How to show orthogonality of associated Laguerre polynomials? $L_p^q(x)$ is associated Laguerre polynomials and defined as below
$$
L_p^q(x)=\frac{x^{-q}e^x}{p!} \frac{d^p}{dx^p}\left( x^{p+q} e^{-x}\right)
$$
I want to show the orthogonality of it. I mean, how to show
$$
\int_0^\infty e^{-x}xL_p^1(x)L_r^1(x) dx = {}_{n+r}P_{k} \ \ \delta_{p,r} 
$$
Now, ${}_n P_k$ is defined as below
$$
{}_a P_b = \frac{a!}{(a-b)!}
$$
 A: Not sure whether it's too late, but here's my answer. I'm a bit confused by your notation (where did $ n$ come from?), so I'll use mine. 
You want to start with Rodrigues form:
    \begin{align}
  L^k_n (x) &= \frac{e^x x^{-k}}{n!} \frac{d^n}{dx^n} \left( x^{n+k} e^{-x} \right)  \\
  &= \frac{e^x x^{-k}}{2 \pi i} \oint \frac{s^{n+k} e^{-s}}{(s-x)^{n+1}} ds
 \end{align} 
    Now, make a substitution: 
    $$
 z = \frac{s-x}{s} 
 $$
$$
 L_n^k (x) =  \frac{1}{2 \pi i} \oint \frac{e^{ - \frac{xz}{1-z} } }{(1-z)^{k+1} z^{n+1}} dz
 $$
    With the help of Cauchy integral formula, we obtain the generating function of associated Laguerre Polynomials:
    $$
 \sum_{ n=0 } ^\infty L_n ^k (x) t^n = \frac{e^{ - \frac{xt}{1-t} } }{(1-t)^{k+1}}
 $$
    Now, to prove the orthogonality of the polynomials, we put two generating functions inside the inner product integral:
    \begin{align}
  & \int _0 ^\infty \frac{e^{- \frac{xt}{1-t}}}{(1-t)^{k+1}} \frac{e^{- \frac{xs}{1-s}}}{(1-s)^{k+1}} e^{-x} x^k dx \\
  &= \frac{1}{(1-t)^{k+1} (1-s)^{k+1}} \int _0 ^\infty e^{ - \frac{1-ts}{(1-t)(1-s)}x } x^{k} dx \\
  &= \frac{1}{(1-ts )^{k+1}} \int _0 ^\infty e^{-x'} x'^{k} dx'
 \end{align} 
    Here, we make another substitution, where
    $$
 x' = \frac{1-ts}{(1-t)(1-s)} x
 $$
    Notice that the RHS is the Gamma function:
    $$
 \int _0 ^\infty e^{-x'} x'^{k} dx' = k!
 $$
    and make use of the differentiation of geometric series:
    $$
 \frac{1}{(1-ts)^{k+1}} = \sum _{i=0} ^\infty \frac{(i+k)!}{i! k!} (ts)^i
 $$
    We get:
    \begin{align}
  & \frac{1}{(1-ts )^{k+1}} \int _0 ^\infty e^{-x'} x'^{k} dx'
  &= k! \sum _{i=0} ^\infty \frac{(i+k)!}{i! k!} (ts)^i \\
  &= \sum_{i=0}^\infty \frac{(i+k)!}{i!} t^i s^i
 \end{align} 
    But this result was deduced from the multiplication of two generating function, therefore:
    \begin{align}
  & \int _0 ^\infty \frac{e^{- \frac{xt}{1-t}}}{(1-t)^{k+1}} \frac{e^{- \frac{xs}{1-s}}}{(1-s)^{k+1}} e^{-x} x^k dx \\
  &= \sum_{n=0}^\infty \sum_{m=0}^\infty \left< L^k_n (x)| L^k _m (x)\right> t^n s^m
 \end{align} 
    Where
    $$
 \left< f(x) | g(x) \right> := \int_0 ^\infty e^{-x} x^k f(x) g(x) dx
 $$
    Compare the coefficients of the two results, note that the innerproduct is 0 whenever $ n \neq m $, we get:
    $$
 \left< L^k_n (x) | L^k_m (x) \right> = \frac{(n+k)!}{n!} \delta_{nm}
 $$
A: I think the following method is conceptually easier than using contour integrals and generating functions. Let us assume that $\alpha>-1$ to have convergent integrals below. Also, let
$$
L_n^\alpha(x)=\frac{1}{n!}x^{-\alpha}e^x\frac{d^n}{dx^n}\bigl(x^{\alpha+n}e^{-x}\bigr).
$$
Assume that $p$ is a polynomial of degree. Then
$$
\langle p,L_n^\alpha\rangle=\int_0^{+\infty} p(x)L_n^\alpha(x)x^\alpha e^{-x}\,dx=\frac{1}{n!}\int_0^{+\infty}p(x)\frac{d^n}{dx^n}\bigl(x^{\alpha+n}e^{-x}\bigr)\,dx.
$$
Integrating by parts $n$ times, we find (the boundary terms vanish at $0$ thanks to the monomial term and at $+\infty$ thanks to $e^{-x}$)
$$
\langle p,L_n^\alpha\rangle
=
\frac{(-1)^n}{n!}\int_0^{+\infty}p^{(n)}(x)x^{\alpha+n}e^{-x}\,dx,\tag{1}
$$
If $p(x)=L_m^\alpha(x)$, $m<n$, we find that its $n$th derivative is constantly equal to zero, and
$$
\langle L_m^{\alpha},L_n^\alpha\rangle=0.
$$
It remains to consider the case $p(x)=L_n^{\alpha}(x)$. We claim that now $p^{(n)}(x)=(-1)^n$, and we use the Leibniz formula to show it,
$$
L_n^\alpha(x)
=x^{-\alpha}e^x\sum_{k=0}^n\frac{1}{k!(n-k)!}\frac{d^ke^{-x}}{dx^k}\frac{d^{n-k}x^{\alpha+n}}{dx^{n-k}}.
$$
To find the coefficient in front of $x^n$ we are lead to look at the term where all derivatives fall on the exponential term, i.e. when $k=n$. We find that this term becomes $\frac{(-1)^n}{n!}x^n$, and it follows that $p^{(n)}(x)=(-1)^n$. Inserting this into (1) above, we find that
$$
\langle L_n^\alpha,L_n^\alpha\rangle
=
\frac{(-1)^n}{n!}\int_0^{+\infty}(-1)^nx^{\alpha+n}e^{-x}\,dx=\frac{\Gamma(\alpha+n+1)}{n!}=\frac{(\alpha+n)!}{n!},
$$
where the last equality only makes sense if $\alpha$ is an integer.
