1
$\begingroup$

While reading the Wikipedia page on Particular values of the Gamma Function, it listed a formula:$$\Gamma\left(\dfrac z2\right)=\sqrt{\pi}\dfrac {(z-2)!!}{2^{(z-1)/2}}\tag{1}$$ Where $z\in\mathbb{Z}$ for positive half integers. $(1)$ can be used to compute $\Gamma\left(\frac 12\right)$ by setting $z=1$ to get$$\Gamma\left(\dfrac 12\right)=\sqrt{\pi}\dfrac {(1-2)!!}{2^{(1-1)/2}}=\sqrt\pi\tag{2}$$ Extending this, I'm wondering

Questions:

  1. Can forumula $(1)$ be generalized to include complex numbers?$$\Gamma\left(a+bi\right)=\text{something}\tag{3}$$
  2. If so, how would you prove such formula?

Running it for WolframAlpha, it says that the Gamma function of a complex number is defined and is possible. But I'm just not sure how to derive a formula for $(3)$.

$\endgroup$
  • $\begingroup$ complex numbers with integer $a$ and $b$.? $\endgroup$ – Nosrati Feb 27 '17 at 4:45
  • $\begingroup$ @MyGlasses Uh... yes. $a,b\in\mathbb{R}$ $\endgroup$ – Crescendo Feb 27 '17 at 15:03
  • $\begingroup$ not sure if this was up when the question was posted, but the Wikipedia page does have some numeric data on complex values of Gamma: en.wikipedia.org/wiki/… no closed forms are given, which may have motivated the question $\endgroup$ – Harry Richman Oct 24 '17 at 13:56
1
$\begingroup$

There is no corresponding formula for complex values of $z$: unlike adding ordinary integers, there is no way to relate $\Gamma(a+i)$ to $\Gamma(a)$ (except that one has absolute value smaller than the other, by trivially bounding the integral definition). One can find the absolute value of $\Gamma(yi)$ for real $y$ by using the reflection formula (which comes out as $\sqrt{\frac{\pi}{y\sinh{(\pi y)}}}$), but there is no equivalent to $\Gamma(1+z)=z\Gamma(z)$. Hence one can relate the values of $\Gamma$ along a horizontal string of points separated by integers, but there is no way to move between these.

This is, in effect, a worse version of the problem with expressing $\Gamma(1/n)$ in finite form for $n>2$: there is no nice formula containing just one of these quantities: you only have things like $$ \Gamma(1/3)\Gamma(2/3) = \frac{2\pi}{\sqrt{3}} $$ from the reflection formula, or $$ \frac{\Gamma(1/3)\Gamma(5/6)}{\Gamma(2/3)} = \sqrt[3]{2}\sqrt{\pi} $$ from the duplication formula.

$\endgroup$

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.