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Is there any $a, b \in \mathbb R, b \ne 0$ such that $\Gamma(a+bi)$ can be evaluated manually? (Like $\Gamma(\frac 12)$)

If there is/are, could you show me how to calculate it?

I found that $\Gamma(i)$ cannot be calculated by hand, but only can be calculated using computer.

Of course I tried

$\Gamma(i+1)=\displaystyle\int_0^\infty x^i~e^{-x}~dx$, and then use $x^i=e^{i\ln x}=\cos\ln x+i\sin\ln x$.

But still no clue..

Thanks.

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    Could anybody comment on the reason of a downvote? Where should I improve? – KYHSGeekCode Sep 22 '18 at 10:01
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    i think this is a good questions –  Sep 22 '18 at 11:00
  • How'd you compute $\Gamma\left(\frac 12 \right)$ by hand exactly? – edmz Sep 22 '18 at 11:12
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    see this https://math.stackexchange.com/questions/215352/why-is-gamma-left-frac12-right-sqrt-pi –  Sep 22 '18 at 11:14
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    @KYHSGeekCode see the following link i hope you find some clues https://math.stackexchange.com/questions/264034/closed-form-expression-for-the-gamma-function-for-complex-numbers-s-aib-with –  Sep 22 '18 at 11:17
  • @VincentLaw Wow, thank you for that good material. – KYHSGeekCode Sep 22 '18 at 11:19
  • you are welcome –  Sep 22 '18 at 11:20
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    Just a thought: In Vincent Law's second link there is a claim that $\arg(\Gamma(z))$ has a closed form. If $\Re(z)=\frac{1}{2}$, then $1-z=\bar{z}$, so $$\frac{\pi}{\sin(\pi z)}=\Gamma(z)\Gamma(1-z)=\Gamma(z)\Gamma(\bar{z})=\Gamma(z)\overline{\Gamma(z)}=|\Gamma(z)|^2,$$ so (if one can find a reference for that particular claim) this might be able to give somewhat of a closed form. – Carl Schildkraut Sep 27 '18 at 06:46
  • @CarlSchildkraut Sounds reasonable. – KYHSGeekCode Sep 27 '18 at 07:05
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    https://en.wikipedia.org/wiki/Riemann%E2%80%93Siegel_theta_function - related – Dark Malthorp Jan 13 '20 at 18:53
  • @DarkMalthorp https://mathoverflow.net/questions/112682/riemann-siegel-function-and-gamma-function/175734#175734 related? – KYHSGeekCode Jan 15 '20 at 00:09

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