Simplify this expression $e^{\psi\left(\frac12+\frac{i}{2\sqrt{3}}\right)+\psi\left(\frac12-\frac{i}{2\sqrt{3}}\right)}$?

Question

Is it possible to simplify this constant expression $e^{\psi\left(\frac12+\frac{i}{2\sqrt{3}}\right)+\psi\left(\frac12-\frac{i}{2\sqrt{3}}\right)}$?

Here $\psi(x)$ is digamma function.

Particularly, the constant is real, so it would be good to get rid of the imaginary numbers here.

A similar question about the following: $$e^{\psi\left(\frac12+\frac{1}{2\sqrt{3}}\right)+\psi\left(\frac12-\frac{1}{2\sqrt{3}}\right)}$$

Answer 1

Maybe one of these is "simpler" $$ e^{\psi\left(\frac12+\frac{i}{2\sqrt{3}}\right)+\psi\left(\frac12-\frac{i}{2\sqrt{3}}\right)}\\ =e^{2\operatorname{Re}\psi\left(\frac12+\frac{i}{2\sqrt{3}}\right)}\\ =\left(e^{\operatorname{Re}\psi\left(\frac12+\frac{i}{2\sqrt{3}}\right)}\right)^2\\ =\left|e^{\psi\left(\frac12+\frac{i}{2\sqrt{3}}\right)}\right|^2\\ $$

Answer 2

The answer is no, there's no way to avoid using digamma functions here, however it's easy enough to show that this expression is real.

$$\psi(1+z)=-\gamma+\sum_{n=1}^\infty \left(\frac{1}{n}-\frac{1}{n+z} \right)$$

$$\psi(1/2+z)=-\gamma+\sum_{n=1}^\infty \left(\frac{1}{n}-\frac{1}{n-1/2+z} \right)$$

$$\psi(1/2+z)+\psi(1/2-z)=-2\gamma+\sum_{n=1}^\infty \left(\frac{2}{n}-\frac{1}{n-1/2+z}-\frac{1}{n-1/2-z} \right)=$$

$$=-2\gamma+2\sum_{n=1}^\infty \left(\frac{1}{n}-\frac{n-1/2}{(n-1/2)^2-z^2} \right)$$

So we have:

$$\color{blue}{\psi\left(\frac{1}{2}+\frac{i}{2\sqrt{3}}\right)+\psi\left(\frac{1}{2}-\frac{i}{2\sqrt{3}}\right)= \\ =-2\gamma+2\sum_{n=1}^\infty \left(\frac{1}{n}-\frac{n-1/2}{(n-1/2)^2+1/12} \right)}$$

I believe Euler–Maclaurin formula could be useful in calculating this series to high precision, as the integral of the general term w.r.t. $n$ has an elementary form.

$\gamma$ is the Euler-Mascheroni constant