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  "body": "\n## Definition\n\nLet $0 \u003c q \u003c 1$, let $a, b \\in \\R$, and let $f$ be a real or complex function defined at the points $a q^k$ and $b q^k$ for all integers $k \\ge 0$.\nThe *Jackson integral* of $f$ from $0$ to $a$ is\n$$\\int_0^a f(x) \\, d_q x = (1 - q)\\, a \\sum_{k=0}^{\\infty} q^k f(a q^k),$$\nwhen the series converges.\nThe Jackson integral from $a$ to $b$ is\n$$\\int_a^b f(x) \\, d_q x = \\int_0^b f(x) \\, d_q x - \\int_0^a f(x) \\, d_q x.$$\n\n## Remarks\n\nExample: for an integer $n \\ge 0$,\n$$\\int_0^1 x^n \\, d_q x = (1 - q) \\sum_{k=0}^{\\infty} q^{k(n+1)} = \\frac{1 - q}{1 - q^{n+1}} = \\frac{1}{[n+1]_q},$$\nthe [[q-analog|GFSDMJHHVA]] of $\\int_0^1 x^n \\, dx = \\frac{1}{n+1}$.\n\nNon-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 $0$.\n\nIf $f$ 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 $q \\to 1^-$.\nIf $f$ is bounded near $0$, then $F(x) = \\int_0^x f(t) \\, d_q t$ satisfies $D_q F = f$ for the [[q-derivative|PMJT3FZE0Y]] $D_q F(x) = \\frac{F(qx) - F(x)}{(q - 1) x}$, $x \\ne 0$.\n",
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