# Ratio of Gamma Functions

Is it possible to show that: \begin{align} \frac{\Gamma\left(\frac{1}{78}\right) \Gamma\left(\frac{29}{78}\right) \Gamma\left(\frac{35}{78}\right) \Gamma\left(\frac{53}{78}\right) \Gamma\left(\frac{55}{78}\right) \Gamma\left(\frac{61}{78}\right)}{\Gamma\left(\frac{2}{78}\right) \Gamma\left(\frac{28}{78}\right) \Gamma\left(\frac{32}{78}\right) \Gamma\left(\frac{44}{78}\right) \Gamma\left(\frac{58}{78}\right) \Gamma\left(\frac{70}{78}\right) } = \sqrt{3} \end{align}

There are other known ratios of Gamma functions, but as always there seems to be a product rule or trick to evaluate the ratios.

• Well yes, else you wouldn't have that particular identity to post. Not to be condescending, but posting a reference, your own work, and motivation would be appreciated. – Zach466920 Jun 21 '15 at 16:00
• $$\frac{\Gamma(1/34)\Gamma(9/34)\Gamma(13/34) \Gamma(15/34)\Gamma(19/34)\Gamma(21/34) \Gamma(25/34)\Gamma(33/34)}{\Gamma(3/34)\Gamma(5/34) \Gamma(7/34)\Gamma(11/34)\Gamma(23/34) \Gamma(27/34)\Gamma(29/34)\Gamma(31/34)} = 1 .$$ See Amer. Math. Monthly, November 2010, page 842 – Dietrich Burde Jun 21 '15 at 16:09

One method to find such formulas is the duplication formula $$\Gamma(z) \Gamma(z+1/2) = 2^{1-2z} \sqrt{\pi} \Gamma(2z),$$ and apply it to nominator and denominator, e.g., to obtain $${\Gamma(1/8) \Gamma(5/8) \Gamma(6/8) \over \Gamma(2/8) \Gamma(3/8) \Gamma(7/8)} = \sqrt{2}.$$ There seem to be more advanced ideas in Deligne, P. Valeurs de fonctions L et périodes d'intégrales, Amer. Math. Soc., Providence, R.I., 1979.