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.


3 Answers 3


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


This isn't a full answer but gives another surprising connection between the two constants. One has $$\gamma = \int_1^\infty \frac{1-\{x\}}{x^2} dx$$ and $$\log \frac{4}{\pi} = \int_1^\infty \frac{\Vert x \Vert}{x^2} dx,$$ where $\{x\} = x-\lfloor x \rfloor$ denotes the fractional part of $x$ and $\Vert x \Vert$ denotes the distance from $x$ to the nearest integer.

One also has $$\int_1^\infty \frac{1-\{x\}}{x^{s+1}} dx = \frac{\zeta(s)-\frac{1}{s-1}}{s}$$ while \begin{align*} \int_1^\infty \frac{\Vert x\Vert}{x^{s+1}}\, dx = \frac{(4-2^{s})\zeta(s-1)+2^{s}-1}{s(s-1)} \end{align*} for all $s \in \mathbb{C}\backslash\{1\}$ with positive real part. These assume the respective limits above at $s = 1$. I found the formulas for $\Vert x \Vert$ by asking, out of curiosity, what $\int_1^\infty \frac{\Vert x\Vert}{x^{s+1}}\, dx$ is. Interestingly, the function $\int_1^\infty \frac{\Vert x\Vert-\tfrac{1}{4}}{x^{s+1}}\, dx$ has analytic continuation to all of $\mathbb{C}$, where $\frac{1}{4} = -3\zeta(-1)$ is the average value of $\Vert x \Vert$ over any of its periods, and one has $$ \int_1^\infty \frac{\Vert x\Vert-\frac{1}{4}}{x}\, dx = -3(\zeta(-1)+\zeta'(-1)) -\frac{13}{12}\log 2 = 3\log A-\frac{13}{12}\log 2 = -0.00464601\ldots,$$ where $A$ is the Glaisher-Kinklein constant. Moreover, the function $ \int_1^\infty \frac{\Vert x\Vert}{x^{s+1}}\, dx$ has meromorphic continuation to $\mathbb{C}$ with a single (simple) pole at $s = 0$ with residue $-3\zeta(-1) = \frac{1}{4}$,

Also, I stumbled on this on Wikipedia: \begin{aligned}\gamma &=\sum _{m=2}^{\infty }(-1)^{m}{\frac {\zeta (m)}{m}}\\&=\log {\frac {4}{\pi }}+\sum _{m=2}^{\infty }(-1)^{m}{\frac {\zeta (m)}{2^{m-1}m}}.\end{aligned}

  • $\begingroup$ Are you sure about this formula? I mean, should not it be just $x$, not $x^2$ in the denominator? $\endgroup$
    – Anixx
    Nov 7, 2023 at 11:11
  • $\begingroup$ Also, check the sign, please $\gamma=-\underset{x\to \infty }{\text{lim}}\left(\int_1^x \frac{1}{t} \, dt-\sum _{k=1}^x \frac{1}{k}\right)$ $\endgroup$
    – Anixx
    Nov 7, 2023 at 11:12
  • $\begingroup$ Yes, I'm sure. The $x^2$ and sign are correct. $\endgroup$ Nov 7, 2023 at 11:54
  • $\begingroup$ Is not integral without the fractional part the same as the sum? $\sum _{k=1}^{\infty } \frac{1}{k^2}=\frac{\pi ^2}{6}$ $\endgroup$
    – Anixx
    Nov 7, 2023 at 12:06
  • $\begingroup$ $\int_1^{\infty } \frac{1}{x^s} \, dx=\frac{1}{s-1}$. How do you get additional $s$ in the denominator? $\endgroup$
    – Anixx
    Nov 7, 2023 at 12:11

This is not an answer per se, but some additional insight. It seems that in certain context there is a meaning in a set of integers with period $2e^{-\gamma}$.

The Chow's EL-numbers are the numbers that can be obtained from $0$ via exponential function, logarithmic function and field operations.

For instance, constants $e,\pi$ and $i$ are all EL-numbers:

\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*}

But what happens if we extend this set with $\lambda=\log(0)$? It is a negatively infinite quantity, equal to the harmonic series with negaive sign:$-\sum_{k=1}^\infty \frac1k$. Its regularized value $-\gamma$.

From the equality $$ \lim_{n\to\infty}\left(\sum_{k=1}^n \frac1k-\int_1^n \frac1tdt\right)=\gamma $$ (or via Laplace transform), $$\int_1^\infty\frac1xdx=-\lambda-\gamma. $$ where the integral $\int_1^\infty\frac1xdx$ is the germ at infinity of the logarithmic function.
If we introduce a new constant $\omega$ as the germ at infinity of the function $f(x)=x$, then (taking into account that $f(x)=\ln x$ has no infinitesimal part at infinity) $$ \int_1^\infty\frac1xdx=\ln\omega $$ and $$ e^{-\lambda}=e^\gamma \omega. $$ Now, $\omega$ is half the numerocity of integers, it corresponds to the numerocity of integers spaced with distance $2$ (like the set of even or odd numbers). So, $e^{-\lambda}=e^\gamma\omega=\omega/(e^{-\gamma})$ is $e^\gamma$ times greater, which corresponds to the numerocity of a set spaced with $2e^{-\gamma}$. Notice in this context that $2\omega/\pi=\omega/(\pi/2)$ is the numerocity of a set spaced with step $\pi$, such as the numerocity of the roots of sine or cosine.

Thus we see that $2e^{-\gamma}$ and $\pi/2$ somehow appear as meaningful periods of a lattice of reals.

Another interesting observation: in $\omega/(\pi/2)$ the denominator is EL-number but the ratio possibly not. But in $\omega/(e^{-\gamma})$ the ratio is EL-number but $e^{-\gamma}$ is not. It turns out that $\omega$ is EL-number iff $\gamma$ is EL-number.


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