Real and imaginary parts of $\ln \Gamma(i b)$

Question

Can one obtain the real and imaginary parts of $\ln \Gamma (i b)$ in terms of simpler functions?

($b$ is a positive number.)

Answer 1

For the derivation we need the formular for $\Im(\psi(i\, z))$ and

\begin{equation} \int_0^{\infty } \psi (1+i\, t) \cos (t \,z) \, dt = -\frac{i}{2} \left(\frac{1}{\frac{2\, (-i \,z)}{2\, \pi }}-\log \left(-\frac{i\, z}{2\, \pi }\right)+\psi\left(-\frac{i\, z}{2 \pi }\right)\right) \end{equation}

\begin{equation} =\frac{i}{2} \,\log \left(\frac{z}{2 \pi }\right)-\frac{\pi }{4} \ \left(\coth \left(\frac{z}{2}\right)-1\right)-\frac{i}{4} \ \left(\psi\left(1-\frac{iz}{2 \pi }\right)+\psi\left(1+\frac{i z}{2 \pi }\right)\right) \end{equation} stocha and A Man Called Old Fashion, where the right hand side is explicitely given in a real - and imaginary part in a closed form. The complete proof of the following proposition is very extensive and time-consuming, we just state the result here.

Propositon: If $z > 0$, then \begin{align} \Re(\log (\Gamma (i\, b)))=\frac{1}{2} (\log (\Gamma (-i\, b))+\log (\Gamma (i\, b)))=\frac{1}{2} \left(\log \left(\frac{\pi }{b}\right)-\log (\sinh(\pi\, b))\right) \end{align} \begin{align} \Im(\log (\Gamma (i\, b)))=\frac{i}{2}\,(\log(\Gamma (1-i\,b))-\log (\Gamma (1+i\,b)))-\frac{\pi }{2} \end{align} \begin{align} =\frac{i}{2}\, (\log (\Gamma (-i\, b))-\log (\Gamma (i\, b))) \end{align}

The second identity of the real part results immediately from \begin{align} \Gamma (z)\, \Gamma (-z)=-\frac{\pi }{z\, \sin (\pi \, z)} \end{align}