I can calculate the values of Gamma function for positive integer arguments using the formula $\Gamma (t) = \displaystyle\int_0^{\infty} e^{-x} x ^ {t-1} \mathrm{d}x $. Which is equal to $ (t-1)! $. But I don't know how to calculate values for $ \Gamma (a+bi) $. Say, for example, how do I calculate $ \Gamma (1+i) $?

  • $\begingroup$ buzzword: analytic continuiation $\endgroup$ – tired Jun 25 '15 at 12:08
  • $\begingroup$ See en.m.wikipedia.org/?title=Gamma_function. $\endgroup$ – Tim Raczkowski Jun 25 '15 at 12:15
  • $\begingroup$ Start with $i!=\displaystyle\int_0^\infty x^i~e^{-x}~dx$, and then use $x^i=e^{i\ln x}=\cos\ln x+i\sin\ln x$. $\endgroup$ – Lucian Jun 25 '15 at 12:44

You can start using the asymptotic development (Stirling series)$$\log\big(\Gamma(x)\big)=x (\log (x)-1)+\frac{1}{2} \left(\log (2 \pi )-\log(x)\right)+\frac{1}{12 x}-\frac{1}{360 x^3}+\frac{1}{1260 x^5}-\frac{1}{1680 x^7}+$$ $$\frac{1}{1188 x^9}-\frac{691}{360360 x^{11}}+\frac{1}{156 x^{13}}-\frac{3617}{122400 x^{15}}+\frac{43867}{244188 x^{17}}+O\left(\left(\frac{1}{x}\right)^{35/2}\right)$$ and use the fact that $x$ is a complex number.

For example, this expansion gives $$\log\big(\Gamma(3+2i)\big)=-0.03163905938061+2.02219319752573 i$$ while the "exact" value (as given by a CAS) would be $$\approx -0.03163905937396 + 2.02219319750133 i$$ So, the expansion would lead to $$\Gamma(3+2i)=-0.422637286329669 + 0.871814255680394 i$$ for an "exact" value (as given by a CAS) $$\approx -0.422637286311202 + 0.871814255696507 i$$ For sure, you could have more terms for the expansion.

  • $\begingroup$ What do you mean by CAS? $\endgroup$ – KYHSGeekCode Sep 22 '18 at 9:41
  • $\begingroup$ @KYHSGeekCode. Computer Algebra System such as Mathematica, Wolfram Alpha and many other. $\endgroup$ – Claude Leibovici Sep 22 '18 at 10:42
  • $\begingroup$ Aha, thank you! $\endgroup$ – KYHSGeekCode Sep 22 '18 at 10:51

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