, we can evaluate

$\displaystyle \int_0^{\infty} x^{\alpha+k} e^{-x} x^i L_n^\alpha(x) dx$

Then combine everything,

$\displaystyle \int_0^{\infty} x^{\alpha+k+i} e^{-x} L_n^\alpha(x) dx = (-1)^n (\alpha+i+k)! \begin{pmatrix} i+k \\ n \end{pmatrix}$

put inside the series expression,

$\displaystyle \int_0^{\infty} x^{\alpha+k} e^{-x} (L_n^\alpha(x))^2 dx = \sum_{i=0}^{n} (-1)^{i+n} \begin{pmatrix} n+\alpha \\ n-i \end{pmatrix} \begin{pmatrix} i+k \\ n \end{pmatrix} \frac{(\alpha+k+i)!}{i!}$

In general, $i + k \geq n || 0 > i+k$.

For $k = 0$, the sum only has 1 term for $i = n$

$\displaystyle \int_0^{\infty} x^{\alpha} e^{-x} (L_n^\alpha(x))^2 dx = \frac{(\alpha+n)!}{n!}$

Using the formula, for $k=1$, we have

$\displaystyle \int_0^{\infty} x^{\alpha+1} e^{-x} (L_n^\alpha(x))^2 dx = \frac{(\alpha+n)!}{n!} (2n+\alpha +1)$

To evaluate the expectation value of radial function

$\displaystyle \langle R_{nl} | \frac{1}{r^m} | R_{nl} \rangle$

We have to calculate the integral,

$\displaystyle \int_0^\infty e^{-x} x^{2l+2-m} \left( L_{n-l-1}^{2l+1} (x) \right)^2 dx$

For $m = 0$, we get the normalization constant,

$\displaystyle A^2 \frac{na_0}{2Z} \int_0^{\infty} e^{-x} x^{2l+2} \left( L_{n-l-1}^{2l+1}(x) \right)^2 dx = 1$

$\displaystyle \langle \frac{1}{r} \rangle = \frac{1}{n^2} \frac{Z}{a_0}$

$\displaystyle \langle \frac{1}{r^2} \rangle = \frac{2}{n^3(2l+1)} \frac{Z^2}{a_0^2}$

In general,

$\displaystyle \langle \frac{1}{r^m} \rangle = \left(\frac{Z}{a_0}\right)^m \frac{2^{m-1}(n-l-1)!}{n^{m+1}(n+l)!} G_{nlm}$

$\displaystyle G_{nlm} = \sum_{i=0}^{n-l-1} (-1)^{n-l-1+i} \begin{pmatrix} n+l \\ n-l-1+i \end{pmatrix} \begin{pmatrix} 1-m+i \\ n-l-1 \end{pmatrix} \frac{(2l+2-m+i)!}{i!}$

For the integral, for $m \neq n$

$\displaystyle \int_0^{\infty} x^{\alpha +k} e^{-x} L_m^{\beta}(x) L_n^\alpha(x) dx$

When $k = 0$, when $m < n$, the integral is zero.

When $k \neq 0, m

$\displaystyle \int_0^{\infty} x^{\alpha +k} e^{-x} L_m^{\beta}(x) L_n^\alpha(x) dx \\ = \sum_{i=0}^{m} \frac{(-1)^i}{i!}\begin{pmatrix} m+\beta \\ m-i \end{pmatrix} \int_0^{\infty} x^{\alpha+k+i} e^{-x} L_n^\alpha(x) dx \\ = \sum_{i=0}^{m} \frac{(-1)^{i+n}}{i!}\begin{pmatrix} m+\beta \\ m-i \end{pmatrix} \begin{pmatrix} k+i \\ n \end{pmatrix} (\alpha+k+i)!$

The sum only non-zero when $m\geq i > n-k$ or $-k > i \geq 0$

This shows that the Laguerre polynomial is orthogonal with weighting $e^{-x} x^{\alpha}$.