I'm trying to solve this integral
$\int_{0}^{\infty} L^n_p L^n_{p'} e^{-x} x^{n-1} dx = \dfrac{1}{n} \dfrac{(p!)^3}{(p-n)!} \delta_{pp'}$
I started with integration by parts where
$u = $ $L^n_p L^n_{p'} e^{-x}$
$dv = x^{n-1} $
After some simplifications, I got the following:$\\$
$ I = \frac{-1}{n} \int_{0}^{\infty} \frac{d}{dx} L^n_p L^n_{p'} e^{-x} x^{n} dx + \frac{-1}{n} \int_{0}^{\infty} \frac{d}{dx} L^n_{p'} L^n_p e^{-x} x^{n} dx + \frac{1}{n} \int_{0}^{\infty} L^n_p L^n_{p'} e^{-x} x^{n} dx$
The integral
$\frac{1}{n} \int_{0}^{\infty} L^n_p L^n_{p'} e^{-x} x^{n} dx$
is already known, which is equal to
$\dfrac{(p!)^3}{(p-n)!} \delta_{pp'}$
after multiply it with $\frac{1}{n}$ it becomes
$\dfrac{1}{n} \dfrac{(p!)^3}{(p-n)!} \delta_{pp'} $
All of that was okay, the issue here is with the rest integrals
$\frac{-1}{n} \int_{0}^{\infty} \frac{d}{dx} L^n_p L^n_{p'} e^{-x} x^{n} dx + \frac{-1}{n} \int_{0}^{\infty} \frac{d}{dx} L^n_{p'} L^n_p e^{-x} x^{n} dx $
I assumed that $p = p'$ so that this integral can be solved, so it becomes:
$\frac{-2}{n} \int_{0}^{\infty} \frac{d}{dx} L^n_p L^n_{p} e^{-x} x^{n} dx $
This integral must be equal to zero, so that the final result is $\dfrac{1}{n} \dfrac{(p!)^3}{(p-n)!} \delta_{pp'} $
But I'm not sure with respect to what should I integrate? It's obvious not with respect to p nor n, there's some other substitution, but I have no idea what should that be.