# Why do we use 'this' Gamma Function. [duplicate]

The Gamma function is a generalization of the factorial defined by Euler as: $$\Gamma(z)=\int\limits_{0}^{\infty}t^{z-1}e^{-t}\,dt$$ for $z\in\mathbb{C}$ with positive real part.

It satisfies, $\Gamma(n)=(n-1)!$ for all $n\in\mathbb{N}$.

My question is why we choose this particular function from all the functions that satisfy the previous property. And, how is the Gamma function deduced?

## marked as duplicate by Start wearing purple, Américo Tavares, MJD, Julian Kuelshammer, Lord_FarinJun 20 '13 at 10:40

• – Martin Jun 20 '13 at 9:56

1. $\Gamma(n+1) = n\Gamma(n)!$ (factorial recurrence)
2. $\Gamma(1) = 1$
3. $n\mapsto \log\Gamma(n)$ is a convex function.
The $\Gamma$ function arises naturally when you work with Mellin transform (an other name for Fourier transform on the group $\mathbb{R}_{>0}$) : if $f : \mathbb{R}_{>0} \rightarrow \mathbb{C}$ is a (suitable) function then its Mellin transform is $$\mathcal{M}(f,s) = \int_0^\infty f(t) t^s \frac{dt}{t}.$$ Let $f(z) = \sum_{n=1}^\infty a_n e^{-nt}$ be a 'Fourier serie' (with $a_n =O(n^C)$), then the its Mellin transform is $$\mathcal{M}(f,s) = \sum_{n \geq 1} a_n \left( \int_0^\infty t^n e^{-t} \frac{dt}{t} \right) n^{-s}.$$ Hence if you define $\Gamma(t) := \int_0^\infty t^n e^{-t} \frac{dt}{t}=\mathcal{M}(\exp,s)$ and $L(f,s)$ the Dirichlet serie $L(f,s):=\sum_{n \geq 1} a_n n^{-s}$, you obtain the nice formula $$\mathcal{M}(f,s) = \Gamma(s) L(f,s).$$ This identity is the key point to get the functional equation of the Zeta function, or the $L$-function of modular forms.
From a practical point of view, the argument of convexity is (imo) rather unnatural. What makes the Gamma function really important and suitable for applications is not that it coincides with factorial at positive integer points, but rather its factorial-type relation $\Gamma(1+z)=z\Gamma(z)$ which holds for any $z\in\mathbb{C}$.
Any other "gamma"-function with this property would differ from the standard one by a periodic function of $z$. Now if this periodic function is non-constant, the alternative gamma-function will have more zeroes and poles than the standard $\Gamma(z)$ for which this set is, in a sense, minimal.