1 Introduction

In recent years, the orthogonal polynomial has had a very important position in the areas of functional neural network and its properties, operators and identification of distributed parameter systems, intelligent instrument automatic calibration, operator and control theory, and so on. And the Gegenbauer polynomial is very important of orthogonal polynomials. It seems to be interesting and important in the area of mathematical physics. Recently, many authors have studied Gegenbauer polynomials related to mathematical physics (see [1–6]).

The information entropy of Gegenbauer polynomials is relevant since it is related to the angular part of the information entropies of certain quantum mechanical systems such as the harmonic oscillator and the hydrogen atom in D dimensions (see [1, 7, 8]). In this paper we will promote the results in [9, 10] to the generalized Gegenbauer polynomials. It will help solve the above general problems.

The generalized Gegenbauer polynomials are given in terms of generating function by

$$ \frac{1}{{{{(\alpha - 2pxt + q{t^{2}})}^{\lambda }}}} = \sum_{n = 0}^{\infty }{C_{\alpha ,p,q,n}^{\lambda }} (x){t^{n}}. $$
(1)

By Newton’s binomial theorem, we get

$$\begin{aligned}& \frac{1}{{{{(\alpha - 2pxt + q{t^{2}})}^{\lambda }}}} = { \biggl(1 - \sqrt{\frac{q}{ \alpha }} t \biggr)^{ - 2\lambda }} { \biggl[\alpha + 2t{ \biggl(1 - \sqrt{\frac{q}{ \alpha }} t \biggr)^{ - 2}}(\sqrt{\alpha q} - px) \biggr]^{ - \lambda }} \\& \hphantom{\frac{1}{{{{(\alpha - 2pxt + q{t^{2}})}^{\lambda }}}} }{} = \sum_{k = 0}^{\infty }{\frac{{{{( - 2)}^{k}}\Gamma (k + \lambda ){{(\sqrt{\alpha q} - px)}^{k}}t^{k}{{(1 - \sqrt{\frac{q}{ \alpha }} t)}^{ - 2k - 2\lambda }}}}{{k!\Gamma (\lambda )}}} \alpha^{-\lambda -k}, \\& {t^{k}} { \biggl(1 - \sqrt{\frac{q}{\alpha }} t \biggr)^{ - 2k - 2\lambda }} = \sum_{m = 0}^{\infty }{ \frac{{\Gamma (m + 2k + 2\lambda )}}{ {m!\Gamma (2k + 2\lambda )}}} {t^{m + k}} { \biggl(\sqrt{\frac{q}{\alpha }} \biggr)^{m}}. \end{aligned}$$

Then

$$\begin{aligned}& \frac{1}{{{{(\alpha - 2pxt + q{t^{2}})}^{\lambda }}}} \\& \quad =\sum_{k = 0}^{\infty }{\frac{{{{( - 2)}^{k}}\Gamma (k + \lambda ){{(\sqrt{\alpha q} - px)}^{k}}}}{{k!\Gamma (\lambda )}}} \sum_{m = 0}^{\infty }{\frac{{\Gamma (m + 2k + 2\lambda )}}{{m! \Gamma (2k + 2\lambda )}}} {t^{m + k}} { \alpha^{ - \lambda - k - \frac{m}{2}}} {q^{\frac{m}{2}}}. \end{aligned}$$

When \(\vert t \vert >0\) is small enough, the double series is absolutely convergent, and the coefficient of \(t^{k}\) is known by the diagonal summation method

$$ C_{\alpha ,p,q,n}^{\lambda }(x) = \sum_{k = 0}^{n} {\frac{ {{{( - 2)}^{k}}\Gamma (k + \lambda )\Gamma (n + k + 2\lambda ){{(\sqrt{ \alpha q} - px)}^{k}}}}{{k!(n - k)!\Gamma (\lambda )\Gamma (2k + 2 \lambda )}}} {\alpha^{ - \lambda - \frac{k}{2} - \frac{n}{2}}} {q^{ \frac{n}{2} - \frac{k}{2}}}. $$

Thus, there are

$$ C_{\alpha ,p,q,n}^{\lambda }(x) = \left ( { \textstyle\begin{array}{@{}c@{}} {n + 2\lambda - 1} \\ n \end{array}\displaystyle } \right ) \sum _{k = 0}^{n} {\frac{{\binom{n}{k} {{(2\lambda + n)}_{k}}}}{{{{(\lambda + \frac{1}{2})}_{k}}}}} { \biggl( \frac{{px - \sqrt{\alpha q} }}{2} \biggr)^{k}} {\alpha^{ - \lambda - \frac{k}{2} - \frac{n}{2}}} {q^{\frac{n}{2} - \frac{k}{2}}}, $$

where \((a)_{k}=a(a+1)(a+2)\cdots(a+k-1)\).

From [11], we have the recursion formula about the generalized Gegenbauer polynomials

$$\begin{aligned}& C_{\alpha ,p,q,0}^{\lambda }(x) = {\alpha^{ - \lambda }},\qquad C_{\alpha ,p,q,1} ^{\lambda }(x) = 2\lambda px{\alpha^{ - \lambda - 1}}, \\& \frac{{n + 1}}{{2\lambda }}C_{\alpha ,p,q,n + 1}^{\lambda }(x) = pxC _{\alpha ,p,q,n}^{\lambda + 1}(x) - qC_{\alpha ,p,q,n - 1}^{\lambda + 1}(x). \end{aligned}$$

By [11], we get

$$ C_{\alpha ,p,q,n}^{\lambda }(x) = \frac{1}{{\Gamma (\lambda )}} \sum _{l = 0}^{ [ {\frac{n}{2}} ] } {\frac{{{{( - 1)} ^{l}}\Gamma (\lambda + n - l)}}{{l!(n - 2l)!}}} {(2xp)^{n - 2l}} {q ^{l}} { \alpha^{ - \lambda - n + l}}. $$

Meanwhile, by the recursion formula about generalized Gegenbauer polynomials and mathematical induction, we get

$$ C_{\alpha ,p,q,n}^{\lambda }(x) = \frac{{ \binom{n + 2\lambda - 1}{n}}}{{\binom{n + \lambda - \frac{1}{2}}{n} }}\sum _{k = 0}^{n} {\frac{{\binom{n + \lambda - \frac{1}{2}}{n - k} \binom{n + \lambda - \frac{1}{2}}{k}{{(\frac{{px - \sqrt{\alpha q} }}{2})}^{k}}}}{{{{(\frac{ {px + \sqrt{\alpha q} }}{2})}^{k - n}}}}{ \alpha^{ - \lambda - n}}}. $$
(2)

It is not difficult to show that \(C_{\alpha ,p,q,n}^{\lambda }(x)\) is a solution of the following Gegenbauer differential equation:

$$ \frac{1}{p} \bigl(\alpha q - {p^{2}} {x^{2}} \bigr){y^{(2)}} - p(2\lambda + 1)xy' + pn(n + 2\lambda )y = 0. $$

From the above equation and mathematical induction, we acquire Rodrigues’ formula for the generalized Gegenbauer polynomials

$$ { \bigl(\alpha q - {p^{2}} {x^{2}} \bigr)^{\lambda - \frac{1}{2}}}C_{\alpha ,p,q,n} ^{\lambda }(x) = \frac{1}{{{\alpha^{\lambda + n}}{p^{n}}}} \frac{{{{( - 2)}^{n}}{{(\lambda )}_{n}}}}{{n!{{(n + 2\lambda )}_{n}}}}{ \biggl( {\frac{d}{{dx}}} \biggr) ^{n}} { \bigl( \alpha q - {p^{2}} {x^{2}} \bigr)^{n + \lambda - \frac{1}{2}}}. $$
(3)

Applying (1) and (2), we get

$$ \int_{-\frac{\sqrt{\alpha q}}{p}}^{\frac{\sqrt{\alpha q}}{p}} \bigl( \alpha q-p^{2}x^{2} \bigr)^{\lambda -\frac{1}{2}}C_{\alpha ,p,q,n}^{\lambda }(x)C_{\alpha ,p,q,m}^{\lambda }(x)\,dx= \frac{q^{n+\lambda }}{p \alpha^{n+\lambda }}\frac{\pi 2^{1-2\lambda }\Gamma (n+2\lambda )}{n!(n+ \lambda )(\Gamma (\lambda ))^{2}}\delta_{m,n}, $$
(4)

where \(\delta_{m,n}\) is the Kronecker symbol and it holds for each fixed \(\lambda\in \mathbf{R}\) with \(\lambda >-\frac{1}{2}\) and \(\lambda \neq 0\).

Equation (5) implies the orthogonality of \(C_{\alpha ,p,q,n}^{\lambda }(x)\), and equation (5) is important in deriving our results in this paper. From (1), we can derive the following derivative of generalized Gegenbauer polynomials \(C_{\alpha ,p,q,n}^{\lambda }(x)\): for \(k\geq 1\),

$$ { \biggl( {\frac{d}{{dx}}} \biggr) ^{k}}C_{p,q,n}^{\lambda }(x) = {2^{k}} {(\lambda )_{k}} {p^{k}}C_{\alpha ,p,q,n - k}^{\lambda + k}(x). $$
(5)

The so-called Euler polynomials \(E_{n}(x)\) are defined by the generating function to be

$$ \frac{2}{{{e^{t}} + 1}}{e^{xt}} = {e^{E(x)t}} = \sum _{n = 0} ^{\infty }{{E_{n}}(x) \frac{{{t^{n}}}}{{n!}}}, $$
(6)

with the usual convention about replacing \(E^{n}(x)\) by \(E_{n}(x)\). In the special case, \(x=0, E_{0}=E_{n}\) are called the nth Euler number.

The Bernoulli polynomials are also defined by the generating function to be

$$ \frac{t}{{{e^{t}} - 1}}{e^{xt}} = {e^{B(x)t}} = \sum _{n = 0} ^{\infty }{{B_{n}}(x) \frac{{{t^{n}}}}{{n!}}}, $$
(7)

with the usual convention about replacing \(B^{n}(x)\) by \(B_{n}(x)\). In the special case, \(x=0, B_{0}=B_{n}\) are called the nth Bernoulli number. From the above equation, we note that

$$ \begin{aligned} &B_{n}(x) = \sum_{k = 0}^{n} {\left ( \textstyle\begin{array}{@{}c@{}} n \\ k \end{array}\displaystyle \right ) {B_{n - k}} {x^{k}}} , \\ & {E_{n}}(x) = \sum_{k = 0}^{n} { \left ( \textstyle\begin{array}{@{}c@{}} n \\ k \end{array}\displaystyle \right ) {E_{n - k}} {x^{k}}}. \end{aligned} $$
(8)

For \(n\in \mathbf{Z}_{+}\), we have

$$ \begin{aligned}&\frac{{d{B_{n}}(x)}}{{dx}} = n{B_{n - 1}}(x), \\ &\frac{{d{E_{n}}(x)}}{ {dx}} = n{E_{n - 1}}(x). \end{aligned} $$
(9)

By the definition of Bernoulli and Euler polynomials, we get

$$\begin{aligned}& {B_{0}} = 1,\qquad {B_{n}}(1) - {B_{n}} = {\delta_{1,n}}, \\& {E_{0}} = 1,\qquad {E_{n}}(1) +{E_{n}} = 2{\delta_{0,n}}. \end{aligned}$$

The Hermite polynomials are defined by the generating function to be

$$ {e^{2xt - {t^{2}}}} = \sum_{n = 0}^{\infty }{{H_{n}}(x) \frac{ {{t^{n}}}}{{n!}}} . $$

For \(n\in \mathbf{Z}_{+}\), \(k\in \mathbf{Z}_{+}\), we have

$$\begin{aligned}& \frac{{d{H_{n}}(x)}}{{dx}} = 2n{H_{n - 1}}(x),\qquad { \biggl( { \frac{d}{{dx}}} \biggr) ^{k}} {H_{n}}(x) = {2^{k}}\frac{{n!}}{{(n - k)!}}{H_{n - k}}(x), \end{aligned}$$
(10)
$$\begin{aligned}& {H_{n}}(x) = \sum_{k = 0}^{n} { \left ( \textstyle\begin{array}{@{}c@{}} n \\ k \end{array}\displaystyle \right ) } {H_{n - k}} {(2x)^{k}}, \end{aligned}$$
(11)

where \(H_{n}\) is the nth Hermite number.

For each fixed λ∈R with \(\lambda >-\frac{1}{2}\) and \(\lambda \neq 0\), let \({\mathrm{\mathbf{P}}_{n}} = \{ p(x) \in \textbf{R}[x]| \deg p(x) \le n\} \) be an inner product space with respect to the generalized inner product

$$ \bigl\langle {{p_{1}}(x),{p_{2}}(x)} \bigr\rangle = \int_{ - \frac{{\sqrt{\alpha q} }}{p}}^{ \frac{{\sqrt{\alpha q }}}{p}} {{ \bigl(\alpha q - p^{2}{x^{2}} \bigr)}^{\lambda - \frac{1}{2}}} {p_{1}}(x){p_{2}}(x)\,dx. $$
(12)

In this paper, we derive some interesting identities involving Gegenbauer polynomials arising from the orthogonality of those for the generalized inner product space \(\mathbf{P}_{n}\) with respect to the weighted inner product. Our methods used in this paper are useful in finding some new identities and relations on the Bernoulli, Euler and Hermite polynomials involving generalized Gegenbauer polynomials.

2 Some identities involving generalized Gegenbauer polynomials

Lemma 1

For \(p(x)\in \mathbf{P}_{n}\), let

$$ p(x) = \sum_{k = 0}^{n} {{d_{k}}C_{\alpha ,p,q,k}^{\lambda }} (x)\quad (d_{k} \in\mathbf{R}). $$
(13)

Then

$$ {d_{k}}=\frac{{{p^{1 - k}}}}{{{q^{k + \lambda }}}}\frac{{(k + \lambda )\Gamma (\lambda )}}{{{{( - 2)}^{k}}\sqrt{\pi }\Gamma (k + \lambda + \frac{1}{2})}} \int_{ - \frac{{\sqrt{\alpha q} }}{p}}^{\frac{{\sqrt{ \alpha q} }}{p}} { \biggl( {\frac{{{d^{k}}}}{{d{x^{k}}}}{{ \bigl( \alpha q - p ^{2}{x^{2}} \bigr)}^{k + \lambda - \frac{1}{2}}}} \biggr) p(x)\,dx}. $$
(14)

Proof

Let us take \(p(x) = \sum_{k = 0}^{n} {{d_{k}}C _{\alpha ,p,q,k}^{\lambda }} (x) \in \mathbf{P}_{n}\), \(d_{k}\) ∈R. Then by (4) and (12), we get

$$\begin{aligned} \bigl\langle {p(x),C_{\alpha ,p,q,k}^{\lambda }(x)} \bigr\rangle =& {d_{k}} \bigl\langle {C_{\alpha ,p,q,k}^{\lambda }(x),C_{\alpha ,p,q,k} ^{\lambda }(x)} \bigr\rangle \\ =& {d_{k}} \int_{ - \frac{{\sqrt{\alpha q }}}{p}}^{\frac{{\sqrt{ \alpha q} }}{p}} {{{ \bigl(\alpha q - p^{2}{x^{2}} \bigr)}^{\lambda - \frac{1}{2}}}C _{\alpha ,p,q,k}^{\lambda }(x)C_{\alpha ,p,q,k}^{\lambda }(x)\,dx} \\ = &{d_{k}}\frac{{{q^{k + \lambda }}}}{p\alpha^{\lambda +k} }\frac{ {\pi {2^{1 - 2\lambda }}\Gamma (k + 2\lambda )}}{{k!(k + \lambda ) {{(\Gamma (\lambda ))}^{2}}}}. \end{aligned}$$

Thus, by the above equation, we get

$$ {d_{k}} = \frac{p\alpha^{\lambda +k} }{{{q^{k + \lambda }}}}\frac{ {k!(k + \lambda ){{(\Gamma (\lambda ))}^{2}}}}{{\pi {2^{1 - 2\lambda }}\Gamma (k + 2\lambda )}} \int_{ - \frac{{\sqrt{\alpha q }}}{p}} ^{\frac{{\sqrt{\alpha q }}}{p}} {{{ \bigl(\alpha q - p^{2}{x^{2}} \bigr)}^{ \lambda - \frac{1}{2}}}p(x)C_{\alpha ,p,q,k}^{\lambda }(x)\,dx} . $$

From the above equation and (3), we have

$$\begin{aligned}& {d_{k}} = \frac{p\alpha^{\lambda +k}}{{{q^{k + \lambda }}}}\frac{ {k!(k + \lambda ){{(\Gamma (\lambda ))}^{2}}}}{{\pi {2^{1 - 2\lambda }}\Gamma (k + 2\lambda )}}\frac{\alpha^{-\lambda -k}}{{{p^{k}}}} \frac{ {{{( - 2)}^{k}}{{(\lambda )}_{k}}}}{{k!{{(k + 2\lambda )}_{k}}}} \int_{ - \frac{{\sqrt{\alpha q }}}{p}}^{\frac{{\sqrt{ \alpha q }}}{p}} { \biggl( {\frac{{{d^{k}}}}{{d{x^{k}}}}{{ \bigl( \alpha q - p ^{2}{x^{2}} \bigr)}^{k + \lambda - \frac{1}{2}}}} \biggr) p(x)\,dx} \\& \quad = \frac{{{p^{1 - k}}}}{{{q^{k + \lambda }}}}\frac{{(k + \lambda ) \Gamma (\lambda )}}{{{{( - 2)}^{k}}\sqrt{\pi }\Gamma (k + \lambda + \frac{1}{2})}} \int_{ - \frac{{\sqrt{\alpha q} }}{p}}^{\frac{{\sqrt{ \alpha q }}}{p}} { \biggl( {\frac{{{d^{k}}}}{{d{x^{k}}}}{{ \bigl( \alpha q - p ^{2}{x^{2}} \bigr)}^{k + \lambda - \frac{1}{2}}}} \biggr) p(x)\,dx}. \end{aligned}$$

This proves Lemma 1. □

Theorem 1

For \(n\in \mathbf{Z}_{+}\), we have

$$ {x^{n}} = \sum_{0 \le k \le n,n - k \equiv 0(\bmod 2)} {\frac{\sqrt{ \alpha }^{n+k+2\lambda }{{{\sqrt{q} }^{n - k}}}}{{{p^{n}}}}} \frac{ {(k + \lambda )n!\Gamma (\lambda )}}{{{2^{n}}(\frac{{n - k}}{2})! \Gamma (\frac{{n + k + 2\lambda + 2}}{2})}}C_{\alpha ,p,q,k}^{\lambda }(x). $$

Proof

Let \(p(x)=x^{n}\in \mathbf{P}_{n}\), from (14) we have

$$\begin{aligned} {d_{k}} =&\frac{{{p^{1 - k}}}}{{{q^{k + \lambda }}}}\frac{{(k + \lambda )\Gamma (\lambda )}}{{{{( - 2)}^{k}}\sqrt{\pi }\Gamma (k + \lambda + \frac{1}{2})}} \int_{ - \frac{{\sqrt{\alpha q }}}{p}}^{\frac{ {\sqrt{\alpha q} }}{p}} { \biggl( {\frac{{{d^{k}}}}{{d{x^{k}}}}{{ \bigl( \alpha q - p^{2}{x^{2}} \bigr)}^{k + \lambda - \frac{1}{2}}}} \biggr) x^{n}\,dx} \\ =&\frac{{{p^{1 - k}}}}{{{q^{k + \lambda }}}}\frac{{(k + \lambda ) \Gamma (\lambda )}}{{{{( - 2)}^{k}}\sqrt{\pi }\Gamma (k + \lambda + \frac{1}{2})}} \int_{ - \frac{{\sqrt{\alpha q }}}{p}}^{\frac{{\sqrt{ \alpha q }}}{p}} {{x^{n}}\,d\biggl( { \frac{{{d^{k - 1}}}}{{d{x^{k - 1}}}} {{ \bigl(\alpha q - {p^{2}} {x^{2}} \bigr)}^{k + \lambda - \frac{1}{2}}}} \biggr) } \\ =&( - n)\frac{{{p^{1 - k}}}}{{{q^{k + \lambda }}}}\frac{{(k + \lambda )\Gamma (\lambda )}}{{{{( - 2)}^{k}}\sqrt{\pi }\Gamma (k + \lambda + \frac{1}{2})}} \int_{ - \frac{{\sqrt{\alpha q} }}{p}}^{\frac{{\sqrt{ \alpha q }}}{p}} {{x^{n - 1}} \biggl( { \frac{{{d^{k - 1}}}}{{d{x^{k - 1}}}} {{ \bigl(\alpha q - {p^{2}} {x^{2}} \bigr)}^{k + \lambda - \frac{1}{2}}}} \biggr) }\,dx \\ =&\cdots \\ =&\frac{{{p^{1 - k}}}}{{{q^{k + \lambda }}}}\frac{{(k + \lambda ) \Gamma (\lambda )\frac{{n!}}{{(n - k)!}}}}{{{{( 2)}^{k}}\sqrt{ \pi }\Gamma (k + \lambda + \frac{1}{2})}} \int_{ - \frac{{\sqrt{\alpha q} }}{p}}^{ \frac{{\sqrt{\alpha q }}}{p}} {{x^{n - k}} {{ \bigl(\alpha q - {p^{2}} {x^{2}} \bigr)} ^{k + \lambda - \frac{1}{2}}}\,dx} \\ =& \bigl(1 + {( - 1)^{n - k}} \bigr)\frac{{{p^{1 - k}}}}{{{q^{k + \lambda }}}}\frac{ {(k + \lambda )\Gamma (\lambda )\frac{{n!}}{{(n - k)!}}}}{{{{2}^{k}}\sqrt{ \pi }\Gamma (k + \lambda + \frac{1}{2})}} \int_{0}^{\frac{{\sqrt{ \alpha q} }}{p}} {{x^{n - k}} {{ \bigl(\alpha q - {p^{2}} {x^{2}} \bigr)}^{k + \lambda - \frac{1}{2}}}\,dx} \\ =& \bigl(1 + {( - 1)^{n - k}} \bigr)\frac{\sqrt{\alpha }^{n+k+2\lambda }{{{\sqrt{q} }^{n - k}}}}{ {{p^{n}}}}\frac{{(k + \lambda )\Gamma (\lambda ) \frac{{n!}}{{(n - k)!}}}}{{{2^{k}}\sqrt{\pi }\Gamma (k + \lambda + \frac{1}{2})}} \int_{0}^{1} {{y^{n - k}} {{ \bigl(1 - {y^{2}} \bigr)}^{k + \lambda - \frac{1}{2}}}\,dy}. \end{aligned}$$

Let us assume that \(n-k\equiv 0(\bmod 2)\) and \(y=\sqrt{x}\). Then we get

$$\begin{aligned} {d_{k}} =& \frac{\sqrt{\alpha }^{n+k+2\lambda }{{{\sqrt{q} }^{n - k}}}}{{{p^{n}}}}\frac{{(k + \lambda )\Gamma (\lambda )\frac{{n!}}{ {(n - k)!}}}}{{{2^{k}}\sqrt{\pi }\Gamma (k + \lambda + \frac{1}{2})}} \int_{0}^{1} {{x^{\frac{{n - k - 1}}{2}}} {{(1 - x)}^{k + \lambda - \frac{1}{2}}}\,dx} \\ =& \frac{\sqrt{\alpha }^{n+k+2\lambda }{{{\sqrt{q} }^{n - k}}}}{ {{p^{n}}}}\frac{{(k + \lambda )\Gamma (\lambda ) \frac{{n!}}{{(n - k)!}}}}{{{2^{k}}\sqrt{\pi }\Gamma (k + \lambda + \frac{1}{2})}}B \biggl(k + \lambda + \frac{1}{2},\frac{{n - k + 1}}{2} \biggr) \\ =& \frac{\sqrt{\alpha }^{n+k+2\lambda }{{{\sqrt{q} }^{n - k}}}}{ {{p^{n}}}}\frac{{(k + \lambda )\Gamma (\lambda ) \frac{{n!}}{{(n - k)!}}}}{{{2^{k}}\sqrt{\pi }\Gamma (k + \lambda + \frac{1}{2})}}\frac{{\Gamma (\frac{{n - k + 1}}{2})\Gamma (k + \lambda + \frac{1}{2})}}{{\Gamma (\frac{{n + k + 2\lambda + 2}}{2})}}, \end{aligned}$$

where \(B(\alpha ,\beta )\) is the beta function which is defined by \(B(\alpha ,\beta )=\frac{\Gamma (\alpha )\Gamma (\beta )}{\Gamma ( \alpha +\beta )}\).

It is not difficult to show that

$$ \Gamma \biggl(\frac{n-k+1}{2} \biggr)=\frac{(n-k)!\sqrt{\pi }}{2^{n-k}( \frac{n-k}{2})!}. $$

Therefore, from (13), we get

$$ {x^{n}} = \sum_{0 \le k \le n,n - k \equiv 0(\bmod 2)} {\frac{\sqrt{ \alpha }^{n+k+2\lambda }{{{\sqrt{q} }^{n - k}}}}{{{p^{n}}}}} \frac{ {(k + \lambda )n!\Gamma (\lambda )}}{{{2^{n}}(\frac{{n - k}}{2})! \Gamma (\frac{{n + k + 2\lambda + 2}}{2})}}C_{\alpha ,p,q,k}^{\lambda }(x). $$

 □

Theorem 2

For \(n\in \mathbf{Z}_{+}\), we have the identities

$$\begin{aligned}& \frac{{{B_{n}}(x)}}{{n!}} = \Gamma (\lambda )\sum_{k = 0} ^{n} {\left ( {\frac{{(k + \lambda )}}{{{2^{k}}(n - k)!}} \sum _{0 \le l \le n - k,l \equiv 0(\bmod 2)} { \frac{{\frac{\sqrt{ \alpha }^{2k+2\lambda +l}{{{\sqrt{q} }^{l}}}}{{{p^{l + k}}}} \binom{n - k}{l} {B_{n - k - l}}l!}}{{{2^{l}}(\frac{l}{2})!\Gamma (\frac{ {2k + 2\lambda + l + 2}}{2})}}} } \right ) } \\& \hphantom{\frac{{{B_{n}}(x)}}{{n!}} = }{}\times C_{\alpha ,p,q,k}^{\lambda }(x), \\& \frac{{{E_{n}}(x)}}{{n!}} = \Gamma (\lambda )\sum_{k = 0} ^{n} {\left ( {\frac{{(k + \lambda )}}{{{2^{k}}(n - k)!}} \sum _{0 \le l \le n - k,l \equiv 0(\bmod 2)} { \frac{{\frac{\sqrt{ \alpha }^{2k+2\lambda +l}{{{\sqrt{q} }^{l}}}}{{{p^{l + k}}}} \binom{n - k}{l} {E_{n - k - l}}l!}}{{{2^{l}}(\frac{l}{2})!\Gamma (\frac{ {2k + 2\lambda + l + 2}}{2})}}} } \right ) } \\& \hphantom{\frac{{{B_{n}}(x)}}{{n!}} = }{} \times C_{\alpha ,p,q,k}^{\lambda }(x). \end{aligned}$$

Proof

Let us take \(p(x)=B_{n}(x)\in \mathbf{P}_{n}\). Then, applying (9) and (14), we get

$$\begin{aligned} {d_{k}} &=\frac{{{p^{1 - k}}}}{{{q^{k + \lambda }}}}\frac{{(k + \lambda )\Gamma (\lambda )}}{{{{( - 2)}^{k}}\sqrt{\pi }\Gamma (k + \lambda + \frac{1}{2})}} \int_{ - \frac{{\sqrt{\alpha q }}}{p}} ^{\frac{{\sqrt{\alpha q }}}{p}} { \biggl( {\frac{{{d^{k}}}}{{d{x^{k}}}} {{ \bigl(\alpha q - p^{2}{x^{2}} \bigr)}^{k + \lambda - \frac{1}{2}}}} \biggr) B _{n}(x)\,dx} \\ &=(-n)\frac{{{p^{1 - k}}}}{{{q^{k + \lambda }}}}\frac{{(k + \lambda )\Gamma (\lambda )}}{{{{( - 2)}^{k}}\sqrt{\pi }\Gamma (k + \lambda + \frac{1}{2})}} \\ &\quad {}\times \int_{ - \frac{{\sqrt{\alpha q} }}{p}}^{\frac{{\sqrt{ \alpha q} }}{p}} { \biggl( {\frac{{{d^{k - 1}}}}{{d{x^{k - 1}}}}{{ \bigl( \alpha q - {p^{2}} {x^{2}} \bigr)}^{k + \lambda - \frac{1}{2}}}} \biggr) {B_{n - 1}}(x)\,dx} \\ &=\cdots \\ &=\frac{{{p^{1 - k}}}}{{{q^{k + \lambda }}}}\frac{{(k + \lambda ) \Gamma (\lambda )\frac{{n!}}{{(n - k)!}}}}{{{{(2)}^{k}}\sqrt{\pi } \Gamma (k + \lambda + \frac{1}{2})}} \int_{ - \frac{{\sqrt{\alpha q }}}{p}}^{\frac{{\sqrt{\alpha q} }}{p}} {{{ \bigl( \alpha q - {p^{2}} {x^{2}} \bigr)}^{k + \lambda - \frac{1}{2}}} {B_{n - k}}(x)\,dx}. \end{aligned}$$

From (8), we have

$$\begin{aligned}& \int_{ - \frac{{\sqrt{\alpha q} }}{p}}^{ \frac{{\sqrt{\alpha q }}}{p}} {{{ \bigl(\alpha q - {p^{2}} {x^{2}} \bigr)}^{k + \lambda - \frac{1}{2}}} {B_{n - k}}(x)\,dx} \\& \quad = \sum_{l = 0}^{n - k} {\left ( { \textstyle\begin{array}{@{}c@{}} {n - k} \\ l \end{array}\displaystyle } \right ) } {B_{n - k - l}} \int_{ - \frac{{\sqrt{\alpha q} }}{p}} ^{\frac{{\sqrt{\alpha q }}}{p}} {{{ \bigl(\alpha q - {p^{2}} {x^{2}} \bigr)}^{k + \lambda - \frac{1}{2}}} {x^{l}}\,dx} \\& \quad = \sum_{0 \le l \le n - k,l = 0(\bmod 2)} {\left ( { \textstyle\begin{array}{@{}c@{}} {n - k} \\ l \end{array}\displaystyle } \right ) {B_{n - k - l}}} \frac{{{(\alpha q)^{k + \lambda + \frac{l}{2}}}}}{{{p^{l + 1}}}} \int_{0}^{1} {{{(1 - x)}^{k + \lambda - \frac{1}{2}}} {x^{\frac{{l - 1}}{2}}}\,dx} \\& \quad = \sum_{0 \le l \le n - k,l = 0(\bmod 2)} {\left ( { \textstyle\begin{array}{@{}c@{}} {n - k} \\ l \end{array}\displaystyle } \right ) {B_{n - k - l}}} \frac{{{(\alpha q)^{k + \lambda + \frac{l}{2}}}}}{{{p^{l + 1}}}}\frac{{\Gamma (k + \lambda + \frac{1}{2})\Gamma (\frac{{l + 1}}{2})}}{{\Gamma (\frac{{2k + 2\lambda + l + 2}}{2})}}. \end{aligned}$$

It is easy to show that

$$ \Gamma \biggl(\frac{l+1}{2} \biggr)= \biggl(\frac{l-1}{2} \biggr) \biggl( \frac{l-3}{2} \biggr)\cdots \biggl(\frac{1}{2} \biggr) \Gamma \biggl( \frac{1}{2} \biggr)= \frac{(\frac{l}{2})^{l}l!\Gamma (\frac{1}{2})}{( \frac{l}{2})!}=\frac{l!\sqrt{\pi }}{2^{l}(\frac{l}{2})!}. $$

So, we get

$$ {d_{k}} = \frac{{n!(k + \lambda )\Gamma (\lambda )}}{{{2^{k}}(n - k)!}}\sum_{0 \le l \le n - k,l = 0(\bmod 2)} { \frac{\sqrt{\alpha } ^{2k+2\lambda +l}{{{\sqrt{q} }^{l}}}}{{{p^{l + k}}}}} \frac{{ \binom{n - k}{l} {B_{n - k - l}}l!}}{{{2^{l}}(\frac{l}{2})!\Gamma (\frac{ {2k + 2\lambda + l + 2}}{2})}}. $$

By the same method, we have

$$ {d_{k}} = \frac{{n!(k + \lambda )\Gamma (\lambda )}}{{{2^{k}}(n - k)!}}\sum_{0 \le l \le n - k,l=0(\bmod 2)} { \frac{\sqrt{\alpha }^{2k+2 \lambda +l}{{{\sqrt{q} }^{l}}}}{{{p^{l + k}}}}} \frac{{ \binom{n - k}{l} {E_{n - k - l}}l!}}{{{2^{l}}(\frac{l}{2})!\Gamma (\frac{ {2k + 2\lambda + l + 2}}{2})}}. $$

Now Theorem 2 follows from (13). □

Theorem 3

For \(n,k\in \mathbf{Z}_{+}\) with \(n\geq k\), we have

$$\begin{aligned} C_{\alpha ,p,q,n - k}^{\lambda }(x)C_{\alpha ,p,q,k}^{\lambda }(x) &= {2^{\lambda + 1}}\left ( { \textstyle\begin{array}{@{}c@{}} {n - k + 2\lambda - 1} \\ {n - k} \end{array}\displaystyle } \right ) \left ( { \textstyle\begin{array}{@{}c@{}} {k + 2\lambda - 1} \\ k \end{array}\displaystyle } \right ) \\ &\quad {}\times \sum_{r = 0}^{n} {\sum _{i = r}^{n} {\sum_{m = 0}^{i} { \biggl\{ {{( - 1)}^{i + r}}} } } \times (r + \lambda ) \\ &\quad {}\times \frac{{\binom{n - k}{i - m} \binom{k}{m} {{(2\lambda + k)}_{m}}{{(2\lambda + n - k)}_{i - m}}i!{{( \lambda + \frac{1}{2})}_{i}}}}{{{{(\lambda + \frac{1}{2})}_{m}}{{( \lambda + \frac{1}{2})}_{i - m}}(i - r)!{{(2\lambda )}_{r + i + 1}}}} \biggr\} \\ &\quad {}\times \frac{\sqrt{\alpha }^{-2\lambda -n+r}\sqrt{q}^{n-r}}{p ^{r}} C_{\alpha ,p,q,r}^{\lambda }(x). \end{aligned}$$

Proof

From the front expression, we get

$$\begin{aligned} C_{\alpha ,p,q,n - k}^{\lambda }(x)C_{\alpha ,p,q,k}^{\lambda }(x) &= \left ( { \textstyle\begin{array}{@{}c@{}} {n - k + 2\lambda - 1} \\ {n - k} \end{array}\displaystyle } \right ) \left ( { \textstyle\begin{array}{@{}c@{}} {k + 2\lambda - 1} \\ k \end{array}\displaystyle } \right ) \\ & \quad {}\times \sum_{i = 0}^{n} {\left ( { \sum_{m = 0}^{i} {\frac{{\binom{n - k}{i - m} \binom{k}{m} {{(2\lambda + k)}_{m}}{{(2\lambda + n - k)}_{i - m}}}}{{{{( \lambda + \frac{1}{2})}_{m}}{{(\lambda + \frac{1}{2})}_{i - m}}}}} } \right ) } \\ &\quad {}\times { \biggl(\frac{{px - \sqrt{\alpha q} }}{2} \biggr)^{i}} {q^{ \frac{{n - i}}{2}}} \alpha^{-2\lambda -\frac{n}{2}-\frac{i}{2}}, \end{aligned}$$

\(p(x)=C_{\alpha ,p,q,n - k}^{\lambda }(x)C_{\alpha ,p,q,k}^{\lambda }(x) \in \mathbf{P}_{n}\), \(p(x) = \sum_{r = 0}^{n} {{d_{r}}C_{ \alpha ,p,q,r}^{\lambda }(x)}\), by (14)

$$\begin{aligned} {d_{r}} &= \frac{{{p^{1 - r}}}}{{{q^{r + \lambda }}}}\frac{{(r + \lambda )\Gamma (\lambda )}}{{{{( - 2)}^{r}}\sqrt{\pi }\Gamma (r + \lambda + \frac{1}{2})}} \\ &\quad \times \int_{ - \frac{{\sqrt{\alpha q} }}{p}}^{\frac{{\sqrt{ \alpha q} }}{p}} { \biggl( {\frac{{{d^{r}}}}{{d{x^{r}}}}{{ \bigl( \alpha q - {p^{2}} {x^{2}} \bigr)}^{r + \lambda - \frac{1}{2}}}} \biggr) C_{\alpha ,p,q,n - k}^{\lambda }(x)C_{\alpha ,p,q,k}^{\lambda }(x)\,dx} \\ &= \frac{{{p^{1 - r}}}}{{{q^{r + \lambda }}}}\frac{{(r + \lambda ) \Gamma (\lambda )}}{{{{( - 2)}^{r}}\sqrt{\pi }\Gamma (r + \lambda + \frac{1}{2})}}\left ( { \textstyle\begin{array}{@{}c@{}} {n - k + 2\lambda - 1} \\ {n - k} \end{array}\displaystyle } \right ) \left ( { \textstyle\begin{array}{@{}c@{}} {k + 2\lambda - 1} \\ k \end{array}\displaystyle } \right ) \\ &\quad \times \sum_{i = 0}^{n} {\left ( { \sum_{m = 0}^{i} {\frac{{\binom{n - k}{i - m} \binom{k}{m} {{(2\lambda + k)}_{m}}{{(2\lambda + n - k)}_{i - m}}}}{{{{( \lambda + \frac{1}{2})}_{m}}{{(\lambda + \frac{1}{2})}_{i - m}}}}} } \right ) } {q^{\frac{{n - i}}{2}}} \alpha^{-2\lambda -\frac{n}{2}-\frac{i}{2}} \\ &\quad {}\times \int_{ - \frac{{\sqrt{\alpha q }}}{p}}^{\frac{{\sqrt{ \alpha q }}}{p}} { \biggl( {\frac{{{d^{r}}}}{{d{x^{r}}}}{{ \bigl( \alpha q - {p^{2}} {x^{2}} \bigr)}^{r + \lambda - \frac{1}{2}}}} \biggr) {{ \biggl(\frac{{px - \sqrt{ \alpha q} }}{2} \biggr)}^{i}}\,dx} \\ &= \frac{{{p^{1 - r}}}}{{{q^{r + \lambda }}}}\frac{{(r + \lambda ) \Gamma (\lambda )}}{{{{( - 2)}^{r}}\sqrt{\pi }\Gamma (r + \lambda + \frac{1}{2})}}\left ( { \textstyle\begin{array}{@{}c@{}} {n - k + 2\lambda - 1} \\ {n - k} \end{array}\displaystyle } \right ) \left ( { \textstyle\begin{array}{@{}c@{}} {k + 2\lambda - 1} \\ k \end{array}\displaystyle } \right ) \\ &\quad {}\times \sum_{i = r}^{n} {\left ( { \sum_{m = 0}^{i} {\frac{{\binom{n - k}{i - m} \binom{k}{m} {{(2\lambda + k)}_{m}}{{(2\lambda + n - k)}_{i - m}}}}{{{{( \lambda + \frac{1}{2})}_{m}}{{(\lambda + \frac{1}{2})}_{i - m}}}}} } \right ) } {q^{\frac{{n - i}}{2}}} \alpha^{-2\lambda -\frac{n}{2}-\frac{i}{2}} \\ &\quad \times \int_{ - \frac{{\sqrt{\alpha q }}}{p}}^{\frac{{\sqrt{ \alpha q }}}{p}} { \biggl( {\frac{{{d^{r}}}}{{d{x^{r}}}}{{ \bigl( \alpha q - {p^{2}} {x^{2}} \bigr)}^{r + \lambda - \frac{1}{2}}}} \biggr) {{ \biggl(\frac{{px - \sqrt{ \alpha q} }}{2} \biggr)}^{i}}\,dx}. \end{aligned}$$

We can show that

$$\begin{aligned} &\int_{ - \frac{{\sqrt{\alpha q} }}{p}}^{ \frac{{\sqrt{\alpha q }}}{p}} { \biggl( {\frac{{{d^{r}}}}{{d{x^{r}}}} {{ \bigl(\alpha q - {p^{2}} {x^{2}} \bigr)}^{r + \lambda - \frac{1}{2}}}} \biggr) {{ \biggl(\frac{{px - \sqrt{\alpha q} }}{2} \biggr)}^{i}}\,dx} \\ &\quad = \frac{{{{( - 1)}^{r}}i!}}{{{2^{i}}(i - r)!}} \int_{ - \frac{{\sqrt{\alpha q} }}{p}}^{ \frac{{\sqrt{\alpha q }}}{p}} {{{ \bigl(\alpha q - {p^{2}} {x^{2}} \bigr)}^{r + \lambda - \frac{1}{2}}} {{(\sqrt{\alpha q} - px)}^{i - r}} {{( - 1)} ^{i - r}}\,dx} \\ &\quad = \frac{{{{( - 1)}^{i}}i!}}{{{2^{i}}(i - r)!}} \int_{ - \frac{{\sqrt{\alpha q} }}{p}}^{ \frac{{\sqrt{\alpha q }}}{p}} {{{(\sqrt{\alpha q} + px)}^{r + \lambda - \frac{1}{2}}} {{(\sqrt{\alpha q} - px)}^{i + \lambda - \frac{1}{2}}}\,dx} \\ &\quad = \frac{{{{( - 1)}^{i}}i!}}{{{2^{i}}(i - r)!}}\frac{{{{\sqrt{ \alpha q }}^{i + r + 2\lambda }}}}{p} \int_{0}^{1} {{{(2x)}^{r + \lambda - \frac{1}{2}}} {{(2 - 2x)}^{i + \lambda - \frac{1}{2}}}2\,dx} \\ &\quad = \frac{{{{( - 1)}^{i}}{2^{r + 2\lambda }}i!}}{{(i - r)!}}\frac{ {{{\sqrt{\alpha q }}^{i + r + 2\lambda }}}}{p}\frac{{\Gamma (i + \lambda + \frac{1}{2})\Gamma (r + \lambda + \frac{1}{2})}}{{\Gamma (r + i + 2\lambda + 1)}}. \end{aligned}$$

Applying some identities about gamma function, we get

$$\begin{aligned} {d_{r}} &= {2^{\lambda + 1}}(r + \lambda )\left ( { \textstyle\begin{array}{@{}c@{}} {n - k + 2\lambda - 1} \\ {n - k} \end{array}\displaystyle } \right ) \left ( { \textstyle\begin{array}{@{}c@{}} {k + 2\lambda - 1} \\ k \end{array}\displaystyle } \right ) \\ &\quad {}\times \sum_{i = r}^{n} {\left ( { \sum_{m = 0}^{i} {( - 1)^{r + i} \frac{{\binom{n - k}{i - m} \binom{k}{m} {{(2\lambda + k)}_{m}}{{(2\lambda + n - k)}_{i - m}}i!{{( \lambda + \frac{1}{2})}_{i}}}}{{{{(\lambda + \frac{1}{2})}_{m}}{{( \lambda + \frac{1}{2})}_{i - m}}(i - r)!{{(2\lambda )}_{r + i + 1}}}}} } \right ) } \\ &\quad {}\times \frac{\sqrt{\alpha }^{-2\lambda -n+r}{{{\sqrt{q} }^{n - r}}}}{ {{p^{r}}}}. \end{aligned}$$

This proves Theorem 3. □

Theorem 4

For \(n\in \mathbf{Z}_{+}\), we have

$$\begin{aligned} {C_{\alpha ,p,q,n}}(x) &=\sum_{k = 0}^{n} { \Biggl\{ \frac{{{{( \lambda + 1)}_{k - 1}}(k + \lambda ){2^{2k}} \binom{n + k + 2\lambda - 1}{n - k} }}{{\binom{n + \lambda - \frac{1}{2}}{n - k} }}} {\sqrt{q} ^{n - k}}\sqrt{\alpha }^{3k-n} \\ &\quad {}\times \sum_{l = 0}^{n - k} {\left ( { \textstyle\begin{array}{@{}c@{}} {n + \lambda - \frac{1}{2}} \\ {n - k - l} \end{array}\displaystyle } \right ) } \left ( { \textstyle\begin{array}{@{}c@{}} {n + \lambda - \frac{1}{2}} \\ l \end{array}\displaystyle } \right ) {( -1)^{l}} \frac{{ \binom{k + \lambda + l - \frac{1}{2}}{l} \binom{\lambda + n - l - \frac{1}{2}}{n - l} }}{{ \binom{n}{l} \binom{k + 2\lambda + n}{n + k} \binom{n + k}{k} k!}} \Biggr\} C_{\alpha ,p,q,k}^{\lambda }(x). \end{aligned}$$

Proof

By (2), (5) and (14), we get

$$\begin{aligned} {d_{k}} &= \frac{{{p^{1 - k}}}}{{{q^{k + \lambda }}}}\frac{{(k + \lambda )\Gamma (\lambda )}}{{{{( - 2)}^{k}}\sqrt{\pi }\Gamma (k + \lambda + \frac{1}{2})}} \int_{ - \frac{{\sqrt{\alpha q} }}{p}}^{\frac{ {\sqrt{\alpha q} }}{p}} { \biggl( {\frac{{{d^{k}}}}{{d{x^{k}}}}{{ \bigl( \alpha q - {p^{2}} {x^{2}} \bigr)}^{k + \lambda - \frac{1}{2}}}} \biggr) C _{\alpha ,p,q,n}^{\lambda }(x)\,dx} \\ &=\frac{(\lambda )_{k}(k+\lambda )\Gamma (\lambda )}{\sqrt{\pi } \Gamma (k+\lambda +\frac{1}{2})}\frac{p}{q^{k+\lambda }} \int_{-\frac{\sqrt{\alpha q}}{p}}^{\frac{\sqrt{\alpha q}}{p}} \bigl( \alpha q-p^{2}x^{2} \bigr)^{k+\lambda -\frac{1}{2}}C_{\alpha ,p,q,n-k}^{ \lambda +k}(x)\,dx \\ &=\frac{{{{(\lambda )}_{k}}(k + \lambda )\Gamma (\lambda )}}{{\sqrt{ \pi }\Gamma (k + \lambda + \frac{1}{2})}} \frac{p}{{{q^{k + \lambda }}}}\frac{{ \binom{n + k + 2\lambda - 1}{n - k} }}{{\binom{n + \lambda - \frac{1}{2}}{n - k} }}\sum _{l = 0}^{n - k} {\left ( { \textstyle\begin{array}{@{}c@{}} {n + \lambda - \frac{1}{2}} \\ {n - k - l} \end{array}\displaystyle } \right ) } \\ &\quad {}\times \left ( { \textstyle\begin{array}{@{}c@{}} {n + \lambda - \frac{1}{2}} \\ l \end{array}\displaystyle } \right ) {( - 1)^{l}} { \biggl( \frac{1}{2} \biggr)^{n - k}}\alpha^{-\lambda -n+k} \\ &\quad {}\times \int_{ - \frac{{\sqrt{\alpha q }}}{p}}^{\frac{{\sqrt{ \alpha q }}}{p}} {{{(\sqrt{\alpha q} - px)}^{k + \lambda - \frac{1}{2} + l}}} {(\sqrt{\alpha q} + px)^{\lambda + n - \frac{1}{2} - l}}\,dx. \end{aligned}$$

It is not difficult to show that

$$\begin{aligned}& \int_{ - \frac{{\sqrt{\alpha q} }}{p}}^{\frac{{\sqrt{\alpha q }}}{p}} {{{(\sqrt{\alpha q} - px)}^{k + \lambda - \frac{1}{2} + l}}} {(\sqrt{\alpha q } + px)^{\lambda + n - \frac{1}{2} - l}}\,dx \\& \quad = \frac{{{{\sqrt{\alpha q} }^{k + 2\lambda + n}}}}{p} \int_{0}^{1} {{{(2 - 2x)}^{{k + \lambda - \frac{1}{2} + l}}} {{(2x)}^{{\lambda + n - \frac{1}{2} - l}}}}\,dx \\& \quad = \frac{{{{\sqrt{\alpha q }}^{k + 2\lambda + n}}}}{p}{2^{k + 2 \lambda + n}}\frac{{\Gamma (k + \lambda + l + \frac{1}{2})\Gamma ( \lambda + n - l + \frac{1}{2})}}{{\Gamma (k + 2\lambda + n + 1)}}. \end{aligned}$$

Applying some identities involving gamma function, we get

$$\begin{aligned}& \Gamma \biggl(k + \lambda + l + \frac{1}{2} \biggr) = \left ( { \textstyle\begin{array}{@{}c@{}} {k + \lambda + l - \frac{1}{2}} \\ l \end{array}\displaystyle } \right ) l!\Gamma \biggl(k + \lambda + \frac{1}{2} \biggr), \\& \Gamma \biggl(\lambda + n - l + \frac{1}{2} \biggr) = \left ( { \textstyle\begin{array}{@{}c@{}} {\lambda + n - l - \frac{1}{2}} \\ {n - l} \end{array}\displaystyle } \right ) (n - l)!\Gamma \biggl(\lambda + \frac{1}{2} \biggr), \end{aligned}$$

and

$$ \Gamma (k + 2\lambda + n + 1) = \left ( { \textstyle\begin{array}{@{}c@{}} {k + 2\lambda + n} \\ {n + k} \end{array}\displaystyle } \right ) (n + k)!\Gamma (2\lambda + 1). $$

As we all know, the duplication formula for the gamma function is given by

$$ \Gamma (z)\Gamma \biggl(z + \frac{1}{2} \biggr) = {2^{1 - 2z}} \sqrt{\pi }\Gamma (2z). $$

From this formula, we get

$$\begin{aligned}& \int_{ - \frac{{\sqrt{\alpha q}}}{p}}^{ \frac{{\sqrt{\alpha q} }}{p}} {{{(\sqrt{\alpha q } - px)}^{k +\lambda - \frac{1}{2}}} {{(\sqrt{\alpha q} + px)}^{\lambda + n - l -\frac{1}{2}}}}\,dx \\& \quad=\frac{\sqrt{\alpha q }^{k + 2\lambda + n}}{p}{2^{ k + n}} \frac{\binom{k + \lambda + l - \frac{1}{2}}{l} \binom{\lambda + n - l - \frac{1}{2}}{n - l} \Gamma (k + \lambda + \frac{1}{2}) }{\binom{n}{l} \binom{k + 2\lambda + n}{n + k} \binom{n + k}{k} k!2\lambda \Gamma (\lambda )} \sqrt{\pi }. \end{aligned}$$

So, we get

$$\begin{aligned} d_{k} &= (\lambda + 1)_{k - 1}(k + \lambda ){2^{2k}}\frac{\binom{n + k + 2\lambda - 1}{n - k}}{\binom{n + \lambda - \frac{1}{2}}{n - k}} \sum_{l = 0}^{n - k} {\left ( { \textstyle\begin{array}{@{}c@{}} {n + \lambda - \frac{1}{2}} \\ {n - k - l} \end{array}\displaystyle } \right ) } \left ( { \textstyle\begin{array}{@{}c@{}} {n + \lambda - \frac{1}{2}} \\ l \end{array}\displaystyle } \right ) \\ &\quad {}\times {( - 1)^{l}} {\sqrt{q} ^{n - k}}\sqrt{\alpha }^{3k-n}\frac{ \binom{k + \lambda + l - \frac{1}{2}}{l} \binom{\lambda + n - l - \frac{1}{2}}{n - l}\binom{n}{l}}{\binom{k + 2\lambda + n}{n + k}\binom{n + k}{k}k!}. \end{aligned}$$

This completes the proof of Theorem 4. □

Theorem 5

For \(n\in \mathbf{Z}_{+}\), we have

$$\begin{aligned}& \frac{{{B_{n + 1}}(x + 1) - {B_{n + 1}}(x)}}{{(n + 1)!}} = \sum_{0 \le k \le n,n - k \equiv 0(\bmod 2)} { \frac{\sqrt{\alpha }^{2\lambda +k+n}{{{\sqrt{q} }^{n - k}}}}{{{p^{n}}}}} \\& \hphantom{\frac{{{B_{n + 1}}(x + 1) - {B_{n + 1}}(x)}}{{(n + 1)!}} =}{}\times \frac{{(k + \lambda )n!\Gamma (\lambda )}}{{{2^{n}}(\frac{ {n - k}}{2})!\Gamma (\frac{{n + k + 2\lambda + 2}}{2})}}C_{\alpha ,p,q,k} ^{\lambda }(x), \\& \frac{{{E_{n}}(x + 1) + {E_{n}}(x)}}{{2(n)!}} = \sum_{0 \le k \le n,n - k \equiv 0(\bmod 2)} { \frac{\sqrt{\alpha }^{2\lambda +k+n}{{{\sqrt{q} }^{n - k}}}}{{{p^{n}}}}} \\& \hphantom{\frac{{{E_{n}}(x + 1) + {E_{n}}(x)}}{{2(n)!}} = }{}\times \frac{{(k + \lambda )n!\Gamma (\lambda )}}{{{2^{n}}(\frac{ {n - k}}{2})!\Gamma (\frac{{n + k + 2\lambda + 2}}{2})}}C_{\alpha ,p,q,k} ^{\lambda }(x). \end{aligned}$$

Proof

Applying Theorem 1, we get

$$ {x^{n}} = \sum_{0 \le k \le n,n - k \equiv 0(\bmod 2)} { \frac{\sqrt{ \alpha }^{2\lambda +k+n}{{{\sqrt{q} }^{n - k}}}}{{{p^{n}}}}} \frac{ {(k + \lambda )n!\Gamma (\lambda )}}{{{2^{n}}(\frac{{n - k}}{2})! \Gamma (\frac{{n + k + 2\lambda + 2}}{2})}}C_{\alpha ,p,q,k}^{\lambda }(x). $$

From (6) and (7), we have

$$\begin{aligned}& {e^{xt}} = \frac{1}{t}\frac{t}{{{e^{t}} - 1}}{e^{xt}} \bigl({e^{t}} - 1 \bigr) = \frac{1}{t}\frac{t}{{{e^{t}} - 1}} \bigl({e^{(x + 1)t}} - {e^{xt}} \bigr) \\& \hphantom{{e^{xt}} }= \frac{1}{t} \Biggl(\sum_{n = 0}^{\infty }{{B_{n}}(x + 1)} \frac{ {{t^{n}}}}{{n!}} - \sum_{n = 0}^{\infty }{{B_{n}}(x)} \frac{ {{t^{n}}}}{{n!}} \Biggr) \\& \hphantom{{e^{xt}} }= \sum_{n = 0}^{\infty }{\frac{{{B_{n + 1}}(x + 1) - {B_{n + 1}}(x)}}{{n + 1}} \frac{{{t^{n}}}}{{n!}}}, \\& {e^{xt}} = \frac{2}{{{e^{t}} + 1}}{e^{xt}} \biggl( \frac{{{e^{t}} + 1}}{2} \biggr) = \frac{1}{2} \biggl(\frac{2}{{{e^{t}} + 1}}{e^{(x + 1)t}} + \frac{2}{{{e^{t}} + 1}}{e^{xt}} \biggr) \\& \hphantom{{e^{xt}} }= \frac{1}{2} \Biggl(\sum_{n = 0}^{\infty }{{E_{n}}(x + 1)} \frac{ {{t^{n}}}}{{n!}} + \sum_{n = 0}^{\infty }{{E_{n}}(x)} \frac{ {{t^{n}}}}{{n!}} \Biggr). \end{aligned}$$

So, we get

$$ {x^{n}} = \frac{{{B_{n + 1}}(x + 1) - {B_{n + 1}}(x)}}{{n + 1}} = \frac{1}{2} \bigl({E_{n}}(x + 1) + {E_{n}}(x) \bigr). $$

We can complete the proof of Theorem 5. □

Theorem 6

For \(n\in \mathbf{Z}_{+}\), we have

$$\begin{aligned} \frac{{{H_{n}}(x)}}{n!} &= \Gamma (\lambda )\sum_{k = 0}^{n} {\frac{{(k + \lambda )}}{{(n - k)!}}\left ( {\sum_{0 \le l \le n - k,l \equiv 0(\bmod 2)} { \frac{\sqrt{\alpha }^{2\lambda +2k+l}{{{\sqrt{q} }^{l}}}}{{{p^{l + k}}}}\frac{{ \binom{n - k}{l} l!{H_{n - k - l}}}}{{(\frac{l}{2})!\Gamma (\frac{{2k + 2 \lambda + l + 2}}{2})}}} } \right ) } \\ &\quad {}\times C_{\alpha ,p,q,k}^{\lambda }(x). \end{aligned}$$

Proof

From (10), (11) and (14), we get

$$\begin{aligned} {d_{k}} &= \frac{{{p^{1 - k}}}}{{{q^{k + \lambda }}}}\frac{{(k + \lambda )\Gamma (\lambda )}}{{{{( - 2)}^{k}}\sqrt{\pi }\Gamma (k + \lambda + \frac{1}{2})}} \int_{ - \frac{{\sqrt{\alpha q }}}{p}}^{\frac{ {\sqrt{\alpha q} }}{p}} { \biggl( {\frac{{{d^{k}}}}{{d{x^{k}}}}{{ \bigl( \alpha q - {p^{2}} {x^{2}} \bigr)}^{k + \lambda - \frac{1}{2}}}} \biggr) {H_{n}}(x)\,dx} \\ &= \frac{{{p^{1 - k}}}}{{{q^{k + \lambda }}}}\frac{{(k + \lambda ) \Gamma (\lambda )}}{{\sqrt{\pi }\Gamma (k + \lambda + \frac{1}{2})}}\frac{ {n!}}{{(n - k)!}} \int_{ - \frac{{\sqrt{\alpha q} }}{p}}^{\frac{ {\sqrt{\alpha q }}}{p}} {{{ \bigl(\alpha q - {p^{2}} {x^{2}} \bigr)}^{k + \lambda - \frac{1}{2}}} {H_{n - k}}(x)\,dx} \\ &=\frac{{{p^{1 - k}}}}{{{q^{k + \lambda }}}}\frac{{(k + \lambda ) \Gamma (\lambda )}}{{\sqrt{\pi }\Gamma (k + \lambda + \frac{1}{2})}}\frac{ {n!}}{{(n - k)!}} \\ &\quad {}\times \sum_{l = 0}^{n - k} {\left ( { \textstyle\begin{array}{@{}c@{}} {n - k} \\ l \end{array}\displaystyle } \right ) } {H_{n - k - l}} {2^{l}} \int_{ - \frac{{\sqrt{\alpha q }}}{p}}^{\frac{{\sqrt{\alpha q} }}{p}} {{{ \bigl( \alpha q - {p^{2}} {x^{2}} \bigr)}^{k + \lambda - \frac{1}{2}}} {x^{l}}\,dx}. \end{aligned}$$

Let us assume that \(l\equiv 0(\bmod 2)\), first let \(x=\frac{ \sqrt{q}}{p}y\), then \(y=\sqrt{x}\), we have

$$\begin{aligned} {d_{k}} & = \frac{{(k + \lambda )\Gamma (\lambda )}}{{\sqrt{\pi } \Gamma (k + \lambda + \frac{1}{2})}}\frac{{n!}}{{(n - k)!}} \\ &\quad {}\times \sum_{l = 0}^{n - k} {\left ( { \textstyle\begin{array}{@{}c@{}} {n - k} \\ l \end{array}\displaystyle } \right ) } {H_{n - k - l}} {2^{l}} \frac{\sqrt{\alpha }^{2\lambda +2k+l} {{{\sqrt{q} }^{l}}}}{{{p^{l + k}}}} \int_{0}^{1} {{{(1 - x)}^{k + \lambda - \frac{1}{2}}} {x^{\frac{{l - 1}}{2}}}\,dx} \\ &= \frac{{(k + \lambda )\Gamma (\lambda )}}{{\sqrt{\pi }\Gamma (k + \lambda + \frac{1}{2})}}\frac{{n!}}{{(n - k)!}} \\ &\quad {}\times \sum_{0 \le l \le n - k,l \equiv 0(\bmod 2)} {\left ( { \textstyle\begin{array}{@{}c@{}} {n - k} \\ l \end{array}\displaystyle } \right ) } {H_{n - k - l}} {2^{l}}\frac{\sqrt{\alpha }^{2\lambda +2k+l} {{{\sqrt{q} }^{l}}}}{{{p^{l + k}}}} \frac{{\Gamma (k + \lambda + \frac{1}{2})\Gamma (\frac{{l + 1}}{2})}}{{\Gamma (\frac{{2k + 2\lambda + l + 2}}{2})}} \\ &= \frac{{(k + \lambda )\Gamma (\lambda )n!}}{{(n - k)!}}\sum_{0 \le l \le n - k,l \equiv 0(\bmod 2)} { \frac{{ \binom{n - k}{l} {H_{n - k - l}}l!}}{{(\frac{l}{2})!\Gamma (\frac{{2k + 2 \lambda + l + 2}}{2})}}} \frac{\sqrt{\alpha }^{2\lambda +2k+l}{{{\sqrt{q} }^{l}}}}{{{p^{l + k}}}}. \end{aligned}$$

 □