The contour integral representation of Laguerre polynomials is
$$L_n^{(\alpha)}(x)=\frac{1}{2\pi i}\oint_C\frac{e^{-\frac{x t}{1-t}}}{(1-t)^{\alpha+1}\,t^{n+1}} \; dt$$

How can one get here starting from the orthogonality condition

$$\int_0^\infty x^\alpha e^{-x} L_n^{(\alpha)}(x)L_m^{(\alpha)}(x)dx=\frac{\Gamma(n+\alpha+1)}{n!} \delta_{n,m}$$

without using other properties (such as Rodrigues' formula, recurrence relation... etc.)?

  • $\begingroup$ anything unclear ? $\endgroup$
    – reuns
    Jun 7, 2019 at 21:26
  • $\begingroup$ Thanks for your answer. Sorry for the late reply. How would it work in more general setting such as $\int_a^b w(x)P_m(x)P_n(x)dx = h_n \delta_{n,m}$? $\endgroup$
    – ukar
    Jun 23, 2019 at 18:03

1 Answer 1


There are a lot of basis $\left(\frac{b_n}{\sqrt{\langle b_n,b_n\rangle}}\right)_{n \ge 0}$ orthonormal for the $\langle u,v \rangle = \int_0^\infty u(x)v(x) x^\alpha e^{-x}dx$ inner product, but (up to sign) there is only one such basis such that $b_n$ is a real polynomial of degree $n$, which is $b_0 = 1$, $b_{n+1} = x^{n+1} - \sum_{m=0}^n \langle x^{n+1}, \frac{b_m}{\langle b_m,b_m\rangle}\rangle b_m$.

Let $f(x,t) = \frac{e^{-\frac{x t}{1-t}}}{(1-t)^{\alpha+1}}= \sum_{n=0}^\infty a_n(x) t^n$ then $a_n$ is a real polynomial of degree $n$ and $$\sum_{n,m} \langle a_n,a_m\rangle t^nu^m=\int_0^\infty f(x,t)f(x,u)x^\alpha e^{-x}dx\\ = (\frac{t}{1-t}+\frac{t}{1-u}+1)^{-a-1}((1-t)(1-u))^{-a-1}\Gamma(a+1)=(t(1-u)+u(1-t)+(1-u)(1-t))^{-a-1}\Gamma(a+1) = \Gamma(a+1) (1-tu)^{-a-1}\\= g(tu)$$

Which means $\langle a_n,a_m\rangle = 0$ whenever $n \ne m$, thus $\frac{a_n}{\sqrt{\langle a_n,a_n\rangle}} = \pm \frac{b_n}{\sqrt{\langle b_n,b_n\rangle}}$.


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