Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let us consider the famous gamma function.

My question is:

Is it possible to derive a closed expression for the gamma function for complex numbers $s=a+ib$ with $a≠1/2$, $b≠0$.

I know that such a closed form exists for integer values of $x=n$ for which $G(n)=(n-1)!$. Also, some non integers values have a closed form. Also if $a=1/2$ then a closed form exists. See this link:

share|cite|improve this question
You may find expressions for the module for $a=0, a=\frac 12$ or $a=1$ here ($(6.1.29)$ and more) (btw your link shows rather a relation of the argument of zeta with $\zeta(1/2+it)$ : $\Gamma$ is the 'simpler' function!). – Raymond Manzoni Dec 23 '12 at 9:16
@RaymondManzoni: Thank you very much. – ZE1 Dec 23 '12 at 9:48
You are welcome ! (sorry for the typos 'module' $\to$ modulus and the MO link gives a relation of the argument of zeta with $\Gamma(1/2+it)$ : zeta is the complicated function and the relation just shows that its argument is simple since it may be written as a function of $\Gamma$ and other elementary functions). $\Gamma$ itself has only trivial closed forms as opposed to the derivative of $\log(\Gamma)$ i.e. the digamma or $\psi$ function that admits a closed form at every rational value. – Raymond Manzoni Dec 23 '12 at 10:14
I don't know of an impossibility proof. Proofs exist that the gamma function can't be written as an elementary function but this doesn't exclude specific closed forms as at $\frac 12$. – Raymond Manzoni Dec 23 '12 at 10:32
The closest thing I could find is the Hölder theorem from the discussion here. See too the more recent paper1 in french and paper2. – Raymond Manzoni Dec 23 '12 at 11:03
up vote 1 down vote accepted

The $\Gamma$ function lacks a closed form containing only elementary functions. Here are the equivalent formulas of the $\Gamma$ function however:

$$\Gamma(z)=\int_{0}^{+\infty}t^{z-1}e^{-t}dt$$ $$\Gamma(z)=\frac1z\prod_{n=1}^{\infty}\frac{(1+\frac1n)^z}{1+\frac zn}$$ $$\Gamma(z)=\frac{e^{-\gamma z}}{z}\prod_{n=1}^{\infty}\frac{e^{\frac zn}}{1+\frac zn}$$ where $\gamma$ is the Euler–Mascheroni constant. Of course we can relate the $\Gamma$ function with elementary functions via the following indentity: $$\Gamma(1-z)\Gamma(z)=\frac{\pi}{\sin \pi z}$$ We also have Riemann's functional equation $$\zeta(s)=2^s\pi^{s-1}\sin\frac{\pi s}2\Gamma(1-s)\zeta(1-s)$$ which relates the $\Gamma$ and the $\zeta$ functions.

If by closed form we mean an expression containing only elementary functions then no, $\Gamma$ has no such form. For more information read this

share|cite|improve this answer
I mean this case: $$\Gamma(s)=f(a,b)$$ such that the expression of $f$ is well know. I find one but I want to use it as contraduction in a proof. – ZE1 Dec 23 '12 at 9:19
The expression of $f$ is given in term of $a$ and $b$ in an explicit manner. – ZE1 Dec 23 '12 at 9:21
I find this explicit formula: $arg(Γ(α+iβ))=λ(α,β)-σ(α,β)$ with $λ(α,β)$ and $σ(α,β)$ are given in term of $cos,sinh,sin, cosh$. – ZE1 Dec 23 '12 at 9:29
@user53124 And so? Even if $\arg \Gamma$ is elementary, $\Gamma$ is not elementary so any attempt to find such a form wil be futile. – Nameless Dec 23 '12 at 9:31
"If by closed form we mean an expression containing only elementary functions then no, $\Gamma$ has no such form" : does not exists or impossible. – ZE1 Dec 23 '12 at 9:31

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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