I have tried to redefine some special functions in the most "natural" way, that is the way which allows to simplify the relations the most. I would call these functions "parelementary". The definitions are as follows:
$$\operatorname{pg }\,\,x =\frac 1\pi \psi\left(\frac x\pi\right)$$
$$\operatorname{cpg }\,\,x=-\frac 1\pi \psi\left(1-\frac x\pi\right)=-\operatorname{pg }(\pi-x)$$
$$\operatorname{gn}\,\,x=\frac 1{\sqrt{\pi}}\Gamma\left(\frac x \pi \right)$$
$$\operatorname{lgn}\,\,x=\ln \left|\frac 1{\sqrt{\pi}}\Gamma\left(\frac x\pi\right)\right|=\ln|\operatorname{gn}(x)|$$
$$\operatorname{zn}\,\,x=\frac {2x}\pi\left(\frac {2x}\pi-1\right)\zeta\left(\frac{2x}{\pi}\right)$$
$$\operatorname{expi}\,\,x=\pi^{\frac{x}\pi}$$
$$\operatorname{czn}\,\,x=\frac {2x}\pi\left(\frac {2x}\pi-1\right)\zeta\left(1-\frac{2x}{\pi}\right)\pi^{\frac{2x}\pi}=\operatorname{zn}\left(\frac\pi2-x\right)\operatorname{expi}(2x)$$
Here is a graphic of pg (orange), cpg (blue), gn (green), zn (red), czn (violet).
They exhibit the following relations:
$$\operatorname{pg}\,\, x + \operatorname{cpg}\,\, x = - \cot x$$
$$\operatorname{pg}\,\, (-x) + \operatorname{cpg}\,\, x = \frac 1x$$
$$\operatorname{gn}\,\, x\operatorname{gn}(\pi-x)=\csc x$$
$$(\operatorname{gn}\,\, x)'=\operatorname{gn}\,\, x\, \operatorname{pg}\,\, x $$
$$(\operatorname{lgn}\,\, x)'=\operatorname{pg}\,\, x$$
But also there are relations which involve expi:
$$\operatorname{zn}\left(x + \frac \pi4\right) \operatorname{expi}(-x) \operatorname{gn}\left(x + \frac \pi4\right)=\operatorname{zn}\left(-x + \frac \pi4\right) \operatorname{expi}(x) \operatorname{gn}\left(-x + \frac \pi4\right)$$
$$\operatorname{czn}\left(-x + \frac \pi4\right) \operatorname{expi}\left(x-\frac\pi2\right) \operatorname{gn}\left(x + \frac \pi4\right)=\operatorname{czn}\left(x + \frac \pi4\right) \operatorname{expi}\left(-x-\frac\pi2\right) \operatorname{gn}\left(-x + \frac \pi4\right)$$
$$D^{s/\pi}[\operatorname{cpg }\,\,x]|_{x=0}=\operatorname{expi}\left(\frac\pi2-s\right)\operatorname{gn}\, s\operatorname{zn} \frac s2$$
Thus I wonder what is the special role of exponential function with base $\pi^{1/\pi}$.
Can the above functions be re-defined such a way so to get rid of the exponent while keeping the relations, thus simplifying them?
Can the above functions be re-defined so to replace the exponent with base $\pi^{1/\pi}$ with exponent with base $e$ everywhere but still keeping the identities?
Does exponent with base $\pi^{1/\pi}$ exhibit any duality with exponent with base $e$?
Does exponent with base $\pi^{1/\pi}$ satisfy any simple and unique differential or functional equation?