Is the Jackson integral of $e_q(x)$ , $e_q(x)$ itself? Some info-

*

*Jackson integral (q-analog of standard integration) simply defined as -$$\int f(x){\mathrm d}_qx=(1-q)x\sum_{k=0}^\infty q^kf(q^kx)$$

*q-exponential defined as $$e_q(x)=\sum_{n=0}^\infty \frac{x^n}{[n]_q!} = 
\sum_{n=0}^\infty \frac{x^n (1-q)^n}{(q;q)_n} = 
\sum_{n=0}^\infty x^n\frac{(1-q)^n}{(1-q^n)(1-q^{n-1}) \cdots (1-q)}$$
My question - $$\text{is}\int e_q(x){\mathrm d}_qx = e_q(x) ?$$
My approach- $$\int e_q(x){\mathrm d}_qx=(1-q)x\sum_{k=0}^\infty q^ke_q(q^kx) \\ =(1-q)x\sum_{k=0}^\infty q^k\sum_{n=0}^\infty(q^kx)^n\frac{(1-q^k)^n}{(q^k;q^k)_n} \\ $$
How should I proceed?

Write out $e_q(q^kx)$ explicitly and then interchange the summations:
$$ \begin{array}{ll} \displaystyle \int e_q(x)\,\mathrm{d}_qx & \displaystyle =(1-q)x\sum_{k=0}^\infty q^k e_q(q^kx) \\[5pt] & \displaystyle  =(1-q)x\sum_{k=0}^\infty q^k \sum_{n=0}^\infty \frac{(q^kx)^n}{[n]_q!} \\[5pt] & \displaystyle = (1-q)x\sum_{n=0}^\infty \frac{x^n}{[n]_q!} \sum_{k=0}^\infty (q^{n+1})^k \\[5pt] & \displaystyle =\sum_{n=0}^\infty \frac{x^{n+1}}{[n]_q!} \frac{1-q~~~~~}{1-q^{n+1}} \\[5pt] & \displaystyle = \sum_{n=0}^\infty \frac{x^{n+1}}{[n+1]_q!} \\[5pt] & = e_q(x)-1. \end{array} $$
In particular, $\displaystyle \int x^n\,\mathrm{d}_qx=\frac{x^{n+1}}{[n+1]_q} $.
