Evaluate \[ \Gamma(1/3)\Gamma(2/3). \]
Find a formula for \(\Gamma(3z)\) in terms of \[ \Gamma(z),\qquad \Gamma(z+1/3),\qquad \Gamma(z+2/3). \]
Solution
The required formulas are \[ \boxed{\Gamma(1/3)\Gamma(2/3)=\frac{2\pi}{\sqrt3}} \] and \[ \boxed{ \Gamma(3z) =\frac{3^{\,3z-1/2}}{2\pi} \Gamma(z)\Gamma(z+1/3)\Gamma(z+2/3).} \]
1. Euler’s reflection formula gives\[ \Gamma(1/3)\Gamma(2/3)=\frac{2\pi}{\sqrt3}. \]
Euler’s reflection formula is \[ \Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)}. \] Substituting \(z=1/3\) yields \[ \Gamma(1/3)\Gamma(2/3) =\frac{\pi}{\sin(\pi/3)} =\frac{\pi}{\sqrt3/2} =\frac{2\pi}{\sqrt3}. \]
2. Define, for \(\operatorname{Re}w>0\),\[ F(w) =\frac{3^w}{2\pi\sqrt3} \Gamma(w/3) \Gamma((w+1)/3) \Gamma((w+2)/3). \]Then \(F(w+1)=wF(w)\).
Using \(\Gamma(u+1)=u\Gamma(u)\), \[ \begin{aligned} F(w+1) &=\frac{3^{w+1}}{2\pi\sqrt3} \Gamma((w+1)/3) \Gamma((w+2)/3) \Gamma(w/3+1)\\ &=\frac{3^{w+1}}{2\pi\sqrt3} \Gamma((w+1)/3) \Gamma((w+2)/3) \frac w3\Gamma(w/3)\\ &=wF(w). \end{aligned} \]
3. One has \(F(1)=1\), and \(F\) is bounded on the strip\(1\le\operatorname{Re}w\le2\).
By 1, \[ \begin{aligned} F(1) &=\frac{3}{2\pi\sqrt3} \Gamma(1/3)\Gamma(2/3)\Gamma(1)\\ &=\frac{3}{2\pi\sqrt3}\cdot\frac{2\pi}{\sqrt3} =1. \end{aligned} \]
If \(1\le\operatorname{Re}w\le2\), then the real parts of \[ \frac w3,\qquad \frac{w+1}{3},\qquad \frac{w+2}{3} \] lie in fixed compact subintervals of \((0,\infty)\). The bound \(|\Gamma(u)|\le\Gamma(\operatorname{Re}u)\) for \(\operatorname{Re}u>0\) therefore bounds each Gamma factor there, and \(|3^w|=3^{\operatorname{Re}w}\) is bounded as well. Hence \(F\) is bounded on the required strip.
4. Wielandt’s characterization implies \(F(w)=\Gamma(w)\).
Wielandt’s theorem states that a holomorphic function on the right half-plane satisfying \[ F(w+1)=wF(w),\qquad F(1)=1, \] and bounded on \(1\le\operatorname{Re}w\le2\) must equal \(\Gamma(w)\). The hypotheses were verified in 2 and 3, so \[ \Gamma(w) =\frac{3^w}{2\pi\sqrt3} \Gamma(w/3) \Gamma((w+1)/3) \Gamma((w+2)/3). \]
5. Replacing \(w\) by \(3z\) gives the triplication formula.
From 4, \[ \Gamma(3z) =\frac{3^{3z}}{2\pi\sqrt3} \Gamma(z)\Gamma(z+1/3)\Gamma(z+2/3). \] Since \(\sqrt3=3^{1/2}\), \[ \Gamma(3z) =\frac{3^{\,3z-1/2}}{2\pi} \Gamma(z)\Gamma(z+1/3)\Gamma(z+2/3). \] Initially this identity holds where the displayed Gamma factors are holomorphic; both sides are meromorphic functions, so it extends to all \(z\) by analytic continuation.