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 $$ \sum_{k=0}^\infty \frac{1}{(2k+1)^2}= \int_0^1 \frac{-\log y}{1 - y^2} = \mathbb{P}(Y \geq 1)= \frac{\pi^2}{8}$$
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 $\{ x+iy: |x| < a \pi/2 \}$ and evaluating the left and right sides. The Brownian motion exit time is $\tau = a^2$ \pi^2/4$ and $f(z) $f(z) = \log(\frac{1-z}{1+z})$ log(\frac{1-z}{1+z}) = -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.
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.
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.
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 $K / \mathbb{Q}$.
