the same object.

The $q$-analogues of an integer $n$ ($q$-integer), a factorial of $n$ ($q$-factorial of $n$)$,$ and the binomial coefficient $\binom{n}{k}$($q$-binomial coefficient) are respectively given as$$\left[ n\right] _{q}=\frac{1-q^{n}}{1-q},\text{ }\left( \left[ n\right]
_{q}\right) !=\left\{ 
\begin{array}{ll}
\prod\limits_{1}^{n}\left[ k\right] _{q}\text{ }, & n=1,2,3,... \\ 
1\text{ \ \ \ \ \ \ \ \ }, & \text{ }n=0\end{array}\right. \text{ and }\left[ 
\begin{array}{c}
n \\ 
k\end{array}\right] _{q}=\dprod\limits_{1}^{n}\frac{1-q^{n-k+1}}{1-q^{k}}.$$

Clearly, $$\lim_{q\rightarrow 1}\left[ n\right] _{q}=n,\text{ }\lim_{q\rightarrow
1}\left( \left[ n\right] _{q}\right) !=n!\text{ and }\lim_{q\rightarrow 1}\left[ 
\begin{array}{c}
n \\ 
k\end{array}\right] _{q}=\binom{n}{k}.$$If $\alpha \in 
\mathbb{C}
,$ then the $q$-analogue of $\alpha $ is given as $\dfrac{1-q^{\alpha }}{1-q}
$ and, it sometimes makes sense when $\alpha $ is not, $\left[ \infty \right]
_{q}=\dfrac{1}{1-q}.$

If $n\in 
\mathbb{N}
,$ then the $q$-analogue of $\left( x+a\right) ^{n}$ and the derivative are respectively given as $$\left. \left( x+a\right) _{q}^{n}=\prod\limits_{j=0}^{n-1}\left(
x+q^{j}a\right) \text{\ and }D_{q}\left( x+a\right) _{q}^{n}=\left[ n\right]
\left( x+a\right) _{q}^{n-1},\text{ }\left( x+a\right) _{q}^{0}=1\right\} . 
\tag{3}$$If $x=1$ and $a=x,$ then the above formula makes sense for $n=\infty ,$ giving$$\left( 1+x\right) _{q}^{\infty }=\prod\limits_{0}^{\infty }\left(
1+q^{k}x\right) .  \tag{4}$$The $q$-Jackson integral from $0$ to $a$ is given by Jackson $\left[ 11\right] $ as$$\int\nolimits_{0}^{a}f\left( x\right) d_{q}x=\left( 1-q\right)
a\sum_{0}^{\infty }f\left( aq^{k}\right) q^{k},  \tag{5}$$provided the sum converges absolutely.

The $q$-Jackson integral in a 