This question is based on Chow - What is a closed-form number?.

The author of the linked paper had proposed a plausible definition of "elementary numbers" (which he calls "EL-numbers") concept, building it analogously to the concept of elementary function (author admits, "elementary number" would be a better name for his proposal, but the term is already occupied).

So, the author defines a set of $\mathbb{E}$ of "EL numbers" which stands for "elementary" and well as "exponentially-logarithmic".

He defines the set as any numbers that can be produced by applying finite number of field operations, exponential and logarithmic functions to the number $0$.

For instance, in his system \begin{gather*} 1=\exp(0) \\ e=\exp(\exp(0)) \\ i=\exp\left(\frac{\log(-1)}2\right)=\exp\left(\frac{\log(0-\exp(0))}{\exp(0)+\exp(0)}\right) \\ \pi=-i\log(-1)=-\exp\left(\frac{\log(0-\exp(0))}{\exp(0)+\exp(0)}\right)\log(0-\exp(0)). \end{gather*}

It turns out that any root of a polynomial with rational coefficients, expressible in the radicals, is also in $\mathbb{E}$.

So, my question is, does the Euler–Mascheroni constant $\gamma$ belong to the EL-numbers? I think no, but that it is "nearly-elementary" in the same way as the digamma function is a nearly-elementary function.

My thoughts on this revolve around these points:

  1. There is symmetry between $\pi/4$ and $e^{-\gamma}$

  2. $\gamma=\psi(1)$, but $\psi(x)$ is antidifference of $1/x$ while logarithm is antiderivative. $\psi(x)$ relates to $\log x$ the same way as Bernoulli polynomials relate to monomials (both Bernoulli polynomials and $\psi(x)$ are slices of Hurwitz Zeta function).

  3. Many divergent integrals and their logarithms regularize to $\gamma$, particularly, $\operatorname{reg} \int_0^1 \frac1x dx=\gamma$ (which makes it in some sense the regularized value of logarithm at zero).

  • $\begingroup$ It seems odd to write "in his system"—the equalities you propose are just equalities (upon making suitable choices of branch for the logarithm function), not dependent on working in any particular system. It seems like you might mean instead "For instance, … show that $1$, $e$, $i$, and $\pi$ are all EL-numbers." $\endgroup$
    – LSpice
    Mar 27, 2021 at 16:31
  • $\begingroup$ Another relevant link: cp4space.hatsya.com/2020/10/17/closed-form-numbers $\endgroup$
    – Anixx
    Mar 27, 2021 at 16:34
  • 3
    $\begingroup$ Doesn't answer the question but is related: using the theory of exponential motives Fresan and Jossen argue that an analogue of Grothendieck period conjecture for exponential periods implies that $\gamma$ is algebraically independent of $2\pi i$. See Corollary 12.8.8 here. $\endgroup$
    – Wojowu
    Mar 27, 2021 at 17:21

1 Answer 1


Despite the simplicity and elegance of the definition of EL numbers, it is very hard to prove that an explicit number is not EL, as Chow points out throughout the paper.

In particular, I am certain that as of right now, nobody knows if the Euler-Mascheroni constant is or isn’t EL. It most likely isn’t, but we can’t even prove that it is irrational.

  • 1
    $\begingroup$ My take on this is that the divergent integral $\int_0^1\frac1xdx$ seems to regularize to $\gamma$, and at the same time $\int_0^1\frac1xdx=\gamma+\int_1^\infty\frac1xdx$, so in a very fuse sense, $\gamma=-\log 0 - \log\infty$ $\endgroup$
    – Anixx
    Mar 27, 2021 at 17:03
  • 2
    $\begingroup$ Of course, the both operations are prohibited in the EL-numbers, though :-) $\endgroup$
    – Anixx
    Mar 27, 2021 at 17:04
  • $\begingroup$ The question is mostly about special algebraic roles of $\gamma$. Of course, if $\gamma$ is rational (which is highly unlikely) it is in EL, but I am interested in other possibilities to express it. $\endgroup$
    – Anixx
    Mar 27, 2021 at 17:07
  • 1
    $\begingroup$ Particularly, given there is symmetry between $\pi/4$ and $e^{-\gamma}$ mathoverflow.net/questions/341470/… $\endgroup$
    – Anixx
    Mar 27, 2021 at 17:08
  • 10
    $\begingroup$ @Anixx Alon Amit is correct. There is a "proof by authority" that the answer to your question is unknown. The proof splits into two cases. The first case is that $\gamma$ has been proved to not be in EL. This would yield a proof that $\gamma$ is irrational, which is unknown; contradiction. The second case is that $\gamma$ has been proved to be in EL. This would imply a finite relation between $\gamma$ and $e$, which would rank as among the most astounding mathematical results of all time. We would all have heard about it. But we haven't; contradiction. The proof is complete. $\endgroup$ Mar 27, 2021 at 22:41

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