Can anyone explain me how to find the value of $\left(\frac{1}{3}\right)!$?

Question

I know how to calculate the value of $\left(\dfrac{1}{2}\right)!$ using the gamma function, but I don't know how to find the value of $\left(\dfrac{1}{3}\right)!$ or $\left(\dfrac{1}{6}\right)!$ using the gamma function.

Answer 1

According to Wikipedia, "It has been proved that $\Gamma(n+r)$ is a transcendental number and algebraically independent of $\pi$ for any integer $n$ and each of the fractions $r=\frac16,\frac14,\frac13,\frac23,\frac34,\frac56$," so it isn't possible to do what you ask.

Answer 2

$$ \Gamma\left(\frac{1}{3}\right)=\frac{2^{7/9}\pi^{2/3}}{3^{1/12}{\rm agm}(2,\sqrt{2+\sqrt{3}})^{1/3}} $$ where $\rm agm$ is the arithmetic-geometric mean. One can then deduce that $$ \Gamma\left(\frac{1}{6}\right)=\frac{2^{14/9}3^{1/3}\pi^{5/6}}{{\rm agm}(1+\sqrt{3},\sqrt{8})^{2/3}} $$ See : Upper bound on integral: $\int_1^\infty \frac{dx}{\sqrt{x^3-1}} < 4$.

To get the values of $\Gamma\left(\frac{4}{3}\right)$ and $\Gamma\left(\frac{7}{6}\right)$, you can use the formula $$\Gamma\left(n+\frac{1}{p}\right)=\Gamma\left(\frac{1}{p}\right)\frac{(pn-p+1)!^{(p)}}{p^n}$$