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

($b$ is a positive number.)

  • I would more likely expect a relationship with the complex conjugate, not the imaginary part. What makes you think such a relationship would exist? – Brevan Ellefsen Oct 09 '19 at 09:35
  • @BrevanEllefsen I don't know that if such a thing exists. I just needed that to simplify my calculations. –  Oct 09 '19 at 10:21
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    "your calculations" ? We fully understand $|\Gamma(it)|$ from $\Gamma(s)\Gamma(1-s)= \pi / \sin \pi s$, to understand $\Im \log \Gamma(it)$ we need more, the explicit formula for $\Gamma'/\Gamma$ and its approximation, ie. Stirling approximation. – reuns Oct 09 '19 at 13:47
  • @reuns Thanks for your comment. I am trying to solve a problem in physics which resulted in some terms including $\ln \Gamma(i b)$. To solve the resulting equations I need to separate the real and imaginary parts of the terms. My problem is that I cannot figure this out for $\ln \Gamma(i b)$. –  Oct 09 '19 at 16:37
  • @reuns I'm trying to figure out your hints. How is the imaginary part of $\log \Gamma(it)$ related to $\Gamma^' / \Gamma$? –  Oct 09 '19 at 18:27
  • Reuns said the absolute value is related. There is no obvious relationship between $\Gamma(ib)$ and $\Gamma(b)$ so your best bet is expanding via simpler functions, e.g. the Stirling Approximation – Brevan Ellefsen Oct 10 '19 at 05:48
  • @BrevanEllefsen Thanks for your clarification. –  Oct 10 '19 at 19:01

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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}

stocha
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  • Great! I understand the real part, but the imaginary part is not completely reduced, and what is $y$? –  Oct 12 '19 at 20:07
  • The real part is beautiful and concise. But, the imaginary part still has $\ln \Gamma$. Is it possible to reduce that? –  Oct 12 '19 at 20:11
  • For b = 1/2 I got -1.81485 for the imaginary part on both sides. This is the result of the already derivated equations, I think that furher reduction is difficult. – stocha Oct 12 '19 at 20:15
  • Do you have any idea how theoretically one can reduce it? –  Oct 12 '19 at 20:18
  • Maybe use the result of the real part, by extending the argument of the $\log\Gamma$ – stocha Oct 12 '19 at 20:18
  • I have to thing about it – stocha Oct 12 '19 at 20:19
  • Perhaps, $\Gamma(z+1) = z \Gamma(z)$? –  Oct 12 '19 at 20:19
  • One has to express $\Im(\log (\Gamma (i, b)))$ by $\Re(\log (\Gamma (i, b)))$ to get a reduction – stocha Oct 12 '19 at 20:23
  • I'm trying to reduce it by using $\Gamma(z+1) = z \Gamma(z)$, if you could come up with something, please share it! –  Oct 12 '19 at 20:35
  • The ansatz just gives, $\Im(\log (\Gamma (i, b)))=\frac{i}{2} (\log (\Gamma (-i, b))-\log (\Gamma (i, b)))$ no more reduction – stocha Oct 12 '19 at 20:41
  • but I think the real part is not posted until now – stocha Oct 12 '19 at 20:46
  • I think that's true! –  Oct 12 '19 at 20:47
  • Somehow $ \frac{\Gamma (-i,b)} {\Gamma (b)}$ cannot be simplified, because of the complex argument. I think the imaginary part cannot be reduced more. – stocha Oct 12 '19 at 22:25