Definitions
- $\gamma$ is the classical Euler-Mascheroni constant.
- $\zeta(s)$ is the Riemann zeta function.
- $\Gamma(s)$ is the Gamma function.
Question
From the Claim Results below for $n=2$, I was surprised to find $\gamma_2 = 2\gamma$.
Thus, the Euler-Mascheroni constant $\large \gamma=\frac{\gamma_2}{2}$ can be expressed as
$ \large \bbox[10px,border:2px solid #0000ff]{\gamma =\frac{\pi^{\frac{1}{2}}}{2 \, \Gamma\left(\frac{3}{2}\right)} \lim\limits_{s\to 2}\left(\zeta(s-1)-\dfrac{1}{s-2}\right)} $
Is there a way to numerically verify that expression correctly approximates the classical Euler-Mascheroni constant $\gamma \approx 0.5772156649...$?
Claim
The $n$-dimensional Euler-Mascheroni constant, $\gamma_n$, is given by:
$$ \gamma_n = \begin{cases} \displaystyle \gamma, & \quad n = 1 \\[12pt] \displaystyle \frac{\pi^{\frac{n - 1}{2}}}{\Gamma\left( \dfrac{n + 1}{2} \right)} \lim\limits_{s \to n} \left( \zeta(s - 1) - \frac{1}{s - 2} \right), & \quad n \ge 2 \end{cases} $$
Proof
Case: $n \ge 3$
The volume of an $n$-dimensional hypersphere is
$$ V_n(R) = \frac{\pi^{n/2} R^n}{\Gamma\left(1 + \frac{n}{2}\right)} $$
For $n-1$ dimensions with radius $R=\frac{1}{k}$ is
$$ V_{n-1}(\frac{1}{k}) = \frac{\pi^{\frac{n-1}{2}}}{k^{n-1}\Gamma\left( \frac{n+1}{2} \right)} $$
Total volume of the Stack of Hyperspherical Slices $V_s$
$$ V_s = \sum_{k=1}^\infty V_{n-1}\left( \dfrac{1}{k} \right) = \frac{\pi^{\frac{n - 1}{2}}}{\Gamma\left( \dfrac{n + 1}{2} \right)} \sum_{k=1}^\infty \dfrac{1}{k^{n - 1}} = \frac{\pi^{\frac{n - 1}{2}} \, \zeta(n - 1)}{\Gamma\left( \dfrac{n + 1}{2} \right)} $$
Total volume of the $n$-dimensional Gabriel's Horn $V_h$
$$ V_h = \int_1^\infty V_{n-1}\left( \dfrac{1}{x} \right) \, dx = \frac{\pi^{\frac{n - 1}{2}}}{\Gamma\left( \dfrac{n + 1}{2} \right)} \int_1^\infty \dfrac{1}{x^{n - 1}} \, dx = \frac{\pi^{\frac{n - 1}{2}}}{(n - 2) \, \Gamma\left( \dfrac{n + 1}{2} \right)} $$
Define $\gamma_n$ as
$$ \gamma_n = V_s - V_h = \frac{\pi^{\frac{n - 1}{2}}}{\Gamma\left( \dfrac{n + 1}{2} \right)} \left( \zeta(n - 1) - \dfrac{1}{n - 2} \right) $$
Case: $n = 2$
To avoid the problematic terms $\zeta(1)$ and $\frac{1}{0}$, reformulate $\gamma_n$ as
$$ \gamma_n=\frac{\pi^{\frac{n-1}{2}}}{\Gamma\left(\dfrac{n+1}{2}\right)} \lim\limits_{s\to n}\left(\zeta(s-1)-\dfrac{1}{s-2}\right) $$
Case: $n = 1$
Define the $1$-dimensional Euler-Mascheroni constant as $\gamma_1 = \gamma$.
Claim Results
\begin{array}{|c|c|c|} \hline \textbf{n-dimension} & \mathbf{\gamma_n} & \mathbf{\approx} \\ \hline 1 & \hspace{5pt} \gamma \hspace{5pt} & 0.5772156649... \\ \hline 2 & \hspace{5pt} 2\gamma \hspace{5pt} & 1.1544313298... \\ \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}
Interestingly, $0$-dimensional $\gamma_0 = \frac{5}{12\pi}$
