# Is there any deep philosophy or intuition behind the similarity between $\pi/4$ and $e^{-\gamma}$?

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.

• – Lucian Apr 23 at 15:58

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 $$\gamma(i)=\frac{\pi}{4}-\ln\frac{\Gamma(1/4)^2}{\pi\sqrt{2\pi}}+i\ln\frac{8\sqrt\pi}{\Gamma(1/4)^2}.$$
• 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 $$\operatorname{reg}\frac1{\pi }\ln \left(\frac{\omega _--\frac{z}{\pi }}{\omega _++\frac{z}{\pi }}\right)=\cot z$$. – Anixx Sep 13 '19 at 12:58
• Another consideration here is that $$\operatorname{reg}\left( \log \left(\frac{a+\omega _-}{a+\omega _+}\right)\right)=-\frac1a$$, so the formula becomes even simplier: $$\gamma(a)=\operatorname{reg}\left( \omega _- \log \left(\frac{a+\omega _-}{a+\omega _+}\right)\right)$$ – Anixx Sep 13 '19 at 13:05
• 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... – Anixx Sep 13 '19 at 13:24