Special Functions – The gamma and the beta function – The gamma function

The gamma function \(\Gamma(z)\) is defined by:

Definition: \(\Gamma(z)=\displaystyle\int_0^{\infty}e^{-t}t^{z-1}\,dt,\quad\text{Re}(z)>0.\)

From this definition it is clear that \(\Gamma(z)\) is analytic for \(\text{Re}(z)>0\). Using integration by parts we find that

\begin{align*} \Gamma(z+1)=\int_0^{\infty}e^{-t}t^z\,dt=-\int_0^{\infty}t^z\,de^{-t}=-e^{-t}t^z\Big|_0^{\infty}+\int_0^{\infty}e^{-t}\,dt^z =z\int_0^{\infty}e^{-t}t^{z-1}\,dt=z\Gamma(z),\quad\text{Re}(z)>0. \end{align*}

Hence we have

Theorem: \(\Gamma(z+1)=z\Gamma(z),\quad\text{Re}(z)>0\).

Further we have: \(\Gamma(1)=\displaystyle\int_0^{\infty}e^{-t}\,dt=-e^{-t}\Big|_0^{\infty}=1\). This implies that \(\Gamma(n+1)=n!\) for \(n=0,1,2,\ldots\).

Analytic continuation

For \(-1<\text{Re}(z)\leq0\) we define: \(\Gamma(z)=\displaystyle\frac{\Gamma(z+1)}{z}\) for \(z\neq0\). Then the gamma function \(\Gamma(z)\) is analytic for \(\text{Re}(z)>-1\) except for \(z=0\). For \(z=0\) we obtain

\[\lim\limits_{z\to0}z\Gamma(z)=\lim\limits_{z\to0}\Gamma(z+1)=\Gamma(1)=1.\]

This implies that \(\Gamma(z)\) has a single pole at \(z=0\) with residue \(1\). This process can be repeated for \(-2<\text{Re}(z)\le -1\), \(-3<\text{Re}(z)\le -2\), etcetera. Then the gamma function turns out to be an analytic function on \(\mathbb{C}\) except for single poles at \(z=0,-1,-2,\ldots\). The residue at \(z=-n\) equals

\begin{align*} \lim\limits_{z\to -n}(z+n)\Gamma(z)&=\lim\limits_{z\to -n}(z+n)\frac{\Gamma(z+1)}{z} =\lim\limits_{z\to -n}(z+n)\frac{1}{z}\frac{1}{z+1}\cdots\frac{1}{z+n-1}\frac{\Gamma(z+n+1)}{z+n}\\[2.5mm] &=\frac{\Gamma(1)}{(-n)(-n+1)\cdots(-1)}=\frac{(-1)^n}{n!},\quad n=0,1,2,\ldots. \end{align*}

Definition: The shifted factorial is defined by \((a)_0=1\) and \((a)_k=a(a+1)(a+2)\cdots(a+k-1)\) for \(k=1,2,3,\ldots\).

Lemma: \(\displaystyle\int_0^1(1-t)^nt^{z-1}\,dt=\frac{n!}{(z)_{n+1}}\) for \(\text{Re}(z)>0\) and \(n=0,1,2,\ldots\).

Proof: For \(n=0\) we have for \(\text{Re}(z)>0\) that \(\displaystyle\int_0^1t^{z-1}\,dt=\frac{t^z}{z}\bigg|_0^1=\frac{1}{z}=\frac{0!}{z^1}\). Now suppose that the formula holds for certain value of \(n\), then

\begin{align*} \int_0^1(1-t)^{n+1}t^{z-1}\,dt&=\int_0^1(1-t)(1-t)^nt^{z-1}\,dt=\int_0^1(1-t)^nt^{z-1}\,dt-\int_0^1(1-t)^nt^z\,dt\\[2.5mm] &=\frac{n!}{(z)_{n+1}}-\frac{n!}{(z+1)_{n+1}}=\frac{n!}{(z)_{n+2}}\left(z+n+1-z\right)=\frac{(n+1)!}{(z)_{n+2}}. \end{align*}

This proves, by mathematical induction, that the lemma holds for all \(n=0,1,2,\ldots\).

Now we set \(t=u/n\) to find that

\[\frac{1}{n^z}\int_0^n\left(1-\frac{u}{n}\right)^nu^{z-1}\,du=\frac{n!}{(z)_{n+1}}\quad\Longrightarrow\quad \int_0^n\left(1-\frac{u}{n}\right)^nu^{z-1}\,du=\frac{n!\,n^z}{(z)_{n+1}}.\]

Since we have \(\displaystyle\lim\limits_{n\rightarrow\infty}\left(1-\frac{u}{n}\right)^n=e^{-u}\), we conclude that \(\Gamma(z)=\displaystyle\int_0^{\infty}e^{-u}u^{z-1}\,du=\lim\limits_{n\rightarrow\infty}\frac{n!\,n^z}{(z)_{n+1}}\).

Theorem: \(\displaystyle\int_0^{\infty}\frac{x^{s-1}}{e^x-1}\,dx=\Gamma(s)\zeta(s),\quad\text{Re}(s)>1\). Here \(\zeta(s):=\displaystyle\sum_{n=1}^{\infty}n^s\) denotes the Riemann zeta function.

Proof: We use that \(\displaystyle\frac{1}{1-e^{-x}}=\displaystyle\sum_{n=0}^{\infty}\left(e^{-x}\right)^n\) for \(x>0\) and the substitution \((n+1)x=t\) and obtain that

\begin{align*} \int_0^{\infty}\frac{x^{s-1}}{e^x-1}\,dx&=\int_0^{\infty}x^{s-1}e^{-x}\cdot\frac{1}{1-e^{-x}}\,dx=\sum_{n=0}^{\infty}\int_0^{\infty}x^{s-1}e^{-(n+1)x}\,dx\\[2.5mm] &=\sum_{n=0}^{\infty}\int_0^{\infty}\frac{t^{s-1}}{(n+1)^{s-1}}e^{-t}\cdot\frac{1}{n+1}\,dt=\sum_{n=0}^{\infty}\frac{1}{(n+1)^s}\int_0^{\infty}t^{s-1}e^{-t}\,dt\\[2.5mm] &=\Gamma(s)\zeta(s),\quad\text{Re}(s)>1. \end{align*}

For instance this implies that \(\displaystyle\int_0^{\infty}\frac{x}{e^x-1}\,dx=\Gamma(2)\zeta(2)=\tfrac{1}{6}\pi^2\) and \(\displaystyle\int_0^{\infty}\frac{x^2}{e^x-1}\,dx=\Gamma(3)\zeta(3)=2\zeta(3)\).

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Last modified on April 20, 2024