# Q-gamma function

In q-analog theory, the $q$-gamma function, or basic gamma function, is a generalization of the ordinary gamma function closely related to the double gamma function. It was introduced by Jackson (1905). It is given by $\Gamma _{q}(x)=(1-q)^{1-x}\prod _{n=0}^{\infty }{\frac {1-q^{n+1}}{1-q^{n+x}}}=(1-q)^{1-x}\,{\frac {(q;q)_{\infty }}{(q^{x};q)_{\infty }}}$ when $|q|<1$, and $\Gamma _{q}(x)={\frac {(q^{-1};q^{-1})_{\infty }}{(q^{-x};q^{-1})_{\infty }}}(q-1)^{1-x}q^{\binom {x}{2}}$ if $|q|>1$. Here $(\cdot ;\cdot )_{\infty }$ is the infinite $q$-Pochhammer symbol. The $q$-gamma function satisfies the functional equation $\Gamma _{q}(x+1)={\frac {1-q^{x}}{1-q}}\Gamma _{q}(x)=[x]_{q}\Gamma _{q}(x)$ In addition, the $q$-gamma function satisfies the q-analog of the Bohr–Mollerup theorem, which was found by Richard Askey (Askey (1978)).
For non-negative integers $n$, $\Gamma _{q}(n)=[n-1]_{q}!$ where $[\cdot ]_{q}$ is the $q$-factorial function. Thus the $q$-gamma function can be considered as an extension of the $q$-factorial function to the real numbers.
The relation to the ordinary gamma function is made explicit in the limit $\lim _{q\to 1\pm }\Gamma _{q}(x)=\Gamma (x).$ There is a simple proof of this limit by Gosper. See the appendix of (Andrews (1986)).

## Transformation properties

The $q$-gamma function satisfies the q-analog of the Gauss multiplication formula (Gasper & Rahman (2004)): $\Gamma _{q}(nx)\Gamma _{r}(1/n)\Gamma _{r}(2/n)\cdots \Gamma _{r}((n-1)/n)=\left({\frac {1-q^{n}}{1-q}}\right)^{nx-1}\Gamma _{r}(x)\Gamma _{r}(x+1/n)\cdots \Gamma _{r}(x+(n-1)/n),\ r=q^{n}.$

### Integral representation

The $q$-gamma function has the following integral representation (Ismail (1981)): ${\frac {1}{\Gamma _{q}(z)}}={\frac {\sin(\pi z)}{\pi }}\int _{0}^{\infty }{\frac {t^{-z}\mathrm {d} t}{(-t(1-q);q)_{\infty }}}.$

### Stirling formula

Moak obtained the following q-analogue of the Stirling formula (see Moak (1984)): $\log \Gamma _{q}(x)\sim (x-1/2)\log[x]_{q}+{\frac {\mathrm {Li} _{2}(1-q^{x})}{\log q}}+C_{\hat {q}}+{\frac {1}{2}}H(q-1)\log q+\sum _{k=1}^{\infty }{\frac {B_{2k}}{(2k)!}}\left({\frac {\log {\hat {q}}}{{\hat {q}}^{x}-1}}\right)^{2k-1}{\hat {q}}^{x}p_{2k-3}({\hat {q}}^{x}),\ x\to \infty ,$ ${\hat {q}}=\left\{{\begin{aligned}q\quad \mathrm {if} \ &0<q\leq 1\\1/q\quad \mathrm {if} \ &q\geq 1\end{aligned}}\right\},$ $C_{q}={\frac {1}{2}}\log(2\pi )+{\frac {1}{2}}\log \left({\frac {q-1}{\log q}}\right)-{\frac {1}{24}}\log q+\log \sum _{m=-\infty }^{\infty }\left(r^{m(6m+1)}-r^{(3m+1)(2m+1)}\right),$ where $r=\exp(4\pi ^{2}/\log q)$, $H$ denotes the Heaviside step function, $B_{k}$ stands for the Bernoulli number, $\mathrm {Li} _{2}(z)$ is the dilogarithm, and $p_{k}$ is a polynomial of degree $k$ satisfying $p_{k}(z)=z(1-z)p'_{k-1}(z)+(kz+1)p_{k-1}(z),p_{0}=p_{-1}=1,k=1,2,\cdots .$

## Raabe-type formulas

Due to I. Mező, the q-analogue of the Raabe formula exists, at least if we use the $q$-gamma function when $|q|>1$. With this restriction, $\int _{0}^{1}\log \Gamma _{q}(x)dx={\frac {\zeta (2)}{\log q}}+\log {\sqrt {\frac {q-1}{\sqrt[{6}]{q}}}}+\log(q^{-1};q^{-1})_{\infty }\quad (q>1).$ El Bachraoui considered the case $0<q<1$ and proved that $\int _{0}^{1}\log \Gamma _{q}(x)dx={\frac {1}{2}}\log(1-q)-{\frac {\zeta (2)}{\log q}}+\log(q;q)_{\infty }\quad (0<q<1).$

## Special values

The following special values are known. $\Gamma _{e^{-\pi }}\left({\frac {1}{2}}\right)={\frac {e^{-7\pi /16}{\sqrt {e^{\pi }-1}}{\sqrt[{4}]{1+{\sqrt {2}}}}}{2^{15/16}\pi ^{3/4}}}\,\Gamma \left({\frac {1}{4}}\right),$ $\Gamma _{e^{-2\pi }}\left({\frac {1}{2}}\right)={\frac {e^{-7\pi /8}{\sqrt {e^{2\pi }-1}}}{2^{9/8}\pi ^{3/4}}}\,\Gamma \left({\frac {1}{4}}\right),$ $\Gamma _{e^{-4\pi }}\left({\frac {1}{2}}\right)={\frac {e^{-7\pi /4}{\sqrt {e^{4\pi }-1}}}{2^{7/4}\pi ^{3/4}}}\,\Gamma \left({\frac {1}{4}}\right),$ $\Gamma _{e^{-8\pi }}\left({\frac {1}{2}}\right)={\frac {e^{-7\pi /2}{\sqrt {e^{8\pi }-1}}}{2^{9/4}\pi ^{3/4}{\sqrt {1+{\sqrt {2}}}}}}\,\Gamma \left({\frac {1}{4}}\right).$ These are the analogues of the classical formula $\Gamma \left({\frac {1}{2}}\right)={\sqrt {\pi }}$.
Moreover, the following analogues of the familiar identity $\Gamma \left({\frac {1}{4}}\right)\Gamma \left({\frac {3}{4}}\right)={\sqrt {2}}\pi$ hold true: $\Gamma _{e^{-2\pi }}\left({\frac {1}{4}}\right)\Gamma _{e^{-2\pi }}\left({\frac {3}{4}}\right)={\frac {e^{-29\pi /16}\left(e^{2\pi }-1\right){\sqrt[{4}]{1+{\sqrt {2}}}}}{2^{33/16}\pi ^{3/2}}}\,\Gamma \left({\frac {1}{4}}\right)^{2},$ $\Gamma _{e^{-4\pi }}\left({\frac {1}{4}}\right)\Gamma _{e^{-4\pi }}\left({\frac {3}{4}}\right)={\frac {e^{-29\pi /8}\left(e^{4\pi }-1\right)}{2^{23/8}\pi ^{3/2}}}\,\Gamma \left({\frac {1}{4}}\right)^{2},$ $\Gamma _{e^{-8\pi }}\left({\frac {1}{4}}\right)\Gamma _{e^{-8\pi }}\left({\frac {3}{4}}\right)={\frac {e^{-29\pi /4}\left(e^{8\pi }-1\right)}{16\pi ^{3/2}{\sqrt {1+{\sqrt {2}}}}}}\,\Gamma \left({\frac {1}{4}}\right)^{2}.$

## Matrix version

Let $A$ be a complex square matrix and positive-definite matrix. Then a $q$-gamma matrix function can be defined by $q$-integral: $\Gamma _{q}(A):=\int _{0}^{\frac {1}{1-q}}t^{A-I}E_{q}(-qt)\mathrm {d} _{q}t$ where $E_{q}$ is the q-exponential function.

## Other q-gamma functions

For other $q$-gamma functions, see Yamasaki 2006.

## Numerical computation

An iterative algorithm to compute the q-gamma function was proposed by Gabutti and Allasia.