In this section, we establish general symmetry identities for the generalized Hermite–Euler polynomials \({}_HE_{n}^{(\alpha )}(x,y)\), of order\(\alpha \) and alternate power sum. Throughout this section \(\alpha \) will be taken as an arbitrary real or complex parameter.
Theorem 1
For integers
\(n\ge 0\), \(a\ge 1\)
and
\(b\ge 1,\) if a
and
b
have
the same parity, then the following identity holds true:
$$\begin{aligned}&\sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) a^{n-k}b^{k}{}_HE_{n-k}^{(\alpha )}(bx,b^{2}z)\sum \limits _{i=0}^{k}\left( \begin{array} {lll}k\\ i\end{array}\right) T_i(a-1)E_{k-i}^{(\alpha -1)}(ay)\nonumber \\&\quad =\sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) a^{k}b^{n-k} {}_HE_{n-k}^{(\alpha )}(ax,a^{2}z)\sum \limits _{i=0}^{k}\left( \begin{array}{lll}k\\ i\end{array}\right) T_i(b-1)E_{k-i}^{(\alpha -1)}(by) \end{aligned}$$
(20)
Proof
Let \(G(t)=:\frac{2^{2{\alpha }-1}(1-(-e^{bt})^a)e^{ab(x+y)t+a^2b^2zt^2}}{(e^{at}+1)^{\alpha }(e^{bt}+1)^{\alpha }}\)
$$\begin{aligned}G(t)&={\left( \frac{2}{e^{at}+1}\right) ^{\alpha }}e^{abxt+a^2b^2zt^2}{\left( \frac{1-{(-e^{bt})}^a}{1+e^{bt}}\right) }{\left( \frac{2}{e^{bt}+1}\right) ^{{\alpha }-1}}e^{abyt} \\& ={\left( \sum \limits _{n=0}^{\infty }{}_HE_{n}^{(\alpha )}(bx,b^{2}z)\frac{(at)^n}{n!}\right) }{\left( \sum \limits _{i=0}^{\infty }T_i{(a-1)}\frac{(bt)^i}{i!}\right) } \left( \sum \limits _{k=0}^{\infty }E_k^{(\alpha -1)}{(ay)\frac{(bt)^k}{k!}}\right) \\&={\left( {\sum \limits _{n=0}^{\infty }{}_HE_{n}^{(\alpha )}(bx,b^{2}z)\frac{(at)^n}{n!}}\right) }{\left( \sum \limits _{k=0}^{\infty }\sum \limits _{i=0}^{\infty } {b^{i+k}}T_i{(a-1)}E_k^{(\alpha -1)}(ay)\frac{t^{i+k}}{i!k!}\right) } \\&={\sum \limits _{n=0}^{\infty }{}_HE_{n}^{(\alpha )}(bx,b^{2}z)\frac{(at)^n}{n!}}{\sum \limits _{k=0}^{\infty }{b^k}\sum \limits _{i=0} ^{k}\left( \begin{array}{lll}k\\ i\end{array}\right) T_i{(a-1)}E_{k-i}^{(\alpha -1)}(ay)\frac{t^k}{k!}} \\ G(t)&={\sum \limits _{n=0}^{\infty }\sum \limits _{k=0}^{\infty }{a^n}{b^k}{}_HE_{n}^{(\alpha )}(bx,b^{2}z)\sum \limits _{i=0}^{k} \left( \begin{array}{lll}k\\ i\end{array}\right) T_i{(a-1)}E_{k-i}^{(\alpha -1)}(ay)\frac{t^{n+k}}{k!n!}} \end{aligned}$$
Replacing n by \(n-k\) in the R.H.S. of above equation, we get
$$\begin{aligned} G(t)={\sum \limits _{n=0}^{\infty }\left[ \sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) {a^{n-k}}{b^k}{}_HE_{n-k} ^{(\alpha )}(bx,b^{2}z)\sum \limits _{i=0}^{k}\left( \begin{array}{lll}k\\ i\end{array}\right) T_i{(a-1)}E_{k-i}^{(\alpha -1)}(ay)\right] }\frac{t^{n}}{n!} \end{aligned}$$
(21)
Using a similar plan, we get
$$\begin{aligned} G(t)={\sum \limits _{n=0}^{\infty }\left[ \sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) {b^{n-k}}{a^k}{}_HE_{n-k} ^{(\alpha )}(ax,a^{2}z)\sum \limits _{i=0}^{k}\left( \begin{array}{lll}k\\ i\end{array}\right) T_i{(b-1)}E_{k-i}^{(\alpha -1)}(by)\right] }\frac{t^{n}}{n!} \end{aligned}$$
(22)
By comparing the coefficients of \(\frac{t^n}{n!}\) in the R.H.S of above Eqs. (21) and (22), we arrive at the desired result.
Remark 1
For \(z=0\) in Theorem (1), the result reduces to known result of Liu and Wang [2009, p. 3348, (2.1)].
Corollary 1
For integers
\( n\ge 0\), \(a\ge 1\)
and
\( b\ge 1 \) , if
a
and
b
have the same parity, then the following identity holds true:
$$\begin{aligned}&\sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) a^{n-k}b^{k}E_{n-k}^{(\alpha )}(bx)\sum \limits _{i=0}^{k}\left( \begin{array} {lll}k\\ i\end{array}\right) T_i(a-1)E_{k-i}^{(\alpha -1)}(ay) \nonumber \\&\quad =\sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) a^{k}b^{n-k} E_{n-k}^{(\alpha )}(ax)\sum \limits _{i=0}^{k}\left( \begin{array}{lll}k\\ i\end{array}\right) T_i(b-1)E_{k-i}^{(\alpha -1)}(by) \end{aligned}$$
(23)
Theorem 2
For integers
\( n\ge 0\), \(a\ge 1\)
and
\( b\ge 1 \) , if
a
is odd and
b
is even, then the following identity holds true:
$$\begin{aligned}&\sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) a^{n-k}b^{k}{}_HE_{n-k}^{(\alpha )}(bx,b^{2}z)\sum \limits _{i=0}^{k}\left( \begin{array}{lll}k\\ i\end{array}\right) T_i(a-1)E_{k-i}^{(\alpha -1)}(ay)\nonumber \\&\quad =-2\sum \limits _{l=0,l\ne n}^{n+1}\left( \begin{array}{lll}{n+1}\\ ~l~\end{array}\right) \frac{B_{n+1-l}}{n+1} a^{n-l}\sum \limits _{k=0}^{l} \left( \begin{array}{lll}l\\ k\end{array}\right) b^{n-k}a^{k}{}_HE_{l-k}^{(\alpha )}(ax,a^{2}z)\nonumber \\&\qquad \times \sum \limits _{i=0}^{k}\left( \begin{array}{lll}k\\ i\end{array}\right) T_i(b-1)E_{k-i}^{(\alpha -1)}(by) \end{aligned}$$
(24)
Proof
Let \(G(t)=:\frac{2^{2{\alpha }-1}(1-(-e^{bt})^a)e^{ab(x+y)t+a^2b^2zt^2}}{(e^{at}+1)^{\alpha }(e^{bt}+1)^{\alpha }}\)
$$\begin{aligned}G(t)&={\left( \frac{2}{e^{at}+1}\right) ^{\alpha }}e^{abxt+a^2b^2zt^2}{\left( \frac{1-{(-e^{bt})}^a}{1+e^{bt}}\right) }{\left( \frac{2}{e^{bt}+1}\right) ^{{\alpha }-1}}e^{abyt}\\ & ={\left( \sum \limits _{n=0}^{\infty }{}_HE_{n}^{(\alpha )}(bx,b^{2}z)\frac{(at)^n}{n!}\right) }{\left( \sum \limits _{i=0}^{\infty }T_i{(a-1)}\frac{(bt)^i}{i!}\right) } \left( \sum \limits _{k=0}^{\infty }E_k^{(\alpha -1)}{(ay)\frac{(bt)^k}{k!}}\right) \\& ={\left( \sum \limits _{n=0}^{\infty }{}_HE_{n}^{(\alpha )}(bx,b^{2}z)\frac{(at)^n}{n!}\right) }{\left( \sum \limits _{k=0}^{\infty }{b^k}\sum \limits _{i=0} ^{k}\left( \begin{array}{lll}k\\ i\end{array}\right) T_i{(a-1)}E_{k-i}^{(\alpha -1)}(ay)\frac{t^k}{k!}\right) } \end{aligned}$$
Replacing n by \(n-k\) in the R.H.S. of above equation, we get
$$\begin{aligned} G(t)={\sum \limits _{n=0}^{\infty }\left[ \sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) {a^{n-k}}{b^k}{}_HE_{n-k} ^{(\alpha )}(bx,b^{2}z)\sum \limits _{i=0}^{k}\left( \begin{array}{lll}k\\ i\end{array}\right) T_i{(a-1)}E_{k-i}^{(\alpha -1)}(ay)\right] }\frac{t^{n}}{n!} \end{aligned}$$
(25)
Since G(t) is not symmetric in a and b, thus G(t) can also be expanded as
$$\begin{aligned}G(t)&={\left( \frac{2}{e^{bt}+1}\right) ^{\alpha }}e^{abxt+a^2b^2zt^2}{\left( \frac{1-{(-e^{at})}^b}{1+e^{at}}\right) }{\left( \frac{1-{(-e^{bt})}^a}{1-{(-e^{at})}^b} \right) }{\left( \frac{2}{e^{at}+1}\right) ^{{\alpha }-1}}e^{abyt}\nonumber \\& ={\left( \sum \limits _{l=0}^{\infty }{}_HE_{l}^{(\alpha )}(ax,a^{2}z)\frac{(bt)^l}{l!}\right) }{\left( \sum \limits _{i=0}^{\infty }T_i{(b-1)}\frac{(at)^i}{i!}\right) } \left( -\sum \limits _{n=0}^{\infty }\frac{B_n+B_n(1)}{n!}(abt)^{n-1}\right) \nonumber \\&\qquad \times \left( \sum \limits _{k=0}^{\infty }E_k^{(\alpha -1)}{(by)\frac{(at)^k}{k!}}\right) \nonumber \\G(t)&=\left( -\sum \limits _{n=0}^{\infty }\frac{B_n+B_n(1)}{n!}(abt)^{n-1}\right) {\left( \sum \limits _{l=0}^{\infty }{}_HE_{l}^{(\alpha )}(ax,a^{2}z)\frac{(bt)^l}{l!} \right) }\nonumber \\&\qquad \times {\left( \sum \limits _{k=0}^{\infty }\sum \limits _{i=0}^{k}a^k\left( \begin{array}{lll}k\\ i\end{array}\right) T_i{(b-1)}E_{k-i}^{(\alpha -1)}(by)\frac{t^k}{k!} \right) } \end{aligned}$$
(26)
Using identity (7) and comparing the coefficients of \(\frac{t^n}{n!}\) in the R.H.S. of Eqs. (25) and (26), we get the desired result.
Remark 2
For \( z=0\) in Theorem (2), the result reduces to known result of Liu and Wang [2009, p. 3349, Theorem (4)].
Corollary 2
For integers
\( n\ge 0\), \(a\ge 1\)
and
\( b\ge 1 \) , if
a
is odd and
b
is even, then the following identity holds true:
$$\begin{aligned}&\sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) a^{n-k}b^{k}E_{n-k}^{(\alpha )}(bx)\sum \limits _{i=0}^{k}\left( \begin{array}{lll}k\\ i\end{array}\right) T_i(a-1)E_{k-i}^{(\alpha -1)}(ay)\nonumber \\&\quad =-2\sum \limits _{l=0,l\ne n}^{n+1}\left( \begin{array}{lll}{n+1}\\ l\end{array}\right) \frac{B_{n+1-l}}{n+1} a^{n-l}\sum \limits _{k=0}^{l} \left( \begin{array}{lll}l\\ k\end{array}\right) b^{n-k}a^{k}E_{l-k}^{(\alpha )}(ax)\nonumber \\&\qquad \times \sum \limits _{i=0}^{k}\left( \begin{array}{lll}k\\ i\end{array}\right) T_i(b-1)E_{k-i}^{(\alpha -1)}(by) \end{aligned}$$
(27)
Theorem 3
For integers
\( n\ge 1\) , \(a\ge 1\)
and
\( b\ge 1\), if
a
is even and
b
is odd, then the following identity holds true:
$$\begin{aligned}&\sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) a^{n-k} b^k{}_HE_{n-k}^{(\alpha )}(bx,b^2 z)\sum \limits _{i=0}^{k}\left( \begin{array} {lll}k\\ i\end{array}\right) T_i(a-1)E_{k-i}^{(\alpha -1)}(ay)\nonumber \\&\quad =\sum \limits _{l=0}^{n-1}\left( \begin{array}{lll}n\\ l\end{array}\right) E_{n-l}(0)a^{n-l} \sum \limits _{k=0}^{l}\left( \begin{array}{lll}l\\ k\end{array}\right) b^{n-k} a^k {}_HE_{l-k}^{(\alpha )} (ax,a^2 z)\nonumber \\&\qquad \times \sum \limits _{i=0}^k\left( \begin{array}{lll}k\\ i\end{array}\right) T_i (b-1)E_{k-i}^{(\alpha -1)}(by) \end{aligned}$$
(28)
Proof
Let
$$\begin{aligned}G(t)&=:\frac{2^{2{\alpha }-1}(1-(-e^{bt})^a)e^{ab(x+y)t+a^2b^2zt^2}}{(e^{at}+1)^{\alpha }(e^{bt}+1)^{\alpha }}\nonumber \\G(t)&={\left( \frac{2}{e^{at}+1}\right) ^{\alpha }}e^{abxt+a^2b^2zt^2}{\left( \frac{1-{(-e^{bt})}^a}{1+e^{bt}}\right) }{\left( \frac{2}{e^{bt}+1}\right) ^{{\alpha }-1}}e^{abyt}\nonumber \\&={\left( \sum \limits _{n=0}^{\infty }{}_HE_{n}^{(\alpha )}(bx,b^{2}z)\frac{(at)^n}{n!}\right) }{\left( \sum \limits _{i=0}^{\infty }T_i{(a-1)}\frac{(bt)^i}{i!}\right) } \left( \sum \limits _{k=0}^{\infty }E_k^{(\alpha -1)}{(ay)\frac{(bt)^k}{k!}}\right) \nonumber \\ \end{aligned}$$
(29)
Another expansion of G(t) is as follows :
$$\begin{aligned}G(t)&={\left( \frac{2}{e^{bt}+1}\right) ^{\alpha }}e^{abxt+a^2b^2zt^2}{\left( \frac{1-{(-e^{at})}^b}{1+e^{at}}\right) } {\left( \frac{1-{(-e^{bt})}^a}{1-{(-e^{at})}^b}\right) }{\left( \frac{2}{e^{at}+1}\right) ^{{\alpha }-1}}e^{abyt}\nonumber \\ &=\left( \sum \limits _{l=0}^{\infty }{}_HE_{l}^{(\alpha )}(ax,a^{2}z)\frac{(bt)^l}{l!}\right) \left( -\sum \limits _{n=0}^{\infty } \frac{B_n+B_n(1)}{n!}(abt)^{n-1}\right) \left( \sum \limits _{i=0}^{\infty }T_i{(b-1)}\frac{(at)^i}{i!}\right) \nonumber \\&\quad \times \left( \sum \limits _{k=0}^{\infty }E_k^{(\alpha -1)} {(by)\frac{(at)^k}{k!}}\right) \nonumber \\G(t)&=\left( \sum \limits _{l=0}^{\infty }{}_HE_{l}^{(\alpha )}(ax,a^{2}z)\frac{(bt)^l}{l!}\right) \left( -\sum \limits _{n=0}^{\infty } \frac{B_n+B_n(1)}{n!}(abt)^{n-1}\right) \nonumber \\&\quad \times \sum \limits _{k=0}^{\infty }\sum \limits _{i=0}^{k}\left( \begin{array}{lll}k\\ i\end{array}\right) a^kT_i{(b-1)}E_{k-i}^{(\alpha -1)}(by)\frac{t^k}{k!} \end{aligned}$$
(30)
Now using Eq. (17) and identity (7) and comparing the coefficients of \(\frac{t^n}{n!}\) in the R.H.S. of Eqs. (29) and (30), we arrive at the desired result.
Remark 3
For \( z=0 \) in Theorem (3), the result reduces to known result of Liu and Wang [2009, p. 3350, (2.11)].
Corollary 3
For integers
\( n\ge 1,\)
\(a\ge 1\)
and
\( b\ge 1\), if
a
is even and
b
is odd, then the following identity holds true:
$$\begin{aligned}&\sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) a^{n-k} b^kE_{n-k}^{(\alpha )}(bx)\sum \limits _{i=0}^{k}\left( \begin{array} {lll}k\\ i\end{array}\right) T_i(a-1)E_{k-i}^{(\alpha -1)}(ay)\nonumber \\&\quad =\sum \limits _{l=0}^{n-1}\left( \begin{array}{lll}n\\ l\end{array}\right) E_{n-l}(0)a^{n-l} \sum \limits _{k=0}^{l}\left( \begin{array}{lll}l\\ k\end{array}\right) b^{n-k} a^k E_{l-k}^{(\alpha )}(ax)\nonumber \\&\qquad \times \sum \limits _{i=0}^k\left( \begin{array}{lll}k\\ i\end{array}\right) T_i (b-1)E_{k-i}^{(\alpha -1)}(by) \end{aligned}$$
(31)
Theorem 4
For integers
\(n\ge 0\), \(a\ge 1 \)
and
\( b\ge 1\), if
a
and
b
have same parity, then the following identity holds true:
$$\begin{aligned}&\sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) {a^k}{b^{n-k}}\sum \limits _{i=0}^{a-1}{(-1)^i}{}_HE_k^{(\alpha )}(bx+{\frac{b}{a}}i,b^2z) E_{n-k}^{(\alpha -1)}(ay)\nonumber \\&\quad =\sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) {a^{n-k}}{b^k}\sum \limits _{i=0}^{b-1}{(-1)^i}{}_HE_k^{(\alpha )} (ax+{\frac{a}{b}}i,a^2z)E_{n-k}^{(\alpha -1)}(by) \end{aligned}$$
(32)
Proof
Let
$$\begin{aligned} G(t)=:\frac{2^{2{\alpha }-1}(1-(-e^{bt})^a)e^{ab(x+y)t+a^2b^2zt^2}}{(e^{at}+1)^{\alpha }(e^{bt}+1)^{\alpha }} \end{aligned}$$
We expand G(t) as follows :
$$\begin{aligned}G(t)&={\left( \frac{2}{e^{at}+1}\right) ^{\alpha }}e^{abxt+a^2b^2zt^2}{\left( \frac{1-{(-e^{bt})}^a}{1+e^{bt}}\right) }{\left( \frac{2}{e^{bt}+1}\right) ^{{\alpha }-1}}e^{abyt}\\& ={\left( \frac{2}{e^{at}+1}\right) ^{\alpha }}e^{abxt+a^2b^2zt^2}{\left( \sum \limits _{i=0}^{a-1}(-1)^ie^{bti}\right) }{\left( \sum \limits _{n=0}^{\infty }E_n^ {(\alpha -1)}(ay)\frac{(bt)^n}{n!}\right) }\\&={\left( \sum \limits _{i=0}^{a-1}(-1)^i{\left( \frac{2}{e^{at}+1}\right) }^{\alpha }e^{(bx+\frac{b}{a}i)at+a^2b^2zt^2}\right) }{\left( \sum \limits _{n=0}^{\infty } E_{n}^{(\alpha -1)}(ay)\frac{(bt)^n}{n!}\right) }\\G(t)&={\left( \sum \limits _{i=0}^{a-1}(-1)^i\sum \limits _{k=0}^{\infty }{}_HE_k^{(\alpha )}(bx+\frac{b}{a}i;b^2z)\frac{(at)^k}{k!}\right) }{\left( \sum \limits _{n=0} ^{\infty }E_{n}^{(\alpha -1)}(ay)\frac{(bt)^n}{n!}\right) } \end{aligned}$$
Replacing n by \(n-k\) in the R.H.S. of above equation, we get
$$\begin{aligned} G(t)=\sum \limits _{n=0}^{\infty }\left[ \sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) a^kb^{n-k}\sum \limits _{i=0}^ {a-1}(-1)^i{}_HE_k^{(\alpha )}(bx+\frac{b}{a}i;b^2z)E_{n-k}^{(\alpha -1)}(ay)\right] \frac{t^n}{n!} \end{aligned}$$
(33)
Using a similar plan, we obtain
$$\begin{aligned} G(t)=\sum \limits _{n=0}^{\infty }\left[ \sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) b^ka^{n-k}\sum \limits _{i=0}^ {b-1}(-1)^i{}_HE_k^{(\alpha )}(ax+\frac{a}{b}i;a^2z)E_{n-k}^{(\alpha -1)}(by)\right] \frac{t^n}{n!} \end{aligned}$$
(34)
By comparing the coefficients of \(\frac{t^n}{n!}\) in the R.H.S. of last two Eqs. (33) and (34), we arrive at the desired result.
Remark 4
For \(z=0\) in Theorem (4), the result reduces to known result of Liu and Wang [2009, Theorem 2.10].
Corollary 4
For integers
\(n\ge 0\), \(a\ge 1 \)
and
\( b\ge 1\), if
a
and
b
have same parity, then the following identity holds true:
$$\begin{aligned}&\sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) {a^k}{b^{n-k}}\sum \limits _{i=0}^{a-1}{(-1)^i}E_k^{(\alpha )}(bx+{\frac{b}{a}}i) E_{n-k}^{(\alpha -1)}(ay)\nonumber \\&\quad =\sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) {a^{n-k}}{b^k}\sum \limits _{i=0}^{b-1}{(-1)^i}E_k^{(\alpha )} (ax+{\frac{a}{b}}i)E_{n-k}^{(\alpha -1)}(by) \end{aligned}$$
(35)
Theorem 5
For integers
\(n\ge 0,\)
\(a\ge 1\)
and
\(b\ge 1\), if
a
is odd and
b
is even, then the following identity holds true:
$$\begin{aligned}&\sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) {a^k}{b^{n-k}}\sum \limits _{i=0}^{a-1}{(-1)^i}{}_HE_k^{(\alpha )}(bx+{\frac{b}{a}}i,b^2z) E_{n-k}^{(\alpha -1)}(ay)\nonumber \\&\quad =-2\sum \limits _{l=0,l\ne n}^{n+1}\left({n+1}\right) \frac{B_{n+1-l}}{n+1} b^{n-l} \sum \limits _{k=0}^{l}{}\left( \begin{array}{lll}l\\ k\end{array}\right) a^{n-k}b^{k}\nonumber \\&\qquad \times \sum \limits _{i=0}^{b-1}(-1)^i{}_HE_k^{(\alpha )}(ax+\frac{a}{b}i,a^{2}z) E_{l-k}^{(\alpha -1)}(by) \end{aligned}$$
(36)
Proof
Let
$$\begin{aligned} G(t)=:\frac{2^{2{\alpha }-1}(1-(-e^{bt})^a)e^{ab(x+y)t+a^2b^2zt^2}}{(e^{at}+1)^{\alpha }(e^{bt}+1)^{\alpha }} \end{aligned}$$
In view of definition (1.15), G(t) has the following expansion:
$$\begin{aligned}&G(t)={\left( \frac{2}{e^{bt}+1}\right) ^{\alpha }}e^{abxt+a^2b^2zt^2}{\left( \frac{1-{(-e^{at})}^b}{1+e^{at}}\right) }{\left( \frac{1-{(-e^{bt})}^a}{1-{(-e^{at})}^b} \right) }{\left( \frac{2}{e^{at}+1}\right) ^{{\alpha }-1}}e^{abyt}\nonumber \\&\quad ={\left( -\sum \limits _{n=0}^{\infty }\frac{B_{n}+B_{n}(1)}{n!}abt^{n-1}\right) }{\left( \sum \limits _{i=0}^{b-1}{(-1)^i}\sum \limits _{n=0}^{\infty } {}_HE_n^{(\alpha )}(ax+{\frac{a}{b}}i,a^2z)\frac{(bt)^n}{n!}\right) }\nonumber \\&\qquad \times {\left( \sum \limits _{n=0}^{\infty }E_{n}^{(\alpha -1)}(by)\frac{(bt)^n}{n!}\right) } \end{aligned}$$
(37)
Using identity (7) and comparing the coefficients of \(\frac{t^n}{n!}\) in the R.H.S. of Eqs. (33) and (37), we get the result (36).
Remark 5
For \(z =0\) in Theorem (5), the result reduces to known result of Liu and Wang [2009, Theorem (2.13)].
Corollary 5
For integers
\(n\ge 0,\)
\(a\ge 1\)
and
\(b\ge 1\), if
a
is odd and
b
is even, then the following identity holds true:
$$\begin{aligned}&\sum \limits _{k=0}^{n}\left( \begin{array}{lll}n\\ k\end{array}\right) {a^k}{b^{n-k}}\sum \limits _{i=0}^{a-1}{(-1)^i}E_k^{(\alpha )}(bx+{\frac{b}{a}}i) E_{n-k}^{(\alpha -1)}(ay) \nonumber \\&\quad =-2\sum \limits _{l=0,l\ne n}^{n+1}\left( \begin{array}{lll}{n+1}\\ l \end{array}\right) \frac{B_{n+1-l}}{n+1} b^{n-l} \sum \limits _{k=0}^{l}{}\left( \begin{array}{lll}l\\ k\end{array}\right) a^{n-k}b^{k}\nonumber \\&\qquad \times \sum \limits _{i=0}^{b-1}(-1)^iE_k^{(\alpha )}(ax+\frac{a}{b}i) E_{l-k}^{(\alpha -1)}(by) \end{aligned}$$
(38)