The imaginary part of the digamma function when its argument is pure imaginary is known as


$$\Im\psi(\mathrm{i}b)=\frac{1}{2}b^{-1}+\frac{1}{2}\pi\coth{\pi b},$$ and its real part is much more involved.


My question is: Can one obtain the real and imaginary parts of $\ln \Gamma (i b)$ in terms of simpler functions, or in terms of $\ln \Gamma (b)$?


($b$ is a positive number.)



Real and imaginary parts of $\ln \Gamma(i b)$
It is helpful to use $\Gamma(1+ib)=ib\Gamma(ib)$ and evaluate $\Gamma(1+ib)$. Using equations 6.1.25 and 6.1.27 of [Abramowitz & Stegun](http://people.math.sfu.ca/~cbm/aands/page_256.htm),
$$\ln\Gamma(1+ib)=\ln|\Gamma(1+ib)|+i\,{\rm Arg}\,\Gamma(1+ib),$$
$$\ln\left|\Gamma\left(1+ib\right)\right|=-\tfrac{1}{2}\sum\_{n=1}^\infty\ln(1+b^2/n^2),$$
$${\rm Arg}\,\Gamma(1+ib)=b\gamma\_E+\sum\_{n=1}^\infty\left(\frac b{n}-\arctan\frac b{n}\right),$$
with $\gamma\_E$ Euler's constant.
These sums are rapidly converging, so should be useful. A closed form expression in terms of elementary functions does not exist, as far as I know.


