Gamma of 3z using triplication formula:

Question

I have to demostrate the gamma function for 3z as you see below: Using the multiplication formula demostrate gamma(3z)

Gamma functions of argument $3z$ can be expressed using a triplication formula $$\Gamma (3z)=(2 \pi)^{-1}3^{3z-1/2}\Gamma(z)\Gamma \left(z+\frac{1}{3}\right)\Gamma\left(z+\frac{2}{3}\right)$$

Answer 1

Hint:

$$(3n)!~=~\prod_{k=0}^{n-1}(3k+1)(3k+2)(3k+3)~=~\prod_{k=0}^{n-1}(3k+1)\cdot\prod_{k=0}^{n-1}(3k+2)\cdot\prod_{k=0}^{n-1}(3k+3)=$$

$$=~3^n\cdot\prod_{k=0}^{n-1}\bigg(k+\frac13\bigg)\cdot3^n\cdot\prod_{k=0}^{n-1}\bigg(k+\frac23\bigg)\cdot3^n\cdot\prod_{k=0}^{n-1}\big(k+1\big)~=$$

$$=~3^{3n}\cdot\frac{\big(n-2/3\big)!}{\big(1/3\big)!}\cdot\frac{\big(n-1/3\big)!}{\big(2/3\big)!}\cdot n!$$

Now use the reflection formula.

Answer 2

$\Gamma(z)$ is log-convex (by the Bohr-Mollerup theorem), continuous and non-vanishing over $\mathbb{R}^+$.
By defining $$ g(z)=\frac{\Gamma(3z)}{\Gamma(z)\,\Gamma\left(z+\tfrac{1}{3}\right)\,\Gamma\left(z+\tfrac{2}{3}\right)} $$ from the functional identity $\Gamma(s+1)=s\,\Gamma(s)$ we get $$ \frac{g(z+1)}{g(z)} = \frac{(3z+2)(3z+1)3z}{z\left(z+\tfrac{1}{3}\right)\left(z+\tfrac{2}{3}\right)} = 27$$ hence $g(z)=C\cdot 3^{3z}$ for some constant $C$. From $\Gamma(s)\,\Gamma(1-s)=\frac{\pi}{\sin(\pi s)}$ we get: $$ g(1)=\frac{2}{\Gamma\left(\tfrac{4}{3}\right)\,\Gamma\left(\tfrac{5}{3}\right)}=\frac{9}{\Gamma\left(\tfrac{1}{3}\right)\,\Gamma\left(\tfrac{2}{3}\right)}=\frac{9}{\pi}\,\sin\frac{\pi}{3}$$ and $C=\frac{1}{2\pi\sqrt{3}}$, finishing the proof of the given identity over $\mathbb{R}^+$.