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 $${\Gamma }_q(x)=(1-q{)}^{1-x}\prod _{n=0}^{\mathrm{\infty }}\frac{1-q^{n+1}}{1-q^{n+x}}=(1-q{)}^{1-x}\frac{(q;q{)}_{\mathrm{\infty }}}{(q^x;q{)}_{\mathrm{\infty }}}$$ when $|q|<1$, and $${\Gamma }_q(x)=\frac{(q^{-1};q^{-1}{)}_{\mathrm{\infty }}}{(q^{-x};q^{-1}{)}_{\mathrm{\infty }}}(q-1{)}^{1-x}{q}^{(\genfrac{}{}{0ex}{}{x}{2})}$$ if $|q|>1$. Here $(\cdot ;\cdot {)}_{\mathrm{\infty }}$ 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 $$\underset{q\to 1\pm }{\lim }{\Gamma }_q(x)=\Gamma (x).$$ 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)): $$\frac{1}{{\Gamma }_q(z)}=\frac{\sin (\pi z)}{\pi }\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{t^{-z}\mathrm{d}t}{(-t(1-q);q{)}_{\mathrm{\infty }}}.$$

Stirling formula

Moak obtained the following q-analogue of the Stirling formula (see Moak (1984)): $$\log {\Gamma }_q(x)\sim (x-1/2)\log [x{]}_q+\frac{{\mathrm{L}\mathrm{i}}_2(1-q^x)}{\log q}+C_{\hat{q}}+\frac{1}{2}H(q-1)\log q+\sum _{k=1}^{\mathrm{\infty }}\frac{B_{2k}}{(2k)!}{\left(\frac{\log \hat{q}}{{\hat{q}}^x-1}\right)}^{2k-1}{\hat{q}}^x{p}_{2k-3}({\hat{q}}^x),x\to \mathrm{\infty },$$ $$\hat{q}=\left\{\begin{array}{cc}q\mathrm{i}\mathrm{f}& 0<q\le 1\\ 1/q\mathrm{i}\mathrm{f}& q\ge 1\end{array}\right\},$$ $$C_q=\frac{1}{2}\log (2\pi )+\frac{1}{2}\log \left(\frac{q-1}{\log q}\right)-\frac{1}{24}\log q+\log \sum _{m=-\mathrm{\infty }}^{\mathrm{\infty }}(r^{m(6m+1)}-r^{(3m+1)(2m+1)}),$$ where $r=\exp (4{\pi }^2/\log q)$, $H$ denotes the Heaviside step function, $B_k$ stands for the Bernoulli number, ${\mathrm{L}\mathrm{i}}_2(z)$ is the dilogarithm, and $p_k$ is a polynomial of degree $k$ satisfying $$p_k(z)=z(1-z)p{\text{'}}_{k-1}(z)+(kz+1)p_{k-1}(z),p_0=p_{-1}=1,k=1,2,\cdots .$$

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, $$\underset{0}{\overset{1}{\int }}\log {\Gamma }_q(x)dx=\frac{\zeta (2)}{\log q}+\log \sqrt{\frac{q-1}{\sqrt[6]{q}}}+\log (q^{-1};q^{-1}{)}_{\mathrm{\infty }}(q>1).$$ El Bachraoui considered the case $0<q<1$ and proved that $$\underset{0}{\overset{1}{\int }}\log {\Gamma }_q(x)dx=\frac{1}{2}\log (1-q)-\frac{\zeta (2)}{\log q}+\log (q;q{)}_{\mathrm{\infty }}(0<q<1).$$

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: $${\Gamma }_{e^{-2\pi }}\left(\frac{1}{4}\right){\Gamma }_{e^{-2\pi }}\left(\frac{3}{4}\right)=\frac{e^{-29\pi /16}(e^{2\pi }-1)\sqrt[4]{1+\sqrt{2}}}{2^{33/16}{\pi }^{3/2}}\Gamma {\left(\frac{1}{4}\right)}^2,$$ $${\Gamma }_{e^{-4\pi }}\left(\frac{1}{4}\right){\Gamma }_{e^{-4\pi }}\left(\frac{3}{4}\right)=\frac{e^{-29\pi /8}(e^{4\pi }-1)}{2^{23/8}{\pi }^{3/2}}\Gamma {\left(\frac{1}{4}\right)}^2,$$ $${\Gamma }_{e^{-8\pi }}\left(\frac{1}{4}\right){\Gamma }_{e^{-8\pi }}\left(\frac{3}{4}\right)=\frac{e^{-29\pi /4}(e^{8\pi }-1)}{16{\pi }^{3/2}\sqrt{1+\sqrt{2}}}\Gamma {\left(\frac{1}{4}\right)}^2.$$

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: $${\Gamma }_q(A):=\underset{0}{\overset{\frac{1}{1-q}}{\int }}{t}^{A-I}{E}_q(-qt){\mathrm{d}}_{q}t$$ 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.