1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Use some Riemann sum to approximate it value. Probably the error can be evaluated (not sure anyway). $\endgroup$ – Masacroso Feb 13 '17 at 1:55
0
$\begingroup$

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 $z$.

$\endgroup$
1
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.