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The classical Euler-Mascheroni Constant is

$$ \gamma = \lim_{n \to \infty} \left( \sum_{k=1}^n \frac{1}{k} - \ln(n) \right) $$

I'm trying to extend $\gamma$ to higher dimensions denoted by $\gamma_n$ and am curious if a higher dimensional generalization of the Euler-Mascheroni Constant is valid?

$2D$ Representation (Classical $\gamma$)

Graphical representation of $\gamma$

gamma

  • Green Bars: Represent the harmonic series $\sum_{k=1}^n \frac{1}{k}$

  • Orange Area Under Curve: Represents $\int_{1}^{n} \frac{1}{x} \, dx = \ln(n)$

  • Blue Dotted Area: Represents the difference to give the Euler—Mascheroni constant

$$ \underbrace{\gamma}_{\text{blue area}} = \lim_{n \to \infty} \left( \underbrace{\sum_{k=1}^n \frac{1}{k}}_{\text{green area}} - \underbrace{\ln(n)}_{\text{orange area}} \right) $$

Applying similar geometric methods, we can extend it to higher dimensions.

For example, consider the following $3D$ representation.

$3D$ Representation ($\gamma_3$)

Visualize revolving the bars around the $x$-axis to form Gabriel's Wedding Cake (an infinite stack of cylinders with volume $\frac{\pi^3}{6}$) and revolve the curve $y=\frac{1}{x}$ to form Gabriel's Horn (volume $\pi$). The difference between these two volumes represents the Euler—Mascheroni constant $\gamma_3=\frac{\pi^3}{6} - \pi$.

Gabriel's Cake

gcake

Gabriel's Horn

ghorn

Furthermore, each $n$-dimension has its own distinct Euler-Mascheroni constant:

\begin{array}{|c|c|c|} \hline \textbf{n-dimension} & \mathbf{\gamma_n} & \mathbf{\approx} \\ \hline 3 & \hspace{5pt} \pi \left( \zeta(2) - 1 \right) \hspace{5pt} & 2.0261201264...\\ \hline 4 & \hspace{5pt} \frac{2}{3} \pi \left( 2 \, \zeta(3) - 1 \right) \hspace{5pt} & 2.9407690791... \\ \hline 5 & \hspace{5pt} \frac{1}{2} \pi^2 \left( \zeta(4) - \frac{1}{3} \right) \hspace{5pt} & 3.6961170085... \\ \hline 6 & \hspace{5pt} \frac{8}{15} \pi^2 \left( \zeta(5) - \frac{1}{4} \right) \hspace{5pt} & 4.1422216722... \\ \hline 7 & \hspace{5pt} \frac{1}{6} \pi^3 \left( \zeta(6) - \frac{1}{5} \right) \hspace{5pt} & 4.2237941871... \\ \hline 8 & \hspace{5pt} \frac{16}{105} \pi^3 \left( \zeta(7) - \frac{1}{6} \right) \hspace{5pt} & 3.9767533569... \\ \hline \end{array}


Definitions

  • $\gamma$ - classical Euler-Mascheroni constant.
  • $\zeta(s)$ - Riemann zeta function.
  • $\Gamma(s)$ - Gamma function.
  • $V_c$ - Volume of Gabriel's Wedding Cake in $n$-dimensions.
  • $V_h$ - Volume of Gabriel's Horn in $n$-dimensions.
  • $\gamma_n$ - Euler-Mascheroni constant in $n$-dimensions.

Conjecture

For dimension $n \ge 3$, the $n$-dimensional Euler-Mascheroni constant is:

$$ \gamma_n=\frac{\pi^{\frac{n-1}{2}}}{\Gamma\left(\dfrac{n+1}{2}\right)} \left(\zeta(n-1)-\dfrac{1}{n-2}\right) $$

Proof

The volume of an $(n-1)$-dimensional Hyper-sphere with radius $R$ is

$$ V_{n-1}(R) = \frac{\pi^{\frac{n-1}{2}} R^{n-1}}{\Gamma\left( \frac{n+1}{2} \right)} $$

The volume of an $n$-dimensional Hyper-cylinder is

$$ V_{n}(R) = \underbrace{V_{n-1}(R)}_{\text{base}} \times \underbrace{h}_{\text{height}} $$

The volume of Gabriel's Wedding Cake $V_{c}$ with $R_k = \frac{1}{k}$ and $h=1$ is

\begin{align*} V_c &= \sum_{k=1}^{\infty} V_n(R_k) \\ &= \sum_{k=1}^{\infty} \frac{\pi^{\frac{n-1}{2}}}{\Gamma\left(\frac{n+1}{2}\right)} \left(\frac{1}{k}\right)^{n-1} \\ &= \frac{\pi^{\frac{n-1}{2}}}{\Gamma\left(\frac{n+1}{2}\right)} \sum_{k=1}^{\infty} \frac{1}{k^{n-1}} \\ &= \frac{\pi^{\frac{n-1}{2}}}{\Gamma\left(\frac{n+1}{2}\right)} \zeta(n-1) \end{align*}

The volume of the $n$-dimensional Gabriel's Horn $V_h$

\begin{align*} V_h &= \int_1^\infty V_{n-1}\left( \frac{1}{x} \right) \, dx \\ &= \frac{\pi^{\frac{n - 1}{2}}}{\Gamma\left( \frac{n + 1}{2} \right)} \int_1^\infty \frac{1}{x^{n - 1}} \, dx \\ &= \frac{\pi^{\frac{n - 1}{2}}}{\, \Gamma\left( \frac{n + 1}{2} \right)} \left(\frac{1}{n - 2} \right) \end{align*}

The $n$-dimensional Euler-Mascheroni constant $\gamma_n$ is defined as

\begin{align*} \gamma_n &= V_c - V_h \\ &= \frac{\pi^{\frac{n-1}{2}}}{\Gamma\left( \frac{n+1}{2} \right)} \zeta(n-1) - \frac{\pi^{\frac{n - 1}{2}}}{\, \Gamma\left( \frac{n + 1}{2} \right)} \left(\frac{1}{n - 2} \right) \\ &= \frac{\pi^{\frac{n - 1}{2}}}{\Gamma\left( \dfrac{n + 1}{2} \right)} \left( \zeta(n - 1) - \dfrac{1}{n - 2} \right) \quad \blacksquare \end{align*}


References

vengy
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  • What's the reason to choose cylinder in $3$ dimensions? Understood the revolution of the bars, just wondering if the choice of the $3$D shape is unique... – Srini Oct 29 '24 at 00:45
  • Rethinking...this method isn't really linked to the Euler-Mascheroni constant. As noted by @Srini, why use a cylinder shape when a cone/sphere would work too, right? Overall, the method is basically computing the difference between Gabriel's Wedding Cake and Gabriel's Horn in n-dimensions. ;) Anyhow, I learned a lot along the way and thanks for your help! – vengy Oct 29 '24 at 01:57
  • In your figure for the 3D case, I believe the horn should pass through the left-side of the cylinders instead of the right-side to be consistent with your calculation. The picture makes it look like the volume of the horn is larger than the volume of the cylinders which is backwards. I think its natural to extend the rectangles to cylinders since the intersection along the plane of the x-y axis where $z=0$ is consistent with the 2D case. – Steven Clark Oct 29 '24 at 02:12
  • https://math.stackexchange.com/questions/46713/euler-mascheroni-constant-expression-further-simplification?noredirect=1&lq=1#comment3319970_46718 – Mats Granvik Nov 22 '24 at 16:55

1 Answers1

1

Your generalization is valid once you adjust the 2D case to be consistent with the intersection of the x-y plane with Gabriel's horn and the cones in the 3D case as illustrated in Figure (1) below.

Illustration of corrected 2D case

Figure (1): Illustration of corrected 2D case


Since $$\int\limits_1^n \frac{1}{x}-\left(-\frac{1}{x}\right) \, dx=2 \int_1^n \frac{1}{x} \, dx=2 \log(n)\tag{1}$$

the 2D case should be evaluated as

$$\underset{n\to \infty }{\text{lim}}\left(\sum _{k=1}^n \frac{2}{k}-2 \log (n)\right)=2 \gamma\tag{2}$$

which is consistent with your formula result

$$\gamma_2=\underset{n\to 2}{\text{lim}} \left(\frac{\pi^{\frac{n-1}{2}}}{\Gamma\left(\frac{n+1}{2}\right)} \left(\zeta(n-1)-\frac{1}{n-2}\right)\right)=2 \gamma\tag{3}$$

for the case $n=2$.


I believe the function

$$f_\gamma(s)=\frac{\pi^{\frac{s-1}{2}}}{\Gamma\left(\frac{s+1}{2}\right)} \left(\zeta(s-1)-\frac{1}{s-2}\right)\tag{4}$$

is analytic with a removable singularity at $s=2$ and $\lim\limits_{s\to\pm\infty} f_\gamma(s)=0$ both of which are suggested by Figure (1) below.

Illustration of f_gamma(s) for -10<s<30

Figure (1): Illustration of $f_\gamma(s)$ for $-10<s<30$


The function $f_\gamma(s)$ has trivial zeros at the negative odd integers as illustrated in Figure (2) below, but the existence and location of non-trivial zeros is more of a mystery to me.

Illustration of f_gamma(s) for -10<s<0

Figure (2): Illustration of $f_\gamma(s)$ for $-10<s<0$


Steven Clark
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  • So, the $N$-dimensional constant converges to the $2D$ Euler-Mascheroni constant: $$ \lim_{n \to 2^+} \frac{\pi^{\frac{n-1}{2}}}{2, \Gamma\left(\frac{n+1}{2}\right)} \left( \zeta(n-1) - \frac{1}{n-2} \right) = \gamma $$ – vengy Dec 22 '24 at 17:33
  • @vengy The N-dimensional constant evaluated at $N=2$ converges to $\gamma_2=2 \gamma$ per formula (3) above. You've divided formula (3) by 2. It seems to me you're still attempting to evaluate the 2-dimensional case inconsistent with the evaluations for the higher dimensional cases. If you want the 2-dimensional case to evaluate to $\gamma$ instead of $2 \gamma$, then you should also multiply your evaluations for the higher dimensional cases by $\frac{1}{2}$. – Steven Clark Dec 22 '24 at 23:56