What is the value of $\psi_2^{(0)}(1)$, where $\psi_q^{(0)}(z)$ is the $q$-digamma function?
My attempt:
\begin{align*} \psi_2^{(0)}(z) &=\frac{1}{\Gamma_2(z)}\frac{d\Gamma_2(z)}{dz} \\&=\frac{(2^z\ ;2)_\infty}{(2\ ;2)_\infty(1-2)^{1-z}}\cdot\frac{d}{dz}\left((1-2)^{1-z}\frac{(2\ ;2)_\infty}{(2^z\ ;2)_\infty}\right) \\ &=\frac{1}{(-1)^{1-z}}\prod_{n=0}^\infty\frac{1-2^{n+z}}{1-2^{n+1}}\cdot\frac{d}{dz}(-1)^{1-z}\prod_{n=1}^\infty\frac{1-2^{n+1}}{1-2^{n+z}}, \end{align*}
where $\Gamma_q(z)$ is the q-gamma function, and $(a\ ;q)_k$ is the q-Pochhammer symbol. Evaluating at $1$, we see
$$\begin{align*} \psi_2^{(0)}(1)&=\frac{1}{(-1)^0}\prod_{n=0}^\infty\frac{1-2^{n+1}}{1-2^{n+1}}\cdot\frac{d}{dz}(-1)^{1-z}\prod_{n=1}^\infty\frac{1-2^{n+1}}{1-2^{n+z}}\Bigr|_{z=1} \\&=\frac{d}{dz}(-1)^{1-z}\prod_{n=1}^\infty\frac{1-2^{n+1}}{1-2^{n+z}}\Bigr|_{z=1} \\&=\frac{d}{dz}\exp\left((\ln(-1)^{1-z}+\sum_{n=0}^\infty\ln\left(\frac{1-2^{n+1}}{1-2^{n+z}}\right)\right)\Bigr|_{z=1} \\&=\exp\left(\ln(-1)^{1-z}+\sum_{n=0}^\infty\ln\left(\frac{1-2^{n+1}}{1-2^{n+z}}\right)\right)\left(i\pi+\sum_{n=0}^\infty\frac{2^{n+z}\ln2}{1-2^{n+z}}\right)\Bigr|_{z=1} \\&=i\pi+\ln2\sum_{n=0}^\infty\frac{2^{n+1}}{1-2^{n+1}} \end{align*}$$
which clearly diverges.... and even if it converged, I would have an imaginary part. Wolfram Alpha gives only an approximation of about $-0.7671026$, so what am I missing here?
Consider $D_{x} a^{x}$: \begin{align} D_{x} a^{x} &= D_{x} e^{x \, ln(a)} \\ &= \ln(a) \, e^{x \, ln(a)} \\ &= a^{x} \, \ln(a). \end{align} Now, \begin{align} \frac{d}{dz}\left((1-q)^{1-z}\frac{(q\ ;q)_\infty}{(q^z\ ;q)_\infty}\right) &= (1-q)^{1-x} \, \frac{d}{dx} \left( \frac{(q\ ;q)_\infty}{(q^z\ ;q)_\infty}\right) - (1-q)^{1-x} \, \ln(1-q) \, \frac{(q\ ;q)_\infty}{(q^z\ ;q)_\infty} \end{align} This is the process presented by the proposer. It is correct. What must be considered now is $0 < q <1$, $q > 1$. For these cases see formulas (1.4) and (1.5) of Some inequalities of the q-Digamma function