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Theory and Modern Applications

# 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

(1.1)

From (1.1), we note that

${L}_{n}^{\alpha }\left(x\right)=\sum _{r=0}^{n}\frac{{\left(-1\right)}^{r}\left(\genfrac{}{}{0}{}{n+\alpha }{n-r}\right){x}^{r}}{r!}\phantom{\rule{1em}{0ex}}\left(\text{see [1–3]}\right).$
(1.2)

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

${L}_{n}^{\alpha }\left(x\right)=\frac{{x}^{-\alpha }{e}^{x}}{n!}\left(\frac{{d}^{n}}{d{x}^{n}}\left({e}^{-x}{x}^{n+\alpha }\right)\right)\phantom{\rule{1em}{0ex}}\left(\text{see [1–3]}\right).$
(1.3)

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

${\int }_{0}^{\mathrm{\infty }}{x}^{\alpha }{e}^{-x}{L}_{m}^{\alpha }\left(x\right){L}_{n}^{\alpha }\left(x\right)\phantom{\rule{0.2em}{0ex}}dx=\frac{\mathrm{\Gamma }\left(\alpha +n+1\right)}{n!}{\delta }_{m,n},$
(1.4)

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

$\frac{t}{{e}^{t}-1}{e}^{xt}={e}^{B\left(x\right)t}=\sum _{n=0}^{\mathrm{\infty }}{B}_{n}\left(x\right)\frac{{t}^{n}}{n!}\phantom{\rule{1em}{0ex}}\left(\text{see [1–29]}\right),$
(1.5)

with the usual convention about replacing ${B}^{n}\left(x\right)$ by ${B}_{n}\left(x\right)$.

In the special case, $x=0$, ${B}_{n}\left(0\right)={B}_{n}$ are called the n th Bernoulli numbers. By (1.5), we get

${B}_{n}\left(x\right)=\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){B}_{n-l}{x}^{l}\phantom{\rule{1em}{0ex}}\left(\text{see [1–29]}\right).$
(1.6)

The Euler numbers are defined by

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

${E}_{n}\left(x\right)={\left(E+x\right)}^{n}=\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){E}_{n-l}{x}^{l}\phantom{\rule{1em}{0ex}}\left(\text{see [11–24]}\right).$
(1.8)

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

$\frac{2}{{e}^{t}+1}{e}^{xt}={e}^{E\left(x\right)t}=\sum _{n=0}^{\mathrm{\infty }}{E}_{n}\left(x\right)\frac{{t}^{n}}{n!}\phantom{\rule{1em}{0ex}}\left(\text{see [15–29]}\right).$
(1.9)

By (1.5) and (1.9), we get

$\frac{2}{{e}^{t}+1}{e}^{xt}=\frac{1}{t}\left(2-2\frac{2}{{e}^{t}+1}\right)\left(\frac{t{e}^{xt}}{{e}^{t}-1}\right)=-2\sum _{n=0}^{\mathrm{\infty }}\left(\sum _{l=0}^{n}\frac{{E}_{l+1}}{l+1}\left(\genfrac{}{}{0}{}{n}{l}\right){B}_{n-l}\left(x\right)\right)\frac{{t}^{n}}{n!}.$
(1.10)

Thus, by (1.10), we see that

${E}_{n}\left(x\right)=-2\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right)\frac{{E}_{l+1}}{l+1}{B}_{n-l}\left(x\right).$
(1.11)

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

$\frac{t}{{e}^{t}-1}{e}^{xt}=\frac{t}{2}\left(\frac{2{e}^{xt}}{{e}^{t}+1}\right)+\left(\frac{t}{{e}^{t}-1}\right)\left(\frac{2{e}^{x}t}{{e}^{t}+1}\right).$
(1.12)

Thus, by (1.12), we see that

${B}_{n}\left(x\right)=\sum _{k=0,k\ne 1}^{n}\left(\genfrac{}{}{0}{}{n}{k}\right){B}_{k}{E}_{n-k}\left(x\right).$
(1.13)

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

$〈p\left(x\right),q\left(x\right)〉={\int }_{0}^{\mathrm{\infty }}{x}^{\alpha }{e}^{-x}p\left(x\right)q\left(x\right)\phantom{\rule{0.2em}{0ex}}dx,$

where $p\left(x\right),q\left(x\right)\in {\mathbb{P}}_{n}$. From (1.4), we note that $\left\{{L}_{0}^{\alpha }\left(x\right),{L}_{1}^{\alpha }\left(x\right),\dots ,{L}_{n}^{\alpha }\left(x\right)\right\}$ 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 $\left\{{L}_{0}^{\alpha }\left(x\right),{L}_{1}^{\alpha }\left(x\right),\dots ,{L}_{n}^{\alpha }\left(x\right)\right\}$ for ${\mathbb{P}}_{n}$.

## 2 Some identities on Bernoulli and Euler polynomials

Let $p\left(x\right)\in {\mathbb{P}}_{n}$. Then $p\left(x\right)$ can be generated by $\left\{{L}_{0}^{\alpha }\left(x\right),{L}_{1}^{\alpha }\left(x\right),\dots ,{L}_{n}^{\alpha }\left(x\right)\right\}$ in ${\mathbb{P}}_{n}$ to be

$p\left(x\right)=\sum _{k=0}^{n}{C}_{k}{L}_{k}^{\alpha }\left(x\right),$
(2.1)

where

$\begin{array}{rcl}〈p\left(x\right),{L}_{k}^{\alpha }\left(x\right)〉& =& {C}_{k}〈{L}_{k}^{\alpha }\left(x\right),{L}_{k}^{\alpha }\left(x\right)〉\\ =& {C}_{k}{\int }_{0}^{\mathrm{\infty }}{x}^{\alpha }{e}^{-x}{L}_{k}^{\alpha }\left(x\right){L}_{k}^{\alpha }\left(x\right)\phantom{\rule{0.2em}{0ex}}dx\\ =& {C}_{k}\frac{\mathrm{\Gamma }\left(\alpha +k+1\right)}{k!}.\end{array}$
(2.2)

From (2.2), we note that

$\begin{array}{rcl}{C}_{k}& =& \frac{k!}{\mathrm{\Gamma }\left(\alpha +k+1\right)}〈p\left(x\right),{L}_{k}^{\alpha }\left(x\right)〉\\ =& \frac{k!}{\mathrm{\Gamma }\left(\alpha +k+1\right)}\frac{1}{k!}{\int }_{0}^{\mathrm{\infty }}\left(\frac{{d}^{k}}{d{x}^{k}}{x}^{k+\alpha }{e}^{-x}\right)p\left(x\right)\phantom{\rule{0.2em}{0ex}}dx\\ =& \frac{1}{\mathrm{\Gamma }\left(\alpha +k+1\right)}{\int }_{0}^{\mathrm{\infty }}\left(\frac{{d}^{k}}{d{x}^{k}}{x}^{k+\alpha }{e}^{-x}\right)p\left(x\right)\phantom{\rule{0.2em}{0ex}}dx.\end{array}$
(2.3)

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

$\begin{array}{rcl}{C}_{k}& =& \frac{1}{\mathrm{\Gamma }\left(\alpha +k+1\right)}{\int }_{0}^{\mathrm{\infty }}\left(\frac{{d}^{k}}{d{x}^{k}}{x}^{k+\alpha }{e}^{-x}\right)\sum _{m=0,m\ne 1}^{n}\left(\genfrac{}{}{0}{}{n}{m}\right){B}_{n}{E}_{n-m}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx\\ =& \frac{{\left(-1\right)}^{k}}{\mathrm{\Gamma }\left(\alpha +k+1\right)}\sum _{m=0,m\ne 1}^{n-k}\sum _{l=k}^{n-m}\left(\genfrac{}{}{0}{}{n}{m}\right)\left(\genfrac{}{}{0}{}{n-m}{l}\right){B}_{m}{E}_{n-m-l}\frac{l!}{\left(l-k\right)!}{\int }_{0}^{\mathrm{\infty }}{x}^{l+\alpha }{e}^{-x}\phantom{\rule{0.2em}{0ex}}dx\\ =& \frac{{\left(-1\right)}^{k}}{\mathrm{\Gamma }\left(\alpha +k+1\right)}\sum _{m=0,m\ne 1}^{n-k}\sum _{l=k}^{n-m}\left(\genfrac{}{}{0}{}{n}{m}\right)\left(\genfrac{}{}{0}{}{n-m}{l}\right){B}_{m}{E}_{n-m-l}\frac{l!}{\left(l-k\right)!}\mathrm{\Gamma }\left(l+\alpha +1\right)\\ =& {\left(-1\right)}^{k}\sum _{m=0,m\ne 1}^{n-k}\sum _{l=k}^{n-m}\left(\genfrac{}{}{0}{}{n}{m}\right)\left(\genfrac{}{}{0}{}{n-m}{l}\right){B}_{m}{E}_{n-m-l}\frac{l!}{\left(l-k\right)!}\frac{\left(l+\alpha \right)\left(l+\alpha -1\right)\cdots \alpha }{\left(\alpha +k\right)\left(\alpha +k-1\right)\cdots \alpha }\\ =& {\left(-1\right)}^{k}n!\sum _{m=0,m\ne 1}^{n-k}\sum _{l=k}^{n-m}\frac{{B}_{m}}{m!}\frac{{E}_{n-m-l}}{\left(n-m-l\right)!}\left(\genfrac{}{}{0}{}{l+\alpha }{l-k}\right).\end{array}$
(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

${B}_{n}\left(x\right)=n!\sum _{k=0}^{n}{\left(-1\right)}^{k}\left(\sum _{m=0,m\ne 1}^{n-k}\sum _{l=k}^{n-m}\frac{{B}_{m}}{m!}\frac{{E}_{n}-m-l}{\left(n-m-l\right)!}\left(\genfrac{}{}{0}{}{l+\alpha }{l-k}\right)\right){L}_{k}^{\alpha }\left(x\right).$

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

$\begin{array}{rcl}{C}_{k}& =& \frac{1}{\mathrm{\Gamma }\left(\alpha +k+1\right)}{\int }_{0}^{\mathrm{\infty }}\left(\frac{{d}^{k}}{d{x}^{k}}{x}^{k+\alpha }{e}^{-x}\right)\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right)\frac{{E}_{l+1}}{l+1}{B}_{n-l}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx\\ =& \frac{1}{\mathrm{\Gamma }\left(\alpha +k+1\right)}\sum _{l=0}^{n-k}\sum _{m=0}^{n-l}\left(\genfrac{}{}{0}{}{n}{l}\right)\left(\genfrac{}{}{0}{}{n-l}{m}\right)\frac{{E}_{l+1}}{l+1}{B}_{n-l-m}{\int }_{0}^{\mathrm{\infty }}\left(\frac{{d}^{k}}{d{x}^{k}}{x}^{k+\alpha }{e}^{-x}\right){x}^{m}\phantom{\rule{0.2em}{0ex}}dx\\ =& \frac{{\left(-1\right)}^{k}}{\mathrm{\Gamma }\left(\alpha +k+1\right)}\sum _{l=0}^{n-k}\sum _{m=k}^{n-l}\left(\genfrac{}{}{0}{}{n}{l}\right)\left(\genfrac{}{}{0}{}{n-l}{m}\right)\frac{{E}_{l+1}}{l+1}{B}_{n-l-m}\frac{m!}{\left(m-k\right)!}\mathrm{\Gamma }\left(m+\alpha +1\right)\\ =& {\left(-1\right)}^{k}\sum _{l=0}^{n-k}\sum _{m=k}^{n-l}\left(\genfrac{}{}{0}{}{n}{l}\right)\left(\genfrac{}{}{0}{}{n-l}{m}\right)\frac{m!}{\left(m-k\right)!}\frac{{E}_{l+1}}{\left(l+1\right)}{B}_{n-l-m}\frac{\left(\alpha +m\right)\left(\alpha +m-1\right)\cdots \alpha }{\left(\alpha +k\right)\left(\alpha +k-1\right)\cdots \alpha }\\ =& {\left(-1\right)}^{k}n!\sum _{l=0}^{n-k}\sum _{m=k}^{n-l}\left(\genfrac{}{}{0}{}{\alpha +m}{m-k}\right)\frac{{E}_{l+1}}{\left(l+1\right)!}\frac{{B}_{n-l-m}}{\left(n-l-m\right)!}.\end{array}$
(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

$-\frac{{E}_{n}\left(x\right)}{2}=n!\sum _{k=0}^{n}{\left(-1\right)}^{k}\left(\sum _{l=0}^{n-k}\sum _{m=k}^{n-l}\left(\genfrac{}{}{0}{}{\alpha +m}{m-k}\right)\frac{{E}_{m+1}}{\left(m+1\right)!}\frac{{B}_{n-m-l}}{\left(n-m-l\right)!}\right){L}_{k}^{\alpha }\left(x\right).$

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

$\begin{array}{rcl}{B}_{n-m}\left(x\right){B}_{m}\left(x\right)& =& \sum _{r}\left\{\left(\genfrac{}{}{0}{}{n-m}{2r}\right)m+\left(\genfrac{}{}{0}{}{m}{2r}\right)\left(n-m\right)\right\}\frac{{B}_{2r}{B}_{n-2r}\left(x\right)}{n-2r}\\ +{\left(-1\right)}^{m+1}\frac{\left(n-m\right)!m!}{n!}{B}_{n}\in {\mathbb{P}}_{n}\phantom{\rule{1em}{0ex}}\left(\text{see [8]}\right).\end{array}$
(2.6)

Let us take $p\left(x\right)={B}_{n-m}\left(x\right){B}_{m}\left(x\right)\in {\mathbb{P}}_{n}$. Then $p\left(x\right)$ can be generated by an orthogonal basis $\left\{{L}_{0}^{\alpha }\left(x\right),{L}_{1}^{\alpha }\left(x\right),\dots ,{L}_{n}^{\alpha }\left(x\right)\right\}$ in ${\mathbb{P}}_{n}$ to be

$p\left(x\right)=\sum _{k=0}^{n}{C}_{k}{L}_{k}^{\alpha }\left(x\right).$
(2.7)

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

$\begin{array}{rcl}{C}_{k}& =& \frac{1}{\mathrm{\Gamma }\left(\alpha +k+1\right)}{\int }_{0}^{\mathrm{\infty }}\left(\frac{{d}^{k}}{d{x}^{k}}{x}^{k+\alpha }{e}^{-x}\right)p\left(x\right)\phantom{\rule{0.2em}{0ex}}dx\\ =& \frac{1}{\mathrm{\Gamma }\left(\alpha +k+1\right)}\sum _{r=0}^{\left[\frac{n}{2}\right]}\left\{\left(\genfrac{}{}{0}{}{n-m}{2r}\right)m+\left(\genfrac{}{}{0}{}{m}{2r}\right)\left(n-m\right)\right\}\\ ×\frac{{B}_{2r}}{n-2r}{\int }_{0}^{\mathrm{\infty }}\left(\frac{{d}^{k}}{d{x}^{k}}{x}^{k+\alpha }{e}^{-x}\right){B}_{n-2r}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx\\ =& \frac{1}{\mathrm{\Gamma }\left(\alpha +k+1\right)}\sum _{r=0}^{\left[\frac{n}{2}\right]}\left\{\left(\genfrac{}{}{0}{}{n-m}{2r}\right)m+\left(\genfrac{}{}{0}{}{m}{2r}\right)\left(n-m\right)\right\}\frac{{B}_{2r}}{n-2r}\\ ×\sum _{l=0}^{n-2r}\left(\genfrac{}{}{0}{}{n-2r}{l}\right){B}_{n-2r-l}{\int }_{0}^{\mathrm{\infty }}\left(\frac{{d}^{k}}{d{x}^{k}}{x}^{k+\alpha }{e}^{-x}\right){x}^{l}\phantom{\rule{0.2em}{0ex}}dx\\ =& \frac{1}{\mathrm{\Gamma }\left(\alpha +k+1\right)}\sum _{r=0}^{\left[\frac{n}{2}\right]}\sum _{l=0}^{n-2r}\left\{\left(\genfrac{}{}{0}{}{n-m}{2r}\right)m+\left(\genfrac{}{}{0}{}{m}{2r}\right)\left(n-m\right)\right\}\left(\genfrac{}{}{0}{}{n-2r}{l}\right)\\ ×\frac{{B}_{2r}{B}_{n-2r-l}}{n-2r}{\int }_{0}^{\mathrm{\infty }}\left(\frac{{d}^{k}}{d{x}^{k}}{x}^{k+\alpha }{e}^{-x}\right){x}^{l}\phantom{\rule{0.2em}{0ex}}dx\\ =& \frac{{\left(-1\right)}^{k}}{\mathrm{\Gamma }\left(\alpha +k+1\right)}\sum _{r=0}^{\left[\frac{n-k}{2}\right]}\sum _{l=k}^{n-2r}\left\{\left(\genfrac{}{}{0}{}{n-m}{2r}\right)m+\left(\genfrac{}{}{0}{}{m}{2r}\right)\left(n-m\right)\right\}\left(\genfrac{}{}{0}{}{n-2r}{l}\right)\\ ×\frac{{B}_{2r}{B}_{n-2r-l}l!}{\left(n-2r\right)\left(l-k\right)!}\mathrm{\Gamma }\left(\alpha +l+1\right).\end{array}$
(2.8)

It is easy to show that

$\begin{array}{rcl}\frac{\mathrm{\Gamma }\left(\alpha +l+1\right)}{\mathrm{\Gamma }\left(\alpha +k+1\right)\left(l-k\right)!}& =& \frac{\left(\alpha +l\right)\left(\alpha +l-1\right)\cdots \alpha \mathrm{\Gamma }\left(\alpha \right)}{\left(\alpha +k\right)\left(\alpha +k-1\right)\cdots \alpha \mathrm{\Gamma }\left(\alpha \right)\left(l-k\right)!}\\ =& \frac{\left(\alpha +l\right)\left(\alpha +l-1\right)\cdots \left(\alpha +k+1\right)}{\left(\alpha -k\right)!}=\left(\genfrac{}{}{0}{}{\alpha +l}{l-k}\right).\end{array}$
(2.9)

By (2.8) and (2.9), we get

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

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

It is easy to show that

$\frac{{t}^{2}{e}^{t\left(x+y\right)}}{{\left({e}^{t}-1\right)}^{2}}=\left(x+y-1\right)\frac{{t}^{2}{e}^{t\left(x+y-1\right)}}{{e}^{t}-1}-\frac{{t}^{2}\phantom{\rule{0.2em}{0ex}}d}{dt}\left(\frac{{e}^{t\left(x+y-1\right)}}{{e}^{t}-1}\right).$
(2.11)

From (2.11), we have

$\sum _{k=0}^{n}\left(\genfrac{}{}{0}{}{n}{k}\right){B}_{k}\left(x\right){B}_{n-k}\left(y\right)=\left(1-n\right){B}_{n}\left(x+y\right)+\left(x+y-1\right)n{B}_{n-1}\left(x+y\right)\phantom{\rule{1em}{0ex}}\left(\text{see [11]}\right).$
(2.12)

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

$\sum _{k=0}^{n}\left(\genfrac{}{}{0}{}{n}{k}\right){B}_{k}\left(x\right){B}_{n-k}\left(x\right)=\left(1-n\right){B}_{n}\left(2x\right)+\left(2x-1\right){B}_{n-1}\left(2x\right).$
(2.13)

Let us take $p\left(x\right)={\sum }_{k=0}^{n}\left(\genfrac{}{}{0}{}{n}{k}\right){B}_{k}\left(x\right){B}_{n-k}\left(x\right)\in {\mathbb{P}}_{n}$. Then $p\left(x\right)$ can be generated by an orthogonal basis $\left\{{L}_{0}^{\alpha }\left(x\right),{L}_{1}^{\alpha }\left(x\right),\dots ,{L}_{n}^{\alpha }\left(x\right)\right\}$ in ${\mathbb{P}}_{n}$ to be

$p\left(x\right)=\sum _{k=0}^{n}\left(\genfrac{}{}{0}{}{n}{k}\right){B}_{k}\left(x\right){B}_{n-k}\left(x\right)=\sum _{k=0}^{n}{C}_{k}{L}_{k}^{\alpha }\left(x\right).$
(2.14)

From (2.3), (2.13), and (2.14), we can determine the coefficients ${C}_{k}$’s to be

$\begin{array}{rcl}{C}_{k}& =& \frac{1}{\mathrm{\Gamma }\left(\alpha +k+1\right)}{\int }_{0}^{\mathrm{\infty }}\left(\frac{{d}^{k}}{d{x}^{k}}{x}^{k+\alpha }{e}^{-x}\right)p\left(x\right)\phantom{\rule{0.2em}{0ex}}dx\\ =& \frac{1}{\mathrm{\Gamma }\left(\alpha +k+1\right)}\left\{\left(1-n\right){\int }_{0}^{\mathrm{\infty }}\left(\frac{{d}^{k}}{d{x}^{k}}{x}^{k+\alpha }{e}^{-x}\right){B}_{n}\left(2x\right)\phantom{\rule{0.2em}{0ex}}dx\\ +n{\int }_{0}^{\mathrm{\infty }}\left(\frac{{d}^{k}}{d{x}^{k}}{x}^{k+\alpha }{e}^{-x}\right)\left(2x-1\right){B}_{n-1}\left(2x\right)\phantom{\rule{0.2em}{0ex}}dx\right\}.\end{array}$
(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

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## Acknowledgements

The authors would like to express their deep gratitude to the referees for their valuable suggestions and comments.

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Correspondence to Taekyun Kim.

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Kim, T., Rim, SH., Dolgy, D. et al. Some identities on Bernoulli and Euler polynomials arising from the orthogonality of Laguerre polynomials. Adv Differ Equ 2012, 201 (2012). https://doi.org/10.1186/1687-1847-2012-201

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### Keywords

• Differential Equation
• Generate Function
• Partial Differential Equation
• Ordinary Differential Equation
• Functional Analysis