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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.)

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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, $$\ln\Gamma(1+ib)=\ln|\Gamma(1+ib)|+i\,{\rm Arg}\,\Gamma(1+ib),$$ $$\ln\left|\Gamma\left(1+ib\right)\right|=-\tfrac{1}{2}\sum_{n=1}^\infty\ln(1+b^2/n^2),$$ $${\rm Arg}\,\Gamma(1+ib)=b\gamma_E+\sum_{n=1}^\infty\left(\frac b{n}-\arctan\frac b{n}\right),$$ 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.

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  • $\begingroup$ It's really helpful. Thanks! $\endgroup$
    – user147095
    Commented Oct 11, 2019 at 17:19
  • $\begingroup$ Also a question: How should we define the branch cut of Arg $\Gamma (1 + i b)$? Because $\tan^{-1} z$ has a branch cut at $(-i \infty, -i]$ and $[i, i \infty)$, however here the argument of $\tan^{-1}$ is variable. (Here, we have assumed that $b$ is complex.) $\endgroup$
    – user147095
    Commented Oct 11, 2019 at 17:35
  • $\begingroup$ these equations are for real $b$; if $b$ is complex you will want to use the more general formulas in A&S, valid for any complex argument of the Gamma function $\endgroup$ Commented Oct 11, 2019 at 19:12

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