The Gamma Function & Analytic Continuation
The Gamma Function & Analytic Continuation
Analytic continuation extends an analytic function beyond its original domain of definition — if $f$ and $g$ are analytic on domains $D_1\subset D_2$ and agree on $D_1$, then $g$ is the unique analytic continuation of $f$ to $D_2$. The Gamma function $\Gamma(z)=\int_0^\infty t^{z-1}e^{-t}\,dt$ (originally defined for $\text{Re}(z)>0$) is analytically continued to $\mathbb{C}\setminus\{0,-1,-2,\ldots\}$ and satisfies the reflection formula and Stirling's approximation.
Gamma Function & Functional Equation
The Gamma function for $\text{Re}(z)>0$: $\Gamma(z)=\int_0^\infty t^{z-1}e^{-t}\,dt$. Functional equation (integration by parts): $\Gamma(z+1)=z\Gamma(z)$. Since $\Gamma(1)=1$, this gives $\Gamma(n)=(n-1)!$ for positive integers. Analytic continuation to $\mathbb{C}\setminus\{0,-1,-2,\ldots\}$ using $\Gamma(z)=\Gamma(z+n)/z(z+1)\cdots(z+n-1)$; simple poles at $z=0,-1,-2,\ldots$ with $\text{Res}(\Gamma,-n)=(-1)^n/n!$. Reflection formula: $\Gamma(z)\Gamma(1-z)=\pi/\sin(\pi z)$.
Identity Theorem & Analytic Continuation
If two analytic functions $f$ and $g$ agree on a set $E$ with an accumulation point in a connected domain $D$, then $f=g$ on all of $D$. This uniqueness theorem implies that an analytic function is completely determined by its values on any convergent sequence. Analytic continuation: if $f_1$ is analytic on $D_1$ and $f_2$ on $D_2$ with $D_1\cap D_2\neq\emptyset$ and $f_1=f_2$ on $D_1\cap D_2$, then $f_2$ is the unique analytic continuation of $f_1$ to $D_2$. The Riemann zeta function $\zeta(s)=\sum n^{-s}$ ($\text{Re}(s)>1$) continues to $\mathbb{C}\setminus\{1\}$ with a simple pole at $s=1$.
Example 1
Compute $\Gamma(1/2)$ using the reflection formula.
Solution: $\Gamma(1/2)\Gamma(1-1/2)=\Gamma(1/2)^2=\pi/\sin(\pi/2)=\pi$. So $\Gamma(1/2)=\sqrt{\pi}$. This gives $\int_0^\infty t^{-1/2}e^{-t}\,dt=\sqrt{\pi}$, or equivalently $\int_0^\infty e^{-x^2}\,dx=\sqrt{\pi}/2$ (the Gaussian integral, via $t=x^2$).
Example 2
Find the analytic continuation of $f(z)=\sum_{n=0}^\infty z^n$ beyond $|z|<1$.
Solution: $\sum z^n=1/(1-z)$ for $|z|<1$. The function $g(z)=1/(1-z)$ is analytic on $\mathbb{C}\setminus\{1\}$ and agrees with $f$ on $|z|<1$. By the Identity Theorem, $g$ is the unique analytic continuation. The 'natural boundary' of $f$ is the circle $|z|=1$ (dense set of singularities) — but $g$ continues through it except at $z=1$. This shows the original series representation misses the actual domain of the function.
Practice
- Prove $\Gamma(n+1)=n!$ for positive integers using the functional equation.
- Show $\Gamma(z)\Gamma(1-z)=\pi/\sin(\pi z)$ by computing $B(z,1-z)=\Gamma(z)\Gamma(1-z)/\Gamma(1)$ as a Beta integral and evaluating via residues.
- State Stirling's formula $\Gamma(n+1)\approx\sqrt{2\pi n}(n/e)^n$ and use it to estimate $100!$.
- Show that the Riemann zeta function $\zeta(s)=\sum n^{-s}$ has a functional equation $\zeta(s)=2^s\pi^{s-1}\sin(\pi s/2)\Gamma(1-s)\zeta(1-s)$.
Show Answer Key
1. $\Gamma(1)=\int_0^\infty e^{-t}dt=1=0!$. $\Gamma(n+1)=\int_0^\infty t^n e^{-t}dt=n\int_0^\infty t^{n-1}e^{-t}dt=n\Gamma(n)$. By induction, $\Gamma(n+1)=n!$.
2. $B(z,1-z)=\int_0^1 t^{z-1}(1-t)^{-z}dt$. Substituting $t=u/(1+u)$: $B(z,1-z)=\int_0^\infty\frac{u^{z-1}}{1+u}du=\frac{\pi}{\sin(\pi z)}$ (by the keyhole contour result). Since $B(z,1-z)=\Gamma(z)\Gamma(1-z)/\Gamma(1)=\Gamma(z)\Gamma(1-z)$, we get the reflection formula.
3. $\Gamma(101)=100!\approx\sqrt{200\pi}(100/e)^{100}\approx9.33\times10^{157}$. The exact value is $100!=9.3326\ldots\times10^{157}$.
4. The functional equation is $\zeta(s)=2^s\pi^{s-1}\sin(\tfrac{\pi s}{2})\Gamma(1-s)\zeta(1-s)$. It relates values at $s$ and $1-s$, extending $\zeta$ to all $\mathbb{C}$ (except $s=1$). Proof uses the Mellin transform of the theta function $\vartheta(t)=\sum e^{-\pi n^2 t}$ and its modularity $\vartheta(1/t)=\sqrt{t}\,\vartheta(t)$.