# Q-Gamma function

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Let us define the following

$$\displaystyle \Gamma_q(x) = \frac{(q;\, q)_{\infty}}{(q^x;\,q)_{\infty}}(1-q)^{1-x}$$

Prove that

$$\displaystyle \lim_{q \to 1^-}\Gamma_q(x)=\Gamma(x)$$​

Naturally the q-beta function is defined as

$$\displaystyle B_q(x,y) = \frac{\Gamma_q(x)\Gamma_q(x)}{\Gamma_q(x+y)}$$

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
We need to prove that

$$\displaystyle \lim_{q \to 1^-} \frac{(q;\, q)_{\infty}}{(q^x;\,q)_{\infty}}(1-q)^{1-x}=\Gamma(x)$$

But first we start by working on the factorial then we extend the result by analytic continuation . So we need to prove

$$\displaystyle \lim_{q \to 1^-} \frac{(q;\, q)_{\infty}}{(q^k;\,q)_{\infty}}(1-q)^{1-k}=(k-1)!$$

Start by

$$\displaystyle \frac{(q;\, q)_{\infty}}{(q^k;\,q)_{\infty}} = \prod_{n\geq 0}\frac{(1-q^{n+1})}{(1-q^{k+n})}= (1-q)(1-q^2)\cdots(1-q^{k-1})$$

So we get

$$\displaystyle \lim_{q \to 1^-}\frac{(1-q)(1-q^2)\cdots(1-q^{k-1})}{(1-q)^{k-1}} =\lim_{q \to 1^-} \frac{(1-q)(1-q^2)\cdots(1-q^{k-1})}{(1-q)(1-q)\cdots(1-q)}=(k-1)!$$