Unexpected appearances of $\pi^2 /~6$.

Question

"The number $\frac 16 \pi^2$ turns up surprisingly often and frequently in unexpected places." - Julian Havil, Gamma: Exploring Euler's Constant.

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It is well-known, especially in 'pop math,' that $$\zeta(2)=\frac1{1^2}+\frac1{2^2}+\frac1{3^2}+\cdots = \frac{\pi^2}{6}.$$ Euler's proof of which is nice. I would like to know where else this constant appears non-trivially. This is a bit broad, so here are the specifics of my question:

1. We can fiddle with the zeta function at arbitrary even integer values to eek out a $\zeta(2)$. I would consider these 'appearances' of $\frac 16 \pi^2$ to be redundant and ask that they not be mentioned unless you have some wickedly compelling reason to include it.
2. By 'non-trivially,' I mean that I do not want converging series, integrals, etc. where it is obvious that $c\pi$ or $c\pi^2$ with $c \in \mathbb{Q}$ can simply be 'factored out' in some way such that it looks like $c\pi^2$ was included after-the-fact so that said series, integral, etc. would equal $\frac 16 \pi^2$. For instance, $\sum \frac{\pi^2}{6\cdot2^n} = \frac 16 \pi^2$, but clearly the appearance of $\frac 16\pi^2$ here is contrived. (But, if you have an answer that seems very interesting but you're unsure if it fits the 'non-trivial' bill, keep in mind that nobody will actually stop you from posting it.)

I hope this is specific enough. This was my attempt at formally saying 'I want to see all the interesting ways we can make $\frac 16 \pi^2$.' With all that being said, I will give my favorite example as an answer below! :$)$

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There used to be a chunk of text explaining why this question should be reopened here. It was reopened, so I removed it.

Answer 1

Let $I(n)$ be the probability that two integers chosen randomly from $[1,n]$ are coprime. Then, $$\lim_{n \to \infty} I(n)=\frac{6}{\pi^2}.$$ So, you could say the odds that two randomly-chosen positive integers are coprime is $1$ in $\frac{\pi^2}6$.

Answer 2

Define a continuous analog of the binomial coefficient as

$$\binom{x}{y}=\frac{\Gamma(x+1)}{\Gamma(y+1)\Gamma(x-y+1)}.$$

While exploring integrals of the form

$$\int_{-\infty}^\infty\prod_{n=1}^m\binom{x_n}{t}\,\mathrm dt$$

I was surprised the first time I saw

$$\int_{-\infty}^\infty\binom{1}{t}^3\,\mathrm dt=\frac{3}{2}+\frac{6}{\pi^2}$$

show up.

Answer 3

Unexpected at first glance is $$2\sum_{m\ge1}\sum_{n\ge1}\frac{(-1)^n}{n^3}\sin(\tfrac{n}{m^{2k}})=\frac{1}{6}\zeta(6k)-\frac{\pi^2}{6}\zeta(2k).$$ A generalization may be found here.

Perhaps more unexpected is $$\sqrt3 \int_0^\infty \frac{\arctan x}{x^2+x+1} \, dx=\frac{\pi^2}{6},$$ which is proven here.

Even nicer is $$\frac1{12}\int_0^{2\pi}\frac{x\,dx}{\phi-\cos^2 x}=\frac{\pi^2}6,$$ which can be seen here. Here $\phi=\frac{1+\sqrt5}2$ is the golden ratio.

A pleasing logarithmic integral is $$\frac83\int_1^{1+\sqrt2}\frac{\ln x}{x^2-1}dx=\frac{\pi^2}6-\frac23\ln^2(1+\sqrt2),$$ proven here.

Another nice trigonometric integral: $$2\int_0^{\pi/2}\cot^{-1}\sqrt{1+\csc x}\, dx=\frac{\pi^2}{6},$$ from here.

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Edit: as was stated in the comments of this answer, it's the $\pi^2$ that counts, though un-scaled integrals evaluating to $\pi^2/6$ are best. With this in mind, I present a nice $\zeta$-quotient integral involving $\pi^2$: $$\int_0^\infty \left(\frac{\tanh(x)}{x^3}-\frac{1}{x^2\cosh^2(x)}\right)\, dx=\frac{7\zeta(3)}{\pi^2}=\frac{7\zeta(3)}{6\zeta(2)},$$ shown here.

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I just derived another identity: $$\int_0^\infty\frac{(x^2+1)\arctan x}{x^4+\tfrac14x^2+1}dx=\frac{\pi^2}{6}.$$ Since I just found this identity I present the proof. In the link I provided after the second identity it is shown that $$f(a)=\int_0^\infty \frac{\arctan x}{x^2+2ax+1}dx=\frac{\pi}{4\sqrt{1-a^2}}\left(\frac\pi2-\phi(a)\right)\qquad |a|<1$$ where $\phi(a)=\arctan\frac{a}{\sqrt{1-a^2}}$. First off, notice that $\phi(-a)=-\phi(a)$. Thus $$j(a)=\frac12(f(a)+f(-a))=\int_0^\infty\frac{(x^2+1)\arctan x}{x^4+2(1-2a^2)x^2+1}dx=\frac{\pi^2}{8\sqrt{1-a^2}}.$$ Hence $$j(\sqrt{7}/4)=\int_0^\infty\frac{(x^2+1)\arctan x}{x^4+\tfrac14x^2+1}dx=\frac{\pi^2}{6}.$$

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Expect more nice examples as I gather the best ones.

Answer 4

Problem 11953 from AMM (January 2017) asked for the evaluation of the following double integral whose value turns out to be equal to $\frac{\pi^2}{6}$. $$\int_0^\infty \!\!\!\int_0^\infty \frac{\sin x \sin y \sin (x + y)}{xy(x + y)} \, dx \, dy = \frac{\pi^2}{6}.$$

Problem 2074 from Mathematics Magazine (June 2019) asked for the following evaluation of a limit of a sum whose value turns out to be equal to $\frac{\pi^2}{6}$. $$\lim_{n \to \infty} \sum_{k = 1}^n \frac{(-1)^{k + 1}}{k} \binom{n}{k} H_k = \frac{\pi^2}{6}.$$ Here $H_n = \sum_{k = 1}^n \frac{1}{k}$ denotes the $n$th Harmonic number.

And here are a few sums: $$\sum_{n = 1}^\infty \frac{H_n}{n2^{n - 1}} = \frac{\pi^2}{6}.$$

$$\sum_{n = 1}^\infty \frac{H_n}{n (n + 1)} = \frac{\pi^2}{6}.$$

$$\frac{3}{2} \sum_{n = 0}^\infty \left (\frac{1}{(6n + 1)^2} + \frac{1}{(6n + 5)^2} \right ) = \frac{\pi^2}{6}.$$ and $$\sum_{n = 1}^\infty \frac{3}{n^2 \binom{2n}{n}} = \frac{\pi^2}{6}.$$

And a few more sums, this time involving the variant harmonic number term $\Lambda_n$ were $$\Lambda_n = 1 + \frac{1}{3} + \cdots + \frac{1}{2n - 1} = \sum_{k = 1}^n \frac{1}{2k - 1} = H_{2n} - \frac{1}{2} H_n.$$

$$\sum_{n = 1}^\infty \frac{\Lambda_n}{n(2n - 1)} = \frac{\pi^2}{6},$$ $$\sum_{n = 1}^\infty (-1)^{n+ 1} \left (\frac{2n + 1}{n(n+ 1)} \right )^2 \Lambda_n = \frac{\pi^2}{6},$$ and $$2\sum_{n = 1}^\infty \frac{(-1)^{n + 1} \Lambda_n}{3^{n - 1} n} = \frac{\pi^2}{6}.$$

Some function values: $$\zeta (2) = \operatorname{Li}_2 (1) = \frac{\pi^2}{6},$$ where $\zeta$ denotes the Riemann zeta function while $\operatorname{Li}_2 (x)$ is the dilogarithm. $$6 \operatorname{Li}_2 \left (\frac{1}{2} \right ) - 6 \operatorname{Li}_2 \left (\frac{1}{4} \right ) - 2 \operatorname{Li}_2 \left (\frac{1}{8} \right ) + \operatorname{Li}_2 \left (\frac{1}{64} \right ) = \frac{\pi^2}{6}.$$

And some strange integrals: $$\int_0^1 (x^{-x})^{{{(x^{-x})}^{(x^{-x})}}^\cdots} \, dx = \frac{\pi^2}{6},$$ and $$\int_0^\infty \frac{dx}{\operatorname{Ai}^2 (x) + \operatorname{Bi}^2(x)} = \frac{\pi^2}{6},$$ where $\operatorname{Ai}(x)$ and $\operatorname{Bi}(x)$ denote the Airy functions of the first and second kinds, respectively.

Answer 5

Here is the simplest,

$$\int_0^1 \frac{\ln x}{x-1}\ \mathrm{d}x=\frac{\pi^2}6$$

Answer 6

In terms of the two real branches of the Lambert W function

$\require{begingroup} \begingroup$ $\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}$

\begin{align} \int_0^1 \frac{\Wp(-\tfrac t{\mathrm e})\,(\Wp(-\tfrac t{\mathrm e})-\Wm(-\tfrac t{\mathrm e}))} {t\,(1+\Wp(-\tfrac t{\mathrm e}))\,(1+\Wm(-\tfrac t{\mathrm e}))}\, dt &=\frac{\pi^2}6 \tag{1}\label{1} . \end{align}

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Edit

And another one, with different integrand curve:

\begin{align} \int_0^1 \frac{\Wp(-\tfrac t\e)+t\,(1+\Wm(-\tfrac t\e))}{t\,(1+\Wm(-\tfrac t\e))} \, dt &=\frac{\pi^2}6 \tag{2}\label{2} . \end{align}

The intersection point of the integrands in \eqref{1} and \eqref{2} can be found exactly at $t=\tfrac1\Omega-1\approx 0.763222834$, where $\Omega$ is Omega constant, $\Omega \e^{\Omega }=1,\ \Omega=\W(1)\approx 0.56714329$ (thanks, @omegadot).

Also, one more:

\begin{align} \int_0^1 \ln\left(\frac{-\Wm(-t\,\exp(-t))}t\right) \, dt &= \int_0^1 -t-\Wm(-t\,\exp(-t)) \, dt =\frac{\pi^2}6 \tag{3}\label{3} . \end{align}

$\endgroup$

Answer 7

Related, but certainly not in an immediately obvious way, to $\zeta(2)$ is the density of the squarefree numbers.

Call a natural number squarefree if no square larger than $1$ divides it (e.g. 12 is not squarefree because 4 divides it, but 30 is squarefree). Let $S$ be the set of squarefree numbers. Then $$ \lim_{n\to\infty}\frac{\#([1..n]\cap S)}{n} = \frac{6}{\pi^2}. $$

See here: http://mathworld.wolfram.com/Squarefree.html

Answer 8

I'll give you three cute examples from the book, (Almost) Impossible Integrals, Sums, and Series.

A particular case of the generalization from Section $1.11$, page $7$ $$i)\ 1- \int_0^1 \left(2 x+ 2^2 x^{2^2-1}+2^3 x^{2^3-1}+2^4 x^{2^4-1}+\cdots\right) \frac{\log(x)}{1+x} \textrm{d}x=\frac{\pi^2}{6}.$$

A particular case of the generalization from Section $1.38$, page $25$ $$ii) \ \frac{1}{2}\int_0^{ \infty} \int_0^{\infty}\frac{x -y}{e^x-e^y} \textrm{d}x \textrm{d}y=\int_0^{ \infty} \int_0^y\frac{x -y}{e^x-e^y} \textrm{d}x \textrm{d}y=\frac{\pi^2}{6}.$$ The first example from Section $1.17$, page $10$ $$\frac{6}{7\zeta(3)}\int _0^1 \int _0^1 \frac{\displaystyle \log \left(\frac{1}{x}\right)-\log \left( \frac{1}{y} \right)}{\displaystyle \log \left(\log \left(\frac{1}{x}\right)\right)-\log \left(\log \left(\frac{1}{y}\right)\right)} \textrm{d}x \textrm{d}y =\frac{6}{\pi^2}.$$

Another curious sum of (crazy) integrals leading to the same value which was proposed by the author of the mentioned book is

$$\frac{\pi^2}{6}=\frac{4}{3}\int_0^{\pi/2} \log \left(\frac{\left(x^2\sin^2(x)+ \pi ^2/4 \cos ^2(x)\right)^{x/2}}{x^x}\right)\sec ^2(x) \textrm{d}x$$ $$-\frac{2}{3} \int_0^1 \frac{\log \left(\left(x^2+\left(1-x^2\right) \cos (\pi x)+1\right)/2\right)}{x-x^3} \textrm{d}x,$$

but also the amazing $\zeta(2)\zeta(3)$ product in the harmonic series (with zeta tail) form

$$\frac{1}{2\zeta(3)}\sum_{n=1}^{\infty} \frac{H_n^2}{n}\left(\zeta(2)-1-\frac{1}{2^2}-\cdots-\frac{1}{n^2}\right)=\frac{\pi^2}{6},$$

or

$$\frac{\pi^2}{6}=4\sum_{n=1}^{\infty}\biggr(2n\biggr(1-\frac{1}{2^{2n+1}}\biggr)\zeta(2n+1)-2\log(2)\biggr(1-\frac{1}{2^{2n}}\biggr)\zeta(2n)$$ $$-\frac{1}{2^{2n}}\sum_{k=1}^{n-1}(1-2^{k+1})\zeta(k+1)(1-2^{2n-k})\zeta(2n-k)\biggr).$$

Answer 9

Here is a crazy-looking integral, which I believe I originally saw on the (now abandoned) integrals and series forum:

$$\int_{0}^{1}\frac{1}{\sqrt{1-x^2}}\arctan\left(\frac{88\sqrt{21}}{36x^2+215}\right)dx=\frac{\pi^2}{6}$$

Answer 10

A somewhat surprising occurence, which can be seen immediatelly via the Euler product, appears in the study of visible points of lattice.

Given a lattice $\Gamma \subset \mathbb R^d$, meaning $\Gamma=\mathbb Z v_1 \oplus ... \oplus\mathbb Z v_d$ for some $\mathbb R$ basis $v_1,.., v_d$ of $\mathbb R^d$, the visible points of $\Gamma$ are defined as $$V:= \{ z=n_1v_1+...+n_dv_d : n_1,.., n_d \in \mathbb Z , \mbox{ gcd } (n_1,.., n_d)=1 \}$$

The, we have the following result, (see Prop.~6 in Diffraction from visible lattice points and k-th power free integers)

Proposition The natural density of $V$ is $$ \mbox{dens}(V)=\frac{1}{ \det(A) \zeta(d) } $$ where $A$ is the matrix with columns $v_1,v_2,...,v_d$. Here, natural density means the density calculated with respect to the sequence $A_n=[-n,n]^d$, note that this set can have a different density with respect to other sequences.

In particular, the visible sets of $\mathbb Z^2$, given by $$V=\{ (n,m) \in \mathbb Z^2 : \mbox{gcd}(m,n) =1 \}$$ have natural density $\frac{1}{\zeta(2)}$.

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The so called "cut and project" formalism establises a connection between the above example and some sets in compact groups, which appeared in my research area recently.

Consider the group $$\mathbb K:= \prod_{p \in P} \left( \mathbb Z^2 / p \mathbb Z^2 \right)$$ where $p$ denotes the set of all primes. $\mathbb K$ is a compact Abelian group, and hence has a probability Haar measure $\theta_{\mathbb K}$.

Now, $\phi(m,n) := \left( (m,n)+p \mathbb Z^2 \right)_{p \in P}$ defines an embedding of $\mathbb Z^2$ into $\mathbb K$.

Define the set $$W:= \prod_{p \in P} \left( \bigl(\mathbb Z^2 / p \mathbb Z^2\bigr) \backslash \{ (0,0) + p\mathbb Z^2 \} \right)$$

Then, the visible points of $\mathbb Z^2$ are exactly $$V= \phi^{-1}(W)$$

The set $W$, which is used in the study of diffraction of $V$, has the following properties:

• $W$ is closed and hence compact.
• $W$ has empty interior (hence is fractal shape).
• $\theta_{\mathbb K}(W) = \frac{1}{\zeta(2)}$

The last property is where I was going to, and it is intuitively not that hard to see once you identify $\theta_{\mathbb K}(W)$ as the product of the counting measures on $\mathbb Z^2 / p \mathbb Z^2$: this immediatelly gives $$\theta_{\mathbb K}(W) = \prod_{p \in P}\frac{p^2-1}{p^2}$$

P.S. There are similar appearences of $\zeta(n)$ in the study of $k$th power free integers, that is all the integers $n \in \mathbb Z$ which are not divisible by the $k$th power of any prime, for a fixed $k$.

Answer 11

Consider the following picture:, centered at the origin of $\mathbf{R}^{2}.$ It is a concentric arrangement of circles $\color{red}{\text{(- this should be discs ?)}}$; each circle has radius $1/n.$ We can think of it as an infinite bulls-eye. The sum of the areas shaded in red is equal to $\frac{1}{2}\pi\zeta(2).$ In particular $$\sum_{k=1}^\infty \int_0^1 \frac{\sin (\pi (2k - 1)/ r)}{2k - 1} r \, dr = \frac{\pi}{8}\left(1-\zeta(2)\right), $$ Surprisingly if you take this arrangement and rotate it about the $x-$axis then you have a similar arrangement with circles being replaced by $3-$balls each with radius $1/n.$ In this case the sum of volumes shaded in red is equal to $\pi\zeta(3).$

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Update: It dawned on me that I can in fact extend the notion "volume shaded in red" to higher dimensions.

Let $K_{i}$ be the $n-$ ball at the center of the origin of Euclidean $n-$space, $\mathbf{E}^{n},$ with radius $\frac{1}{i}$ and whose volume I denote by $\mu\left(K_{i}\right).$ Consider $$ \sum\limits_{i=1}^{\infty}(-1)^{i+1}\mu\left(K_{i}\right). $$ A closed form for this quantity is known whenever $n$ is an even number: $$ (-1)^{1+\frac{n}{2}}{\left(2^{n-1}-1 \right)B_{n}\above 1.5pt \Gamma(1+\frac{n}{2})\Gamma(1+n)}\pi^{\frac{3}{2}n}. $$ Inspections shows the numerator of the rational part is the sequence A036280(n/2). You can check in the case that $n=2$ the quantity computes to $\frac{1}{2}\pi\zeta(2).$

Answer 12

Two simple trigonometric integrals are $$\frac{4}{3}\int_0^1 \frac{\arctan{x}}{1+x^2}dx =\frac{\pi^2}{6}$$ and

$$\frac{4}{3}\int_0^1 \frac{\arcsin{x}}{\sqrt{1-x^2}}dx = \frac{\pi^2}{6}$$

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Using inverse hyperbolic functions:

$$\frac{10}{3} \int_0^1\frac{\operatorname{arcsinh}\left({\frac{x}{2}}\right)}{x}dx=\frac{\pi^2}{6}$$

$$\frac{4}{3} \int_0^1 \frac{\operatorname{arctanh}{x}}{x} dx = \frac{\pi^2}{6}$$

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From series $$ \sum_{k=0}^\infty \frac{1}{((k+1)(k+2))^2} = \frac{\pi^2}{3}-3 $$

and

$$ \sum_{k=0}^\infty \frac{k}{((k+1)(k+2))^2} = 5- \frac{\pi^2}{2} $$

$\frac{\pi^2}{6}$ arises directly when cancelling out the integer terms:

$$ \sum_{k=0}^\infty \frac{5+3k}{((k+1)(k+2))^2} =\frac{\pi^2}{6}$$

Similarly, $$\frac{8}{3}\sum_{k=0}^\infty \frac{4k+5}{(2k+1)^2(2k+3)^2} = \frac{\pi^2}{6}$$

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More series and integrals are available at http://oeis.org/A013661

Answer 13

This paper gives a polynomial-time approximation algorithm for the Minimum Equivalent Digraph (MEG) problem, with approximation ratio $\pi^2/6$.

The problem is, given a directed graph, to find a min-size subset $S$ of the edges that preserves all reachability relations between pairs of vertices. (That is, for every pair $u, v$ of vertices, if there is a path from $u$ to $v$ in the original graph, then there is such a path that uses only edges in $S$.) The problem is NP-hard. This was the first poly-time algorithm with approximation ratio less than 2.

Answer 14

Integral representations are given by

$$2\int_0^1 x \left \lfloor{\frac1x}\right \rfloor \ dx=2\int_1^\infty\frac{1}{t}\lfloor{t\rfloor}\frac{dt}{t^2}=2\sum_{n=1}^{\infty}\int_n^{\infty}t^{-3}\,dt=\sum_{n=1}^{\infty}n^{-2}=\frac{\pi^2}{6}$$

and

$$\frac{4}{3}\int_0^1\int_0^1\frac{1}{1-x^2y^2}\,dxdy=\frac{\pi^2}{6}$$

Then, for every $-1 < \alpha \le 1$,

$$\int_0^\infty\frac{\log(1+\alpha x)}{x(1+x)}\,dx= \log(\alpha)\log(1-\alpha)+\text{Li}_2(\alpha)$$

and when $\alpha=1$, this integral becomes

$$\int_0^\infty\frac{\log(1+ x)}{x(1+x)}\,dx= \frac{\pi^2}{6}$$

Answer 15

Here is one, $$-\sum_{n=0}^{\infty}\left[\zeta(2n)-\zeta(2n+2)-\zeta(2n+3)+\zeta(2n+4)\right]=\frac{\pi^2}{6}$$

Answer 16

This is one of those amazing series for $1/\pi^2$. You can find them in this paper by G. Almkvist and J. Guillera.

$$\left(\frac{2}{5}\right)^{3}\sum_{n=0}^{\infty}\frac{(6n)!}{n!^{6}10^{6n}}(532n^{2}+126n+9)=\frac{6}{\pi^{2}}$$

Answer 17

How about

$$\int_0^1 dx \, \log{x} \, \log{(1-x)} = 2 - \frac{\pi^2}{6} $$

Answer 18

This I found neat

$$\int_0^\pi \frac{\log(\frac{\cos x}{2}+1)}{\cos x} dx=\frac{\pi^2}{6}$$

Answer 19

For $0< x<1$, \begin{align}\text{Li}_2(x)+\text{Li}_2(1-x)+\ln x\ln(1-x)=\dfrac{\pi^2}{6}\end{align}

And, for $0\leq x\leq 1$, $\displaystyle \text{Li}_2(x)=\sum_{n=1}^\infty \dfrac{x^n}{n^2}$

Answer 20

In physics, $\pi^2/6$ appears as a proportionality constant between a metal's internal energy (or at least the contribution of the electrons to that energy) on the one hand and the density of states $\times$ the Fermi temperature on the other hand. It appears there as another manifestation of the identity $\zeta(2)=\pi^2/6$, i.e. its derivation has not really an independent character here compared to the installments of the other answers.

Answer 21

I was quite pleasantly surprised to derive the continued fraction identity $$\frac{\pi^2}{6}=-1+\cfrac{1}{1-\cfrac{1}{2-\cfrac{1}{5-\cfrac{16}{13-\cfrac{81}{25-\cfrac{256}{41-\cdots}}}}}},$$ which comes from the identity for the polylogarithm $\mathrm{Li}_s(z)$: $$\mathrm{Li}_s(z)+1=\cfrac{1}{1-\cfrac{z}{1+z-\cfrac{z}{2^s+z-\cfrac{2^{2s}z}{3^s+2^sz-\cfrac{3^{2s}z}{4^s+3^sz-\cfrac{4^{2s}z}{5^s+4^sz-\cdots}}}}}},$$ which is valid for $\Re(s)>1$ and $|z|\le1$.

Answer 22

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ With Abel-Plana Formula: \begin{align} {\pi^{2} \over 6\phantom{^{2}}} & = {3 \over 2} + 2\int_{0}^{\infty}{\sin\pars{2\arctan\pars{t}} \over \pars{1 + t^{2}}\pars{\expo{2\pi{\large t}} - 1}}\,\dd t \end{align}

Answer 23

You have $$\int\limits_{-\infty}^{+\infty}\int\limits_{-\infty}^{+\infty}\frac{\sin(x^2+y^2)}{x^2+y^2}dxdy=\frac{\pi^2}{2}.$$ Of course, dividing by 3 you have the expected value $\pi^2/6$.

Answer 24

$$\sum_{n=0}^{\infty}\frac{C_n^5}{2^{10n}}\left(\frac{n+1}{2n-1}\right)^4(4n-1)\sum_{k=0}^{n}(-1)^k{n \choose k}\frac{(n-k)[4k^2-2k+1]}{(2k-1)(2k+1)}=\frac{4}{\pi^2}$$

Where $C_n$; Catalan numbers