Geometric Interpretations of $\frac{\pi^3}{12}$?

Question

If $n$ is a non zero positive number I write $K^{1/n}$ for the disc centered at the origin of $\mathbb{R}^2$ with radius $1/n.$ I construct an infinite arrangement of concentric discs centered at the origin using $K^{1/n}$ for each $n.$ I denote this arrangement by $BK.$ I set $A_{BK}=\sum_{n=1}^\infty(-1)^{n+1}\text{area}(K^{1/n}).$ I believe I have correctly that:

\begin{align} A_{BK}&=\sum_{n=1}^\infty(-1)^{n+1}\text{area}(K^{1/n})\\ &=\pi\sum_{n=1}^\infty(-1)^{n+1}\bigg(\frac{1}{n}\bigg)^2\\ &=\pi\eta(2)\\ &=\frac{\pi^3}{12} \end{align}

where $\eta$ is the Dirichlet eta function. $A_{BK}$ is the infinite analogue of the "area between two discs" and so geometrically we can think of ${\pi^3 \above 1.5pt 12}$ as the "area between infinitely many discs."

Questions: Can we construct an integral (possibly a multiple integral) to calculate $A_{BK}$ ?

The picture below should help. We are adding up the red shaded areas.

Answer 1

Well, the simplest thing to do would be to write $$ f(r) = \begin{cases} 1 & \lfloor 1/r \rfloor \text{ odd} \\ 0 & \lfloor 1/r \rfloor \text{ even} \end{cases}. $$ This function is equal to 1 on $(1/2,1/2]$, equal to 0 on $(1/3, 1/2]$, equal to 1 on $(1/4,1/3]$, and so forth. The area of your bullseye is then $$ A = \int_0^{2 \pi} \int_0^1 f(r) r \, dr \, d\theta = 2 \pi \int_0^1 f(r) r \, dr . $$

Now, we also know that the square wave function (oscillating between -1 and 1, with period 1) is given by a Fourier sine series: $$ \frac{4}{\pi} \sum_{k = 1}^\infty \frac{\sin (2 \pi (2k - 1) x)}{2k - 1} $$ So one could try writing $f(r)$ in terms of this series: $$ f(x) = \frac{1}{2} - \frac{2}{\pi} \sum_{k = 1}^\infty \frac{\sin (\pi (2k - 1)/ x)}{2k - 1}. $$ The area then becomes $$ A = 2 \pi \left[\frac{1}{4} - \frac{2}{\pi} \int_0^1 \sum_{k=1}^\infty \frac{\sin (\pi (2k - 1)/ r)}{2k - 1} r \, dr \right]\\ $$ But since the partial sums here obey the conditions of the dominated convergence theorem, we have $$ A = \frac{\pi^3}{12} = \frac{\pi}{2} - 4 \sum_{k=1}^\infty \int_0^1 \frac{\sin (\pi (2k - 1)/ r)}{2k - 1} r \, dr $$ or $$\sum_{k=1}^\infty \int_0^1 \frac{\sin (\pi (2k - 1)/ r)}{2k - 1} r \, dr = \frac{\pi}{8}-\frac{\pi^3}{48}, $$ which is not a result I would have expected.

Answer 2

Since by the (inverse) Laplace transform $$ \frac{1}{n^2} = \int_{0}^{+\infty} x e^{-nx}\,dx $$ $\eta(2)$ admits the following integral representation $$ \int_{0}^{+\infty}\sum_{n\geq 1}(-1)^{n+1} x e^{-nx}\,dx = \int_{0}^{+\infty}\frac{x}{e^x+1}\,dx\stackrel{x\mapsto -\log u}{=}\int_{0}^{1}\frac{-\log u}{1+u}\,du$$ which is turned into $\int_{0}^{1}\frac{\log(1+u)}{u}\,du$ by integration by parts. There are a good number of (equivalent) alternatives provided by Fourier or Fourier-Legendre series.