$\int_{0}^{\infty}e^{-x}p_n(x)p_m(x)=0$ when $p_n(x)=(-1)^nn!\sum_{k=0}^{n}\binom{n}{k}\frac{(-x)^k}{k!}$

Question

Let $n \in \mathbb N_0$. Consider the polynomials $p_n$ defined by

$$p_n(x)=(-1)^nn!\sum_{k=0}^{n}\binom{n}{k}\frac{(-x)^k}{k!}$$

I want to show that $\int_{0}^{\infty}e^{-x}p_n(x)p_m(x)=0$ for $n \neq m$.

The hint is that for $m,n \in \mathbb N_0$ we have:

$$\sum_{k=0}^{n}\sum_{j=0}^{m}(-1)^{j+k}\binom{j+k}{k}\binom{n}{k}\binom{m}{j}=\delta_{m,n}$$

$p_n(x)p_m(x)=(-1)^{n+m}n!m!\sum_{k=0}^{n}\sum_{j=0}^{m}(-1)^{j+k}\binom{n}{k}\binom{m}{j}\frac{x^{j+k}}{j!k!}$. But I don't know how to continue or how I can bring this into a form to use the given hint

Answer 1

We obtain with $p_n(x)=(-1)^nn!\sum_{k=0}^n\binom{n}{k}\frac{(-x)^k}{k!}$:

\begin{align*} \color{blue}{\int_{0}^\infty}&\color{blue}{ e^{-x}p_n(x)p_m(x)\,dx}\\ &=\int_{0}^\infty e^{-x}\left((-1)^nn!\sum_{k=0}^n\binom{n}{k}\frac{(-x)^k}{k!}\right)\left((-1)^mm!\sum_{j=0}^m\binom{m}{j}\frac{(-x)^j}{j!}\right)\,dx\\ &=(-1)^{n+m}n!m!\sum_{k=0}^n\sum_{j=0}^m\binom{n}{k}\binom{m}{j}\frac{(-1)^{k+j}}{k!j!}\int_{0}^\infty e^{-x}x^{k+j}\,dx\tag{1}\\ &=(-1)^{n+m}n!m!\sum_{k=0}^n\sum_{j=0}^m\binom{j+k}{k}\binom{n}{k}\binom{m}{j}\frac{(-1)^{k+j}}{(k+j)!}\int_{0}^\infty e^{-x}x^{k+j}\,dx\tag{2}\\ &=(-1)^{n+m}n!m!\sum_{k=0}^n\sum_{j=0}^m\binom{j+k}{k}\binom{n}{k}\binom{m}{j}(-1)^{k+j}\tag{3}\\ &\,\,\color{blue}{=(-1)^{n+m}n!m!\delta_{m,n}}\tag{4}\\ \end{align*}

and the claim follows.

Comment:

• In (1) we do some rearrangements and use the linearity of the integral- and sigma-operator.

• In (2) we use the binomial identity $\binom{j+k}{j}=\frac{(j+k)!}{j!k!}$.

• In (3) we use the identity $\int_{0}^\infty e^{-x}x^n\,dx=n!$ which can be shown by integration by parts and induction.

• In (4) we apply the given hint.

Answer 2

The polynomials correspond to the Laguerre polynomials. The identity is the result of their orthogonality: See https://en.wikipedia.org/wiki/Laguerre_polynomials#Orthogonality.

Answer 3

In order to show the orthogonality of $p_n$ and $p_m$ (with respect to the inner product $\langle f,g\rangle = \int_{0}^{+\infty}f(x)g(x)e^{-x}\,dx$) for $n\neq m$ it is enough to prove the orthogonality of $p_n(x)$ and $x^j$ for $j<\deg(p_n)=n$. Now

$$\langle p_n(x), x^j \rangle = (-1)^n n!\sum_{k=0}^{n}\binom{n}{k}\frac{(-1)^k}{k!}\int_{0}^{+\infty}x^{k+j}e^{-x}\,dx $$ equals $$ (-1)^n n!j!\sum_{k=0}^{n}\binom{n}{k}(-1)^k \binom{k+j}{j} $$ where the sum is the $n$-th power of the forward difference operator $\delta:q(x)\mapsto q(x+1)-q(x)$ applied to the polynomial $q(x)=\binom{x+j}{j}$. Since $\deg q = j<n$, the RHS is zero.