# Asymptotic behavior of the generalized polygamma function

The generalized polygamma function^{}$$\!^{} is defined as$$\psi^{(\nu)}(z)=e^{-\gamma\!\;\nu}\;\partial_\nu\!\left(\frac{e^{\gamma\!\;\nu}\;\zeta(\nu+1,z)}{\Gamma(-\nu)}\right),\tag1$$where the order \nu can be an arbitrary complex number. For a nonnegative integer order we have to take a limit \nu\to n, and then we get the usual polygamma function$$\psi^{(n)}(z)=\partial_z^{n+1}\ln\Gamma(z),\quad n\in\mathbb N.\tag2$$I am trying to understand asymptotic behavior of the generalized polygamma function for large imaginary orders. Based on numerical calculations I conjecture that$$\lim\limits_{x\to\infty}\frac{\ln\left|\psi^{(ix)}(1)\right|}x\stackrel?=\frac\pi2.\tag3$$Could you suggest and ideas how to prove (or refute) this conjecture? Some sources^{} use a different generalization of the usual polygamma function to complex orders :$$\begin{align}\pmb\psi^{(\nu)}(z)&=\frac{1+z\,\psi^{(0)}(z)+\nu\,(\ln z-\psi^{(0)}(-\nu)-\gamma)}{z^{1+\nu}\,\Gamma(1-\nu)}\\&+\frac{\nu\,z^{1-\nu}}{\Gamma(2-\nu)}\,\sum_{k=1}^\infty\frac{{_2F_1}\left(\begin{array}{c}1,\,1\\2-\nu\end{array}\middle|-\!{\Large\frac z k}\right)}{k\,(k+z)}.\end{align}\tag4$$For example, this is how PolyGamma is implemented in Mathematica. Again, for non-negative integer orders we have to take a limit, and it also gives us back the usual polygamma function. But for other (fractional and complex) orders this generalization behaves differently. Still, it seems that for large imaginary orders it has a similar behavior and I conjecture that$$\lim\limits_{x\to\infty}\frac{\ln\left|\pmb\psi^{(ix)}(1)\right|}x\stackrel?=\frac\pi2.\tag5$$Update: There is an equivalent definition for \pmb\psi^{(\nu)}(z) using an integral representation that looks somewhat simpler than formula (4) above, but is valid only for \Re(\nu)<0; it can be extended to \Re(\nu)\ge0 using analytic continuation:$$\pmb\psi^{(\nu)}(z)=\frac1{\Gamma(-\nu)\,z^\nu}\left(\frac{\psi^{(0)}(-\nu)+\gamma-\ln z}{z}+\int_0^1\frac{\psi^{(0)}(1+x\!\;z)}{(1-x)^{1+\nu}}\,dx\right).\tag6\$