Here is a couple of examples of the similarity from Wikipedia, in which the expressions differ only in signs. I encountered other analogies as well.

$${\begin{aligned}\gamma &=\int _{0}^{1}\int _{0}^{1}{\frac {x-1}{(1-xy)\ln xy}}\,dx\,dy\\&=\sum _{n=1}^{\infty }\left({\frac {1}{n}}-\ln {\frac {n+1}{n}}\right).\end{aligned}}$$

$${\begin{aligned}\ln {\frac {4}{\pi }}&=\int _{0}^{1}\int _{0}^{1}{\frac {x-1}{(1+xy)\ln xy}}\,dx\,dy\\&=\sum _{n=1}^{\infty }\left((-1)^{n-1}\left({\frac {1}{n}}-\ln {\frac {n+1}{n}}\right)\right).\end{aligned}}$$

$${\begin{aligned}\gamma &=\sum _{n=1}^{\infty }{\frac {N_{1}(n)+N_{0}(n)}{2n(2n+1)}}\\\ln {\frac {4}{\pi }}&=\sum _{n=1}^{\infty }{\frac {N_{1}(n)-N_{0}(n)}{2n(2n+1)}},\end{aligned}}$$

(where $N_1(n)$ and $N_0(n)$ are the number of 1's and 0's, respectively, in the binary expansion of $n$).

I wonder whether is there any algebraic system where $4e^{-\gamma}$ would play a role similar to what $\pi$ plays, say in complex numbers, or a geometric system where $4e^{-\gamma}$ would play some special role, like $\pi$ in Euclidean and Riemannian geometries.


1 Answer 1


The intuition may be helped by considering the generalized Euler constant function $$\gamma(z)=\sum_{n=1}^\infty z^{n-1}\left(\frac{1}{n}-\ln\frac{n+1}{n}\right),\;\;|z|\leq 1.$$ Its values include the Euler constant $\gamma=\gamma(1)$ and the "alternating Euler constant" $\ln 4/\pi=\gamma(-1)$. So any general integral formula or recursion relation for $\gamma(z)$ will establish a connection of the type noted in the OP.

The properties of the function $\gamma(z)$ have been studied in The generalized-Euler-constant function and a generalization of Somos's quadratic recurrence constant (2007). Somos's constant $\sigma=\sqrt{1\sqrt{2\sqrt{3\cdots}}}$ is obtained as $\gamma(1/2)=2\ln(2/\sigma)$.

Another special value $$\gamma(i)=\frac{\pi}{4}-\ln\frac{\Gamma(1/4)^2}{\pi\sqrt{2\pi}}+i\ln\frac{8\sqrt\pi}{\Gamma(1/4)^2}.$$

  • $\begingroup$ Interesting! I googled this paper: projecteuclid.org/euclid.bbms/1290608199 Following that formula and the rules of the theory I am currently working on extended.fandom.com/wiki/Extended_Wiki the euler's constant function can be represented as $$\gamma(a)-1=\operatorname{reg}\left( \left(a+\omega _-\right) \log \left(\frac{a+\omega _-}{a+\omega _+}\right)\right)$$, which is to be compared to another identity $$\operatorname{reg}\frac1{\pi }\ln \left(\frac{\omega _--\frac{z}{\pi }}{\omega _++\frac{z}{\pi }}\right)=\cot z$$. $\endgroup$
    – Anixx
    Sep 13, 2019 at 12:58
  • $\begingroup$ Another consideration here is that $$\operatorname{reg}\left( \log \left(\frac{a+\omega _-}{a+\omega _+}\right)\right)=-\frac1a$$, so the formula becomes even simplier: $$\gamma(a)=\operatorname{reg}\left( \omega _- \log \left(\frac{a+\omega _-}{a+\omega _+}\right)\right)$$ $\endgroup$
    – Anixx
    Sep 13, 2019 at 13:05
  • 1
    $\begingroup$ Thus, $$\gamma = \operatorname{reg}\left( \omega _- \ln \left(\frac{\omega _+}{\omega _++1}\right)\right)$$ and $$\ln \pi/4=\operatorname{reg}\left( \omega _- \ln \left(\frac{\omega _-}{\omega _--1}\right)\right)$$ But at the same time, $$\operatorname{reg} \ln \omega_+ =-\gamma$$ I wonder if there is a similarly simple formula for $\pi$ though... $\endgroup$
    – Anixx
    Sep 13, 2019 at 13:24
  • $\begingroup$ I still would like to see something else, maybe something geometric. $\endgroup$
    – Anixx
    Sep 23, 2019 at 20:29

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