Wikipedia mentions that the number $$a = \dfrac{\Gamma\left(\dfrac{1}{4}\right)}{\pi^{1/4}}$$ is transcendental. Since $\Gamma(1/2) = \sqrt{\pi}$ the above number $a$ seems to connected to a combination of $\Gamma(1/4)$ and $\Gamma(1/2)$. On squaring the connection is clear as we can see that $$a^{2} = \frac{\Gamma^{2}(1/4)}{\Gamma(1/2)} = B(1/4, 1/4)$$ where $B$ represents the beta function.

Are there any other known values of beta function which are transcendental? Any reference to a proof of the transcendence of number $a$ given above would also be helpful.

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    $\begingroup$ To mention the obvious: It is clear from the reflection formula that at least one of $q!$ or $(-q)!$ is transcendental, for $q\in\mathbb Q$. $\endgroup$ – Lucian Jun 28 '15 at 11:01
  • $\begingroup$ @Lucian: Nice remark to deal with certain values of $\Gamma$. $\endgroup$ – Paramanand Singh Jun 28 '15 at 11:09

You have the following result.

Theorem. Let $a$ and $b$ be rational numbers which are not integers and such that $a+b$ is not an integer. Then the number $$B(a,b) = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}=\int_{0}^{1}u^{a-1}(1-u)^{b-1} \,du$$ is transcendental.

Here is, among others, an interesting reference from Michel Waldschmidt: Transcendence of Periods (see p. 6).

For example, $$ \frac{\Gamma\!\left(\dfrac{1}{3}\right)}{\pi^{1/3}}\quad \text{and}\quad \frac{\Gamma\!\left(\dfrac{1}{3}\right)}{\pi^{2/3}}$$ are transcendental (see here).


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