Product of Gamma functions II

Question

What is the value of the product of Gamma functions \begin{align} \prod_{k=1}^{10} \Gamma\left(\frac{k}{10}\right) \end{align} and can it be shown that \begin{align} \prod_{k=1}^{20} \Gamma\left(\frac{k}{10}\right) \approx \frac{\pi^{9}}{54} \end{align} and \begin{align} \prod_{k=1}^{40} \Gamma\left(\frac{k}{10}\right) \approx \left( 6 + \frac{625}{4501}\right) \pi^{18}. \end{align}

Answer 1

Original Question

As shown at the end of this answer, $$ \Gamma(x)\Gamma(1-x)=\pi\csc(\pi x)\tag{1} $$ As shown at the beginning of this answer, $$ \prod_{k=1}^{n-1}\sin(k\pi/n)=\frac{n}{2^{n-1}}\tag{2} $$ Using $(1)$ and $(2)$, we get $$ \begin{align} \left[\prod_{k=1}^{10}\Gamma\left(\frac{k}{10}\right)\right]^2 &=\prod_{k=1}^9\Gamma\left(\frac{k}{10}\right)\Gamma\left(1-\frac{k}{10}\right)\\ &=\prod_{k=1}^9\pi\csc(k\pi/10)\\ &=\frac{2^9\pi^9}{10}\tag{3} \end{align} $$ Thus, $$ \prod_{k=1}^{10}\Gamma\left(\frac k{10}\right)=16\pi^4\sqrt{\frac\pi5}\tag{4} $$

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Gauss's Multiplication Formula

Define $$ f(x)=\prod_{k=0}^{n-1}\Gamma\left(x+\frac kn\right)\tag{5} $$ then $f$ is log-convex and $$ \begin{align} x\,f(x) &=x\,\prod_{k=0}^{n-1}\Gamma\left(x+\frac kn\right)\\ &=x\,\Gamma(x)\,\prod_{k=1}^{n-1}\Gamma\left(x+\frac kn\right)\\ &=\Gamma(x+1)\,\prod_{k=0}^{n-2}\Gamma\left(x+\frac{k+1}n\right)\\ &=\prod_{k=1}^{n-1}\Gamma\left(x+\frac1n+\frac kn\right)\\[6pt] &=f\left(x+\frac1n\right)\tag{6} \end{align} $$ Plugging $\frac xn$ into $(6)$ gives $$ \frac xnf\left(\frac xn\right)=f\left(\frac{x+1}n\right)\tag{7} $$ $(7)$ and log-convexity implies that $$ f\left(\frac xn\right)=C_n\frac{\Gamma(x)}{n^x}\tag{8} $$ Using $(1)$ and $(2)$ yield $$ \begin{align} f\left(\frac1n\right)^2 &=\prod_{k=1}^{n-1}\Gamma\left(\frac kn\right)\Gamma\left(1-\frac kn\right)\\ &=\prod_{k=1}^{n-1}\pi\csc(k\pi/n)\\ &=\frac1n2^{n-1}\pi^{n-1}\tag{9} \end{align} $$ $(8)$ and $(9)$ yield $$ C_n=\sqrt{n2^{n-1}\pi^{n-1}}\tag{10} $$ Therefore, $(8)$ and $(10)$ give Gauss's Multiplication Formula $$ \prod_{k=0}^{n-1}\Gamma\left(x+\frac kn\right) =\sqrt{n2^{n-1}\pi^{n-1}}\frac{\Gamma(nx)}{n^{nx}}\tag{11} $$

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Second Question

Using $(11)$, $$ \begin{align} \prod_{k=1}^{20}\Gamma\left(\frac k{10}\right) &=\prod_{k=0}^{9}\Gamma\left(\frac1{10}+\frac k{10}\right)\prod_{k=0}^{9}\Gamma\left(\frac{11}{10}+\frac k{10}\right)\\ &=5120\pi^9\frac{\Gamma(1)}{10^1}\frac{\Gamma(11)}{10^{11}}\\ &=\frac{\pi^9\,9!}{10\cdot5^9}\tag{12} \end{align} $$

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Third Question

Using $(11)$, $$ \begin{align} &\prod_{k=1}^{40}\Gamma\left(\frac k{10}\right)\\ &=\small\prod_{k=0}^{9}\Gamma\left(\frac1{10}+\frac k{10}\right)\prod_{k=0}^{9}\Gamma\left(\frac{11}{10}+\frac k{10}\right)\prod_{k=0}^{9}\Gamma\left(\frac{21}{10}+\frac k{10}\right)\prod_{k=0}^{9}\Gamma\left(\frac{31}{10}+\frac k{10}\right)\\ &=\left(5120\pi^9\right)^2\frac{\Gamma(1)}{10^1}\frac{\Gamma(11)}{10^{11}}\frac{\Gamma(21)}{10^{21}}\frac{\Gamma(31)}{10^{31}}\\ &=\frac{2^{18}\pi^{18}}{10^{62}}10!\,20!\,30!\tag{13} \end{align} $$

Answer 2

Since @robjohn has done an excellent job at providing an answer in detail I will add my comments as an additional solution. The results presented here follow the numbering in robjohn's work.

In robjohn's equation (12) the factor \begin{align} \frac{9! \ \pi^{9}}{2 \cdot 5^{10}} \end{align} has been provided. Numerically this can be seen as $.018579456\cdots \ \pi^{9}$. By comparing this to that of $1/54 = .0185185\cdots$ one can make a good approximation by \begin{align} \prod_{k=1}^{20} \Gamma\left(\frac{k}{10}\right) = \frac{9! \ \pi^{9}}{2 \cdot 5^{10}} \approx \frac{\pi^{9}}{54}. \tag{14} \end{align} Another possible value is $20/1077 = .0185701021\cdots$ which leads to the statement \begin{align} \prod_{k=1}^{20} \Gamma\left(\frac{k}{10}\right) = \frac{9! \ \pi^{9}}{2 \cdot 5^{10}} \approx \frac{20 \ \pi^{9}}{1077}. \tag{15} \end{align}

From equation (13) the factor is \begin{align} \frac{2^{18} \ (10)!(20)!(30)!}{10^{62}} = 6.138858832\cdots = 6 + .138858832\cdots . \end{align} Since $625/4501 = .1388580315\cdots$ then to a fair approximation it can be stated \begin{align} \prod_{k=1}^{40} \Gamma\left(\frac{k}{10}\right) \approx \left( 6 + \frac{625}{4501} \right) \pi^{18}. \tag{16} \end{align}