# How justify that $\Gamma\left(\frac{1}{6}\right)$ is a transcendental number

Let $\Gamma(x)$ the Gamma function. See, if you need its definiton, for example in this Wikipedia.

While I was evaluating an integral with the help of Wolfram Alpha, I have known that $$\Gamma\left(\frac{1}{6}\right)$$ is a transcendental number.

Question. What's the reasoning to know this fact, that $\Gamma\left(\frac{1}{6}\right)$ is a trancendental number? Thanks in advance.

I presume that it is consequence of a theorem or computational method based on the definition of transcendental numbers, and that the Gamma function is a factorial. How justify that previous real number is trascendental?

• Wikipedia claims that $\Gamma(x)$ is transcendental for $x=\frac16,\frac14,\frac13,\frac23,\frac34,\frac56$ and that they are algebraically independent from $\pi$ (unlike $\Gamma(1/2)$) – Simply Beautiful Art Jul 16 '17 at 13:17
• Many thanks @SimplyBeautifulArt I am going to wait if some user want provide more details. I didn't read Wikipedia's claim. My thoughts were that should be an easy consequence of some proposition. Any case I hope that some user tell us what about it, if it is a difficult question or can be deduced from the standard theory. – user243301 Jul 16 '17 at 13:20
• webusers.imj-prg.fr/~michel.waldschmidt/articles/pdf/… is a survey of transcendentality proofs. – Chappers Jul 16 '17 at 13:28
• Many thanks @Chappers – user243301 Jul 16 '17 at 19:51

$$\Gamma\left(\frac{1}{6}\right) = {\frac{2^{\frac{14}{9}}\cdot 3^{\frac{1}{3}}\cdot \pi^{\frac{5}{6}} }{\text{AGM}\left(1+\sqrt{3},\sqrt{8}\right)^{\frac{2}{3}}}}\tag{1}$$
is proved. The trascendence of $\Gamma\left(\frac{1}{6}\right)$ then follows from recalling classical results (Theorem $7$ here) about the trascendence of periods / complete elliptic integrals.