What is the real part of $\ln\Gamma(-i z)$

Question

I knew $$\text{Re}(\Gamma(-iz)) =\frac{1}{z}\int_{0}^{\infty}e^{-t} \sin(z\ln t)\text{d}t $$ But $$ \text{Re}(\ln\Gamma(-iz))=? $$ Is it has any expressions using integrals? Thank you very much.

Answer 1

It looks like you're assuming $z$ is real, because otherwise the integral you gave would be divergent. This makes things much easier, because that means $\Gamma(-iz)^* = \Gamma(iz)$. Since we have $\mathrm{Re}[\ln(-iz)] = \frac{1}{2}\ln|\Gamma(-iz)|^2$, we can use the Gamma reflection formula to find $$ |\Gamma(-iz)|^2 = \Gamma(iz)\Gamma(-iz) = \frac{i}{z}\Gamma(iz)\Gamma(1 - iz) = \frac{i}{z}\frac{\pi}{\sin(i\pi z)} = \frac{\pi}{z\sinh(\pi z)}. $$ So $$ \mathrm{Re}[\ln\Gamma(-iz)] = \frac{1}{2}\ln\left[\frac{\pi}{z\sinh(\pi z)}\right] = \frac{\ln\pi}{2} - \frac{\ln[z\sinh(\pi z)]}{2}. $$