Geometric / physical / probabilistic interpretations of Riemann zeta(n>1)?

Question

What are some physical, geometric, or probabilistic interpretations of the values of the Riemann zeta function at the positive integers greater than one?

I've found some examples:

1) In MO-Q111339 on a Tamagawa number, GH states

$$\mathrm{vol}(\mathrm{SL}_2(\mathbb{R})/\mathrm{SL}_2(\mathbb{Z}))=\zeta(2).$$

2) In "Quantum Gauge Theories in Two Dimensions," Edward Witten derives

$$\mathrm{vol}(\mathcal M)=\frac{2}{(\sqrt{2}\:\pi)^{2g-2}}\zeta(2g-2)$$

from a volume form for the moduli space $\mathcal M$ of flat connections on a gauge group ($G=SU(2)$) bundle over a compact two-dimensional manifold, a Riemann surface of genus $g$, and, for a connected sum of an orientable surface of genus $g$ with $k$ Klein bottles and $r$ copies of the projective plane $RP^2$, he derives

$$\mathrm{vol}(\mathcal M)=\frac{2(1-2^{1-(2g-2+2k+r)})}{(\sqrt{2}\:\pi)^{2g-2+2k+r}} \zeta(2g-2+2k+r).$$

3) In Wikipedia on the Stefan-Boltzmann law, the black body irradiance (total energy radiated per unit surface area of a black body per unit time) is given as

$$j^{*}=2\pi\:3!\zeta(4)\:\frac{(kT)^{4}}{c^{2}h^{3}}.$$

(In n-dimensional space, it's proportional to $n!\zeta(n+1)$, and Planck's law for the electromagnetic energy density inside the 3-D black body has an extra factor of $4/c$.)

4) In "Feynman's Sunshine Numbers," David Broadhurst gives the rate per unit surface area at which a black body at temperature $T$ emits photons as

$$2\pi\:2!\zeta(3)\:\frac{(kT)^{3}}{c^{2}h^{3}}.$$

(And the density of photons inside the body has an extra factor of $4/c$.)

Motivation: I'm motivated not only by general interest, but also by MO-Q111165 and MO-Q111770. Determinants (volumes) of adjacency matrices and, therefore, the cycle index polynomials (CIPs) for the symmetric group pop up in statistical physics, e.g., in Potts q-color field theory and scaling random cluster model, and the CIPS can be "rescaled" to obtain the complete Bell polynomials (OEIS-A036040) which are related to the cumulant expansion polynomials (OEIS-A127671), both of which are related to statistical correlations and their diagrammatics (see references in OEIS-A036040).

5) The $p_n(z)$ of MO-Q111165 seem formally related to the Chern classes $c_{k}(V)$ of a direct (infinite) sum of line bundles $\:\:\:\: V=L_{1}\oplus L_2\oplus ...\:.$ :

With $x_{i}=c_{1}(L_i)$, the first Chern classes,

$$p_k(z)=k!\:c_{k}(V)=k!\:e_{k}(x_{1},x_{2}, ...),$$

where $e_{k}$ are elementary symmetric polynomials. The $\zeta(n)$ can be identified as the power sums of the first Chern classes, and then, for example,

$$3!\:c_{3}(V)=p_3(z)=(z+\gamma)^3-3\zeta(2)(z+\gamma)+2\zeta(3)$$ $$4!\:c_{4}(V)=p_4(z)=(z+\gamma)^4-6\zeta(2)(z+\gamma)^2+8\zeta(3)(z+\gamma)+3[\zeta^2(2)-2\zeta(4)].$$

Update (Nov. 16, 2012): Just found the sequence in a thesis by R. Lu, "Regularized Equivariant Euler Classes and Gamma Functions," which discusses the relationship to Chern and Pontrjagin classes.

See also "An integral lift of the Gamma-genus" and "The motivic Thom isomorphism" by Jack Morava and "Hodge theoretic aspects of mirror symmetry" by L. Katzarkov, M. Kontsevich, and T. Pantev.

Answer 1

Bourgade, Fujita & Yor shows to get Zeta functions from Cauchy Random Variables for even values and the $\chi_4$ L-functions for odd values. For some reason they always come in this pair.

This proof is simplified by Luigi Pace for $\zeta(2)$. The Cauchy Random variable is $$ p_X (x) = \frac{2}{1+x^2}$$

when we look at the ration of two such random variables $Y = X/X'$. $$ p_Y(y) = \frac{4}{\pi^2} \frac{\log y}{y^2-1}$$ Then observe $\mathbb{P}(Y \geq 1) = \mathbb{P}(X < X') = \frac{1}{2}$. So they compute $$ \sum_{k=0}^\infty \frac{1}{(2k+1)^2}= \int_0^1 \frac{-\log y}{1 - y^2} = \mathbb{P}(Y \geq 1)= \frac{\pi^2}{8}$$

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I learned through a blog a proof using 2D Brownian motion at least for the case $\zeta(2)$.

Suppose that $f: \mathbb{C} \to \mathbb{C}$ is an analytic function on the neighbourhood of the unit disk. This
function maps the unit disk to with boundary where . A two dimensional brownian motion started at $f(0)$ takes on average time $$ \mathbb{E}[\tau] = \sum_{k \geq 1} |a_k|^2 $$ to exit domain $f(\mathbb{D})$ where $f(z) = \sum_{k \geq 0} a_k z^k$ and $\tau = \inf \{ t > 0: B_t \in \partial f(\mathbb{D}) \}$ is the hitting time of the boundary .

You can get $\zeta(2)$ by considering Brownian motion on the strip $\{ x+iy: |x| < \pi/2 \}$ and evaluating the left and right sides. The Brownian motion exit time is $\tau = \pi^2/4$ and $$f(z) = \log\left(\frac{1-z}{1+z}\right) = -2\left(z + \frac{z^3}{3} + \frac{z^5}{5} + \dots \right)$$ maps the strip to the unit disk.

This style is traced to the arXiv article by Greg Markowsky.

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Also check out this paper by Noam Elkies who relates them to Alternating permutations. One can show:

\begin{eqnarray*} \sum_{k=0}^\infty \frac{1}{(2k+1)^2} &=& \sum_{k= 0}^\infty \int_0^1 \int_0^1 (xy)^{2k}dx\, dy \\\\ &=& \int_0^1 \int_0^1 \left( \sum_{k= 0}^\infty(xy)^{2k} \right)dx \, dy = \int_0^1 \int_0^1 \frac{ dx \, dy}{1 - (xy)^2} \end{eqnarray*} Then he does the strange Calabi substitution: \[ x = \frac{\sin u}{\cos v} ,y = \frac{\sin v }{\cos u} \]

and recovers a calculus identity: \[ \int_0^1 \int_0^1 \frac{ dx \, dy}{1 - (xy)^2} = \int_{u+v < \pi/2} 1 \, du \, dv = \frac{\pi^2}{8} \]

This proof is extended to higher dimensions in Elkies' paper.

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You can then study the transform $T: L^2[0,\pi/2] \to L^2[0,\pi/2]$, the characteristic function of a triangle.

\[ (Tf)(x)=\int_0^{\pi/2 -x} f(t) \, dt \]

and ask when does $Tf = \lambda f$. The spectrum of this operator is

\[ \lambda = \frac{1}{4k+1} , f_\lambda(x) = cos (4k+1)u \]

Then one can take the trace of $T^n$ and compare to the volume of a polytope:

\begin{eqnarray} \sum_{k=-\infty}^\infty \frac{1}{(4k+1)^k}&=& \sum_\lambda \langle f |T^n | f \rangle \\\\ & =& \mathrm{Vol}\bigg(\{0 < x_1 > x_2 < x_3 > \dots < x_{n-1} > x_n > \frac{\pi}{2}\}\bigg) \end{eqnarray} The volume of this polytope can be expressed in terms of alternating permutations.

I first learned of this iterated integral idea in Stanley's survey on Alternating Permutations, but also in some papers by Chebikin on Parking Functions, this seems to be an example of a chain polytope.

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What other L-functions can take neat values like $L(k) \in \mathbb{Q}\pi^k$ where $k \in \mathbb{Z}$ ? Possibly need an algebraic extension $K / \mathbb{Q}$.

Answer 2

Here's a physical interpretation of ζ(2): Suppose you're stuck in stopped traffic on a long road, and that you're standing just in front of your car. Suppose further that all the cars behind you are blowing their horns at you. Then the noise you hear is ζ(2) times louder than the sound of the first car's horn. Of course, there's a similar visual interpretation, maybe more realistic, using the illumination given by a long string of Christmas lights.

Answer 3

Here is a generalization of GH's answer concerning Tamagawa numbers:

Reference Paul Garett's notes: http://www.math.umn.edu/~garrett/m/v/volumes.pdf

$$ vol(SL(n,\mathbb{Z}) \backslash SL(n, \mathbb{R})) = \zeta(2) \zeta(3) \zeta(4) \zeta(5) \cdots \zeta(n).$$

$$vol(Sp(n,\mathbb{Z})\backslash Sp(n, \mathbb{R})) = \zeta(2) \zeta(4) \zeta(6) \zeta(8) \cdots \zeta(2n)$$

This works for number field and global function fields analogous. There you have to replace Riemann zeta by the Dedekind zeta or Hasse-Weil zeta function and additionally root numbers show up.

Answer 4

I wrote an article about this very subject titled Zeta Values in Geometry and Topology three years ago. My thinking on the points in the article has evolved, in particular, I'm fairly convinced that Questions 0.1-0.4 aren't fruitful lines of inquiry. Still, the material therein is fascinating to me.

Answer 5

This isn't about the Riemann zeta function, but instead has to do with the zeta function of a number field $K$. But I found it unexpected. Let $\zeta_K(s)=\sum_{\mathfrak a\ne0} N\mathfrak a^{-s}$ be the usual zeta function, where we sum over all non-zero integral ideals, and let $\zeta_K^{\text{prin}}(s) = \sum_{(\alpha)\ne0} N\alpha^{-s}$ be the same, except now we sum only over principal ideals. Then the probability that an elliptic curve over $K$ has a global minimal Weierstrass equation is more-or-less equal to $$\frac{\zeta_K^{\text{prin}}(10)}{\zeta_K(10)},$$ where the probability is counted by taking elliptic curve $y^2=x^3+Ax+B$ with $A$ and $B$ in appropriate boxes. See:

• The density of elliptic curves having a global minimal Weierstrass equation, Ebru Bekyel, Journal of Number Theory, Volume 109, Issue 1, November 2004, Pages 41–58 doi:10.1016/j.jnt.2004.06.003

Answer 6

You asked for a geometric interpretation. Let $\gamma_\pm(a)$ be the two geometric shapes described

by the transcendental equation $e^{-y}\pm~e^{-\large x^a}=1$. Then their areas for $~a=\dfrac1n~$ are

$$ \begin{align} A_+&=-\int_0^\infty\ln\Big(1-e^{-\large x^a}\Big)dx=\Gamma(1+n)\cdot\zeta(1+n),\\ A_-&=-\int_0^\infty\ln\Big(1+e^{-\large x^a}\Big)dx=\Gamma(1+n)\cdot\zeta(1+n)\cdot(1-2^{-n}). \end{align} $$

Since the various polynomial curves, such as circles $(x^2+y^2=r^2)$, ellipses & hyperbolae

$\bigg(\dfrac xa\bigg)^2\pm\bigg(\dfrac yb\bigg)^2=1$, superellipses $\bigg(\dfrac xa\bigg)^n+\bigg(\dfrac yb\bigg)^m=1$, etc., have been studied for

centuries, it seemed only natural to expand this line of thought by inquiring about the properties

of exponential ones, such as $a^{\large x^n}\pm b^{\large y^m}=1$.

Answer 7

Elaborating on Nash's comment:

Oliver, special case of Zipf's law, right? That leads to the Zipf–Mandelbrot law that has a probability mass function of $$f(k;N,1,s)=\displaystyle\frac{\frac{1}{(k+1)^s}}{\sum_{i=1}^{N}\frac{1}{(i+1)^s}}$$ and then back to $\mathrm{vol}(\mathcal M)$ for the Klein bottles and particle statistics through $$(1-2^{1-s})\zeta(s)=\sum_{n=0}^{\infty } \frac{1}{2^{n+1}} \sum_{k=0}^{n}(-1)^k \binom{n}{k}\frac{1}{(k+1)^s}$$ $$=\eta(s)=\int_{0}^{\infty }\frac{1}{\exp(x)+1}\frac{x^{s-1}}{(s-1)!}dx$$

where $\eta(s)$ is the Dirichlet eta function, and so the Klein bottle manifolds seem connected to fermions and Fermi-Dirac statistics (as apropos Möbius twists), whereas the orientable Riemann manifolds seem related to bosons and Bose-Einstein statistics.

And, Alan Gut in "Some remarks on the zeta distribution" defines the random variable $U$ with probability mass function (choose your favorite $\sigma= 2, 3, ...$)

$$P(U_\sigma)=\frac{1}{\zeta(\sigma)n^\sigma}$$

and says, "The main point is that, for $\sigma>1$, one can view the normalized zeta function $\varphi_{\sigma}(t)=\frac{\zeta(\sigma\:+\:i\:t)) }{\zeta(\sigma)}$ as the characteristic function of, as it turns out, a compound Poisson distribution. "

He shows how the moments and cumulants of the distribution (related to OEIS A036040 and A127671) given as functions of $\zeta(\sigma)$ and its derivatives are related to the von Mangoldt and Moebius functions and re-derives (and extends) an identity due to Selberg.

On a tangent, the zeta values can be used to translate the Gamma-genus:

With $$R_z = z+\gamma + \sum_{n=1}^{\infty } (-1)^n\zeta (n+1)(d/dz)^n,$$

then $$\displaystyle \exp(\omega\:R_z)\frac{e^{(t\:z)}}{t!}=\exp{(\omega\:d/dt)}\frac{e^{(t\:z)}}{t!}=\frac{e^{((t+\omega)\:z)}}{(t+\omega)!}$$

Answer 8

Hint: A literally enlightning geometric presentation of the Basel-problem at YouTube is the video

The stunning geometry behind this surprising equation by 3Blue1Brown

uploaded a short time ago. It needs a little physics, a little mathematics and a lot of creativity to derive this famous result completely different than Euler's proof. I also think it is pedagogically valuable and could be presented in many classes.

Answer 9

The probability that a ``random" number is $n$-free (AKA does not have any integer $b$ so that $b^n$ divides it) is given by $1/\zeta(n)$

Answer 10

There is a relation to the famous Veneziano amplitudes of nascent string theory (cf. Kholodenko, New Strings for old Veneziano amplitudes) through the factorials via the Euler beta function integral

$$ B(s,\alpha) = \frac{(s-1)!(\alpha-1)!}{(s+\alpha-1)!} x^{s+\alpha-1} |_{x=1} = \int_0^x t^{s-1} \; (x-t)^{\alpha-1} \; dt |_{x=1}.$$

This can be morphed into the core Riemann-Liouville fractional integroderivative of fractional calculus (analytically continued)

$$ D_x^{-s} x^{\alpha-1} |_{x=1} = \int_0^x \frac{t^{s-1}}{(s-1)!} \; (x-t)^{\alpha-1} \; dt |_{x=1}= \frac{(\alpha-1)!}{(s+\alpha-1)!} x^{s+\alpha-1} |_{x=1}\; , $$

which can be expressed in terms of the infinitesimal generator

$$ R_x = -\log(x) + \psi(1 + xD_x) = -(\log(x) + \gamma) - \sum_{n \ge 1} (-1)^n \zeta(n+1) \; (xD_x)^n \; ,$$

incorporating $\zeta(n>1)$ using well-known formulas for the digamma, or Psi, function $\psi(\beta)$ with $ \gamma =- \frac{d\beta!}{d\beta} |_{\beta=0}= \psi(1)$, the Euler-Mascheroni constant, as

$$ D_x^{-s} x^{\alpha-1} = e^{-sR_x} x^{\alpha-1} \; .$$

$$$$ $$$$ For easy reference, from the paper, Veneziano's 4-particle scattering amplitude is proportional to

$$A(s,t,u) = V(s,t) + V(s,u) + V(t,u) \; ,$$

where $V(s,t) = B(-\alpha(s),-\alpha(t))$, in which the arguments of beta are Regge trajectories.

Answer 11

The partition function $Z(T)$ [of primon gases] is given by the Riemann zeta function:

$${\displaystyle Z(T):=\sum _{n=1}^{\infty }\exp \left({\frac {-E_{n}}{k_{B}T}}\right)=\sum _{n=1}^{\infty }\exp \left({\frac {-E_{0}\log n}{k_{B}T}}\right)=\sum _{n=1}^{\infty }{\frac {1}{n^{s}}}=\zeta (s)} $$ with $s = E_0/k_BT$ where $k_B$ is Boltzmann's constant and $T$ is the absolute temperature.

The divergence of the zeta function at $s = 1$ corresponds to the divergence of the partition function at a Hagedorn temperature of $T_H = E_0/k_B$.

So choose your temperature in units of $T_H$...

Answer 12

I know I am late to the party here, but a couple of people mentioned different Basel Problem solutions by Pace and Calabi These solutions, along with another one by Zagier and Kontsevich all have general versions that induce different probabilistic viewpoints of $\zeta(2k).$ I reference my publication for details: https://www.ams.org/journals/qam/2018-76-03/S0033-569X-2018-01499-3/ or https://arxiv.org/abs/1710.03637.

First, Pace considers i.i.d. nonnegative Cauchy Random Variables $X,Y$ and their quotient $Z=X/Y.$ It turns out $Z$ has density function:

$$f_Z(z)= \frac{4}{\pi^2} \frac{\log(z)}{z^2-1}, \quad z>0.$$ Thus, $$\frac{\pi^2}{4} = \int_{0}^{\infty}\frac{\log(z)}{z^2-1} \ dz=\int_{0}^{1}\frac{\log(z)}{z^2-1} \ dz+\int_{1}^{\infty}\frac{\log(z)}{z^2-1} \ dz.$$ Making the substitution $t=1/z$ allows us to deduce $$\frac{\pi^2}{8} = \int_{0}^{1}\frac{\log(z)}{z^2-1} \ dz.$$ After converting the integrand on the right hand side into a geometric series and using monotone convergence theorem to interchange sum and integral, we obtain $$\frac{\pi^2}{8} = \sum_{n=0}^{\infty} \frac{1}{(2n+1)^2} = \frac{3}{4} \zeta(2).$$

Pace's solution generalizes. It leads to

Interpretation 1: The sums $$S(k)=\sum_{n \geq 0} \frac{(-1)^{nk}}{(2n+1)^k}, \quad k \in \mathbb{N},$$ are obtained by considering $k$ i.i.d. nonnegative Cauchy Random Variables $X_1, \dots, X_k$ and integrating the density function of the quotient $Z=X_1/ \dots / X_k.$ Then, upon splitting the sum $\zeta(2k)$ and rearranging, we obtain $$S(2k)= \frac{2^{2k}-1} {2^{2k}}\zeta(2k).$$

With repeated use of partial fractions and Fubini's Theorem, we can obtain a density function for $f_{Z}(z).$

Interpretation 1 will yield $$f_{Z}(z)= \begin{cases} \left(\frac{2}{\pi} \right)^k \frac{k |B_k|}{4^k-2^k}\frac{\ln^{k-1}(z)}{z^2-(-1)^k} & {k \text{ even}} \\ \left(\frac{2}{\pi} \right)^k |E_{k-1}|\frac{\ln^{k-1}(z)}{z^2-(-1)^k} & {k \text{ odd}} \end{cases}, \quad z>0 $$ where $B_k$ and $E_k$ are the $k$th Bernoulli and Eulerian numbers, respectively.

Replicating Pace's steps on $\int_{0}^{\infty} f(z) \ dz$ i.e. splitting of the region of integration and using the geometric series argument, we can recover $S(k).$

Interpretation 1 also allows us to find the sums $$S(k,a)=\sum_{n \in \mathbb{Z}} \frac{(-1)^{nk}}{(an+1)^k}, \quad a>1, \quad \frac{1}{a} \notin \mathbb{N}.$$ In particular, one considers the independent random variables $X_1,$ having density function $$f_{X_1}(x_1)= \frac{a}{\pi} \sin \left(\frac{\pi}{a} \right) \frac{1}{x_1^a + 1}, \quad x_1 \geq 0,$$ and $X_2, \dots, X_k$ all having density function $$f_{X_i}(x_i) = \frac{a}{\pi} \sin \left(\frac{\pi}{a} \right) \frac{x_i^{1-2/a}}{x_i^2 + 1}, \quad x_i \geq 0.$$ We can obtain $S(k,a)$ by integrating the density function for $Z=X_1/(X_2 \dots X_k)^{2/a}.$ Replicating Pace's arguments for $\int_{0}^{\infty} f(z) \ dz$ allows us to recover $S(k,a).$

In previous literature, Bourgade, Fujita, and Yor evaluate $S(k,a)$ in a slightly different way from here (see https://projecteuclid.org/download/pdf_1/euclid.ecp/1465224952), with the crux being differentiating under the integral sign.

Second, Calabi evaluates the double integral $$\int_{0}^{1} \int_{0}^{1} \frac{1}{1-x^2y^2} \ dx \ dy$$ in two ways. He obtains $S(2)$ with a geometric series argument and then uses the trigonometric change of variables $$x=\frac{\sin(u)}{\cos(v)}, \quad y=\frac{\sin(v)}{\cos(u)}$$ whose Jacobian determinant is implicitly the denominator of the integrand. The transformation diffeomorphically maps $(0,1)^2$ to the open isosceles triangle with base and height both $\pi/2.$ Thus $S(2)$ is geometrically the area of this triangle $\pi^2/8,$ hence solving the Basel Problem. The generalization of this is finding the integral $$\int_{(0,1)^k} \frac{1}{1-(-1)^k x_1^2 \dots x_{k}^2} \ dx_1 \dots \ dx_k, $$ which on one hand happens to be equal to $S(k)$ by a geometric series argument. On the other hand, Calabi's general change of variables $$x_i=\frac{\sin(u_i)}{\cos(u_{i+1})}, \quad \dots \quad , x_k=\frac{\sin(u_k)}{\cos(u_1)}$$ where $1 \leq i \leq k-1,$ has a Jacobian determinant which cancels with the integrand's denominator and diffeomorphically maps $(0,1)^k$ to the open convex polytope $$\Delta^{k}=\lbrace (u_1, \dots, u_k) \in \mathbb{R}^k: u_i +u_{i+1}, u_k +u_1<\pi/2, \quad u_1, \dots, u_k >0, \quad 1 \leq i \leq k-1 \rbrace.$$ Thus $S(k)=\text{Volume}(\Delta^{k}).$ Rescaling $u_i=\frac{\pi}{2} v_i,$ and considering $k$ i.i.d. Uniform Random Variables $V_1, \dots, V_k \in (0,1),$ we get

Interpretation 2: $$S(k)= \left(\frac{\pi}{2} \right)^k \text{Pr} \left(V_1+V_2<1, \dots V_{k-1}+V_k<1, V_k +V_1 <1 \right).$$ That is explicitly $S(k)$ is proportional to the probability that $k$ i.i.d. uniform random variables $V_1, \dots, V_k$ in $(0,1)$ have cyclically pairwise consecutive sums less than $1.$

In previous literature, Beukers, Calabi, and Kolk dissect $\Delta^k$ into congruent pyramids (see https://pdfs.semanticscholar.org/35be/01e63c0bfd32b82c97d58ccc9c35471c3617.pdf), and some users have already mentioned Elkies and Silagadze perform spectral analysis on the characteristic function of $\Delta^k$ (see (https://www.maa.org/sites/default/files/pdf/news_old/Elkies.pdf and http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.749.9234&rep=rep1&type=pdf). $\Delta^k$ turns out to be a chain polytope, which Stanley independently interprets in general (see http://dedekind.mit.edu/~rstan/pubs/pubfiles/66.pdf).

In my paper, the aforementioned probability with the uniform random variables leads to a gargantuan formula that does not involve $B_k$ or $E_k$ in any way. Rather, it is the result of imposing different conditions on $V_1, \dots, V_k.$ See my paper for details.

The result of Interpretation 2 is that $$S(k)= \left(\frac{\pi}{4} \right)^k +\left(\frac{\pi}{4} \right)^k \sum_{n=1}^{ \left \lfloor \frac{k}{2} \right \rfloor} \sum_{\substack{(r_1, \dots, r_n) \in [k]^n: \\ |r_p-r_q| \notin \lbrace 0,1,k-1 \rbrace, \\ p,q \in [n]} } \prod_{i=1}^{n} \frac{1}{i+\sum_{j=1}^{i} \alpha_j},$$ where $[m]:= \lbrace 1, \dots, m \rbrace$ and $$\alpha_j=2- \delta(k,2) - \sum_{m=1}^{j-1} \delta(|r_m-r_j|,2)+\delta(|r_m-r_j|,k-2)$$ and $\delta(a,b)$ is the Kronecker Delta Function. In particular, the inner sum is taken over all tuples $(r_1, \dots, r_n) \in [k]^ n$ having cyclically pairwise nonconsecutive entries.

Lastly, Zagier and Kontsevich reproduce Calabi's $\zeta(2)$ proof by evaluating $$\int_{0}^{1}\int_{0}^{1} \frac{1}{\sqrt{xy}(1-xy)}.$$ On one hand, this integral is $4S(2).$ But on the other hand, the change of variables $$x=\frac{\xi^2(\eta^2+1)}{\xi^2+1}, \quad y=\frac{\eta^2(\xi^2+1)}{\eta^2+1}$$ transforms it into $$\int_{0}^{\infty} \int_{0}^{1/\xi} \frac{4}{(\xi^2+1)(\eta^2+1)} d \eta \ d \xi.$$

The general version of this involves evaluating

$$I=\int_{(0,1)^k} \frac{1}{\sqrt{x_1 \dots x_k}(1-x_1 \dots x_k)} \ dx_1 \dots \ dx_k,$$ which is equal to $2^k S(k).$ But the generalized transformation

$$x_i= \frac{\xi_i^2 (\xi_{i+1}^2+1)}{(\xi_{i}^2+1)}, \quad \dots \quad,x_k= \frac{\xi_k^2 (\xi_{1}^2+1)}{(\xi_{k}^2+1)},$$ and defining $\Xi_1, \dots, \Xi_k$ to be $k$ i.i.d. nonnegative Cauchy random variables, it turns out

Intepretation 3 $$S(k)= \left(\frac{\pi}{2} \right)^k \text{Pr} \left(\Xi_1\Xi_2 <1, \dots \Xi_{k-1} \Xi_k <1, \Xi_k \Xi_1 <1 \right).$$ Explicitly, $S(k)$ is proportional to the probability that $k$ i.i.d. nonnegative Cauchy random variables $\Xi_1, \dots, \Xi_k$ have cyclically pairwise consecutive products less than $1.$

The result of Interpretation 3 is the same as the result of Interpretation 2.