# How do I calculate values for Gamma function with complex arguments?

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 $\Gamma (1+i)$?

• buzzword: analytic continuiation – tired Jun 25 '15 at 12:08
• – Tim Raczkowski Jun 25 '15 at 12:15
• Start with $i!=\displaystyle\int_0^\infty x^i~e^{-x}~dx$, and then use $x^i=e^{i\ln x}=\cos\ln x+i\sin\ln x$. – 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.