# Simplify ratio of Gamma Functions

Is it possible to simplify the following ratio of Gamma functions:

$$r\equiv-\frac{\Gamma \left(-\frac{2}{7}\right) \Gamma \left(\frac{6}{7}\right) \Gamma \left(\frac{10}{7}\right)}{\Gamma \left(-\frac{3}{7}\right) \Gamma \left(\frac{1}{7}\right) \Gamma \left(\frac{16}{7}\right)}$$

?

I managed to reduce it to a somehow more symmetric expression of the form $$r=\frac{49}{36}\frac{\Gamma \left(-\frac{2}{7}\right)}{\Gamma(\frac{2}{7})}\frac{\Gamma\left(\frac{3}{7}\right)}{\Gamma\left(-\frac{3}{7}\right)}\frac{\Gamma\left(\frac{6}{7}\right)}{\Gamma\left(-\frac{6}{7}\right)}$$

but I haven't managed to simplify it further.

Thanks in advance for any help!

• You might enjoy this post. The paper therein gives a lot of tools that might be used to simplify ratios of Gamma values at fractional values. Unfortunately $\Gamma(k/7)$ doesn't appear to be something discussed. – Mason Feb 13 '20 at 4:27

I can't see my way through to something nice. Using the recurrence relation for the Gamma function I get $$-\frac{\operatorname{\Gamma}\left(-\frac27\right)\operatorname{\Gamma}\left(\frac67\right)\operatorname{\Gamma}\left(\frac{10}7\right)}{\operatorname{\Gamma}\left(-\frac37\right)\operatorname{\Gamma}\left(\frac17\right)\operatorname{\Gamma}\left(\frac{16}7\right)}=-\frac74\frac{\operatorname{\Gamma}\left(\frac57\right)\operatorname{\Gamma}\left(\frac67\right)\operatorname{\Gamma}\left(\frac37\right)}{\operatorname{\Gamma}\left(\frac47\right)\operatorname{\Gamma}\left(\frac17\right)\operatorname{\Gamma}\left(\frac27\right)}$$ Quadratic nonresidues $$\pmod{7}$$ on top and residues on the bottom. You could multiply numerator and denominator by what is now the numerator and use the multiplication formula for the Gamma function to get $$-\frac{\operatorname{\Gamma}\left(-\frac27\right)\operatorname{\Gamma}\left(\frac67\right)\operatorname{\Gamma}\left(\frac{10}7\right)}{\operatorname{\Gamma}\left(-\frac37\right)\operatorname{\Gamma}\left(\frac17\right)\operatorname{\Gamma}\left(\frac{16}7\right)}=-\frac{7\sqrt7}{32\pi^3}\left(\operatorname{\Gamma}\left(\frac57\right)\operatorname{\Gamma}\left(\frac67\right)\operatorname{\Gamma}\left(\frac37\right)\right)^2$$ But if you got a pretty result one might anticipate an even prettier one for $$\operatorname{\Gamma}\left(\frac13\right)$$ and there is something with a complete elliptic integral of the first kind but it's not very encouraging.

• @user12588 Agreed. – user5713492 Feb 13 '20 at 4:19