Jackson integral

Remarks

Example: for an integer $n \ge 0$, $$\int_0^1 x^n \, d_q x = (1 - q) \sum_{k=0}^{\infty} q^{k(n+1)} = \frac{1 - q}{1 - q^{n+1}} = \frac{1}{[n+1]_q},$$ the q-analog of $\int_0^1 x^n \, dx = \frac{1}{n+1}$.

Non-example: $f(x) = 1/x$ on $(0, 1]$ has no Jackson integral from $0$ to $1$, since the terms $q^k f(q^k) = 1$ do not tend to $0$.

If $f$ is continuous on $[0, a]$, the sum is a Riemann sum for the partition by the points $a q^k$, and it tends to $\int_0^a f(x)\, dx$ as $q \to 1^-$. If $f$ is bounded near $0$, then $F(x) = \int_0^x f(t) \, d_q t$ satisfies $D_q F = f$ for the q-derivative $D_q F(x) = \frac{F(qx) - F(x)}{(q - 1) x}$, $x \ne 0$.