Let $0<q<1$. Define the $q$-gamma function by the formula $$\Gamma_q(z) = \dfrac{(q;q)_{\infty}}{(q^z;q)_{\infty}}(1-q)^{1-z},$$ where $(\cdot;\cdot)_{\infty}$ denotes the q-Pochhammer symbol. The function $\Gamma_q$ is a $q$-analogue of the gamma function.

Qgamma.png

Graph of $\Gamma_q$ on $[-3,3]$.

Properties

Theorem

The following formula holds: $$\Gamma_q(z+1)=\dfrac{1-q^z}{1-q}\Gamma_q(z),$$ where $\Gamma_q$ denotes the $q$-gamma function and $[z]_q$ denotes the $q$-number of $z$.

Proof

References

Daniel S. Moak: The q-gamma function for q greater than 1 (1980)... (previous)... (next)

Proposition: $\Gamma_q(1)=\Gamma_q(2)=1$

Proof: proof goes here █

Theorem ($q$-analog of Bohr-Mollerup): Let $f$ be a function which satisfies $$f(x+1) = \dfrac{1-q^x}{1-q}f(x)$$ for some $q \in (0,1)$, $$f(1)=1,$$ and $\log f(x)$ is convex for $x>0$. Then $f(x) = \Gamma_q(x)$.

Proof: proof goes here █

Theorem (Legendre Duplication Formula): $\Gamma_q(2x)\Gamma_{q^2}\left(\dfrac{1}{2}\right)=\Gamma_{q^2}(x)\Gamma_{q^2}\left( x +\dfrac{1}{2} \right)(1+q)^{2x+1}$

Proof: proof goes here █

Theorem: ($q$-analog) The following formula holds: $$\displaystyle\lim_{q \rightarrow 1^-} \Gamma_q(z) = \Gamma(z),$$ where $\Gamma_q$ is the q-Gamma function and $\Gamma$ is the gamma function.

Proof: █