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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.

enter image description here

Anthony
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  • Do you mean something as simple as $\pi \int_0^1 \frac{\log (x+1)}{x} , dx$ along with some geometric construction? – James Arathoon Jun 27 '18 at 16:10
  • I believe I am missing something. If we have concentric disks, isn't the area of their union equal to the area of the larger disk? What do you mean by arrangement? – Jack D'Aurizio Jun 27 '18 at 17:30
  • @JackD'Aurizio - No. I am computing the area "between" concentric discs. I am deliberately defining $A_{BK}=\sum_{n=1}^\infty(-1)^{n+1}\text{area}(K^{1/n}).$ – Anthony Jun 27 '18 at 17:36
  • @EricTowers yes I did. For example $K^{1/4}$ would be a disc with a radius equal to $1/4$ and so $\text{area}(K^{1/4})=\frac{\pi}{4^2}=\frac{\pi}{16}.$ I will edit the question to reflect that. – Anthony Jun 28 '18 at 14:59
  • The picture you have is a bit misleading, since the bands of light and dark are evenly spaced. What you should really have (according to the sum) is a solid band for $r = [1/2, 1]$; a light band for $[1/3, 1/2]$; a solid band for $[1/4, 1/3]$; and so on. In other words, the solid bands should get smaller towards the center. – Michael Seifert Jun 28 '18 at 18:25
  • @MichaelSeifert It's not drawn to scale. – Anthony Jun 28 '18 at 18:28
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    Here's a scale drawing if you want one. – Michael Seifert Jun 28 '18 at 18:30

2 Answers2

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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.

Jack D'Aurizio
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  • Sure what you have written is direct substitution of $\eta(2)$ and it requires multiplication by $\pi$ to get the desired result. I am trying to take advantage of the geometric construction itself to construct a new integral that computes $\frac{\pi^3}{12}.$ – Anthony Jun 27 '18 at 18:09
  • @AnthonyHernandez: in such a case you may have a look at here – Jack D'Aurizio Jun 28 '18 at 16:48
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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.

Michael Seifert
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