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

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

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! :$$)$$

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

• This video on youtube gives a novel way to evaluate this sum, essentially by high-school geometry involving triangles and circles. Jan 18, 2020 at 2:36
• @Jam I think the 'natural appearances' of $\pi$ are significantly different from the natural appearances of $\pi^2$. You could square the solutions given to the question you link, though they'd hardly be compelling solutions to this question. For instance, $$\int_{-\infty}^\infty e^{-x^2}~dx = \sqrt \pi$$ is certainly a suitable answer to the question you've linked, but I wouldn't say $$\frac{\left( \int_{-\infty}^\infty e^{-x^2}~dx\right)^4}6 = \frac 16\pi^2$$ belongs here. Likewise, I don't think the sqrt of some of the answers here would be compelling responses to the question you linked. Jan 18, 2020 at 17:12
• I believe that "natural appearances of $\pi$" and "natural appearances of $\pi^2\over6$" are two very different things. Jan 18, 2020 at 20:34
• probability of primality is ${\pi^2\over 6}^{\pi(\sqrt{n})}$ but that' just contrived.
– user645636
Jan 20, 2020 at 2:45
• @RoddyMacPhee Correct me if I'm wrong but wouldn't it be $\left(\frac{6}{\pi^2}\right)^{\pi\sqrt n}$, otherwise it is increasing wrt. $n$?
– Jam
Jan 20, 2020 at 20:34

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 $$\frac{\pi^2}6$$.

• But does the proof of this use: $\sum 1/n^2 =\pi^2/6$? Or is it proved independently of that. Jan 18, 2020 at 1:37
• @GEdgar I am unsure. When I say I want to avoid $\sum \frac 1{n^2}$, I mean that I do not want it to be obvious that it appears in the derivation of a solution. In this case, if the proof does use $\zeta(2)$, it certainly isn't obvious and is therefore interesting enough to include! I will append a source when I find one. Jan 18, 2020 at 1:42
• The probability is $\prod_{p\in\Bbb P}(1-p^{-2})=1/\zeta(2)$, so the proof does use the Basel problem's solution, but it also uses the factorization of $\zeta$ over primes. So there are two "non-obvious" parts to the proof, Basel being the second: the first is that you can change that product into (one over) a sum on $\Bbb N$.
– J.G.
Jan 18, 2020 at 10:47
• So, since $\zeta(2) = \pi^2/6$ is used in the proof of this, it is not an independent case where $\pi^2/6$ arises. I would conjecture this is also true of most (or all?) of the other answers here. Jan 18, 2020 at 12:55
• This is clearly zeta(2) so no upvote from me. Sorry. No downvote either.
– mick
Jan 31, 2020 at 17:09

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.

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.

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.

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 $$j(\sqrt{7}/4)=\int_0^\infty\frac{(x^2+1)\arctan x}{x^4+\tfrac14x^2+1}dx=\frac{\pi^2}{6}.$$

Expect more nice examples as I gather the best ones.

• These are nice examples, but in all of them you are cheating by adding a constant in front of the integral/sum in such a way the final constant multiplying pi^2 in the result is 1/6. Jan 18, 2020 at 8:04
• OP here - I think these answers are all perfectly valid, as the most characteristic part of $\frac 16 \pi^2$ (which would be the $\pi^2$ term) arises naturally. This may kind of blur the focus of my question, but I’d much rather have these integrals than not :) Though if clathratus digs up an integral that evaluates to our desired constant with no scaling needed, that would be all the more interesting! Jan 18, 2020 at 10:24
• Similar to your second integral, $$\frac{4}{3} \int_0^\infty \frac{atan{x}}{1 + x^2} dx = \frac{16}{3} \int_0^1 \frac{atan{x}}{1 + x^2} dx=\frac{\pi^2}{6}$$ Jan 19, 2020 at 8:11
• What a brilliant compilation!(+1) Here's another one of the logatirhmic form: $$\int_0^\phi\frac{\ln x}{x^2-1}dx=\frac{\pi^2}{6}-\frac{3}{4}\ln^2\phi$$ where $\phi=\frac{1+\sqrt{5}}{2}$
– user632577
Jan 26, 2020 at 2:43
• @EdwardH. wow! that one is spectacular! source? Jan 26, 2020 at 2:46

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.

Here is the simplest,

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

• That is simply $\text{Li}_2(0)=\sum_{n=1}^\infty \frac1{n^2}=\frac{\pi^2}{6}$ which is not different from $\zeta(2)$. So, this contributes nothing to the OP. Jan 18, 2020 at 4:48
• Not sure agree. The integral could be obtained with elementary mehods. To write it as an expansion, or in advanced functions, is fine, but unnecessary. Besides, any $\frac{\pi^2}6$-example could be equated to $Li_2(0)$. Jan 18, 2020 at 4:58
• Is there a proof of this not using knowledge of the value of $\zeta(2)$ ? Jan 18, 2020 at 13:21
• @GEdgar Do you mind if I answer you the question instead? All we need is to connect $\displaystyle \int_0^1 \frac{\log(x)}{1-x}\textrm{d}x$ to $\displaystyle \int_0^1 \frac{\log(x)}{1-x^2}\textrm{d}x$, which is trivial to do, and then follow the procedure from the penultimate row of page $56$, going backwards, from (Almost) Impossible Integrals, Sums, and Series. Done. So, no need to use the knowledge of the value of $\zeta(2)$. Jan 18, 2020 at 15:08
• I like that you changed this from 'this may be the simplest' to 'here is the simplest.' After seeing the rest of the responses this question has produced, I would make that change too. Jan 20, 2020 at 1:38

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}

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}

$$\endgroup$$

• That is a very nice example. Do you mind if I ask, where does it come from? Jan 18, 2020 at 9:25
• @omegadot: Well, it's just $\int_0^1 \frac{\ln x}{x-1}$ in disguise, using parametric representation of the real branches of the Lambert W function. Jan 18, 2020 at 12:13
• How we do love the Omega constant: $\operatorname{W}_0(1) = \Omega$. Jan 26, 2020 at 5:06
• @omegadot: "What’s in a name?.." Jan 26, 2020 at 6:04
• @omegadot: Even better, $\mathrm{e}$ is just a shortcut for $\Omega^{-1/\Omega}$. Jan 26, 2020 at 11:00

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 $$\lim_{n\to\infty}\frac{\#([1..n]\cap S)}{n} = \frac{6}{\pi^2}.$$

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

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

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

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

• Yes, that is from I&S. I took a whack at it in the comments there, but the closest I could get at the time was the similar looking integral $$\int_0^1\text{arctan}\left(\frac{88\sqrt{21}}{36x^2+215}\right)\,\mathrm dx=\text{arctan}\left(\frac{88\sqrt{21}}{251}\right)+\frac{\alpha+\bar{\alpha}}{2742}$$ where $\alpha=w_1w_2\text{arctan}\left(\frac{6}{w_1}\right)$, $w_1=\sqrt{215+i88\sqrt{21}}$, and $w_2=i(\bar{w_1})^2.$ Sad the forum has been inactive for so long, it was a fun place to post. Jan 19, 2020 at 1:21

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 $$\frac{1}{\zeta(2)}$$.

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 $$k$$.

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 $$\pi\zeta(3).$$

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 $$\frac{1}{2}\pi\zeta(2).$$

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

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

Using inverse hyperbolic functions:

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

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

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

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

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.

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

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

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

• I found it, by experimental mathematic approach. May 8, 2021 at 21:08
• Are you sure the sum starts from $n=0$? I thought the zeta function is undefined at $n=0$. Jun 11, 2021 at 10:26
• @A-Level Student - $\zeta(0) = -\frac{1}{2}$. Jan 7 at 4:55

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

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

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

• Combining with $$\int_0^1 x(1-x)\log{x}\log(1-x)dx=\frac{1}{108}(37-3\pi^2)$$ we get rid of the 2: $$-\int_0^1(216x^2-216x+37)\log{x}\log(1-x)dx=\frac{\pi^2}{6}$$ Jan 26, 2020 at 8:39

This I found neat

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

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

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.

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 $$|z|\le1$$.

$$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$$ 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}

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 $$\pi^2/6$$.

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

Where $$C_n$$; Catalan numbers

• Sibawayh, Where's $\frac{\pi^2}{6}$ here? May 16, 2021 at 8:58
• @BooleanWick I guess Sibawayh means it's equal to $$\frac{2}{3}\left(\frac{\pi^2}{6}\right)^{-1}$$ Jun 11, 2021 at 10:22