A question on the Gamma function for real argument

Question

The Gamma Function for real argument being defined by $$\Gamma(x)=\int_0^{\infty} t^{x-1} e^{-t}dt$$ Prove the equality $$\Gamma(\frac 23)\Gamma(\frac 76)=\Gamma(\frac 53)\Gamma(\frac 16)$$ Remarking that $\frac 23+\frac 76=\frac {11}{6}=\frac53+\frac 16$ it appears the following question: is it true that $$\Gamma(a)\Gamma(b)=\Gamma(c)\Gamma(d)\space\text{when}\space a+b=c+d?$$

Answer 1

This is not true.

The basic idea is that $\Gamma(x+1) =x\Gamma(x) $.

Therefore, $\Gamma(\frac53) =\frac23 \Gamma(\frac23) $ and $\Gamma(\frac76) =\frac16 \Gamma(\frac16) $ so that

$\begin{array}\\ \Gamma(\frac 23)\Gamma(\frac 76)-\Gamma(\frac 53)\Gamma(\frac 16) &=\Gamma(\frac 23)\frac16 \Gamma(\frac16)-\frac23 \Gamma(\frac23)\Gamma(\frac 16)\\ &=(\frac16-\frac23)\Gamma(\frac 23) \Gamma(\frac16)\\ &=-\frac12\Gamma(\frac 23) \Gamma(\frac16)\\ &\approx -3.768724\\ &\ne 0 \end{array} $