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The gamma function is increases on the interval $(x_0, \infty),$ where $x_0$ denotes the unique zero of the digamma function on the positive half line.

The inverse function of gamma function defined on $(\Gamma(x_0),\infty)$ was shown to have an extension to a Pick-function in the cut plane $\mathbb{C}-(-\infty,\Gamma(x_0)]$ (see [1]).

Do you have information for any similar result for digamma function or polygamma function?

[1] http://www.ams.org/journals/proc/2012-140-04/S0002-9939-2011-11023-2/S0002-9939-2011-11023-2.pdf

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We want to solve for $x$ the equation $$\psi ^{(0)}(x)=a$$

Just four months before your question, this paper gave tight bounds $$x_{\text{min}}=\frac{1}{\log \left(1+e^{-x}\right)} \lt \psi^{-1} (x) \lt e^{x}+\frac{1}{2}=x_{\text{max}}$$

What I recently found very interesting is that is is much better to consider instead the equation $$e^{\psi ^{(0)}(x)}=e^a$$ the lhs being very close to linearity.

Use the series expansion around $x=\alpha$ $$e^{\psi ^{(0)}(x)}=e^{\psi ^{(0)}(\alpha )}\left(1+\sum_{n=1}^\infty \frac{A_n}{n!}\,(x-\alpha)^n \right)$$ where $$A_1=\psi ^{(1)}(\alpha )$$ $$A_2=\psi ^{(1)}(\alpha )^2+\psi ^{(2)}(\alpha )$$ $$A_3=\psi ^{(1)}(\alpha )^3+3 \psi ^{(2)}(\alpha ) \psi ^{(1)}(\alpha )+\psi ^{(3)}(\alpha )$$ and so forth.

Truncate to some order $p$ and use power series reversion to get $$x=\alpha +t+\sum_{n=2}^p B_n\, t^n \qquad \text{where} \qquad t=\frac{e^{a-\psi ^{(0)}(\alpha )}-1}{\psi ^{(1)}(\alpha )}$$ the very first coefficients being $$B_2=\frac{-\psi ^{(1)}(\alpha )^2-\psi ^{(2)}(\alpha )}{2 \psi ^{(1)}(\alpha )}$$ $$B_3=\frac{2 \psi ^{(1)}(\alpha )^4+3 \psi ^{(2)}(\alpha ) \psi ^{(1)}(\alpha )^2-\psi ^{(3)}(\alpha ) \psi ^{(1)}(\alpha )+3 \psi ^{(2)}(\alpha )^2}{6 \psi ^{(1)}(\alpha )^2}$$ and so forth.

Trying for $a=\pi$ for different orders of expansion starting at the midpoint

$$\left( \begin{array}{cc} p & x_{(p)} \\ 1 & 23.6388933011739 \\ 2 & 23.6388924116017 \\ 3 & 23.6388924311562 \\ 4 & 23.6388924307264 \\ 5 & 23.6388924307358 \\ 6 & 23.6388924307356 \\ \end{array} \right)$$

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