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Q-Gamma function

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
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
Jan 17, 2013
1,667
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

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