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  "body": "\n## Definition\n\nLet $0 \u003c q \u003c 1$, and let $(a; q)_\\infty = \\prod_{k=0}^\\infty (1 - a q^k)$ be the infinite [[q-Pochhammer symbol|Y1X8ZZQ7VM]].\nFor a real number $x$ that is not in $\\set{0, -1, -2, \\ldots}$, the *$q$-gamma function* is\n$$\\Gamma_q(x) = (1 - q)^{1 - x} \\frac{(q; q)_\\infty}{(q^x; q)_\\infty}.$$\n\n## Remarks\n\nThe products converge because $\\sum q^k \u003c \\infty$, and $(q^x; q)_\\infty \\ne 0$ because $q^{x+k} \\ne 1$ for all $k \\ge 0$.\n\nExample: $\\Gamma_q(1) = 1$, since the two products are equal.\nAlso $\\Gamma_q(2) = (1-q)^{-1} \\frac{(q;q)_\\infty}{(q^2;q)_\\infty} = \\frac{1 - q}{1 - q} = 1$ and $\\Gamma_q(3) = \\frac{(1-q)(1-q^2)}{(1-q)^2} = 1 + q$.\nIn general $\\Gamma_q(x + 1) = [x]_q\\, \\Gamma_q(x)$ with $[x]_q = \\frac{1 - q^x}{1 - q}$, so $\\Gamma_q(n + 1) = [1]_q [2]_q \\cdots [n]_q$.\n\nNon-example of a valid argument: for $x = -1$ the factor $1 - q^{x+1}$ of $(q^x; q)_\\infty$ is $0$, so $\\Gamma_q$ has a pole there.\n\nFor each fixed $x$, $\\Gamma_q(x) \\to \\Gamma(x)$ as $q \\to 1^-$, where $\\Gamma$ is the [[gamma function|0H5SYBSED9]].\nA related definition exists for $q \u003e 1$.\n",
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