# Manually evaluating the gamma function

How would I find the gamma function of i+1 or any other complex number?

Euler integral for i

I tried using integration by parts, but it just went on forever.

• Use some Riemann sum to approximate it value. Probably the error can be evaluated (not sure anyway). – Masacroso Feb 13 '17 at 1:55

There is not a simpler closed form for $\Gamma(i+1)$ other than just writing $\Gamma(i+1)$. It does have a decimal expansion $0.498015668118356042713691117462...$ though.

If you want a numerical answer, Riemann Sums will do, or even Integration by Parts like you tried with error estimates. Stirling's Approximation would also help you for large values of $z$.

Here is the most basic solution:

\begin{align}\Gamma(i+1)&=\lim_{n\to\infty}\frac{n!e^{i\ln(n+1)}}{(1+i)(2+i)(3+i)\dots(n+i)}\\&=\lim_{n\to\infty}\frac{n!(\cos(\ln(n+1))+i\sin(\ln(n+1)))}{(1+i)(2+i)(3+i)\dots(n+i)}\end{align}

Pretty basic and doable with a basic calculator.

• No particular reason it seems. – Simply Beautiful Art Jun 22 '18 at 13:07
• What basic calculator do you mean..? – KYHSGeekCode Sep 15 '18 at 9:21