# Transcendence of $e^{\frac{\pi^2}{12 \log 2}}$

Is it known whether $e^{\frac{\pi^2}{12 \log 2}}$ is transcendental or algebraic?

This number showed up in this other question.

• exp(pi) is transcendental by Gelfond-Schneider: en.wikipedia.org/wiki/Gelfond%E2%80%93Schneider_theorem – Adam P. Goucher Jan 19 '17 at 11:34
• What is the motivation behind this question ? – Sylvain JULIEN Jan 19 '17 at 11:49
• @Sylvain, there was a question maybe yesterday about algebraic non-examples of a result of Levy on continued fractions, and this question came up there. (But OP should have linked to it) – Gerry Myerson Jan 19 '17 at 11:52
• $e^\pi=(-1)^{-i}$ so it's transcendental. The open one is $\pi^e$ but I don't think that $\pi^e$ has some profound meaning, it's just something that looks as silly as $e^\pi$. – Kevin Buzzard Jan 19 '17 at 11:54
• =3.275822918721811159787681882453843863608475525982374149405198924190723215644960355... Clearly trascendental! :) – Yaakov Baruch Jan 19 '17 at 13:01

This is most likely open, since alredy $e^{\pi^2}$ is not known to be transcendental.
As an added difficulty, I don't think that $\frac{\pi^2}{12 \log 2}$ is known to be transcendental either.
There are very few, very limited, tricks to prove this kind of result: things like taking $(-1)^{-i}$ and $i^i$ and applying Gelfond–Schneider, or building the Weierstrass $\wp$-function of $\mathbb{Q}(\sqrt{-d})$ to get the transcendence of $e^{\pi\sqrt{d}}$ from its invariants.
• If $\pi^2/12\log 2$ were algebraic, its exponential would be transcendental, so I don't see why it's a difficulty :P – Wojowu Jan 20 '17 at 6:02