Some identities on Bernoulli and Euler polynomials arising from the orthogonality of Laguerre polynomials

Abstract

In this paper, we derive some interesting identities on Bernoulli and Euler polynomials by using the orthogonal property of Laguerre polynomials.

1 Introduction

The generalized Laguerre polynomials are defined by

$$\frac{exp(-\frac{xt}{1-t})}{{(1-t)}^{\alpha +1}}=\sum _{n=0}^{\mathrm{\infty }}{L}_n^{\alpha }(x)t^n(\alpha \in \mathbb{Q}\text{ with }\alpha >-1).$$

(1.1)

From (1.1), we note that

$$L_n^{\alpha }(x)=\sum _{r=0}^n\frac{{(-1)}^r\left(\genfrac{}{}{0ex}{}{n+\alpha }{n-r}\right)x^r}{r!}\left(\text{see [1–3]}\right).$$

(1.2)

By (1.2), we see that $L_n^{\alpha }(x)$ is a polynomial with degree n. It is well known that Rodrigues’ formula for $L_n^{\alpha }(x)$ is given by

$$L_n^{\alpha }(x)=\frac{x^{-\alpha }{e}^x}{n!}\left(\frac{d^n}{d{x}^n}\left(e^{-x}{x}^{n+\alpha }\right)\right)\left(\text{see [1–3]}\right).$$

(1.3)

From (1.3) and a part of integration, we note that

$${\int }_0^{\mathrm{\infty }}{x}^{\alpha }{e}^{-x}{L}_m^{\alpha }(x)L_n^{\alpha }(x)dx=\frac{\mathrm{\Gamma }(\alpha +n+1)}{n!}{\delta }_{m,n},$$

(1.4)

where ${\delta }_{m,n}$ is a Kronecker symbol. As is well known, Bernoulli polynomials are defined by the generating function to be

$$\frac{t}{e^t-1}{e}^{xt}=e^{B(x)t}=\sum _{n=0}^{\mathrm{\infty }}{B}_n(x)\frac{t^n}{n!}\left(\text{see [1–29]}\right),$$

(1.5)

with the usual convention about replacing $B^n(x)$ by $B_n(x)$.

In the special case, $x=0$, $B_n(0)=B_n$ are called the n th Bernoulli numbers. By (1.5), we get

$$B_n(x)=\sum _{l=0}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)B_{n-l}{x}^l\left(\text{see [1–29]}\right).$$

(1.6)

The Euler numbers are defined by

$$E_0=1,{(E+1)}^n+E_n=2{\delta }_{0,n}\left(\text{see [27, 28]}\right),$$

(1.7)

with the usual convention about replacing $E^n$ by $E_n$.

In the viewpoint of (1.6), the Euler polynomials are also defined by

$$E_n(x)={(E+x)}^n=\sum _{l=0}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)E_{n-l}{x}^l\left(\text{see [11–24]}\right).$$

(1.8)

From (1.7) and (1.8), we note that the generating function of the Euler polynomial is given by

$$\frac{2}{e^t+1}{e}^{xt}=e^{E(x)t}=\sum _{n=0}^{\mathrm{\infty }}{E}_n(x)\frac{t^n}{n!}\left(\text{see [15–29]}\right).$$

(1.9)

By (1.5) and (1.9), we get

$$\frac{2}{e^t+1}{e}^{xt}=\frac{1}{t}(2-2\frac{2}{e^t+1})\left(\frac{t{e}^{xt}}{e^t-1}\right)=-2\sum _{n=0}^{\mathrm{\infty }}(\sum _{l=0}^n\frac{E_{l+1}}{l+1}\left(\genfrac{}{}{0ex}{}{n}{l}\right)B_{n-l}(x))\frac{t^n}{n!}.$$

(1.10)

Thus, by (1.10), we see that

$$E_n(x)=-2\sum _{l=0}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)\frac{E_{l+1}}{l+1}{B}_{n-l}(x).$$

(1.11)

By (1.7) and (1.8), we easily get

$$\frac{t}{e^t-1}{e}^{xt}=\frac{t}{2}\left(\frac{2e^{xt}}{e^t+1}\right)+\left(\frac{t}{e^t-1}\right)\left(\frac{2e^{x}t}{e^t+1}\right).$$

(1.12)

Thus, by (1.12), we see that

$$B_n(x)=\sum _{k=0,k\ne 1}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right)B_k{E}_{n-k}(x).$$

(1.13)

Throughout this paper, we assume that $\alpha \in \mathbb{Q}$ with $\alpha >-1$. Let ${\mathbb{P}}_n=\{p(x)\in \mathbb{Q}[x]|degp(x)\le n\}$ be the inner product space with the inner product

$$〈p(x),q(x)〉={\int }_0^{\mathrm{\infty }}{x}^{\alpha }{e}^{-x}p(x)q(x)dx,$$

where $p(x),q(x)\in {\mathbb{P}}_n$. From (1.4), we note that $\{L_0^{\alpha }(x),L_1^{\alpha }(x),\dots ,L_n^{\alpha }(x)\}$ is an orthogonal basis for ${\mathbb{P}}_n$.

In this paper, we give some interesting identities on Bernoulli and Euler polynomials which can be derived by an orthogonal basis $\{L_0^{\alpha }(x),L_1^{\alpha }(x),\dots ,L_n^{\alpha }(x)\}$ for ${\mathbb{P}}_n$.

2 Some identities on Bernoulli and Euler polynomials

Let $p(x)\in {\mathbb{P}}_n$. Then $p(x)$ can be generated by $\{L_0^{\alpha }(x),L_1^{\alpha }(x),\dots ,L_n^{\alpha }(x)\}$ in ${\mathbb{P}}_n$ to be

$$p(x)=\sum _{k=0}^n{C}_k{L}_k^{\alpha }(x),$$

(2.1)

where

$$\begin{array}{rcl}〈p(x),L_k^{\alpha }(x)〉& =& C_k〈L_k^{\alpha }(x),L_k^{\alpha }(x)〉\\ =& C_k{\int }_0^{\mathrm{\infty }}{x}^{\alpha }{e}^{-x}{L}_k^{\alpha }(x)L_k^{\alpha }(x)dx\\ =& C_k\frac{\mathrm{\Gamma }(\alpha +k+1)}{k!}.\end{array}$$

(2.2)

From (2.2), we note that

$$\begin{array}{rcl}{C}_k& =& \frac{k!}{\mathrm{\Gamma }(\alpha +k+1)}〈p(x),L_k^{\alpha }(x)〉\\ =& \frac{k!}{\mathrm{\Gamma }(\alpha +k+1)}\frac{1}{k!}{\int }_0^{\mathrm{\infty }}\left(\frac{d^k}{d{x}^k}{x}^{k+\alpha }{e}^{-x}\right)p(x)dx\\ =& \frac{1}{\mathrm{\Gamma }(\alpha +k+1)}{\int }_0^{\mathrm{\infty }}\left(\frac{d^k}{d{x}^k}{x}^{k+\alpha }{e}^{-x}\right)p(x)dx.\end{array}$$

(2.3)

Let us take $p(x)={\sum }_{m=0,m\ne 1}^n\left(\genfrac{}{}{0ex}{}{n}{m}\right)B_m{E}_n-m(x)\in {\mathbb{P}}_n$. Then, from (2.3), we have

$$\begin{array}{rcl}{C}_k& =& \frac{1}{\mathrm{\Gamma }(\alpha +k+1)}{\int }_0^{\mathrm{\infty }}\left(\frac{d^k}{d{x}^k}{x}^{k+\alpha }{e}^{-x}\right)\sum _{m=0,m\ne 1}^n\left(\genfrac{}{}{0ex}{}{n}{m}\right)B_n{E}_{n-m}(x)dx\\ =& \frac{{(-1)}^k}{\mathrm{\Gamma }(\alpha +k+1)}\sum _{m=0,m\ne 1}^{n-k}\sum _{l=k}^{n-m}\left(\genfrac{}{}{0ex}{}{n}{m}\right)\left(\genfrac{}{}{0ex}{}{n-m}{l}\right)B_m{E}_{n-m-l}\frac{l!}{(l-k)!}{\int }_0^{\mathrm{\infty }}{x}^{l+\alpha }{e}^{-x}dx\\ =& \frac{{(-1)}^k}{\mathrm{\Gamma }(\alpha +k+1)}\sum _{m=0,m\ne 1}^{n-k}\sum _{l=k}^{n-m}\left(\genfrac{}{}{0ex}{}{n}{m}\right)\left(\genfrac{}{}{0ex}{}{n-m}{l}\right)B_m{E}_{n-m-l}\frac{l!}{(l-k)!}\mathrm{\Gamma }(l+\alpha +1)\\ =& {(-1)}^k\sum _{m=0,m\ne 1}^{n-k}\sum _{l=k}^{n-m}\left(\genfrac{}{}{0ex}{}{n}{m}\right)\left(\genfrac{}{}{0ex}{}{n-m}{l}\right)B_m{E}_{n-m-l}\frac{l!}{(l-k)!}\frac{(l+\alpha )(l+\alpha -1)\cdots \alpha }{(\alpha +k)(\alpha +k-1)\cdots \alpha }\\ =& {(-1)}^{k}n!\sum _{m=0,m\ne 1}^{n-k}\sum _{l=k}^{n-m}\frac{B_m}{m!}\frac{E_{n-m-l}}{(n-m-l)!}\left(\genfrac{}{}{0ex}{}{l+\alpha }{l-k}\right).\end{array}$$

(2.4)

Therefore, by (2.1) and (2.4), we obtain the following theorem.

Theorem 2.1 For $n\in {\mathbb{Z}}_{+}$, we have

From (1.13), we can derive the following corollary.

Corollary 2.2 For $n\in {\mathbb{Z}}_{+}$, we have

$$B_n(x)=n!\sum _{k=0}^n{(-1)}^k\left(\sum _{m=0,m\ne 1}^{n-k}\sum _{l=k}^{n-m}\frac{B_m}{m!}\frac{E_n-m-l}{(n-m-l)!}\left(\genfrac{}{}{0ex}{}{l+\alpha }{l-k}\right)\right)L_k^{\alpha }(x).$$

Let us take $p(x)={\sum }_{l=0}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)\frac{E_{l+1}}{l+1}{B}_{n-l}(x)$. By the same method, we get

$$\begin{array}{rcl}{C}_k& =& \frac{1}{\mathrm{\Gamma }(\alpha +k+1)}{\int }_0^{\mathrm{\infty }}\left(\frac{d^k}{d{x}^k}{x}^{k+\alpha }{e}^{-x}\right)\sum _{l=0}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)\frac{E_{l+1}}{l+1}{B}_{n-l}(x)dx\\ =& \frac{1}{\mathrm{\Gamma }(\alpha +k+1)}\sum _{l=0}^{n-k}\sum _{m=0}^{n-l}\left(\genfrac{}{}{0ex}{}{n}{l}\right)\left(\genfrac{}{}{0ex}{}{n-l}{m}\right)\frac{E_{l+1}}{l+1}{B}_{n-l-m}{\int }_0^{\mathrm{\infty }}\left(\frac{d^k}{d{x}^k}{x}^{k+\alpha }{e}^{-x}\right)x^{m}dx\\ =& \frac{{(-1)}^k}{\mathrm{\Gamma }(\alpha +k+1)}\sum _{l=0}^{n-k}\sum _{m=k}^{n-l}\left(\genfrac{}{}{0ex}{}{n}{l}\right)\left(\genfrac{}{}{0ex}{}{n-l}{m}\right)\frac{E_{l+1}}{l+1}{B}_{n-l-m}\frac{m!}{(m-k)!}\mathrm{\Gamma }(m+\alpha +1)\\ =& {(-1)}^k\sum _{l=0}^{n-k}\sum _{m=k}^{n-l}\left(\genfrac{}{}{0ex}{}{n}{l}\right)\left(\genfrac{}{}{0ex}{}{n-l}{m}\right)\frac{m!}{(m-k)!}\frac{E_{l+1}}{(l+1)}{B}_{n-l-m}\frac{(\alpha +m)(\alpha +m-1)\cdots \alpha }{(\alpha +k)(\alpha +k-1)\cdots \alpha }\\ =& {(-1)}^{k}n!\sum _{l=0}^{n-k}\sum _{m=k}^{n-l}\left(\genfrac{}{}{0ex}{}{\alpha +m}{m-k}\right)\frac{E_{l+1}}{(l+1)!}\frac{B_{n-l-m}}{(n-l-m)!}.\end{array}$$

(2.5)

Therefore, by (1.11), (2.1), and (2.5), we obtain the following theorem.

Theorem 2.3 For $n\in {\mathbb{Z}}_{+}$, we have

$$-\frac{E_n(x)}{2}=n!\sum _{k=0}^n{(-1)}^k\left(\sum _{l=0}^{n-k}\sum _{m=k}^{n-l}\left(\genfrac{}{}{0ex}{}{\alpha +m}{m-k}\right)\frac{E_{m+1}}{(m+1)!}\frac{B_{n-m-l}}{(n-m-l)!}\right)L_k^{\alpha }(x).$$

For $n\in \mathbb{N}$ with $n\ge 2$ and $m\in {\mathbb{Z}}_{+}$ with $n-m\ge 0$, we have

$$\begin{array}{rcl}{B}_{n-m}(x)B_m(x)& =& \sum _r\{\left(\genfrac{}{}{0ex}{}{n-m}{2r}\right)m+\left(\genfrac{}{}{0ex}{}{m}{2r}\right)(n-m)\}\frac{B_{2r}{B}_{n-2r}(x)}{n-2r}\\ +{(-1)}^{m+1}\frac{(n-m)!m!}{n!}{B}_n\in {\mathbb{P}}_n\left(\text{see [8]}\right).\end{array}$$

(2.6)

Let us take $p(x)=B_{n-m}(x)B_m(x)\in {\mathbb{P}}_n$. Then $p(x)$ can be generated by an orthogonal basis $\{L_0^{\alpha }(x),L_1^{\alpha }(x),\dots ,L_n^{\alpha }(x)\}$ in ${\mathbb{P}}_n$ to be

$$p(x)=\sum _{k=0}^n{C}_k{L}_k^{\alpha }(x).$$

(2.7)

From (2.3), (2.6), and (2.7), we note that

$$\begin{array}{rcl}{C}_k& =& \frac{1}{\mathrm{\Gamma }(\alpha +k+1)}{\int }_0^{\mathrm{\infty }}\left(\frac{d^k}{d{x}^k}{x}^{k+\alpha }{e}^{-x}\right)p(x)dx\\ =& \frac{1}{\mathrm{\Gamma }(\alpha +k+1)}\sum _{r=0}^{[\frac{n}{2}]}\{\left(\genfrac{}{}{0ex}{}{n-m}{2r}\right)m+\left(\genfrac{}{}{0ex}{}{m}{2r}\right)(n-m)\}\\ \times \frac{B_{2r}}{n-2r}{\int }_0^{\mathrm{\infty }}\left(\frac{d^k}{d{x}^k}{x}^{k+\alpha }{e}^{-x}\right)B_{n-2r}(x)dx\\ =& \frac{1}{\mathrm{\Gamma }(\alpha +k+1)}\sum _{r=0}^{[\frac{n}{2}]}\{\left(\genfrac{}{}{0ex}{}{n-m}{2r}\right)m+\left(\genfrac{}{}{0ex}{}{m}{2r}\right)(n-m)\}\frac{B_{2r}}{n-2r}\\ \times \sum _{l=0}^{n-2r}\left(\genfrac{}{}{0ex}{}{n-2r}{l}\right)B_{n-2r-l}{\int }_0^{\mathrm{\infty }}\left(\frac{d^k}{d{x}^k}{x}^{k+\alpha }{e}^{-x}\right)x^{l}dx\\ =& \frac{1}{\mathrm{\Gamma }(\alpha +k+1)}\sum _{r=0}^{[\frac{n}{2}]}\sum _{l=0}^{n-2r}\{\left(\genfrac{}{}{0ex}{}{n-m}{2r}\right)m+\left(\genfrac{}{}{0ex}{}{m}{2r}\right)(n-m)\}\left(\genfrac{}{}{0ex}{}{n-2r}{l}\right)\\ \times \frac{B_{2r}{B}_{n-2r-l}}{n-2r}{\int }_0^{\mathrm{\infty }}\left(\frac{d^k}{d{x}^k}{x}^{k+\alpha }{e}^{-x}\right)x^{l}dx\\ =& \frac{{(-1)}^k}{\mathrm{\Gamma }(\alpha +k+1)}\sum _{r=0}^{[\frac{n-k}{2}]}\sum _{l=k}^{n-2r}\{\left(\genfrac{}{}{0ex}{}{n-m}{2r}\right)m+\left(\genfrac{}{}{0ex}{}{m}{2r}\right)(n-m)\}\left(\genfrac{}{}{0ex}{}{n-2r}{l}\right)\\ \times \frac{B_{2r}{B}_{n-2r-l}l!}{(n-2r)(l-k)!}\mathrm{\Gamma }(\alpha +l+1).\end{array}$$

(2.8)

It is easy to show that

$$\begin{array}{rcl}\frac{\mathrm{\Gamma }(\alpha +l+1)}{\mathrm{\Gamma }(\alpha +k+1)(l-k)!}& =& \frac{(\alpha +l)(\alpha +l-1)\cdots \alpha \mathrm{\Gamma }(\alpha )}{(\alpha +k)(\alpha +k-1)\cdots \alpha \mathrm{\Gamma }(\alpha )(l-k)!}\\ =& \frac{(\alpha +l)(\alpha +l-1)\cdots (\alpha +k+1)}{(\alpha -k)!}=\left(\genfrac{}{}{0ex}{}{\alpha +l}{l-k}\right).\end{array}$$

(2.9)

By (2.8) and (2.9), we get

$$\begin{array}{rcl}{C}_k& =& {(-1)}^k\sum _{r=0}^{[\frac{n-k}{2}]}\sum _{l=k}^{n-2r}\{\left(\genfrac{}{}{0ex}{}{n-m}{2r}\right)m+\left(\genfrac{}{}{0ex}{}{m}{2r}\right)(n-m)\}\\ \times \left(\genfrac{}{}{0ex}{}{n-2r}{l}\right)\left(\genfrac{}{}{0ex}{}{\alpha +l}{l-k}\right)\frac{l!B_{2r}{B}_{n-2r-l}}{(n-2r)}.\end{array}$$

(2.10)

Therefore, by (2.7) and (2.10), we obtain the following theorem.

Theorem 2.4 For $n\in \mathbb{N}$ with $n\ge 2$ and $m\in {\mathbb{Z}}_{+}$ with $n-m\ge 0$, we have

$$\begin{array}{rcl}{B}_{n-m}(x)B_m(x)& =& \sum _{k=0}^n{(-1)}^k\{\sum _{r=0}^{[\frac{n-k}{2}]}\sum _{l=k}^{n-2r}(\left(\genfrac{}{}{0ex}{}{n-m}{2r}\right)m+\left(\genfrac{}{}{0ex}{}{m}{2r}\right)(n-m))\\ \times \left(\genfrac{}{}{0ex}{}{n-2r}{l}\right)\left(\genfrac{}{}{0ex}{}{\alpha +l}{l-k}\right)\frac{l!B_{2r}{B}_{n-2r-l}}{(n-2r)}\}{L}_k^{\alpha }(x).\end{array}$$

It is easy to show that

$$\frac{t^2{e}^{t(x+y)}}{{(e^t-1)}^2}=(x+y-1)\frac{t^2{e}^{t(x+y-1)}}{e^t-1}-\frac{t^{2}d}{dt}\left(\frac{e^{t(x+y-1)}}{e^t-1}\right).$$

(2.11)

From (2.11), we have

$$\sum _{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right)B_k(x)B_{n-k}(y)=(1-n)B_n(x+y)+(x+y-1)n{B}_{n-1}(x+y)\left(\text{see [11]}\right).$$

(2.12)

Let $x=y$. Then by (2.12), we get

$$\sum _{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right)B_k(x)B_{n-k}(x)=(1-n)B_n(2x)+(2x-1)B_{n-1}(2x).$$

(2.13)

Let us take $p(x)={\sum }_{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right)B_k(x)B_{n-k}(x)\in {\mathbb{P}}_n$. Then $p(x)$ can be generated by an orthogonal basis $\{L_0^{\alpha }(x),L_1^{\alpha }(x),\dots ,L_n^{\alpha }(x)\}$ in ${\mathbb{P}}_n$ to be

$$p(x)=\sum _{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right)B_k(x)B_{n-k}(x)=\sum _{k=0}^n{C}_k{L}_k^{\alpha }(x).$$

(2.14)

From (2.3), (2.13), and (2.14), we can determine the coefficients $C_k$’s to be

$$\begin{array}{rcl}{C}_k& =& \frac{1}{\mathrm{\Gamma }(\alpha +k+1)}{\int }_0^{\mathrm{\infty }}\left(\frac{d^k}{d{x}^k}{x}^{k+\alpha }{e}^{-x}\right)p(x)dx\\ =& \frac{1}{\mathrm{\Gamma }(\alpha +k+1)}\{(1-n){\int }_0^{\mathrm{\infty }}\left(\frac{d^k}{d{x}^k}{x}^{k+\alpha }{e}^{-x}\right)B_n(2x)dx\\ +n{\int }_0^{\mathrm{\infty }}\left(\frac{d^k}{d{x}^k}{x}^{k+\alpha }{e}^{-x}\right)(2x-1)B_{n-1}(2x)dx\}.\end{array}$$

(2.15)

By simple calculation, we get

(2.16)

and

(2.17)

Therefore, by (2.13), (2.14), (2.15), (2.16), and (2.17), we obtain the following theorem.

Theorem 2.5 For $n\in {\mathbb{Z}}_{+}$, we get