Curious closed forms of the $q$-Gamma function.

Question

I found on Wikipedia the $q$-Gamma function, defined as $$\Gamma_q(x)=(1-q)^{1-x}\prod_{n\ge0}\frac{1-q^{n+1}}{1-q^{n+x}}$$ for $|q|<1$. There is a definition for $|q|>1$, but we won't need that here (I think).

The page also provided me with some curious results. For brevity, we define $$j(z)=\frac{1}{\Gamma(\frac14)}\Gamma_{e^{-\pi z}}(\tfrac12).$$ Then Wikipedia tells me that $$\begin{align} j(1)&=\frac{\sqrt{e^\pi-1}\sqrt[4]{1+\sqrt2}}{2^{15/16}\pi^{3/4}e^{7\pi/16}}\\ j(2)&=\frac{\sqrt{e^{2\pi}-1}}{2^{9/8}\pi^{3/4}e^{7\pi/8}}\\ j(4)&=\frac{\sqrt{e^{4\pi}-1}}{2^{7/4}\pi^{3/4}e^{7\pi/4}}\\ j(8)&=\frac{1}{\sqrt{1+\sqrt2}}\cdot\frac{\sqrt{e^{8\pi}-1}}{2^{9/4}\pi^{3/4}e^{7\pi/2}}. \end{align}$$

Can anyone prove these? As some of you may know, my experience lies in the evaluation of integrals and infinite series, and I have very little experience or knowledge when it comes to infinite products. That being said, please forgive me when I do not include any attempts, as I genuinely have no idea where to start. The purpose of this question is to not only see proofs for these fascinating identities but to also to learn techniques for evaluating infinite products in general.

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Edit

I found on Wikipedia the series $$\frac{(q;q)_\infty}{(z;q)_\infty}=\sum_{n\ge0}\frac{(-1)^nq^{n(n+1)/2}}{(q;q)_n(1-zq^n)}\qquad |z|<1$$ where $$(a;q)_n=\prod_{k=0}^{n-1}(1-aq^k).$$ Anyway, we can see that $$\Gamma_q(\tfrac12)=(1-q)^{1/2}\frac{(q;q)_\infty}{(q^{1/2};q)_\infty}=(1-q)^{1/2}\sum_{k\ge0}\frac{(-1)^kq^{k(k+1)/2}}{(q;q)_k(1-q^{k+1/2})},$$ but that really doesn't help me that much, to be honest.

Answer 1

Introduction. The $\Gamma_q$ function is a $q$-analog of the $\Gamma$ function defined by $$ \Gamma_q(x) = \frac{(q;q)_\infty}{\left(q^x;q\right)_\infty}(1-q)^{1-x},\quad |q|<1, $$ where $(a;q)_\infty$ is a $q$-Pochhammer symbol.

Ramanujan's $\psi$ function defined by $$ \psi(q)= \sum_{n=0}^{\infty} q^{n(n+1)/2} = \frac{\left(q^2;q^2\right)_\infty}{\left(q;q^2\right)_\infty}, $$ where the infinite product representation arises from the Jacobi triple product identity.

We can express $\Gamma_{q}\left(1/2\right)$ in terms of the $\psi$ function as the following: $$ \Gamma_q\left(\frac12\right) = \psi\left(\sqrt{q}\right)\sqrt{1-q}. $$ The $\psi$ function and related functions $-$ so-called Ramanujan's theta functions $-$ are widely studied in Berndt's Ramanujan's Notebooks. We summarize here the relevant parts, and refer to Berndt for details.

Values of Ramanujan's $\psi$ function. For $0 < x < 1$, let $$ z = {_2}F{_1}\left(\tfrac12,\tfrac12;1;x\right) $$ and $$ y = \pi \frac{{_2}F{_1}\left(\tfrac12,\tfrac12;1;1-x\right)}{{_2}F{_1}\left(\tfrac12,\tfrac12;1;x\right)}, $$ where ${_2}F{_1}$ is the Gaussian hypergeometric function.

We have \begin{align} \psi\left(e^{-y/4}\right) &= \sqrt{z}\left(1+x^{1/4}\right)^{1/2}\left(\tfrac12 \left(1+\sqrt{x}\right)\right)^{1/8}\left(xe^y\right)^{1/32},\tag{1}\\ \psi\left(e^{-y/2}\right) &= \sqrt{z}\left(\tfrac12 \left(1+\sqrt{x}\right)\right)^{1/4}\left(xe^y\right)^{1/16},\tag{2}\\ \psi\left(e^{-y}\right) &= \sqrt{\tfrac12 z}\left(xe^y\right)^{1/8},\tag{3}\\ \psi\left(e^{-2y}\right) &= \tfrac12 \sqrt{z} \left(xe^y\right)^{1/4},\tag{4}\\ \psi\left(e^{-4y}\right) &= \tfrac12 \sqrt{\tfrac12 z}\left(\left(1-\sqrt{1-x}\right)e^y\right)^{1/2},\tag{5}\\ \psi\left(e^{-8y}\right) &= \tfrac14\sqrt{z}\left(1-\left(1-x\right)^{1/4}\right)e^y.\tag{6} \end{align}

The notation for $z$ and $y$ are defined in Berndt [Part $\text{III}$, p. $101$, Entry $6$, $(6.2)$ and $(6.3)$]. The formulas $(1)\!-\!(6)$ are given with proof in Berndt [Part $\text{III}$, p. $123$, Entry $11$.].

Values of $y$ and $z$ for $x = 1/2$. It is clear that $y = \pi$ for $x = 1/2$. To evaluate $z$, we use the following identity, which is given in Berndt [Part $\text{III}$, p. $89$, $(1.4)$]. If $a$ and $b$ are arbitrary, then $$ {_2}F{_1}\left(a,b;\tfrac12 \left(a + b + 1\right);\tfrac12\right) = \frac{\Gamma\left(\tfrac12\right)\Gamma\left(\tfrac12 \left(a + b + 1\right)\right)}{\Gamma\left(\tfrac12 + \tfrac12 a\right)\Gamma\left(\tfrac12 + \tfrac12 b\right)}. $$ In particular, if $c$ is arbitrary, then $$ {_2}F{_1}\left(1 - c,c;1;\tfrac12\right) = \frac{\Gamma\left(\tfrac12\right)}{\Gamma\left(1 - \tfrac12 c\right)\Gamma\left(\tfrac12 + \tfrac12 c\right)}. $$ For $c = 1/2$, we have $$ z = {_2}F{_1}\left(\tfrac12,\tfrac12;1;\tfrac12\right) = \frac{\Gamma\left(\tfrac12\right)}{\Gamma^2\left(\tfrac34\right)} = \frac{\pi^{1/2}}{\Gamma^2\left(\tfrac34\right)} = \frac{\Gamma^2\left(\tfrac14\right)}{2\pi^{3/2}}, $$ where we used the particular value $\Gamma\left(\tfrac12\right) = \sqrt{\pi}$ and the product identity $\Gamma\left(\tfrac14\right)\Gamma\left(\tfrac34\right) = \sqrt{2}\pi$.

Values of $\Gamma_q\left(1/2\right)$. Let $$ a = \frac{\pi^{1/4}}{\Gamma\left(3/4\right)}. $$ Then \begin{align} \Gamma_{e^{-\pi/2}}\left(\tfrac12\right) &= a2^{-11/32}\left(2^{1/4}+1\right)^{1/2}\left(\sqrt{2}+1\right)^{1/8}e^{\pi/32}\sqrt{1-e^{-\pi/2}},\tag{1'}\\ \Gamma_{e^{-\pi}}\left(\tfrac12\right) &= a2^{-7/16}\left(\sqrt{2}+1\right)^{1/4}e^{\pi/16}\sqrt{1-e^{-\pi}},\tag{2'}\\ \Gamma_{e^{-2\pi}}\left(\tfrac12\right) &= a2^{-5/8}e^{\pi/8}\sqrt{1-e^{-2\pi}},\tag{3'}\\ \Gamma_{e^{-4\pi}}\left(\tfrac12\right) &= a2^{-5/4}e^{\pi/4}\sqrt{1-e^{-4\pi}},\tag{4'}\\ \Gamma_{e^{-8\pi}}\left(\tfrac12\right) &= a2^{-2}\left(2-\sqrt{2}\right)^{1/2}e^{\pi/2}\sqrt{1-e^{-8\pi}},\tag{5'}\\ \Gamma_{e^{-16\pi}}\left(\tfrac12\right) &= a2^{-2}\left(1-2^{-1/4}\right)e^{\pi}\sqrt{1-e^{-16\pi}}.\tag{6'} \end{align}

The values of the $\psi$ function that is corresponding to the formulas $(1\text{'})\!-\!(6\text{'})$ are given in Berndt [Part $\text{V}$, p. $325$, Entry $1$.].