Gamma Function

Subjects: Complex Analysis, Special Functions
Links:Related Functions to the Gamma Function, Poles of Analytic Functions, Infinite Product of Functions

What we want $f:U\to \mathbb{C}$ such that

1. $U$ is the biggest region possible
2. $f\in \mathcal{H}(U)$
3. $f(1)=1$ and $zf(z)=f(z+1)$

We see that $0\notin U$ since $1=f(1)=0\cdot f(0)=0$, which brings problems. Similarly we need that $P(f)={\mathbb{Z}}^{\le 0}$, and we get $$f(z+n+1) = f(z)\prod_{k = 1}^n (z+k)$$and we get that the residues at each nonnegative integers is of the form $$ \text{Res}(f, -n) = \frac{(-1)^n}{n!}$$ with $n\in \mathbb{N}$. Let suppose $Z(f)=\varnothing$, and $f\in \mathcal{M}(\mathbb{C})$, with $P(f)={\mathbb{Z}}^{\le 0}$, with $U=\mathbb{C}\setminus {\mathbb{Z}}^{\le 0}$. Then there's $h\in \mathcal{H}(U)$ such that $f=1/h$. Then $Z(h)=P(f)$. Using Weierstrass Factorization Theorem, and consider that

Then we can there's $g\in \mathcal{H}(U)$ such that:$$h(z) = z e^{g(z)} \prod_{n = 1}^\infty E_1(-z/n)= z e^{g(z)} \prod_{n = 1}^\infty\left(\frac{z+n}{n}\right)e^{-z/n}$$
Exploiting the recurrence relation to with the $h$, we can find out who is $g$, the problem is that there are many, but the $g$ must satisfy$$\gamma = g(z+1)-g(z)$$with $\gamma$ being the Euler–Mascheroni Constant, and the 'simplest' is $g(z)=\gamma z$.

Def: Let $h\in \mathcal{H}(\mathbb{C})$ as $$ h(z) = z e^{\gamma z} \prod_{n = 1}^\infty\left(\frac{z+n}{n}\right)e^{-z/n} $$
and we define $\mathrm{\Gamma }:\mathbb{C}\setminus {\mathbb{Z}}^{\le 0}\to \mathbb{C}$ as$$\Gamma(z) = \frac{1}{h(z)} =\frac{e^{\gamma z}}{z}\prod_{n = 1}^\infty \left(\frac{n}{z+n}\right)e^{z/n}$$
This is known as the Weierstrass product Formula of the Gamma Function

Th: $\mathrm{\Gamma }\in \mathcal{M}(\mathbb{C})$ and $P(\mathrm{\Gamma })={\mathbb{Z}}^{\le 0}$ and satisfies this conditions:

• for all $n\in \mathbb{N}$, we get that $\text{Res}(\mathrm{\Gamma },-n)=(-1{)}^n/n!$
• for every $z\in \mathbb{C}\setminus {\mathbb{Z}}^{\le 0}$, $\mathrm{\Gamma }(z+1)=z\mathrm{\Gamma }(z)$ and $\mathrm{\Gamma }(1)=0$.
• $\mathrm{\Gamma }(n+1)=n!$ for every $n\in \mathbb{N}$

Euler’s Reflection Formula

for $z\notin \mathbb{Z}$, then$$ \Gamma(1-z)\Gamma(z) = \frac{\pi}{\sin(\pi z)} $$

Gauss Product Formula

Gauss rewrote it as Gauss’s product formula:$$ \Gamma(z)= \lim_{n \to \infty}\frac{n^z n!}{z(z+1) \cdots (z+n)}= \lim_{n \to \infty}\frac{n^z n!}{z^{\overline{ n+1}}} = \lim_{n\to \infty} \left(n^z \prod_{\nu = 1}^n \frac{\nu}{\nu +z}\right) $$

Bohr-Mollerup Theorem

There’s a unique positive function $f$ defined on $x\ge 0$ satisfying

• $f(1)=1$
• $f(x+1)=xf(x)$
• $\ln (f(x))$ is convex

Now we want to check that the usual definition of the gamma function actually matches with the definition by Legendre as the integral:

but we need a little constraints to make the integral converge, having that $\mathrm{\Re }(z)>0$, then have that $$ \Gamma(z) = \int_0^\infty t^{z-1} e^{-t} ,d t $$
To actually do this, it is done by a couple of tricks, and a lot of steps:

Some notation, given $0<a<M$ we define$$S(a, M) ={z\in \Bbb C\mid a \le \Re(z) \le M} $$
Lemma 1: Let $0<a<M$, then

• for every $\epsilon >0$, there's a $\delta >0$ such that every $z\in S(a,M)$ if $0<\alpha <\beta <\delta$, then $$ \left|\int_\alpha^\beta t^{z-1} e^{-t}, dt \right|<\varepsilon$$
• for every $\epsilon >0$, there's a $N>0$ such that every $z\in S(a,M)$ if $N<p<q$, then $$\left|\int_p^q t^{z-1}e^{-t}, dt \right| < \varepsilon$$
  Lemma 2: Let $U=\{z\in \mathbb{C}\mid \mathrm{\Re }(z)>0\}$. For each $n\in \mathbb{N}$, let $f_n:U\to \mathbb{C}$ as $$f_n(z) = \int_{1/n}^n t^{z-1}e^{-t}, dt $$Then, $f_n$ converges to $$f(z) = \int_0^\infty t^{z-1} e^{-t} ,d t$$
  in $\mathcal{H}(U)$.

Lemma 3: Let $h_n(z)={(1-{\displaystyle \frac{z}{n}})}^n$, with $h_n\in \mathcal{H}(\mathbb{C})$

• $h_n\to e^{-z}$ in $\mathcal{H}(U)$
• if $t\ge 0$, and $n>t$, then ${(1-{\displaystyle \frac{z}{n}})}^n\le e^{-t}$

Lemma 4: For $x>0$, and $n\in \mathbb{N}$, we get that

Cor of Lemma 4: For $x>0$, and $$g_n(x) = \int_0^n \left(1-\frac{t}{n}\right)^n t^{x-1}, dt$$ Then $$\lim\limits_{n \to \infty }g(x) = \Gamma (x) $$
Th: Let $U=\{z\in \mathbb{C}\mid \mathrm{\Re }(z)>0\}$ Then for $z\in U$ we have that $$ \Gamma(z) = \int_0^\infty t^{z-1} e^{-t} ,d t $$

Miscellaneous

We can also have new forms of the gamma function

We also have that another definition namely, the **Euler’s product formula:****$$ \Gamma(z) = \frac{1}{z} \prod_{n = 1}^\infty \left[\frac{1}{1+\frac{z}{n}}\left(1+\frac{1}{n}\right)^z\right] $$

Legendre’s Duplication Formula

Let $z\in \mathbb{C}$, then

Particular Values

for nonnegative integers $n$, we have that

Raabe’s Identity